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1

CHAPTER 1

Basic Considerations

1.1 Conservation of mass — Mass — density Newton’s second law — Momentum — velocity The first law of thermodynamics — internal energy — temperature 1.2 a) density = mass/volume = M L/ 3 b) pressure = force/area = F L ML T L M LT/ / /2 2 2 2= = c) power = force × velocity = F L T ML T L T ML T× = × =/ / / /2 2 3

d) energy = force × distance = ML T L ML T/ /2 2 2× = e) mass flux = ρAV = M/L3 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = L3/T

1.3 a) density = ML

FT LL

FT L3

2

32 4/

/=

b) pressure = F/L2 c) power = F × velocity = F × L/T = FL/T d) energy = F × L = FL

e) mass flux = MT

FT LT

FT L= =2 /

/

f) flow rate = AV = L2 × L/T = L3/T 1.4 (C) m = F/a or kg = N/m/s2 = N.s2/m. 1.5 (B) [µ] = [τ/du/dy] = (F/L2)/(L/T)/L = F.T/L2. 1.6 a) L = [C] T2. ∴[C] = L/T2 b) F = [C]M. ∴[C] = F/M = ML/T2 M = L/T2 c) L3/T = [C] L2 L2/3 . ∴[C] = L T L L L T3 2 2 3 1 3/ / /⋅ ⋅ = Note: the slope S0 has no dimensions. 1.7 a) m = [C] s2. ∴[C] = m/s2 b) N = [C] kg. ∴[C] = N/kg = kg ⋅ m/s2⋅ kg = m/s2 c) m3/s = [C] m2 m2/3. ∴[C] = m3/s⋅m2⋅ m2/3 = m1/3/s 1.8 a) pressure: N/m2 = kg ⋅ m/s2/m2 = kg/m⋅ s2 b) energy: N⋅ m = kg ⋅ m/s2 × m = kg⋅ m2/s2

c) power: N⋅ m/s = kg ⋅ m2/s3

d) viscosity: N⋅ s/m2 =kg m

ss

1m

kg / m s2 2

⋅ ⋅ = ⋅

2

e) heat flux: J/s = N m

skg m

sms

kg m / s22 3⋅ = ⋅ ⋅ = ⋅

f) specific heat: J

kg KN mkg K

kg ms

mkg K

m / K s2

2 2

⋅= ⋅

⋅= ⋅ ⋅

⋅= ⋅

1.9 kgms

ms

m2 + + =c k f . Since all terms must have the same dimensions (units) we

require: [c] = kg/s, [k] = kg/s2 = N s / m s N / m,2 2⋅ ⋅ = [f] = kg m / s N.2⋅ =

Note: we could express the units on c as [c] = kg / s N s / m s N s / m2= ⋅ ⋅ = ⋅ 1.10 a) 250 kN b) 572 GPa c) 42 nPa d) 17.6 cm3 e) 1.2 cm2 f) 76 mm3 1.11 a) 1.25× 108 N b) 3.21× 10−5 s c) 6.7× 108 Pa d) 5.6× 10−12 m3 e) 5.2 × 10−2 m2 f) 7.8 × 109 m3

1.12 (A) 8 92.36 10 23.6 10 23.6 nPa.− −× = × =

1.13 2 2 2

0.068540.225 0.738

0.00194 3.281

m m

d dλ

ρ ρ= =

×

where m is in slugs, ρ in slug/ft3 and d in feet. We used the conversions in the front cover.

1.14 a) 20 cm/hr = 520/3600 5.555 10 m/s

100−= × 520

/3600 5.555 10 m/s100

−= ×

b) 2000 rev/min = 2000 2× π/60 = 209.4 rad/s c) 50 Hp = 50 × 745.7 = 37 285 W d) 100 ft3/min = 100 × 0.02832/60 = 0.0472 m3/s e) 2000 kN/cm2 = 2× 106 N/cm2 × 1002 cm2/m2 = 2× 1010 N/m2 f) 4 slug/min = 4 × 14.59/60 = 0.9727 kg/s g) 500 g/L = 500 × 10−3 kg/10−3 m3 = 500 kg/m3

h) 500 kWh = 500 × 1000× 3600 = 1.8× 109 J 1.15 a) F = ma = 10 × 40 = 400 N. b) F − W = ma. ∴ F = 10 × 40 + 10 × 9.81 = 498.1 N. c) F − W sin 30° = ma. ∴ F = 10 × 40 + 9.81 × 0.5 = 449 N.

1.16 (C) The mass is the same on earth and the moon: [4(8 )] 32 .du

r rdr

τ µ µ µ= = =

1.17 The mass is the same on the earth and the moon:

3

m = 60

32 21863

.. .= ∴ Wmoon = 1.863× 5.4 = 10.06 lb

1.18 (C) shear sin 4200sin30 2100 N.F F θ= = =o

shear4

2100= 84 kPa

250 10

FA

τ −= =×

1.19 a) λρ

= = ×× ×

= ×−

−−. .

.. ( . )

.43225 2254 8 10

184 3 7 10102

26

10 26m

dm or 0.00043 mm

b) λρ

= = ×× ×

= ×−

−−. .

.. ( . )

.225 2254 8 10

00103 3 7 107 7 102

26

10 25m

dm or 0.077 mm

c) 26

2 10 2

4.8 10.225 .225 .0039m

.00002 (3.7 10 )

m

dλ

ρ

−

−×

= = =× ×

or 3.9 mm

1.20 Use the values from Table B.3 in the Appendix. a) 52.3 + 101.3 = 153.6 kPa. b) 52.3 + 89.85 = 142.2 kPa. c) 52.3 + 54.4 = 106.7 kPa (use a straight- line interpolation). d) 52.3 + 26.49 = 78.8 kPa. e) 52.3 + 1.196 = 53.5 kPa.

1.21 a) 101 − 31 = 70 kPa abs. b) 760 − 31

101 × 760 = 527 mm of Hg abs.

c) 14.7 − 31

101 × 14.7 = 10.2 psia. d) 34 −

31101

× 34 = 23.6 ft of H2O abs.

e) 30 − 31

101 × 30 = 20.8 in. of Hg abs.

1.22 p = po e−gz/RT = 101 e−9.81 × 4000/287 × (15 + 273) = 62.8 kPa From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is

% error = 62.8 61.6

61.6−

× 100 = 1.95 %.

1.23 a) p = 973 + 22,560 20,00025,000 20,000

−−

(785 - 973) = 877 psf

T = −12.3 + 22,560 20,00025,000 20,000

−−

(−30.1 + 12.3) = −21.4°F

b) p = 973 + .512 (785 − 973) + .512

2 (−.488) (628 − 2 × 785 + 973) = 873 psf

T = −12.3 + .512 (−30.1 + 12.3) + .512

2 (−.488) (−48 + 2 × 30.1 − 12.3) = −21.4°F

Note: The results in (b) are more accurate than the results in (a). When we use a linear interpolation, we lose significant digits in the result.

4

1.24 T = −48 + 33,000 30,00035,000 30,000

−−

(−65.8 + 48) = −59°F or (−59 − 32) 59

= −50.6°C

1.25 (B)

1.26 p = nFA

= 4

26.5 cos 42

152 10−×

o

= 1296 MN/m2 = 1296 MPa.

1.27 4

n4

t

(120000) .2 10 2 . 4 N

20 .2 10 .0004N

F

F

−

−

= × × =

= × × = F = 2 2

n tF F+ = 2.400 N.

θ = tan−1 .00042.4

=.0095°

1.28 ρ = mV−

= 0 2180 1728

./

= 1.92 slug/ft3. τ = ρg = 1.92 × 32.2 = 61.8 lb/ft3.

1.29 ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3 γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3

% error for ρ = 976 978

978−

× 100 = −.20%

% error for γ = 9560 978 9.81

978 9.81− ×

× × 100 = −.36%

1.30 S = 13.6 − .0024T = 13.6 − .0024 × 50 = 13.48.

% error = 13.48 13.6

13.6−

× 100 = −.88%

1.31 a) m = W Vg

γ=612 400 500 10

9.81g

−× ×= = 0.632 kg

b) m = 612 400 500 10

9.77

−× × = 0.635 kg

c) m = 612 400 500 10

9.83

−× × = 0.631 kg

1.32 S =/

water

m Vρρ

= 10/. 1.2

water

Vρ

= .1.94

∴ V = 4.30 ft3

1.33 (D) 2 2

3water

( 4) (80 4)1000 1000 968 kg/m

180 180Tρ − −= − = − =

5

1.34 τ = µ dudr

= 1.92 × 10−5 2

30(2 1/12)

(1/12)

×

= 0.014 lb/ft2

1.35 T = force × moment arm = τ 2πRL × R = µ dudr

2πR2L = µ 2

0.41000

R +

2πR2L.

∴µ = 2 2

2

0.00260.4 0.41000 2 1000 2 .01 0.2

12

T

R LR

π π=

+ + × ×

= 0.414 N.s/m2.

1.36 Use Eq.1.5.8: T =32 R Lh

π ω µ =

( )2 .5 / 12π π× × × × ×3 2000 260

4 006

01 12

.

. / = 2.74 ft- lb.

Hp = Tω550

2 74 209550

= ×. .4 = 1.04 Hp

1.37 Fbelt = µdudy

A = × −131 1010002

3..

(.6 × 4) = 15.7 N.

Hp = F V× = ×746

15 7 10746.

= 0.210 Hp

1.38 Assume a linear velocity so .du rdy h

ω= Due to the area

element shown, dT = dF × r = τdA × r = µ dudy

2πr dr × r.

dr

r

τ

T = 3

0

2Rr dr

hµω π

∫ = 2

4

2 36 10400 2

603 12

2 08 12

45 4

πµω π π

hR =

× × × × ×

×

−. ( / )

. / = 91 × 10−5 ft- lb.

1.39 2

30(2 1/12)

(1/12)

×

2 20 0[32 / ] 32 / .

dur r r r

drτ µ µ µ= = = ∴τr = 0 = 0,

τr=0.25 = 32 × 1 × 10−3 × 2

.25/100

(.5/100)= 3.2 Pa, τr=0.5 = 32 × 1 × 10−3 × 2

.5/100

(.5/100)= 6.4 Pa

1.40 (A) 3[10 5000 ] 10 10 5000 0.02 1 Pa.du

rdr

τ µ µ −= = × = × × × =

1.41 The velocity at a radius r is rω. The shear stress is τ µ= ∆∆

uy

.

The torque is dT = τrdA on a differential element. We have

6

0.08

0 = = 2

0.0002r

T rdA rdxωτ µ π∫ ∫ ,

2000 2209.4 rad/s

60πω ×= =

where x is measured along the rotating surface. From the geometry 2x = r, so that

0.08 0.08

2 3

0 0

209.4 / 2 329000= 0.1 2 329000 (0.08 )

0.0002 32

x xT dx x dxπ× = =∫ ∫ = 56.1 N . m

1.42 If τ µ= dudy

= cons’t and µ = AeB/T = AeBy/K = AeCy, then

AeCy dudy

= cons’t. ∴ dudy

= De−Cy.

Finally, 0 0

yuCydu De dy−=∫ ∫ or u(y) =

0

yCyDe

C−− = E (e−Cy − 1)

where A, B, C, D, E, and K are constants.

1.43 µ = =

=

Ae Ae

Ae

B T B

B

/ /

/

.

.

001

000357

293

353 ∴A = 2.334 × 10−6, B = 1776.

µ40 = 2.334 × 10−6 e1776/313 = 6.80 × 10−4 N.s/m2 1.44 m = ρV . Then dm = ρd V + V dρ. Assume mass to be constant in a volume

subjected to a pressure increase; then dm = 0. ∴ρd V = − V dρ, or d VV

.d ρρ

= −

1.45 B =V− p

V∆

∆2200 MPa.= V∴∆ V−= 2 10

2200p

B∆ − ×= = −0.00909 m3 or −9090 cm3

1.46 Use c = 1450 m/s. L = c∆t = 1450 × 0.62 = 899 m

1.47 =B V

p∆∆ −V

= −2100 −1320

. = 136.5 MPa

1.48 a) 327,000 144/1.93c = × = 4670 fps b) 327,000 144/1.93c = × = 4940 fps

c) 308,000 144/1.87c = × = 4870 fps 1.49 V∆ =3.8 × 10−4 × −20 × 1 = .0076 m3.

∆p = −B V∆

V.0076

22701

−= − = 17.25 MPa

1.50 p = 2 2 0741

5 10 6

σR

= ×× −

. = 2.96 × 104 Pa or 29.6 kPa. Bubbles: p = 4σ/R = 59.3 kPa

7

1.51 Use Table B.1: σ = 0.00504 lb/ft. ∴p = 4 4 .00504

1/32 12Rσ ×=

× = 7.74 psf or 0.0538 psi

1.52 See Example 1.4: h = 4 cos 4 0.0736 0.866

0.130 m.1000 9.81 0.0002gD

σ βρ

× ×= =× ×

1.53 (D) 6

4 cos 4 0.0736 13 m or 300 cm.

1000 9.81 10 10h

gDσ βρ −

× ×= = =× × ×

1.54 See Example 1.4: h = 4 cos 4 0.032cos130

1.94 13.6 32.2 0.8/12gDσ βρ

×=× × ×

o

= −0.00145 ft or −0.0174 in

1.55 force up = σ × L × 2 cosβ = force down = ρghtL. ∴h = 2σ β

ρcos

.gt

1.56 Draw a free-body diagram: The force must balance:

W = 2σL or π ρ σd

L g L2

42

= .

∴ =dg

8σπρ

W

σL σL

needle

1.57 From the free-body diagram in No. 1.47, a force balance yields:

Is π ρd

g2

4< 2σ?

π(. ). .

0044

7850 9 81 2 07412

× < ×

0.968 < 0.1482 ∴No 1.58 Each surface tension force = σ × π D. There is a force on the outside and one on the inside of the ring. ∴F = 2σπD neglecting the weight of the ring.

F

D

1.59

h(x)h

dW

σdl

From the infinitesimal free-body shown:

cos .d gh xdxσ θ ρ α=l cosθ = dxdl

.

/d d x d

hg xdx g x

σ σρ α ρ α

∴ = =l l

We assumed small α so that the element thickness is αx.

8

1.60 The absolute pressure is p = −80 + 92 = 12 kPa. At 50°C water has a vapor

pressure of 12.2 kPa; so T = 50°C is a maximum temperature. The water would “boil” above this temperature.

1.61 The engineer knew that water boils near the vapor pressure. At 82°C the vapor

pressure from Table B.1 is 50.8 (by interpolation). From Table B.3, the elevation that has a pressure of 50.8 kPa is interpolated to be 5500 m.

1.62 At 40°C the vapor pressure from Table B.1 is 7.4 kPa. This would be the

minimum pressure that could be obtained since the water would vaporize below this pressure.

1.63 The absolute pressure is 14.5 − 11.5 = 3.0 psia. If bubbles were observed to form

at 3.0 psia (this is boiling), the temperature from Table B.1 is interpolated, using vapor pressure, to be 141°F.

1.64 The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming

atmospheric pressure to be 100 kPa, we have 10 000 + 100 = 600 x. ∴x = 16.83 km. 1.65 (C)

1.66 ρ= =× +

=pRT

10130 287 273 15

.. ( )

1.226 kg/m3. γ = 1.226 × 9.81 = 12.03 N/m3

1.67 3in

101.31.226 kg/m .

0.287 (15 273)p

RTρ = = =

× + 3

out85

1.19 kg/m .0.287 248

ρ = =×

Yes. The heavier air outside enters at the bottom and the lighter air inside exits at the top. A circulation is set up and the air moves from the outside in and the inside out: infiltration. This is the “chimney” effect.

1.68 3750 440.1339 slug/ft .

1716 470p

RTρ ×= = =

× m Vρ= 0.1339 15 2.01 slug.= × =

1.69 (C) p V

m = 800 459.95 kg

0.1886 (10 273)RT×= =

× +

1.70 p

W VRT

= 100(10 20 4) 9.81 9333 N.

0.287 293g = × × × × =

×

9

1.71 Assume that the steel belts and tire rigidity result in a constant volume so that m1 = m2:

V 1 V= 1 1 2 22

1 2

22 1

1

or .

150 460(35 14.7) 67.4 psia or 52.7 psi gage.

10 460

m RT m RTp p

Tp p

T

=

+∴ = = + =− +

1.72 The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have

100000 1 9.81. 10200 kg.m m× = × ∴ = Hence,

p V

m =3100 4 / 3

10200. 12.6 m or 25.2 m.0.287 288

rr d

RTπ×= = ∴ = =×

1.73 2 210 ( 10). 20 32.2. 25.4 fps.

2KE PE mV mg V V= ∆ + ∆ = + − ∴ = × ∴ =

2 210 ( 20). 40 32.2. 35.9 fps.

2mV mg V V= + − ∴ = × ∴ =

1.74 2 21-2

1. a) 200 0 5( 10 ). 19.15 m/s.

2 f fW KE V V= ∆ × = × − ∴ =

b) 10

2 2

0

120 15( 10 ).

2 fsds V= × −∫

2

2 210 120 15( 10 ). 15.27 m/s.

2 2 f fV V× = × − ∴ =

c) 10

2 2

0

1200cos 15( 10 ).

20 2 fs

ds Vπ = × −∫

2 220 1200sin 15( 10 ). 16.42 m/s.

2 2 f fV Vπ

π× = × − ∴ =

1.75 21 2 1 2 2 1

1. 10 40 0.2 0 . 40000.

2E E u u u u= × × + = + ∴ − =% % % %

40000

. 55.8 C where comes from Table B.4.717v vu c T T c∆ = ∆ ∴∆ = = o%

The following shows that the units check:

2 2 2 2 2

car2 2 2

air

kg m / s m kg C m kg CC

kg J/(kg C) N m s (kg m/s ) m s

m Vm c

× ⋅ ⋅ ⋅ ⋅ ⋅= = = = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

o oo

o

where we used N = kg.m/s2 from Newton’s 2nd law.

10

1.76 2

22 1 H O

1. .

2E E mV m c T= = ∆

2

61 100 10001500 1000 2000 10 4180 . 69.2 C.

2 3600T T−× × × = × × × ∆ ∴∆ =

o

We used c = 4180 J/kg. Co from Table B.5. (See Problem 1.75 for a units check.)

1.77 water . 0.2 40000 100 4.18 . 19.1 C.f fm h m c T T T= ∆ × = × ∆ ∴∆ = o

The specific heat c was found in Table B.5. Note: We used kJ on the left and kJ on the right.

1.78. (B) ice water ice water water. 320 .E E m m c T∆ = ∆ × = × ∆

6 35 (40 10 ) 1000 320 (2 10 ) 1000 4.18 . 7.66 C.T T− −× × × × = × × × ∆ ∴∆ = o We assumed the density of ice to be equal to that of water, namely 1000 kg/m3. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem.

1.79. W p d V= mRTV

=∫ d Vd V

mRT=∫V

lnV

mRT=∫2

V2

1 1ln

pmRT

p=

since, for the T = const process, 1p V 1 2p V= 2. Finally,

1-24 1

1716 530ln 78,310 ft-lb.32.2 2

W = × × = −

The 1st law states that 0. 78,310 ft-lb or 101 Btu.vQ W u mc T Q W− = ∆ = ∆ = ∴ = = − −%

1.80 If the volume is fixed the reversible work is zero since the boundary does not

move. Also, since V 1 2

1 2,

mRT T Tp p p

= = so the temperature doubles if the

pressure doubles. Hence, using Table B.4 and Eq. 1.7.17, 200 2

a) (1.004 0.287)(2 293 293) 999 kJ0.287 293vQ mc T

×= ∆ = − × − =×

b) 200 2

(1.004 0.287)(2 373 373) 999 kJ0.287 373vQ mc T

×= ∆ = − × − =×

c) 200 2

(1.004 0.287)(2 473 473) 999 kJ0.287 473vQ mc T

×= ∆ = − × − =×

1.81 W p d V= (p V= 2 V− 11). If = const,

Tp

V2

1

TV

= 2 12

so if 2 ,T T=∫

then V 2 2V= 1 and (2W p V= 1 V− 1) p V= 1 1.mRT= a) 2 0.287 333 191 kJW = × × = b) 2 0.287 423 243 kJW = × × =

11

c) 2 0.287 473 272 kJW = × × = 1.82 = 1.4 287 318 357 m/s. 357 8.32 2970 m.c kRT L c t= × × = = ∆ = × =

1/ 0.4/1.4

22 1

1

500(20 273) 151.8 K or 121.2 C

5000

k kp

T Tp

− = = + = − o

1.83 We assume an isentropic process for the maximum pressure:

/ 1 1.4/0.42

2 11

423(150 100) 904 kPa abs or 804 kPa gage.

293

k kT

p pT

− = = + =

Note: We assumed patm = 100 kPa since it was not given. Also, a measured pressure is a gage pressure.

1.84 / 1 1.4/0.4

22 1

1

473100 534 kPa abs.

293

k kT

p pT

− = = =

2 1( ) (1.004 0.287)(473 293) 129 kJ/kg.vw u c T T= − ∆ = − − = − − − = −

We used Eq. 1.7.17 for cv. 1.85 a) 1.4 287 293 343.1 m/sc kRT= = × × =

b) 1.4 188.9 293 266.9 m/sc kRT= = × × =

c) 1.4 296.8 293 348.9 m/sc kRT= = × × =

d) 1.4 4124 293 1301 m/sc kRT= = × × =

e) 1.4 461.5 293 424.1 m/sc kRT= = × × = Note: We must use the units on R to be J/kg.K in the above equations.

1.86 (D) For this high-frequency wave, 287 323 304 m/s.c RT= = × = 1.87 At 10 000 m the speed of sound 1.4 287 223 299 m/s.c kRT= = × × =

At sea level, 1.4 287 288 340 m/s.c kRT= = × × =

340 299

% decrease 100 12.06 %.340−= × =

1.88 a) = 1.4 287 253 319 m/s. 319 8.32 2654 m.c kRT L c t= × × = = ∆ = × =

b) = 1.4 287 293 343 m/s. 343 8.32 2854 m.c kRT L c t= × × = = ∆ = × =

c) = 1.4 287 318 357 m/s. 357 8.32 2970 m.c kRT L c t= × × = = ∆ = × =

12

C HAPTER 2

Fluid Statics 2.1 Σ ∆ ∆

∆ ∆F ma p z p s

y zay y y y= − =: sinα ρ

2

Σ ∆ ∆∆ ∆ ∆ ∆

F ma p y p sy z

a gy z

z z z z= − = +: cosα ρ ρ2 2

Since ∆ ∆s ycosα = and ∆ ∆s z sin α = , we have

p p

yay y− = ρ

∆2

and ( )p pz

a gz z− = +ρ∆2

Let ∆y → 0 and ∆z → 0:

p pp p

y

z

− =− =

00

∴ = =p p py z .

2.2 p = γh. a) 9810 × 10 = 98 100 Pa or 98.1 kPa b) (0.8 × 9810) × 10 = 78 480 Pa or 78.5 kPa c) (13.6 × 9810) × 10 = 1 334 000 Pa or 1334 kPa d) (1.59 × 9810) × 10 = 155 980 Pa or 156.0 kPa e) (0.68 × 9810) × 10 = 66 710 Pa or 66.7 kPa 2.3 h = p/γ. a) h = 250 000/9810 = 25.5 m b) h = 250 000/(0.8 × 9810) = 31.9 m c) h = 250 000/(13.6 × 9810) = 1.874 m d) h = 250 000/(1.59 × 9810) = 16.0 m e) h = 250 000/(0.68 × 9810) = 37.5 m 2.4 (C) (13.6 9810) (28.5 0.0254) 96600 PaHgp hγ= = × × × =

2.5 S = 20 14462.4 20

phγ

×=

× = 2.31. ρ = 1.94 × 2.31 = 4.48 slug/ft3.

2.6 a) p = γh = 0.76 × (13.6 × 9810) = 9810 h. ∴h = 10.34 m. b) (13.6 × 9810) × 0.75 = 9810 h. ∴h = 10.2 m. c) (13.6 × 9810) × 0.01 = 9810 h. ∴h = 0.136 m or 13.6 cm. 2.7 a) p = γ1h1 + γ2h2 = 9810 × 0.2 + (13.6 × 9810) × 0.02 = 4630 Pa or 4.63 kPa. b) 9810 × 0.052 + 15 630 × 0.026 = 916 Pa or 0.916 kPa. c) 9016 × 3 + 9810 × 2 + (13.6 × 9810) × 0.1 = 60 010 Pa or 60.0 kPa.

y

z

pz∆y

py∆zp∆s

∆s

α

∆z

∆y

ρg∆V

13

2.8 ∆p = ρgh = 0.0024 × 32.2 (–10,000) = –773 psf or –5.37 psi. 2.9 (D) 0 84000 1.00 9.81 4000 44760 Pap p ghρ= − = − × × = 2.10

inside

100 9.813 13.51 Pa

.287 253100 9.81

3 11.67 Pa.287 293

outside oo

base

ii

pgp g h h

RTp

pgp g h h

RT

ρ

ρ

× ∆ = ∆ = ∆ = × = × ∴ ∆× ∆ = ∆ = ∆ = × =×

= 1.84 Pa

If no wind is present this ∆pbase would produce a small infiltration since the higher pressure outside would force outside air into the bottom region (through cracks). 2.11 p = ρgdh where h = –z. From the given information S = 1.0 + h/100 since S(0) = 1 and

S(10) = 1.1. By definition ρ = 1000 S, where ρwater = 1000 kg/m3. Then dp = 1000 (1 + h/100) gdh. Integrate:

dp h gdhp

= +∫∫ 1000 1 1000

10

0

( / )

p = × +×

1000 9 81 1010

2 100

2

. ( ) = 103 000 Pa or 103 kPa

Note: we could have used an average S: Savg = 1.05, so that ρavg = 1050 kg/m3.

2.12 v∇ = + +p

px

ipy

jpz

k∂∂

∂∂

∂∂

$ $ $

= – ρa ix$ – ρα y j$ – ρα z k$ – ρgk$ = – ( )ρ a i a j a kx y z

$ $ $+ + – ρgk$

= – ρva – ρvg

∴ ∇ = − +v v vp a gρ( )

2.13 /

0 0[( ) / ]g Ratmp p T z T αα= −

= 100 [(288 − 0.0065 × 300)/288]9.81/.0065 × 287 = 96.49 kPa

p p ghatm= − = −×

× ×ρ 100100

287 2889 81 300 1000

.. / = 96.44 kPa

% error = 96 96

96100

.44 .49.49−

× = −0.052%

The density variation can be ignored over heights of 300 m or less.

14

2.14 /

00

0

g R

atm atmT z

p p p p pT

αα −

∆ = − = −

= 100 288 0065 20

2881

9 81 0065 287− ×

−

×. . /.

= −0.237 Pa or −0.000237 kPa

This change is very small and can most often be ignored.

2.15 Eq. 1.5.11 gives 310,000 144 .dpd

ρρ

× = But, dp = ρgdh. Therefore,

74.464 10

gdh dρ ρρ×

= or 2 732.2

4.464 10

ddh

ρ

ρ=

×

Integrate, using 0ρ = 2.00 slug/ft3:

2 72 0

32.2

4.464 10

hddh

ρ ρ

ρ=∫ ∫

×. ∴

1 12ρ

− −

= 7.21 × 10-7 h or

72

1 14.42 10 hρ

−=

− ×

Now,

77 7

0 0

2 2ln(1 14.42 10 )

1 14.42 10 14.42 10

h h g gp gdh dh h

hρ −

− −= = = − ×∫ ∫

− × − ×

Assume ρ = const: 2.0 32.2 64.4p gh h hρ= = × × =

a) For h = 1500 ft: paccurate = 96,700 psf and pestimate = 96,600 psf. 96,600 96,700

% error 100 0.103 %96,700

−= × = −

b) For h = 5000 ft: paccurate = 323,200 psf and pestimate = 322,000 psf. 322,000 323,200

% error 100 0.371 %323,200

−= × = −

c) For h = 15,000 ft: paccurate = 976,600 psf and pestimate = 966,000 psf. 966,000 976,600

% error 100 1.085 %976,600

−= × = −

2.16 Use the result of Example 2.2: p = 101 e−gz/RT.

a) p = 101 e−9.81 ×10 000/287 ×273 = 28.9 kPa. b) p = 101 e−9.81 ×10 000/287 ×288 = 30.8 kPa. c) p = 101 e−9.81 ×10 000/287 ×258 = 26.9 kPa.

2.17 Use Eq. 2.4.8: p = 9.81

0.0065 287101(1 0.0065 /288) .z ×−

a) z = 3000. ∴p = 69.9 kPa. b) z = 6000. ∴p = 47.0 kPa. c) z = 9000. ∴p = 30.6 kPa. d) z = 11 000. ∴p = 22.5 kPa.

15

2.18 Use the result of Example 2.2: /

0= .gz RTp

ep

−

0

ln .p gz

p RT= −

0.001 32.2ln .

14.7 1716 455z

= −×

∴z = 232,700 ft.

2.19 p = γh = (13.6 × 9810) × 0.25 = 33 350 Pa or 33.35 kPa. 2.20 a) p = γh. 450 000 = (13.6 × 9810) h. ∴h = 3.373 m b) p + 11.78 × 1.5 = (13.6 × 9810) h. Use p = 450 000, then h = 3.373 m % error is 0.000 %. 2.21 Referring to Fig. 2.6a, the pressure in the pipe is p = ρgh. If p = 2400 Pa, then 2400 = ρgh = ρ × 9.81 h.

a) ρ =×

2400981 36. .

= 680 kg/m3. ∴gasoline

b) ρ =×

2400981 272. .

= 899 kg/m3. ∴benzene

c) ρ =×

2400981 245. .

= 999 kg/m3. ∴water

d) ρ =×

24009 81 154. .

= 1589 kg/m3. ∴carbon tetrachloride

2.22 Referring to Fig. 2.6a, the pressure is p = ρwgh = 21.

2 aVρ Then Vghw

a

2 2=

ρρ

.

a) V 2 2 1000 9 81 06123

=× × ×. .

. = 957. ∴V = 30.9 m/s

b) V 2 2 194 32 2 3 1200238

=× × ×. . /

. = 13,124. ∴V = 115 ft/sec

c) V 2 2 1000 9 81 1123

=× × ×. .

. = 1595. ∴V = 39.9 m/s

d) V 2 2 194 32 2 5 1200238

=× × ×. . /

. = 21,870. ∴V = 148 ft/sec

2.23 (C) 0 30000 0.3 9810 0.1 8020 Paw atm x x water wp p h hγ γ= + − = + × − × = 2.24 See Fig. 2.6b: p1 = –γ1h + γ2H.

p1 = –0.86 × 62.4 × 5

12 + 13.6 × 62.4 ×

9 512.

= 649.5 psf or 4.51 psi.

2.25 0 1 1 2 2 3 3 4 4p p gh gh gh ghρ ρ ρ ρ= + + + + = 3200 + 917×9.81×0.2 + 1000×9.81×0.1 + 1258×9.81×0.15 + 1593×9.81×0.18 = 10 640 Pa or 10.64 kPa

16

2.26 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (Use ∆ ∆p g h= ρ ) 40 000 – 16 000 = 1000×9.81(–.2) + 13 600×9.81×H + 920×9.81×.3.

∴H = .1743 m or 17.43 cm 2.27 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (Use ∆ ∆p g h= ρ ) po – pw = 900×9.81(–.2) + 13 600×9.81(–.1) + 1000×9.81×.15

= –12 300Pa or –12.3 kPa 2.28 ( ) ( ) ( ) ( )p p p p p p p p p p1 5 1 2 2 3 3 4 4 5− = − + − + − + −

p1 = 9810(–.02) + 13 600×9.81(.–04) + 9810(–.02)+13 600×9.81×.16 = 15 620 Pa or 15.62 kPa

2.29 pw + 9810 × .15 – 13.6 × 9810 × .1 – .68 × 9810 × .2 + .86 × 9810 × .15 = po. ∴pw – po = 11 940 Pa or 11.94 kPa. 2.30 pw – 9810 × .12 – .68 × 9810 × .1 + .86 × 9810 × .1 = po. With pw = 15 000, po = 14 000 Pa or 14.0 kPa. 2.31 a) p + 9810 × 2 = 13.6 × 9810 × .1. ∴p = –6278 Pa or –6.28 kPa. b) p + 9810 × .8 = 13.6 × 9810 × .2. ∴p = 18 835 Pa or 18.84 kPa. c) p + 62.4 × 6 = 13.6 × 62.4 × 4/12. ∴p = –91.5 psf or –0.635 psi. d) p + 62.4 × 2 = 13.6 × 62.4 × 8/12. ∴p = 441 psf or 3.06 psi. 2.32 p – 9810 × 4 + 13.6 × 9810 × .16 = 0. ∴p = 17 890 Pa or 17.89 kPa. 2.33 (A) (13.6 9810) 0.16 21350 Pa.ap Hγ= − = − × × = − , 21350 10000 11350 13.6 9810 . 0.0851 maafter after afterp H H= − + = − = × ∴ = 2.34 8200 + 9810 × .25 = 1.59 × 9810 × H. ∴H = 0.683 m

Hnew = .683 + .273 = .956 m. ∆H = .273

2 = .1365.

p + 9810 (.25 + .1365) = 1.59 × 9810 × .956. ∴p = 11 120 Pa or 11.12 kPa. 2.35 p + 9810 × .05 + 1.59 × 9810 × .07 – .8 × 9810 × .1 = 13.6 × 9810 × .05. ∴p = 5873 Pa or 5.87 kPa. Note: In our solutions we usually retain 3 significant digits in the answers (if a number

starts with “1” then 4 digits are retained. In most problems a material property is used, i.e., S = 1.59. This is only 3 sig. digits! ∴ only 3 are usually retained in the answer!

H

∆H

∆H

17

2.36 Before pressure is applied the air column on the right is 48" high. After pressure is applied, it is (4 – H/2) ft high. For an isothermal process 1p V 1 2p V= 2 using absolute pressures. Thus,

14.7 × 144 × 4A = p2(4 – H / 2 )A or p2 = 8467

4 2− H /.

From a pressure balance on the manometer (pressures in psf):

30 × 144 + 14.7 × 144 = 13.6 × 62.4 H + 8467

4 2− H /,

or H2 – 15.59 H + 40.73 = 0. ∴H = 12.27 or 3.32 ft. 2.37 a) ( ) ( ) ( ) ( )p p p p p p p p p p1 5 1 2 2 3 3 4 4 5− = − + − + − + −

4000 = 9800(0.16–0.22) + 15 600(0.10–0.16) + 133 400H + 15 600(0.07–H). ∴H = .0376 m or 3.76 cm

b) 0.6×144 = 62.4(–2/12) + 99.5(–2/12) + 849H + 99.5(2.5/12 – H). ∴H = .1236 ft or 1.483 in.

2.38 a) 2 2

2 21 1 2 3 2

2 /

2 2( ) /

H D dp D dγ γ γ γ

∆=

∆ − + + −

2

22(.1/.005)

9800 2 15 600 2(133 400 15 600)(.1/.005)=

− + × + − = × −8 10 6.487 H

∴ = × ×−∆H 8 10 4006.487 = 0.0034 m or 3.4 mm

b) ∆H =− + × + −

×2 4 2

62 2 99 5 2 99 5 4 206 144

2

2

( /. ).4 . (849 . )( /. )

. = 0.01153 ft or 0.138 in.

2.39 ( ) ( ) ( )p p p p p p p p1 4 1 2 2 3 3 4− = − + − + − (poil = 14.0 kPa from No. 2.30) 15 500 – 14 000 = 9800(0.12 + ∆z) + 680(0.1 – 2∆z) + 860(–0.1 – ∆z).

∴∆z = 0.0451 m or 4.51 cm 2.40 a) pair = –6250 + 625 = –5620 Pa. –5620 + 9800(2 + ∆z) – 13 600 × 9.81(0.1 + 2∆z) = 0. ∴∆z = 0.0025. ∴h = 0.1 + 2∆z = .15 m or 15 cm b) pair = 18 800 + 1880 = 20 680 Pa. 20 680 + 9800(0.8 + ∆z) – 13 600 × 9.81(0.2 + 2∆z) = 0. ∴∆z = 0.00715 m ∴h = .2+ 2∆z = .214 or 21.4 cm c) pair = –91.5 + 9.15 = –82.4 psf. –82.4 + 62.4(6 + ∆z) – 13.6 × 62.4(4/12 + 2∆z) = 0. ∴∆z = 0.00558 ft. ∴h = 4/12 + 2 (0.00558) = 0.3445 ft or 4.13 in. d) pair = 441 + 44.1 = 485 psf 485 + 62.4(2 + ∆z) – 13.6 × 62.4(8/12 + 2∆z) = 0. ∴∆z = 0.0267 ft. ∴h = 8/12 + 2 (0.0267) = 0.7205 ft or 8.65 in.

18

2.41 F hA= γ = 9810 × 10 × π × .32/4 = 6934 N.

2.42 5 1 5 5

(2 ) (2 ) [9800 1 3 (2 )]. 32670 N3 3 3 3

P P× × = × × × × × × × ∴ = a) F = pc A = 9800 × 2

× 42 = 313 600 N or 313.6 kN

b) 2 2

9800 1 (2 4) 9800 2 9800 1 98000 N or 98.0 kN3 3cF p A= = × × × + × × + × × =

c) F = pc A = 9800 × 1 × 2 × 4 × 2 = 110 900 N or 110.9 kN d) F = pc A = 9800 × 1 × 2 × 4/.866 = 90 500 N or 90.5 kN 2.43 For saturated ground, the force on the bottom tending to lift the vault is F = pc A = 9800 × 1.5 × (2 × 1) = 29 400 N. The weight of the vault is approximately W gVρ= 2400 9.81walls = × [2(2×1.5×.1) + 2(2×1×.1) + 2(.8×1.3×.1)] = 28 400 N. The vault will tend to rise out of the ground. 2.44 F = pc A = 6660 × 2 × π × 22 = 167 400 N or 167.4 kN Find γ in Table B.5 in the Appendix. 2.45 a) F = pc A = 9800 (10 − 2.828/3) (2.828 × 2/2) = 251 000 N or 251 kN where the height of the triangle is (32 − 12)1/2 = 2.828 m. b) F = pc A = 9800 × 10 (2.828 × 2/2) = 277 100 N or 277.1 kN c) F = pc A = 9800 (10 − 2.828 × .866/3) (2.828 × 2/2) = 254 500 N or 254.5 kN 2.46 a) F hA= = × × =γ 62 27 33 24.4 . 40,930 lb.

y p = +×

×27 33

6 8 3627 33 24

3

./

. = 27.46'. ∴y = 30 – 27.46 = 2.54'.

8/5.46 = 3/x. ∴x = 2.05’. (2.05, 2.54) ft.

b) F = 62.4 × 30 × 24 = 44,930 lb. The centroid is the center of pressure. y = 2.667'. 8/5.333 = 3/x. ∴x = 2.000' (2.000, 2.667) ft. c) F = 62.4 (30 – 2.667 × .707) × 24 = 42,100 lb.

y p = +×

×39 77

6 8 3639 77 24

3

./

. = 39.86'. y = 42.43 – 39.86 = 2.57'

8/5.43 = 3/x. ∴x = 2.04'. (2.04, 2.57) ft. 2.47 (B) The force acts 1/3 the distance from the hinge to the water line:

5 1 5 5(2 ) (2 ) [9800 1 3 (2 )]. 32670 N

3 3 3 3P P× × = × × × × × × × ∴ =

(x, y)

y

x

19

2.48 a) F hA= = × ×γ π9810 6 2 2 = 739 700 N or 739.7 kN.

y yI

Ayp = + = +×

×6

2 44 6

4ππ

/ = 6.167 m. ∴(x, y)p = (0, –0.167) m

b) F hA= = × ×γ π9810 6 2 = 369 800 N or 369.8 kN.

y p = +×

×6

2 82 6

4ππ

/ = 6.167 m. x2 + y2 = 4

x Fx

pdA x y xdy y y dyp = = − = − −∫ ∫∫−−2 2

62

4 62

2

2

2

2γ γ( ) ( )( ) .

∴ × = − − + =−∫x y y y dypγ π

γγ6 2

224 4 6 322

2

23( ) . ∴xp = 0.8488 m

∴(x, y)p = (0.8488, –0.167) m

c) F = 9810 × (4 + 4/3) × 6 = 313 900 N or 313.9 kN.

y p = +×

×5333

3 4 365 333 6

3

./

. = 5.500 m. ∴y = –1.5

4/2.5 = 1.5/x. ∴x = 0.9375. ∴(x,y)p = (0.9375, –1.5) m

d) F = × + ×9810 423

4( sin 36.9°) × 6 = 330 000 N

y p = +×

×5 6

5 2 366 5 6

3

..4 /

. = 5.657 m. ∴y = 0.343 m

3 cos 53.13° = 1.8, 2.5 – 1.8 = 0.7, 2.4/2.057 = . /7 1x . ∴ x1 = 0.6. x = 1.8 + 0.6 = 2.4. ∴(x,y)p = (2.4, 0.343) m. 2.49 F hA= = × × ×γ 62 11 6 10.4 ( ) = 41,180 lb.

y yI

yAp = + = +×

×11

6 10 1211 60

3 / = 11.758'.

(16 – 11.758) 41,180 = 10P. ∴P = 17,470 lb. 2.50 F hA= = × ×γ 9810 6 20 = 1.777 × 106 N, or 1177 kN.

y yI

Ayp = + = +×

×7 5

4 5 127 5 20

3

./

. = 7.778 m.

(10 – 7.778) 1177 = 5 P. ∴P = 523 kN.

y

x(x, y)

dAdy

xy

3 4

53.13o

yp

P

F

20

2.51 F hA= = × ×γ 9810 12 20 = 2.354 × 106 N, or 2354 kN.

y yI

Ayp = + = +×

×15

4 5 1215 20

3 / = 15.139 m.

(17.5 – 15.139) 2354 = 5 P. ∴P = 1112 kN.

2.52 y yI

AyH bH

bH HH H

Hp = + = +×

= + =2

122 2 6

23

3 //

. y p is measured from the surface.

∴From the bottom, H y H H Hp− = − =23

13

.

Note: this result is independent of the angle α, so it is true for a vertical area or a sloped area.

2.53 31sin40 3 . ( 2) sin40 . 2( 2) .

2 3l

F l l F l P l l Pγ γ= × × = + ∴ = +o o

a) 9810× 23 = 2(2 + 2)P. ∴P = 9810 N b) 9810× 43 = 2(4 + 2)P. ∴P = 52 300 N c) 9810× 53 = 2(5 + 2)P. ∴P = 87 600 N

2.54 h = −1 2 2 2. .4 = 1.1314 m. A = 1.2 × 1.1314 + .4 × 1.1314 = 1.8102 m2 Use 2 forces: F h Ac1 1 9800 5657 1 2 11314= = × × ×γ . ( . . ) = 7527 N

F h Ac2 2 980011314

311314= = × × ×γ

.(.4 . ) = 1673 N

y p123

11314= ( . ). y yI

A yp22

2

3113143

11314 3611314 2 11314 3

= + = +×

× ×. .4 . /

.4 . / . / = 0.5657 m

ΣMhinge = 0: 752711314

31673 11314 0 5657 11314× + × − −

.( . . ) . P = 0. ∴P = 3346 N.

2.55 To open, the resultant force must be just above the hinge, i.e., yp must be just less than h. Let yp = h, the condition when the gate is about to open: y h H A h H I h H h H= + = + = + +( ) / , ( ) , [ ( )]( ) /3 2 362 3

∴ =+

++

+ +=

++

+=

+y

h H h Hh H h H

h H h H h Hp 3

2 363 3 6 2

4

2

( ) /( ) ( ) /

a) hh H

=+2

. ∴h = H = 0.9 m

b) h = H = 1.2 m c) h = H = 1.5 m 2.56 The gate is about to open when the center of pressure is at the hinge.

a) y H Hb

H bp = + = + +×

+1 2 1 8 2

1 8 129 1 8

3

. ( . / ). /

(. ) .. ∴H = 0.

21

b) y H Hb

H bp = + = + +×+

1 2 2 0 22 12

1 2

3

. ( . / )/

( ). ∴H = 0.6667 m.

c) y H Hb

H bp = + = + +×

+1 2 2 2 2

2 2 1211 2 2

3

. ( . / ). /

( . ) .. ∴H = 2.933 m.

2.57 (A) The gate opens when the center of pressure in at the hinge:

31.2 11.2 (1.2 ) /12

5. 5 1.2.2 2 (1.2 ) (11.2 ) / 2p

h I h b hy y y

Ay h b h+ + +

= + = + = + = ++ +

This can be solved by trial-and –error, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields h = 1.08 m.

2.58 FH

bH bH12

212

= × =γ γ

F H b b H2 = × =γ γl l

12 3 2

2γ γbHH

b H× = ×ll

. ∴ =H 3l

a) H = ×3 2 = 3.464 m b) H = 1.732 m c) H = 10.39' d) H = 5.196'

Assume 1 m deep 2.59 The dam will topple if the moment about “O” of F1 and F3 exceeds the restoring moment of W and F2. a) W = × × + ×( .4 )( / )2 9810 6 50 24 50 2 = 21.19 × 106 N

dw =× + ×

+300 27 600 16

300 600 = 19.67 m.

F2 = 9810 × 5 × 11.09 = 0.544 × 106 N. d211 09

3=

. = 3.697 m.

F1 9810452

45= × × = 9.933 × 106 N. d1 = 15 m.

F3 981045 10

230= ×

+× = 8.093 × 106 N. d3

2 943 15 5150 202 943 5150

=× + ×

+. .

. . = 18.18 m.

Wd F dF d F d

w + = × ⋅+ = × ⋅

2 26

1 1 3 36

418 8 102961 10

.

.N mN m

∴will not topple.

b) W = (2.4 × 9810) (6 × 65 + 65 × 12) = 27.55 × 106 N.

dw = 390 27 780 16

390 780× + ×

+ = 19.67 m.

F260 54 10≅ ×. N. d2 3 70≅ . m.

F1 = 9810 × 30 × 60 = 17.66 × 106 N. d1 = 20 m.

F3 981060 10

230= ×

+× = 10.3 × 106 N. d3

2 943 15 7358 202 943 7 358

=× + ×

+. .

. . = 18.57 m.

F2

F1

H/3

l/2

F1 F2

F3

WO

22

Wd F dF d F d

w + = × ⋅+ = × ⋅

2 26

1 1 3 36

543 9 10544 5 10

..

N mN m

∴it will topple.

c) Since it will topple for H = 60, it certainly will topple if H = 75 m. assume 1 m deep 2.60 The dam will topple if there is a net clockwise moment about “O.”

a) W W W W= + = × × × ×1 2 1 6 43 1 62 2. ( ) .4 .4 = 38,640 lb. W2 24 43 2 62 2= × × ×( / ) .4 .4 = 77,280 lb.

W3 40 22 33 2 62= × ×( . / ) .4 = 27,870 lb @ 20.89 ft. F1 62 20 40 1= × × ×.4 ( ) = 49,920 lb @ 40/3 ft. F2 62 5 10 1= × × ×.4 ( ) = 3120 lb @ 3.33 ft

1

32

= 18,720 lb @ 15 ft

= 28,080 lb @ 20 ftp

p

FF

F

=

OMΣ : (49,920)(40/3) + (18,720)(15) + (28,080)(20) − (38,640)(3) − (77,280)(14) − (27,870)(20.89) − (3120)(3.33) < 0. ∴won’t tip. b) W1 = 6 × 63 × 62.4 × 2.4 = 56,610 lb. W2 = (24 × 63/2) × 62.4 × 2.4 = 113,220 lb. F1 62 30 60= × ×.4 = 112,300 lb. 3 (60 22.86/2) 62.4W = × × = 42,790 lb. F2 62 5 10= × ×.4 = 3120 lb Fp1 62 10 30= × ×.4 = 18,720 lb. Fp2 62 50 30 2= × ×.4 / = 46,800 lb.

OMΣ : (112,300)(20) + (18,720)(15) + (46,800)(20) − (56,610)(3) − (113,220)(14) − 42,790(21.24) = 799,000 > 0. ∴will tip.

c) Since it will topple for H = 60 ft., it will also topple for H = 80 ft. 2.61 ΣMhinge = 0. 2.5P – dw × W – d1 × F1 = 0.

∴ P = × × × +×

× ××

×

12 5

23

9800 1 84 23

98002

44

2

. ππ

= 62 700 N

Note: This calculation is simpler than that of Example 2.7. Actually, We could have moved the horizontal force FH and a vertical force FV (equal to W) simultaneously to the center of the circle and then 2.5P = 2FH.=2F1. This was outlined at the end of Example 2.7.

2.62 Since all infinitesimal pressure forces pass thru the center, we can place the resultant forces at the center. Since the vertical components pass thru the bottom point, they produce no moment about that point. Hence, consider only horizontal forces:

( ) 9810 2 (4 10) 784 800N( ) 0.86 9810 1 20 168 700N

waterH

H oil

FF

= × × × == × × × =

ΣM P: . . . 2 784 8 2 168 7 2= × − × ∴P = 616.1 kN.

F1 F2

F3

W

O

W3

F1

dw

d1

PW

23

2.63 Place the resultant forcev vF FH V+ at the center of the circular arc.

vFH passes thru the

hinge showing that P FV= . a) P FV= = × × + × =9810 6 2 4 4 594( )π 200 N or 594.2 kN.

b) P = FV = 62.4 (20 × 6 × 12 + 9π × 12) = 111,000 lb. 2.64 (D) Place the force

v vF FH V+ at the center of the circular arc. FH passes through the

hinge: 24 1.2 9800 ( 1.2 /4) 9800 300000. 5.16 m.VP F w w wπ∴ = = × × + × × = ∴ =

2.65 Place the resultant

v vF FH V+ at the circular arc center.

vFH passes thru the hinge so that

P FV= . Use the water that could be contained above the gate; it produces the same pressure distribution and hence the same FV .

P FV= = 9810 (6 × 3 × 4 + 9π) = 983 700 N or 983.7 kN. 2.66 Place the resultant

v vF FH V+ at the center.

vFV passes thru the hinge

2 × (9810 × 1 × 10) = 2.8 P. ∴P = 70 070 N or 70.07 kN. 2.67 The incremental pressure forces on the circular quarter arc pass through the hinge so that no moment is produced by such forces. Moments about the hinge gives: 3 P = 0.9 W = 0.9 × 400. ∴P = 120 N. 2.68 The resultant

v vF FH V+ of the unknown liquid acts thru the center of the circular arc.

vFV

passes thru the hinge. Thus we use only ( ) .FH oil Assume 1 m wide.

a) ΣMR R

RR

SR

RR

Rx: . 3

98102

43

98004 2

2

+

=

π

πγ ∴ γ x = 4580 N/m3

b) ΣMR R

RR

SR

RR

Rx: . . . 3

62 42

43

62 44 2

2

+

=

π

πγ ∴ γ x = 29.1 lb/ft3

2.69 The force of the water is only vertical (FV)w, acting thru the center. The force of the oil can also be positioned at the center: a) P FH o= = × × ×( ) ( . ) . .0 8 9810 0 3 3 6 = 8476 N. ΣF W F Fy V o V w= = + −0 ( ) ( )

0 = S × 9810 π × .62 × 6 + ..

36364

−

π× 6 × (.8 × 9810) – 9810 × π × .18 × 6

− × × × −9810 8 2 6 62. . . ∴ =S 0 955. . b) g Vρ .W= = 1996 lb. ΣF W F Fy V o V w= = + −0 ( ) ( )

24

0 = S × 62.4 × π × 22 × 20 + 444

−

π× 20 × .8 × 62.4 – 62.4 × π × 2 × 20

− × × × ×62 4 8 2 2 202. . . ∴ =S 0 955. . 2.70 The pressure in the dome is a) p = 60 000 – 9810 × 3 – 0.8 × 9810 × 2 = 14 870 Pa or 14.87 kPa. The force is F = pAprojected = (π × 32) × 14.87 = 420.4 kN. b) From a free-body diagram of the dome filled with oil: Fweld + W = pA Using the pressure from part (a):

Fweld = 14 870 × π × 32 – (.8 × 9810) × 12

43

33π ×

= –23 400 N

or –23.4 kN

2.71 A free-body diagram of the gate and water is shown.

H

F d W H Pw3+ = × .

a) H = 2 m. F = 9810 × 1 × 4 = 39 240 N.

W xdyy

dy= = =×

∫∫9810 2 9810 22

2 98102

23 2

1 2

0

2

0

2 3 2/ /

/ = 26 160 N.

d x

xxdy

xdy

x dx

x dxw = = = =

∫∫

∫

∫2

12

4

4

12

1 41 3

3

0

1

2

0

1

//

= 0.375 m.

∴ = × + ×P13

390 375

226 160 240

. = 17 980 N or 17.98 kN.

b) H = 8'. F = 62.4 × 4 × 32 = 7,987 lb.

W xdy x dx= = × = × ×∫∫62 4 62 4 4 62 16 2 32

0

23.4 .4 .4 / = 2,662 lb.

d xx dx

x dxw = = =

∫

∫

12

4

4

12

16 48 3

3

0

2

2

0

2

//

= 0.75'.

1 87,987 0.75 2,662 2910 lb

8 3P = × + × =

2.72 (A) W Vγ= 900 9.81 9810 0.01 15 . 6 mw w× = × × ∴ =

W

pAFweld

dA=xdy

y

xF

h/3

25

2.73 W = weight of displaced water. a) 20 000 + 250 000 = 9810 × 3 2(6 /2).d d+ ∴d2 + 12d – 18.35 = 0. ∴d = 1.372 m.

b) 270 000 = 1.03 × 9810 × 3 2(6 /2).d d+ d2 + 12d – 17.81 = 0. ∴d = 1.336 m. 2.74 25 + FB = 100. ∴FB = 75 = 9810 −V . ∴ −V = 7.645 × 10−3 m3 γ × 7.645 × 10−3 = 100. or 7645 cm3 ∴γ = 13 080 N/m3. 2.75 3000 × 60 = 25 × 300 ∆d × 62.4. ∴∆d = 0.3846' or 4.62". 2.76 100 000 × 9.81 + 6 000 000 = (12 × 30 + 8h × 30) 9810 ∴h = 1.465 m. ∴distance from top = 2 – 1.465 = 0.535 m 2.77 T + FB = W. (See Fig. 2.11 c.) T = 40 000 – 1.59 × 9810 × 2 = 8804 N or 8.804 kN. 2.78 The forces acting on the balloon are its weight W, the buoyant force FB, and the weight of the air in the balloon Fa. Sum forces:

FB = W + Fa or 43

100043

3 3π ρ π ρR g R ga= +

43

5100 9 81287 293

100043

5100 981

2873 3π π×

××

= + ××.

..

..

Ta

∴Ta = 350.4 K or 77.4°C

2.79 The forces acting on the blimp are the payload Fp, the weight of the blimp W, the buoyant force FB, and the weight of the helium Fh: FB = Fp + W + Fh

1500 150100 9 81287 288

2π × ×××

..

= Fp + 0.1 Fp + 1500 π × 1502 × 100 9812 077 288

××

..

4o /64.I dπ= . Npeople =

986 10800

8. × = 1.23 × 106

Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add significant weight. 2.80 Neglect the bouyant force of air. A force balance yields FB = W + F = 50 + 10 = 60 = 9800 −V . ∴ −V = .006122 m3 Density: g Vρ .W= ρ × ×9 81 006122. . = 50. ∴ρ = 832.5 kg/m3 Specific wt: γ = ρg = 832.5 × 9.81 = 8167 N/m3 Specific gravity: S = ρ/ρwater = 832.5/1000 = 0.8325

26

2.81 From a force balance FB = W + pA. a) The buoyant force is found as follows (h > 16'):

cos ,θ =− −h R

R15

Area = θR2 – (h – 15 – R) R sinθ

∴FB = 10 × 62.4[πR2 − θR2 + (h – 15 – R) R sinθ]. FB = 1500 + γhA. The h that makes the above 2 FB’s equal is found by trial-and- error: h = 16.5: 1859 ? 1577 h = 16.8: 1866 ? 1858

h = 17.0: 1870 ? 1960 ∴h = 16.82 ft.

b) Assume h > 1613

ft. and use the above equations with R = 1.333':

h = 16.4: 1857 ? 1853 ∴h = 16.4 ft.

c) Assume h < 1623

ft. With R = 1.667',

FB = 10 × 62.4[θR2 − (R – h + 15) R sinθ].

FB = 1500 + γhA. cos θ =− +R h

R15

Trial-and-error for h:

h = 16: 1849 ? 1374 h = 16.2: 1853 ? 1765

h = 16.4: 1857 ? 2170 ∴h = 16.25 ft. 2.82 a) W FB= . [ ]0 01 136 1000 015 4 9 81 98102. . . / . .+ × × × × = −h Vπ

− =×

× +×

× = × −Vπ π.

..

. . .0154

150054

06 2 769 102 2

5 3m ∴h = 7.361 × 10−3 m

∴ =mHg213.6 1000 .015 / 4hπ× × × = 0.01769 kg

b) (.01 + .01769) 9.81 = 9810 π π×

× +×

×

..

.. .

0154

150054

122 2

Sx ∴Sx = 0.959.

c) (.01 + .01769) 9.81 = 9810 π ×

×.

.0154

152

Sx. ∴Sx = 1.045.

2.83 (. ) ..

..

. .01 9 81 98100154

150054

122 2

+ =×

× +×

×

mHg

π π ∴mHg = 0.01886.

a) (.01 + .01886) 9.81 = 9810 π ×

×.

.0154

152

Sx. ∴Sx = 1.089.

b) mHg = 0.01886 kg.

h − 15

θ R

pA

FB

W

h − 15

θ R

27

2.84 a) 4 4

o(10/12)

64 64d

Iπ π ×

= = = 0.02367 ft4.

− = =× × × ×

VW

rH O2

8 62 5 12 12 1262

2. .4 ( / ) /.4

π = 0.4363. depth =

.4363( / )π 5 12 2 = 0.8'

∴ = − −GM . /.4363 (. .4)02367 5 = –0.0457'. ∴It will not float with ends horizontal. b) Io = 0.02367 ft4, −V = 0.3636 ft3, depth = 0.6667' GM = − −. /. ( ) /02367 3636 5 4 12 = –0.01823'. ∴It will not float as given.

c) −V = 0.2909, depth = 6.4", GM = ..

.023672909

4 3 212

−−

= 0.0147. ∴It will float.

2.85 With ends horizontal 4o /64.I dπ= The displaced volume is

− = × = × −V d h dx xγ π γ2 5 34 9800 8 014 10/ . since h = d. The depth the cylinder will sink is

depth = −

= × = ×− −VA

d d dx x8 014 10 4 10 20 105 3 2 5. / / .γ π γ

The distance CG is CGh

dx= − × −

210 2 10 25. /γ . Then

GMd

dd

dx

x=×

− + × >−−π

γγ

4

5 3564

8 014 10 210 2 10 2 0

/.

. / .

This gives (divide by d and multiply by γx): 612.5 – .5 γx + 5.1 × 10-5 γ x

2 > 0. Consequently, γx > 8369 N/m3 or γx < 1435 N/m3

2.86 3

3.water

water water

S dWV S d

γ

γ γ− = = =

33.water

water water

S dWV S d

γ

γ γ− = = = ∴h = Sd.

GMd

Sdd Sd d

SS

= − − = − +4

3

122 2

112

12 2

/( / / ) ( ).

If GM = 0 the cube is neutral and 6S2 – 6S + 1 = 0.

∴ =± −

S6 36 24

12 = 0.7887, 0.2113.

The cube is unstable if 0.2113 < S < 0.7887. Note: Try S = 0.8 and S = 0.1 to see if GM > 0. This indicates stability.

2.87 As shown, y =× + ×

+16 9 16 4

16 16 = 6.5 cm above the bottom edge.

GS

SA

A

=× + × + ×

× + × + ×4 9 5 16 8 5 16 4

5 8 2 8 16γ γ γ

γ γ γ. .

. = 6.5 cm.

C

Gh

28

∴130 + 104 SA = 174 + 64 SA. ∴ SA = 1.1.

2.88 a) y =× + × + ×

+ +16 4 8 1 8 7

16 8 8 = 4. x =

× + × + ×+ +

16 1 8 4 8 416 8 8

= 2.5.

For G: y =× × + × × + × ×

× + × + ×12 16 4 5 8 1 1 5 8 7

1 2 16 5 8 15 8. . .

. . . = 4.682.

x =× + × × + × ×

× + × + ×12 16 5 8 4 15 8 4

12 16 5 8 15 8. . .

. . . = 2.364.

G must be directly under C.

tan..

.θ =136682

∴θ =11.3°.

b) y =× + × + ×

+ +

4 2 212

2 3 5

4 2 2

. = 2. x =

× + × + ×

+ +

412

2 2 2 2

4 2 2 = 1.25

For G:y =× × + × + ×

× + × + ×12 4 2 5 1 15 7

12 4 5 2 15 2. . .

. . . = 2.34. x =

× + × + ×× + × + ×

12 2 5 4 15 41 2 4 5 2 15 2. . .. . .

= 1.182

∆y = 0.34, ∆x = 0.068. tan..

.θ =06834

∴θ = 11.3°.

2.89 The centroid C is 1.5 m below the water surface. ∴ CG = 1.5 m.

Using Eq. 2.4.47: GM =×× ×

− = − = >ll

8 128 3

15 1777 15 0 277 03 /

. . . . .

∴The barge is stable.

2.90 y =× + ×

+8 3 16 97 1

8 16 97.485 .414 .

.485 . = 1.8 m. ∴ CG = −1 8 1 5. . = 0.3 m.

Using Eq. 2.4.47: GM =×

− = − =l

l8 1234 97

3 1 3 1163.485 /

.. .46 . . . ∴Stable.

2.91 (A) 2

520000 20000 6660 (1.2 ) 24070 Pa

9.8124070 0.02 30.25 N

plug

plug plug

p h

F p A

γ

π

= + = + × × =

= = × × =.

2.92 a) tan.

.α = =209 81 4

H ∴H = 8.155 m. pmax = 9810 (8.155 + 2) = 99 620 Pa

b) pmax = ρ(g + az) h = 1000 (9.81 + 20) × 2 = 59 620 Pa c) pmax = 1.94 × 60 (–12) – 1.94 (32.2 + 60) (–6) = 2470 psf or 17.15 psi d) pmax = 1.94 (32.2 + 60) (–6) = 1073 psf or 7.45 psi

0.682C

0.136

G

29

2.93 The air volume is the same before and after.

∴ 0.5 × 8 = hb/2. tan.

.α = =109 81

hb

49 8110

=h

h2

..

∴h = 2.856. ∴Use dotted line.

2 512

2 5 2 4. . .452 .w + × × = ∴w = 0.374 m.

a) pA = –1000 × 10 (0 – 7.626) – 1000 × 9.81 × 2.5 = 51 740 Pa or 51.74 kPa b) pB = –1000 × 10 (0 – 7.626) = 76 260 Pa or 76.26 kPa c) pC = 0. Air fills the space to the dotted line. 2.94 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure.

a) 60 000 = –1000 ax (0–8) – 1000 × 9.81 0 2 589 81

− −

..ax

4 = h

ax

2 9 812× .

60 = 8 ax + 24.52 – 9.81 89 81

ax

.. ax – 4.435 = 1.1074 ax .

ax2 – 10.1 ax + 19.67 = 0 ∴ax = 2.64, 7.46 m/s2

b) 60 000 = –1000 ax (–8) – 1000 (9.81 + 10) − +

2 5

89 81

..

.ax

60 = 8 ax + 49.52 – 19.81 8

19 81ax

.. ax – 1.31 = 1.574 ax .

ax2 – 5.1 ax + 1.44 = 0 ∴ax = 0.25, 4.8 m/s2

c) 60 000 = –1000 ax (–8) – 1000 (9.81 + 5) (–2.5 +8

14 81ax

.).

60 = 8 ax + 37.0 – 14.81 8

14 81ax

.. ax – 2.875 = 1.361 ax .

ax2 – 7.6 ax + 8.266 = 0 ∴ax = 1.32, 6.28 m/s2

2.95 a) ax = 20 × .866 = 17.32 m/s2, az = 10 m/s2. Use Eq. 2.5.2 with the peep hole as

position 1. The x-axis is horizontal passing thru A. We have pA = –1000 × 17.32 (0 – 1.232) – 1000 (9.81 + 10) (0 – 1.866) = 58 290 Pa b) pA = –1000 × 8.66 (0 – 1.848) – 1000 (9.81 + 5) (0 – 2.799) = 57 460 Pa c) The peep hole is located at (3.696, 5.598). Use Eq. 2.5.2: pA = –1.94 × 51.96 (0 – 3.696) – 1.94 (32.2 + 30) (0 – 5.598) = 1048 psf d) The peep hole is located at (4.928, 7.464). Use Eq. 2.5.2: pA = –1.94 × 25.98 (–4.928) – 1.94 (32.2 + 15) (–7.464) = 932 psf 2.96 a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2, p(z) = –1000 × 10 (–7.626) – 1000 × 9.81(z) = 76 260 – 9810 z

α

b

hA

B

z

1xw

C

30

∴ = −∫F z dzAB ( ).

76 260 9810 40

2 5

= 640 000 N or 640 kN

b) The pressure on the bottom BC is p(x) = –1000 × 10 (x – 7.626) = 76 260 – 10 000 x.

∴ = −∫F x dxBC ( ).

76 260 10 000 40

7 626

= 1.163 × 106 N or 1163 kN

c) On the top p(x) = –1000 × 10 (x – 5.174) where position 1 is on the top surface:

∴ = −∫F x dxtop ( ).

51 740 10 000 40

5 174

= 5.35 × 105 N or 535 kN

2.97 a) The pressure at A is 58.29 kPa. At B it is pB = –1000 × 17.32 (1.732–1.232) – 1000 (19.81) (1–1.866) = 8495 Pa. Since the pressure varies linearly over AB, we can use an average pressure times the area:

FAB =+

× ×58 290 8495

215 2

. = 100 200 N or 100.2 kN

b) pD = 0. pC = –1000 × 17.32 (–.5–1.232) − 1000 × 19.81(.866–1.866) = 49 810 Pa.

FCD = × × ×12

49 810 15 2 . = 74 720 N or 74.72 kN.

c) pA = 58 290 Pa. pC = 49 810 Pa. ∴ =+

×FAC58 29 49 81

215

. .. = 81.08 kN.

2.98 Use Eq. 2.5.2 with position 1 at the open end:

a) pA = 0 since z2 = z1.

pB = 1000 × 19.81 × 0.6 = 11 890 Pa. pC = 11 890 Pa. b) pA = –1000 × 10 (.9–0) = –9000 Pa. pB = –000 × 10 (.9)–1000 × 9.81(-.6) = –3114 Pa pC = –1000 × 9.81 × (–.6) = 5886 Pa.

c) pA = –1000×20 (0.9) = –18 000 Pa. pB = –1000 × 20 × 0.9–1000×19.81(−0.6) = –6110 Pa. pC = 11 890 Pa

d) pA = 0. pB = 1.94 × (32.2-60) 2512

= −112 psf. pC = –112 psf.

e) pA = 1.94 × 60 −

37 512

. = −364 psf.

pB = 1.94 × 60 −

37 512

.– 1.94 × 32.2 −

2512

= –234 psf.

x

z

A

BCx

z1

31

pC = –1.94 × 32.2 −

2512

= 130 psf.

f) pA = 1.94 × 30 37 512

.

= 182 psf.

pB = –1.94(–30) 37 512

.

– 1.94 × 62.2 −

2512

= 433 psf.

pC = –1.94 × 62.2 × −

2512

= 251 psf.

2.99 Use Eq. 2.6.4 with position 1 at the open end:

ωπ

=×50 260

= 5.236 rad/s.

a) pA =×

× ×1000 5 236

26 15

22.

(. . ) = 11 100 Pa.

pB =12

× 1000 × 5.2362 × .92 + 9810 × .6 = 16 990 Pa.

pC = 9810 × .6 = 5886 Pa.

b) pA =12

× 1000 × 5.2362 × 0.62 = 4935 Pa.

pB =12

× 1000 × 5.2362 × 0.62 + 9810 × 0.4 = 8859 Pa.

pC = 9810 × 0.4 = 3924 Pa.

c) pA =12

× 1.94 × 5.2362 × 37 512

2.

= 259.7 psf.

pB =12

× 1.94 × 5.2362 × 37 512

622512

2..4

+ × = 389.7 psf.

pC =622512

.4 × = 130 psf.

d) pA =12

× 1.94 × 5.2362 × 22 512

2.

= 93.5 psf.

pB =12

× 1.94 × 5.2362 × 22 512

2.

+ 62

1512

.4 × = 171.5 psf.

pC = 621512

.4 × = 78 psf.

A

BC

z1

r

ω

32

2.100 Use Eq. 2.6.4 with position 1 at the open end.

a) pA =12

× 1000 × 102 (0 – 0.92) = –40 500 Pa.

pB = –40 500 + 9810 × 0.6 = –34 600 Pa. pC = 9810 × 0.6 = 5886 Pa.

b) pA =12

× 1000 × 102 (0 – 0.62) = –18 000 Pa.

pB = –18 000 + 9810 × 0.4 = –14 080 Pa.

pC = 9810 × 0.4 = 3924 Pa.

c) pA =12

× 1.94 × 102 037 5144

2

−

. = –947 psf.

pB = -947 + 62.4 × 2512

= –817 psf. pC = 62.4 × 2512

= 130 psf.

d) pA =12

× 1.94 × 102 −

22 512

2

2

. = –341 psf.

pB = –341 + 62.4 × 1512

= –263 psf. pC = 62.4 × 1512

= 78 psf.

2.101.1Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: p2 = 0.

a) 0 = 12

× 1000 ω2 (0 – 0.452) – 9810 (0 – 0.6). ∴ω = 7.62 rad/s.

b) 0 = 12

× 1000 ω2 (0 – 0.32) – 9810 (0 – 0.4). ∴ω = 9.34 rad/s.

c) 0 = 12

× 1.94 ω2 018 75

12

2

2−

.– 62.4 −

2512

. ∴ω = 7.41 rad/s.

d) 0 = 12

× 1.94 ω2 −

11 2512

2

2

.– 62.4 −

1512

. ∴ω = 9.57 rad/s.

2.102 The air volume before and after is equal.

∴ = × ×12

6 202 2π πr h . . . ∴ r h0

2 = 0.144.

a) Using Eq. 2.6.5: r02 25 2× / = 9.81 h

∴h = 0.428 m

∴pA = 12

× 1000 × 52 × 0.62 – 9810 (–0.372)

= 8149 Pa.

b) r02 27 2× / = 9.81 h. ∴h = 0.6 m.

∴pA =1000

2× 72 × 0.62 + 9810 × 0.2 = 10 780 Pa.

A

BC

z1

r

ω

z

1

rω

2

1

2

h

z

rA

r0

33

c) For ω = 10, part of the bottom is bared.

π π π× × = −. . .6 212

12

202

12

1r h r h

Using Eq. 2.6.5:

ω 2

02

2rg

h= , ω 2

12

2rg

h= 1 .

∴ = −01442 2

22

2 12.

gh

gh

ω ω or

h h212

20 144 102 981

− =×

×.

..

Also, h – h1 = 0.8. 1.6h – 0.64 = .7339. ∴h = 0.859 m, r1 = 0.108 m.

∴pA = 12

× 1000 × 102 (0.62 – 0.1082) = 17 400 Pa.

d) Following part (c): h h212

20144 202 9 81

− =×

×.

.. 1.6h – .64 = 2.936.∴ h = 2.235 m.

∴pA = 12

× 1000 × 202 (0.62 – 0.2652) = 57 900 Pa r1 = 0.265 m

2.103 The answers to Problem 2.102 are increased by 25 000 Pa. a) 33 150 Pa b) 35 780 Pa c) 42 400 Pa d) 82 900 Pa

2.104 p r r g h( ) [ (. )].= − − −12

0 82 2ρω ρ

p r r h( ) (. )= + −500 9810 82 2ω if h < .8. p r r r( ) ( )= −500 2 2

12ω if h > .8.

a) F p rdr r r dr= = +∫∫ 2 2 12 500 36503

0

6

π π ( ).

= 6670 N.

(We used h = .428 m)

b) F p rdr r r dr= = +∫∫ 2 2 24 500 19623

0

6

π π ( ).

= 7210 N. (We used h = 0.6 m)

c) F p rdr r r dr= = −−∫∫ 2 2 50 000 1083

108

62π π ( ( . )

.

.

= 9520 N. (We used r1 = 0.108 m)

d) F p rdr r r dr= = −∫∫ 2 2 200 000 2653

265

62π π ( ( . )

.

.

= 26 400 N. (r1 = 0.265 m)

1

h

z

rA

r0

h1

dr

dA = 2πrdr

34

CHAPTER 3

Introduction to Fluids in Motion 3.1

3.2 Pathlines: Release several at an instant in time and take a time exposure of the subsequent motions of the bulbs. Sreakline: Continue to release the devises at a given location and after the last one is released, take a snapshot of the “line” of bulbs. Repeat this for several different release locations for additional streaklines. 3.3

3.4

streakline

pathline

streamline

streakline

pathline hose

boytime tt = 0

streamlines t = 2 hr

pathline t = 2 hr

streakline at t = 3 hry

x

35

3.5 a)udxdt

t vdydt

t= = + = =2 2 2

x t t c y t c= + + = +21

222

= +y y2

∴ − + =x xy y y2 22 4 ∴parabola.

b) x t t c c c= + + ∴ = − = −21 1 22 8 4. , . and

= + + ± + −y y4 2 4 8( )

∴ − + + − =x xy y x y2 22 8 12 0. ∴parabola.

3.6 ˆˆ ˆ ( )

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ using , .

zV dr udy vdxV ui vj wk

dr dxi dy j dzk i j k j i k

× = −= + +

= + + × = × = −

vv v

v

3.7 Lagrangian: Several college students would be hired to ride bikes around the various roads, making notes of quantities of interest. Eulerian: Several college students would be positioned at each intersection and quantities would be recorded as a function of time. 3.8 a) At t V= = =2 0 0 0 2 22 and m / s( , , ) .

At t V= − = + =2 1 2 0 3 2 3 6062 2 and m / s( , , ) . . b) At t V= =2 0 0 0 0 and ( , , ) .

At t V= − = − + − =2 1 2 0 2 8 8 2462 2 and m / s( , , ) ( ) ( ) . .

c) At t V= = − =2 0 0 0 4 42 and m / s( , , ) ( ) .

At t V= − = + − + − =2 1 2 0 2 4 4 62 2 2 and m / s( , , ) ( ) ( ) . 3.9 (D) 5 ˆ( 51.4 10 )j−− ×

A simultaneous solution yields 4/5 and 3/5.x yn n= = (They must both have the same sign.

3.10 a) cos $ / ( ) / . . .α α= ⋅ = + + = ∴ =

voV i V 1 2 3 2 0 832 33 692 2

v

V n i j n i n jn nn n

n n

n nx y

x y

x y

y x

x x

⋅ = + ⋅ + =+ =+ =

∴= −

+ =$ . ( $ $) ( $ $) .0 3 2 0

3 2 01

32

94

12 2

2 2

∴ = = − = −n n n i jx y

213

313

113

2 3, $ ( $ $). or

39.8o

y

x

streamlines t = 5 s

(27, 21)(35, 25)

36

b) cos $ / / ( ) ( ) . .α α= ⋅ = − − + − = − ∴ =v

oV i V 2 2 8 0 2425 1042 2

vV n i j n i n j

n nn n

n nn nx y

x y

x y

x y

y y⋅ = − − ⋅ + =

− − =+ =

∴= −+ =

$ . ( $ $) ( $ $) .0 2 8 02 8 0

14

16 12 2 2 2

∴ = = − = − +n n n i jy x

117

417

117

4, $ ( $ $). or

c) cos $ / / ( ) . . .α α= ⋅ = + − = ∴ = −

voV i V 5 5 8 0 6202 51 672 2

v

V n i j n i n jn nn n

n n

n nx y

x y

x y

x y

y y

⋅ = − ⋅ + =− =+ =

∴=

+ =$ . ( $ $) ( $ $) .0 5 8 0

5 8 01

85

6425

12 2

2 2

∴ = = = +n n n i jy x589

889

189

8 5, $ ( $ $). or

3.11 a) [ ]v v

V dr x i xtj dxi dyj× = + + × + =0 2 0. ( )$ $ ( $ $) .

∴ + − =+

=( ) .x dy xtdx txdx

xdy2 0

2 or

Integrate: [ ]txdx

xdy t x n x y C

+= − + = +∫∫ 2

2 2. . l

2 1 2 3 2( ) .− = − + ∴ln C C = 0.8028. [ ]t x n x y− + = +2 2 0 8028l .

b) [ ]v v

V dr xyi y j dxi dyj× = − × + =0 2 02. $ $ ( $ $) .

∴ + = = −xydy y dxdxx

dyy

2 022 or .

Integrate: 2 2 1 2l l l lnx n y C n n C= − = − −( / ). ( ) ( / ). ∴ = − = − − ∴ = −C nx n y x y2 2 22 2. ( / ). . l l

c) [ ]v v

V dr x i y tj dxi dyj× = + − × + =0 4 02 2. ( )$ $ ( $ $) .

( ) .x dy y tdxtdx

xdyy

2 22 24 0

4+ + =

+= − or

Integrate: t xC

yC

2 21 2

212

12

1 1tan . tan .− −+

= +

= −

∴ = − −

=−C ytx

0 96362

0 9636 21. . tan .

37

3.12 (C) 2ˆ ˆ ˆ ˆ ˆ ˆ2 (2 ) (2 2 ) 16 8 16 .V V V V

a u v w xy yi y xi yj i i jt x y z

∂ ∂ ∂ ∂= + + + = − − = − − +

∂ ∂ ∂ ∂

v v v vv

2 28 16 17.89 m/sa∴ = + =

3.13 a) DVDt

uVx

vVy

wVz

Vt

v v v v v= + + +

∂∂

∂∂

∂∂

∂∂

=0.

b) u Vx

vVy

wVz

Vt

x i y j xi yj∂∂

∂∂

∂∂

∂∂

v v v v+ + + = + = +2 2 2 2 4 4( $) ( $) $ $ = 8 4$ $i j−

c) u Vx

v Vy

w Vz

Vt

x t xti ytj xyt xtj ztk x i xyj∂∂

∂∂

∂∂

∂∂

v v v v+ + + = + + + + +2 22 2 2 2 2 2( $ $) ( $ $) $

+ = − −2 68 100 54yzk i j k$ $ $

d) u Vx

v Vy

w Vz

Vt

x i yzj xyz xzj tz xyj tk zk∂∂

∂∂

∂∂

∂∂

v v v v+ + + = − − − + − + +($ $) ( $) ( $ $) $2 2 2 2

= xi yz x yz xyzt j zt z k$ ( )$ ( ) $− − + + +2 4 22 2 2

= 2 114 15$ $ $i j k− +

3.14 vΩ = −

+ −

+ −

12

12

12

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

wy

vz

iuz

wx

jvx

uy

k$ $ $.

a) vΩ = − =

12

20∂∂

uy

k yk$ $ = −20 $k

b) vΩ = − + − + −

12

0 012

0 012

0 0( )$ ( )$ ( ) $i j k = 0

c) vΩ = − + − + −

12

2 012

0 012

2 0( )$ ( ) $ ( ) $zt i j yt k = 6 2$ $i k−

d) vΩ = + + − + − −

12

0 212

0 012

2 0( )$ ( )$ ( ) $xy i j yz k = − +2 3$ $i k

3.15 The vorticity v v

ω = 2Ω. Using the results of Problem 3.7: a) v

ω = −40$i b) vω = 0 c) v

ω = 12 4$ $i k− d) vω = − +4 6$ $i k

3.16 a) ε∂∂

ε∂∂

ε∂∂xx yy zz

ux

vy

wz

= = = = = =0 0 0, , .

ε∂∂

∂∂

ε∂∂

∂∂xy xz

uy

vx

yuz

wx

= +

= − = = +

=12

20 2012

0, ,

ε∂∂

∂∂yz

vz

wy

= +

= ∴ =

12

00 20 020 0 00 0 0

. rate - of strain

38

b) ε ε εε ε ε

xx yy zz

xy xz yz

= = == = =

2 2 00 0 0, , ., , .

rate-of strain = 2 0 0

0 2 00 0 0

c) ε ε εxx yy zzxt xt yt= = = = = = −2 8 2 8 2 4, , .

ε ε εxy xz yzyt zt= = − = = = =12

2 212

0 012

2 6( ) , ( ) , ( ) .

rate-of strain = 8 2 0

2 8 60 6 4

−

−−

d) ε ε εxx yy zzxz t= = − = − = =1 2 12 2, , .

ε ε εxy xz yzyz xy= − = = = = − =12

2 312

0 012

2 2( ) , ( ) , ( ) .

rate-of strain = 1 3 0

3 12 20 2 2

−

3.17 a) ar r r r rr = −

− +

−

−1040 80

1040

140

2 3 2 2cos cossin

( sin )θ θθ

θ

− +

= − − −

110

4010 2 1 125 1

2

22

r rsin ( .5)( ) . ( )θ = 9.375 m/s2.

ar r r r rθ θ θ

θθ= −

+ +

+

1040 80

1040

1040

2 3 2 2cos sinsin

cos

− −

1100

16004r r

sin cosθ θ = 0 since sin 180° = 0.

aφ = 0.

b) ω ω ω θ θθr z r r r r= = = − +

− −

−0 01

1040 1

1040

2 2, , sin ( sin ) = 0.

At (4, 180°) vω = 0 since v

ω = 0 everywhere.

3.18 a) ar r r r rr = −

− +

− −

1080 240

1080

1080

3 4 3 3cos cossin

( sin )θ θθ

θ

− +

= − −10

808 75 1 9375 1

3

2 2

r rsin

. ( )(. )( )θ = 8.203 m/s2

aθ = 0 since sin 180° = 0. aφ = 0 since v φ = 0. b) ω r = 0, ω θ = 0, ω φ = 0, since sin 180° = 0.

39

3.19 V Va u

t x∂ ∂∂ ∂

= +v v

v v+V

wy

∂+

∂

v.

V ui

z t∂∂

∂=

∂

v For steady flow ∂ ∂u t a/ .= =0 0 so that v

3.20 Assume u(r,x) and v(r,x) are not zero. Then, replacing z with x in the appropriate

equations of Table 3.1 and recognizing that v θ ∂ ∂θ= =0 0 and / :

r xv v u u

a v u a v ur x r x

∂ ∂ ∂ ∂∂ ∂ ∂ ∂

= + = +

3.21 a) u e tt= − − = = ∞−2 1 0 1 10( )( ) ./ 2 m / s at

( )aut

e txt= = −

= =−∂∂

2 1 01

100 2 010( ) . ./ m / s at 2

b) u e tt= − − = = ∞−2 1 0 12 10( .5 )( ) ./ 1.875 m / s at

a e txt= −

= =−2 1 0 21

100 0125 02 2 10( .5 / ) . ./ m / s at 2

c) u e tt= − − =−2 1 2 2 12 2 10( / )( ) ./ 0 for all

a e txt= −

=−2 1 2 21

1002 2 10( / ) ./ for all

3.22 DT Tu

Dt x∂∂

= v+T

wy

∂∂

+ 220(1 ) sin 0.5878100 100 5

T T ty

z t∂ ∂ π π π∂ ∂

+ = − − = − ×

= −0.3693 °C/s.

3.23 DDt

ux

vy

wz t

eρ ∂ρ

∂∂ρ∂

∂ρ∂

∂ρ∂

= + + + = − × − − × −

10 1 23 10 4 3000 10 4

( . ) = − ×⋅

−9 11 10 4. .kg

m s3

3.24 DDt

ux

vy

wz t

ρ ∂ρ∂

∂ρ∂

∂ρ∂

∂ρ∂

= + + + = −

101000

4= −

⋅2500

kgm s3 .

3.25 DDt

ux

ρ ∂ρ∂

= = ×4 01(. ) = 0.04 kg/m3⋅s

3.26 (D) 22

10[10(4 ) ]

(4 )x

u u u u ua u v w u x

t x y z x xx−∂ ∂ ∂ ∂ ∂ ∂

= + + + = = −∂ ∂ ∂ ∂ ∂ ∂−

3 22

10 10 110( 2)( 1)(4 ) 20 6.25 m/s .

4 8(4 )x

x−= − − − = × × =

−

3.27 DDt

Vt

= ⋅ ∇ +v v ∂

∂ observing that the dot product of two vectors

vA A i A j A kx y z= + +$ $ $

and v v vB B i B j B k A B A B A B A Bx y z x x y y z z= + + ⋅ = + +$ $ $ . is

40

3.28

aut

V u

avt

V v

awt

V w

aVt

V V

x

y

z

= + ⋅∇

= + ⋅∇

= + ⋅∇

∴ = + ⋅∇

∂∂∂∂

∂∂

∂∂

v v

v v

v v

vv

v v v ( )

3.29 Using Eq. 3.2.12:

a) v v

v v v v v vv

vA a

d sdt

V rddt

r= + + × + × × + ×2

2 2Ω Ω ΩΩ

( )

= 2 20 4 20 20 1( $ $) $ ( $ .5$)k i k k i× + × × = 160 600$ $j i− m / s2

b) v v v v v v oA V r k j k k i= × + × × = × − + × ×2 2 20 20 30 20 20 3Ω Ω Ω( ) ( $ cos $) $ ( $ $) = −507$i

3.30 vΩ =

× ×= × −2

24 60 607 272 10 5π $ . $k k rad/s.

vv i k i k= − − = − −5 707 707 3 3( . $ . $) .535$ .535 $ m/s.

v v v v v vA V r= × + × ×2Ω Ω Ω( )

= 2 7 272 10 3 3 7 272 105 5× × × − − + × ×− −. $ ( .535$ .535 $) . $k i k k [ . $ ( . $ . $)]7 272 10 6 10 707 7075 6× × × − +− k i k = − × +−51 4 10 0 02245. $ . $j i m / s 2 .

Note: We have neglected the acceleration of the earth relative to the sun since it is quite small

(it is d s dt2 2v / ). The component 5 ˆ( 51.4 10 )j−− × is the Coriolis acceleration and causes air motions to move c.w. or c.c.w. in the two hemispheres.

3.31 a) two-dimensional (r, z) b) two-dimensional (x, y) c) two-dimensional (r, z) d) two-dimensional (r, z) e) three-dimensional (x, y, z) f) three-dimensional (x, y, z)

g) two-dimensional (r, z) h) one-dimensional (r) 3.32 Steady: a, c, e, f, h Unsteady: b, d, g 3.33 b. It is an unsteady plane flow.

3.34 a) d) e) 3.35 f, h

41

3.36 a) inviscid. b) inviscid. c) inviscid. d) viscous inside the boundary layer.

e) viscous inside the boundary layers and separated regions. f) viscous. g) viscous. h) viscous. 3.37 d and e. Each flow possesses a stagnation point. 3.38 3.39 (C) The only velocity component is u(x). We have neglected v(x) since it is

quite small. If v(x) in not negligible, the flow would be two-dimensional. 3.40 Re = V L / ν = 2 × .015/.77 × 10-6 = 39 000. ∴Turbulent.

3.41 Re = =VLν

.2 × .8/1.4 × 10-5 = 11 400. ∴Turbulent.

3.42 Re.

.= =

×× −

VLν

4 061 7 10 5 = 14 100. ∴Turbulent.

Note: We used the smallest dimension to be safe!

3.43 a) Re. .

.51= =

××

=−

V Dν

1 2 0 011 10

795.5 Always laminar.

b) Re.

.51= =

××

=−

VDν

1 2 11 10

795 500. May not be laminar.

3.44 Re = 3 × 105 = TVxν

. ν µ ρ= / where µ µ= ( ).T

a) T = 223 K or −50°C. ∴ = × ⋅−µ 1 10 5.5 N s / m .2

∴ =×

×= ×

−−ν

1 5 103376 1 23

2 5 105

5.. .

. . m /s2

3 10900 1000

3600 2 105

5× =×

× × −

xT

.5. ∴xT = 0.03 m or 3 cm

b) T = −48°F. ∴µ = 3.3 × 10−7 lb-sec/ft2. ν =×

= ×−

−3 3 1000089

3 7 107

4..

. ft2/sec.

3 10600 5280

3600 3 7 105

4× =×× × −

x T

.. ∴xT = 0.13' or 1.5"

42

3.45 Assume the flow is parallel to the leaf. Then 3 × 105 = Vx T / .ν ∴ = × = × × × =−x VT 3 10 3 10 1 4 10 6 8 175 5 4ν / .5 . / . m . The flow is expected to be laminar.

3.46 a) MVc

= =× ×

=100

14 287 2360 325

.. . For accurate calculations the flow is

compressible. Assume incompressible flow if an error of 4%, or so, is acceptable.

b) MVc

= =× ×

=80

14 287 2880 235

.. . ∴Assume incompressible.

c) MVc

= =× ×

=100

14 287 3730 258

.. . ∴Assume incompressible.

3.47 DDt

ux

vy

wz t

ρ ∂ρ∂

∂ρ∂

∂ρ∂

∂ρ∂

= + + + = 0. For a steady, plane flow

∂ρ ∂/ t = 0 and w = 0. Then

ux

vy

∂ρ∂

∂ρ∂

+ = 0.

3.48 Du

Dt xρ ∂ρ

∂= v

y∂ρ∂

+ w+z t

∂ρ ∂ρ∂ ∂

+ 0.= ∴incompressible.

3.49 (B) 2 9810 0.800

. 113 m/s.2 1.23

water

air

hV pV

γρ ρ

×= = = ∴ =

3.50 V p2

2=

ρ. Use ρ = 0.0021 slug/ft3.

a) v p= = × ×2 2 3 144 0021/ . / .ρ = 203 ft/sec.

b) v p= = × ×2 2 9 144 0021/ . / .ρ = 351 ft/sec.

c) v p= = × ×2 2 09 144 0021/ . / .ρ = 111 ft/sec.

3.51 pV

= =×

ρ

2 2

2123

120 10003600

2. / = 683 Pa.

∴F = pA = 683 π × 0.0752 = 12.1 N.

3.52 V p2

20+ =

ρ. ∴ =

−=

×V

p2 2 2000123ρ .

= 57.0 m/s

43

3.53 (C) 2 2 2

1 2 1. 0.200 0.600. 2 9.81 0.400 2.80 m/s.2 2 2V V Vp

Vg g gγ

+ = + = ∴ = × × =

3.54 (B) The manometer reading h implies:

2 221 1 2 2

2 22

or (60 10.2). 9.39 m/s2 2 1.13

V p V pV V

ρ ρ+ = + = − ∴ = The

temperature (the viscosity of the water) and the diameter of the pipe are not needed.

3.55 a) 22

0

2 2VV p

ρ+ =

22( 10 )

. . 502

o oo

p px pp p x ρ

ρ ρ ρ−

+ + = ∴ = −

b) 22

0

2 2VV p

ρ+ =

22(10 )

. . 502

o oo

p py pp p y ρ

ρ ρ ρ+ + = ∴ = −

3.56 22

2 2U pV p

ρ ρ∞ ∞+ = + .

a) v v U r rr cθ θ= = = − −∞0 180 1 12 2 and o: ( / )( ).

( )∴ = − = −

∞ ∞p U v Urr

rrr

c cρ ρ2 2

22 2 22

2

4

.

b) Let r r p Uc T= = ∞: ρ2

2

c) ( ) [ ]v r r v U p v Ur c= = − ∴ = − = −∞ ∞ ∞0 22 2

1 42 2 2 and = U 2: sin . sinθ θθρ ρ

θ

d) Let θ ρ= = − ∞903290

2o : p U

3.57 22

2 2U pV p

ρ ρ∞ ∞+ = + .

a) v θ θ= =0 180 and o : ( )p U v Urr

rrr

c c= − =

−

∞ ∞

ρ ρ2 2

22 2 23 6

.

b) Let r r p Uc T= = ∞: 12

2ρ .

c) ( ) [ ]v r r p v Ur c= = − = −∞ ∞02 2

1 42 2 2 and = U 2: sinρ ρ

θθ

d) Let θ ρ= = − ∞903290

2o : p U

44

3.58 22

2 2U pV p

ρ ρ∞ ∞+ = + .

a) ( )p U ux x

= − = − +

= − +

∞

ρ ρ ππ

ρ2 2

10 10202

50 1 112 2 2

2 2

= − +

502 1

2ρx x

b) 10 when 1. 50 ( 2 1) 50u x p ρ ρ−= = − = − − + =

c) ( )2 2

2 2 2 60 130 30 450 1 1

2 2 2p U u

x xρ ρ π

ρπ∞

= − = − + = − +

= − +

4502 1

2ρx x

d) 10 when 1. 450 ( 2 1) 450u x p ρ ρ−= = − = − − + =

3.59 V p V p

V p p12

1 22

21 1 22 2

0 20+ = + = − =ρ ρ

. and kPa.

( )V p p V22

1 2 22 2

100020 6 32= − = ∴ =

ρ( . 000) = 40. m / s

3.60 Assume the velocity in the plenum is zero. Then

2 2

21 1 2 22 2

2 or (60 10.2). 9.39 m/s

2 2 1.13V p V p

V Vρ ρ

+ = + = − ∴ =

We found ρ = 113. kg / m 3 in Table B.2.

3.61 Bernoulli from the stream to the pitot probe: pV

pT = +ρ2

2.

Manometer: .T Hgp H H h p hγ γ γ γ+ − − = −

Then, 2

2 HgV

p H H pρ γ γ+ + − = . 2 (2 )HgV Hγ γ

ρ

−∴ =

a) V V2 13 6 1 98001000

2 0 04 3 14=−

× ∴ =( . )

( . ). . m / s

b) V V2 13 6 1 98001000

2 0 1 4 97=−

× ∴ =( . )

( . ). . m / s

c) V V2 13 6 1 62 41 94

2 2 12 11 62=−

× ∴ =( . ) .

.( / ). . fps

d) V V2 13 6 1 62 41 94

2 4 12 16 44=−

× ∴ =( . ) .

.( / ). . fps

45

3.62 The pressure at 90° from Problem 3.56 is 290 3 /2.p Uρ ∞= − The pressure at the

stagnation point is 2 /2.Tp Uρ ∞= The manometer provides: p H pT − =γ 90

2 21 31.204 9800 0.04 1.204 . 12.76 m/s

2 2U U U∞ ∞ ∞× − × = − × ∴ =

3.63 The pressure at 90° from Problem 3.57 is 290 3 /2.p Uρ ∞= − The pressure at the

stagnation point is 2 /2.Tp Uρ ∞= The manometer provides: p H pT − =γ 90

2 21 31.204 9800 0.04 1.204 . 12.76 m/s

2 2U U U∞ ∞ ∞× − × = − × ∴ =

3.64 Assume an incompressible flow with point 1 outside in the room where p1 0= and v 1 0= . The Bernoulli’s equation gives, with p hw2 2= γ ,

2

1

2V 1p

ρ+

22 2 .2

V pρ

= +

a) 02

9800 0 021 204

18 0422

2= +− ×

∴ =V

V.

.. . m / s

b) 02

9800 0 081 204

36 122

2= +− ×

∴ =V

V.

.. . m / s

c) 02

62 4 1 120 00233

66 822

2= +− ×

∴ =V

V. /.

. . fps

d) 02

62 4 4 120 00233

133 622

2= +− ×

∴ =V

V. /.

. . fps

3.65 Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel where p V1 10 0= = and . Bernoulli’s equation gives

02

12

22

22 2

2= + ∴ = −V p

p Vρ

ρ.

a) ρ = =×

= ∴ = − × × = −p

RTp

900 287 253

1 23912

1 239 100 619522

.. . . kg / m Pa3

b) ρ = =×

= ∴ = − × × = −p

RTp

950 287 273

1 21212

1 212 100 606022

.. . . kg / m Pa3

c) ρ = =×

= ∴ = − × × = −p

RTp

920 287 293

1 09412

1 094 100 547022

.. . . kg / m Pa3

d) ρ = =×

= ∴ = − × × = −p

RTp

1000 287 313

1 11312

1113 100 556622

.. . . kg / m Pa3

3.66 (A) 2

1

2V

g

21 2 2

2p V p

gγ γ+ = +

22

2800000

. . 40 m/s.9810 2 9.81

VV= ∴ =

×

46

3.67 a) p h V h hA A A= = × = = =γ 9800 4 39 0 2 200 Pa, Using . ,

2

2AVg

AA

ph

γ+ +

22 2

22V p

hg γ

= + +22

2. 2AV

p pg

γ= −

= −×

× = −3914

2 9 819800 58

2

200 700 Pa.

b) 0 and 0.B Bp V= = Bernoulli’s eq. gives, with the datum through the pipe,

Vg

ph

Vg

ph pB B

B

222

22 2

2

2 24

142 9 81

9800 58+ + = + + = −×

= −

γ γ.

. 700 Pa

3.68 Bernoulli: 2

2 2

2V p

g γ+

21 1

2V p

g γ= +

Manometer: 2

21 22Hg

Vp z H H z p

gγ γ γ γ γ+ + − − = +

Substitute Bernoulli’s into the manometer equation:

( )2

11 1.

2HgV

p H pg

γ γ γ+ − = +

a) Use H = 0.01 m: V

V12

1

98002 9 81

13 6 1 9800 0 01 1××

= − × ∴ =.

( . ) . .572 m / s

Substitute into Bernoulli:

pV V

g122

12 2 2

220 1

2 9 819800 198=

−=

−×

× =γ.572.

600 Pa

b) Use H = 0.05 m: V

V12

1

98002 9 81

13 6 1 9800 0 05 3××

= − × ∴ =.

( . ) . .516 m / s

Substitute into Bernoulli:

pV V

g122

12 2 2

220 3

2 9 819800 193=

−=

−×

× =γ.516.

600 Pa

c) Use H = 0.1 m: V

V12

1

98002 9 81

13 6 1 9800 0 1 4 972××

= − × ∴ =.

( . ) . . m / s

Substitute into Bernoulli:

pV V

g122

12 2 2

220 4 972

2 9 819800 187=

−=

−×

× =γ..

400 Pa

47

3.69 Bernoulli across nozzle: 2

1

2V 2

1 2 2

2p V pρ ρ

+ = + 2 1. 2 /V p ρ∴ =

Bernoulli to max. height: 2

1

2V

g1

1p

hγ

+ +2

2

2V

g= 2p

γ+ 2 2 1. / .h h p γ+ ∴ =

a) V p2 12 2 700 1000 37 42= = × =/ / .ρ 000 m / s

h p2 1 700= =/ γ 000 / 9800 = 71.4 m

b) V p2 12 2 1 1000 52 92= = × =/ / .ρ 400 000 m / s h p2 1= =/ γ 1 400 000 / 9800 = 142.9 m c) V p2 12 2 100 1 94 121 8= = × × =/ / . .ρ 144 fps

h p2 1= = ×/ γ 100 144 / 62.4 = 231 ft d) V p2 12 2 200 1 94 172 3= = × × =/ / . .ρ 144 fps

h p2 1 200= = ×/ γ 144/ 62.4 = 462 ft 3.70 a) Apply Bernoulli’s eq. from the surface to a point on top of the downstream flow:

2

1

2V

g1p

γ+

22 2

1 2V p

hg γ

+ = + 2 2. 2 ( )h V g H h+ ∴ = −

b) Apply Bernoulli’s eq. from a point near the bottom upstream to a point on the bottom of the downstream flow:

2

1

2V

g

21 2 2

1 2. 2

p V ph h

gγ γ+ + = + +

Using p H p h h h V g H h1 2 1 2 2 2= = = = −γ γ, , ( ) and

3.71 2

1

2V 2

1 2 2 .2

p V pρ ρ

+ = + p2 = −100 000 Pa, the lowest possible pressure.

a) 600222 000

1000100 000

1000= −

V. ∴ V2 = 37.4 m/s.

b) 300222 000

1000100 000

1000= −

V. ∴ V2 = 28.3 m/s.

48

c) 80 1441.94

14.7 1441.94

×= −

×V22

2. ∴ V2 = 118.6 ft/sec.

d) 40 1441.94

14.7 1441.94

×= −

×V22

2. ∴ V2 = 90.1 ft/sec.

3.72 A water system must never have a negative pressure, since a leak could ingest impurities. ∴ The least pressure is zero gage.

V p

gzV p

gz12

11

22

222 2

+ + = + +ρ ρ

. V V1 2= . Let z1 0= , and p2 0= .

500 0001000

= 9 81 2. .z ∴ z2 = 51.0 m.

3.73 a) ( ) ( )p V V1 22

12 2 2

21000

22 10= − = −

ρ = −48 000 Pa

b) ( ) ( )2 2 2 21 2 1

9022 10 43300 Pa

2 2p V V

ρ= − = − = −

c) ( ) ( )2 2 2 21 2 1

6802 10 32600 Pa

2 2p V V

ρ= − = − = −

d) ( ) ( )2 2 2 21 2 1

1.232 10 59.0 Pa

2 2p V V

ρ= − = − = −

3.74 V p V p1

21 2

22

2 2+ = +

ρ ρ. ( ) ( )2 2 2 2

1 2 11.23

2 82 2

p V Vρ

= − = − = −36.9 Pa

3.75 (D) ( ) ( )2 2 2 21 2 1

90230 15 304400 Pa

2 2p V V

ρ= − = − =

3.76 Apply Bernoulli’s equation between the exit (point 2) where the radius is R and a point 1 in between the exit and the center of the tube at a radius r less than R:

V p V p

pV V1

21 2

22

122

12

2 2 2+ = + ∴ =

−ρ ρ

ρ. .

Since V V2 1< , we see that p1 is negative (a vacuum) so that the envelope would tend to rise due to the negative pressure over most of its area (except for a small

area near the end of the tube).

3.77 Re .=V Dν

For air ν ≅ × −1 10 5.5 . Use reasonable dimensions from your

experience!

49

a) Re.

.5.=

××

= ×−

20 0 031 10

4 1054 ∴Separate

b) Re.

.5.=

××

=−

20 0 0051 10

67005 ∴Separate

c) Re.5

. .=×

×= ×−

20 21 10

2 7 1056 ∴Separate

d) Re.

.5.=

××

=−

5 0 0021 10

6705 ∴Separate

e) Re.5

. .=×

×= ×−

20 21 10

2 7 1056 ∴Separate

f) Re.5

.=×

×= ×−

100 31 10

2 1057

∴It will tend to separate, except streamlining the components eliminates separation.

3.78 A burr downstream of the opening will create a region that acts similar to a stagnation region thereby creating a high pressure since the velocity will be relatively low in that region.

3.79 ∆ ∆pVR

n= = × =ρ2 2

1000100 05

0 02.

. 40 000 Pa Along AB, we

expect V VA B> <10 10 m / s and m /s.

3.80 The higher pressure at B will force the fluid toward the lower pressure at A, especially in the wall region of slow moving fluid, thereby causing a secondary flow normal to the pipe’s axis. This results in a relatively high loss for an elbow.

3.81 Refer to Bernoulli’s equation: V p V p1

21 2

22

2 2+ = +

ρ ρ

p pA B> since V VA B< p pC D< since V VC D> p pB D> since V VD B>

stagnation region

A

B

VA

VB

50

CHAPTER 4

The Integral Forms of the Fundamental Laws

4.1 a) No net force may act on the system: Σ

vF = 0.

b) The energy transferred to or from the system must be zero: Q - W = 0. c) If 3 3 2

ˆ ˆˆ 10 ( ) 0nV V n i j= ⋅ = ⋅ − =v

is the same for all volume elements then

Σv vF

DDt

V dm= ∫ , or Σv vF

DDt

mV= ( ). Since mass is constant for a system

Σv

vF m

DVDt

= . Since DVDt

a F mav

v v v= =, . Σ

4.2 Extensive properties: Mass, m; Momentum, mVv

; kinetic energy, 12

mV 2 ;

potential energy, mgh; enthalpy, H. Associated intensive properties (divide by the mass): unity, 1; velocity,

vV ;V2/2;

gh; H/m = h (specific enthalpy). Intensive properties: Temperature, T; time, t; pressure, p; density, ρ; viscosity, µ. 4.3 (B) 4.4

System ( )t V= 1

c.v.( )t V= 1

System ( )t t V+ ∆ = 1 V+ 2

c.v.( )t t V+ ∆ = 1

4.5

System ( )t V= 1 V+ 2

c.v.( )t V= 1 V+ 2

System ( )t t V+ ∆ = 2 V+ 3

c.v.( )t t V+ ∆ = 1 V+ 2

1 2

12 3

pump

51

4.6 a) The energy equation (the 1st law of Thermo). b) The conservation of mass. c) Newton’s 2nd law. d) The energy equation. e) The energy equation. 4.7

4.8

4.9

4.10 $ $ $ . ($ $)n i j i j1

12

12

0 707= − − = − + . $ . $ .5 $n i j2 0 866 0= − . $ $n j3 = − .

1 1 1ˆ ˆ ˆˆ 10 [ 0.707( )] 7.07 fpsnV V n i i j= ⋅ = ⋅ − + = −

v

V V n i i jn2 2 2 10 0866 05 866= ⋅ = ⋅ − =v

$ $ ( . $ . $) . fps

3 3 2ˆ ˆˆ 10 ( ) 0nV V n i j= ⋅ = ⋅ − =

v

4.11 flux = ηρ $n VA⋅

v

flux1 = ηρ ηρ[ . ($ $)] $ / .− + ⋅ = −0 707 10 0 707 10i j iA A

flux2 = ηρ ηρ( . $ .5$) $ / .0 866 0 10 0 866 10i j iA A− ⋅ =

flux3 = ηρ( $) $− ⋅ =j iA10 03

nv n

vωn v n

vn

v

n

v nn

n

v vv

v

nn

nn

v

v

v

v

n n vv

52

4.12 ( $) ( .5$ . $) $( )vB n A i j j⋅ = + ⋅ × 15 0 0 866 10 12

= × × =15 0 866 120 1559. cm 3 Volume = 15 60 10 12 1559 sin cm 3o × × = 4.13 The control volume must be independent of time. Since all space coordinates are

integrated out on the left, only time remains; thus, we use an ordinary derivative to differentiate a function of time. But, on the right, we note that ρ and η may be functions of (x, y, z, t); hence, the partial derivative is used.

4.14

4.15

4.16

4.17 If fluid crosses the control surface only on areas A1 and A2, ρ ρ ρ$ $ $

. .

n VdA n VdA n V dAAAc s

⋅ = ⋅ + ⋅ =∫∫∫v v v

021

For uniform flow all quantities are constant over each area: ρ ρ1 1 1 2 2 2 0

21

$ $n V dA n V dAAA

⋅ + ⋅ =∫∫v v

Let A1 be the inlet so $n V V A1 1 1 2⋅ = −v

and be the outlet so $ .n V V2 2 2⋅ =v

Then − + =ρ ρ1 1 1 2 2 2 0V A V A or ρ ρ2 2 2 1 1 1A V A V=

1

2

1system (∆t) is involumes 1 and 2

c.v. (0) = c.v. (∆t) = volume 1

1

23

system (∆t) = V1 + V2 + V3

c.v. (∆t) = V1 + V2

system boundary at (t + ∆t)

53

4.18 Use Eq. 4.4.2 with mV representing the mass in the volume:

0 = + ⋅∫dmdt

n V dAV

c s

ρ $. .

v= + −

dmdt

A V A VV ρ ρ2 2 1 1

= + −dm

dtQ mV ρ & .

Finally,

dmdt

m QV = −& .ρ

4.19 Use Eq. 4.4.2 with mS representing the mass in the sponge:

0 = + ⋅∫dmdt

n VdAS ρ $v

= + + −dmdt

A V A V A VS ρ ρ ρ2 2 3 3 1 1

= + + −dm

dtm A V QS & .2 3 3 1ρ ρ

Finally,

dmdt

Q m A VS = − −ρ ρ1 2 3 3& .

4.20 (D) 22000.04 70 0.837 kg/s

0.287 293p

m AV AVRT

ρ π= = = × × =×

& .

4.21 A1V1 = A2V2. π × 125144

2.× 60 = π × 2 5

144

2. V2. ∴V2 = 15 ft/sec.

& ..

m AV= = ×ρ π1941 25144

602

= 3.968 slug/sec. Q = AV = π 125144

2.× 60 = 2 045. / . ft sec3

4.22 A1V1 = A2V2. π × .0252 × 10 = (2π × .6 × .003)V2. ∴V2 = 1.736 m/s. & .m AV= = × ×ρ π1000 025 102 = 19.63 kg/s. Q = AV=π × .0252 × 10 = 0 01963. . m / s3 4.23 &min = ρA1V1 + ρA2V2. 200 = 1000 π × .0252 × 25 + 1000 Q2. ∴Q2 = 0 1509. . m / s3

4.24 ρ11

1

40 1441716 520

= =××

pRT

= .006455 slug/ft3. ρ27 144

1716 610=

××

= .000963 slug/ft3.

& .& .

( / )..m AV V m

A= ∴ = =

×ρ

ρ π 1

1 12

22 144 006455

∴V1 = 355 fps.

& . . ( / ) .m V2 20 2 000963 2 3 144= = × × ∴V2 = 4984 fps.

54

4.25 ρ ρ ρ ρ1 1 1 2 2 2 11

2500

287 3934

1246287 522

8 317A V A Vp

RT= = =

×= =

×=.

..433 .

..

kgm

kgm3 3

4.433 π × .052 × 600 = 8.317 π × .052 V2. ∴V2 = 319.8 m/s. &m A V= ρ1 1 1 = 20.89 kg/s. Q A V1 1 1= = 4 712. . m / s3 Q2 = 2.512 . m / s3 4.26 ρ ρ1 1 1 2 2 2A V A V=

p

RTA V

pRT

A V1

11 1

2

22 2=

200293

0 05 40120

0 03 1202

2

2π π× × = × ×. . .T

∴ = −T2 189 9 83. . K or Co

4.27 a) A V A V1 1 2 2= . (2 × 1.5 + 1.5 × 1.5) 3 =πd2

2

42× . ∴d2 = 3.167 m

b) (2 × 1.5 + 1.5 × 1.5) 3 =πd2

2

422

× . ∴d2 = 4.478 m

c) (2 × 1.5 + 1.5 × 1.5) 3 =13 2

866 22πRR

R− ×

×. .

∴R = 3.581 m. ∴d2 = 7.162 m

4.28 (A) Refer to the circle of Problem 4.27:

2 375.7 2( 0.4 0.10 0.40 sin75.5 ) 3 0.516 m /s.

360Q AV π

×= = × × − × × × =o

4.29 a) vrr

r V vdArr

rdr rrr

drrr r

= −

= = −

= −

∫∫ ∫10 1 10 1 2 20

002

000

2

00

00 0

. . π π π

∴ = −

=V

rr r202 3

1030

202

02

= 3.333 m/s.

& . .m A V= = × × ×ρ π1000 04 3 332 = 16 75. .kg / s Q = AV = 0 01675. . m / s3

b) vrr

r Vrr

rdrr rr

= −

= −

= −

∫10 1 10 1 2 20

2 4

2

02 0

22

02

0

02

020

. . π π π ∴V = 5 m/s

& .m A V= = × × ×ρ π1000 04 52 = 25 13. kg / s. Q = AV = 0 02513. m / s.3

c) v rr

r Vrr

rdr rr

r

= −

= −

+∫20 1 20 1 2 10 4

002

0202

0

0

. / ./

π π π ∴V = 5.833 m/s

& . .m AV= = × × ×ρ π1000 04 58332 = 29 32. kg / s. Q =0 02932. . m / s3

θ R

cosθ = 1/2 θ = 60o

55

4.30 a) Since the area is rectangular, V = 5 m/s.

& . .m A V= = × × ×ρ 1000 08 8 5 = 320 kg / s. Q = &mρ

= 0 32. . m / s3

b) vyh

yh

= −

40

2

2 with y = 0 at the lower wall.

∴ = −

= ×∫Vhw

yh

yh

wdyh

wh

40 406

2

20

. ∴V = 6.667 m/s.

& . . .m A V= = × × ×ρ 1000 08 8 6 667 = 426 7. kg / s. Q = 0 4267. . m / s3

c) V × .08 = 10 × .04 + 5 × .02 + 5 × .02. ∴V = 7.5 m/s.

& . . .5m A V= = × × ×ρ 1000 08 8 7 = 480 kg / s. & &Q

m=

ρ = 0 48. . m / s3

4.31 a) A V v dA1 1 2= ∫ . π π π×

× = −

=∫

124

6 1 2 24

2

0

2

02

020

vrr

rdr vrr

max max .

With r01

24= , v max = 12 fps. ∴ =v r( ) 12 1 576 2( ) .− r fps

b) A V v dA1 1 2= ∫ . 1

126 1

43

2

2× × = −

=

−∫w v

yh

wdy v wh

h

h

max max .

With h = 124

, v max = 9 fps. ∴ =v y( ) 9 1 576 2( ) .− y fps

c) 1 1 2 .A V v dA= ∫ 0 22

2 0max max2

0 0

0.01 2 1 2 2 .4

r rrv rdr v

rπ π π

× × = − = ∫

With r0 = 0.01 m, maxv = 4 m/s. ( )v r∴ = 24(1 10000 ) m/s.r−

d) n 2

max max24

0.02 2 1 .3

h

h

y hw v wdy v w

h−

× × = − = ∫

With h = 0.01 m, maxv = 3 m/s. ( )v y∴ = 23(1 10000 ) m/s.y−

4.32 If dm dt/ ,= 0 then ρ ρ ρ1 1 1 2 2 2 3 3 3A V A V A V= + . In terms of &m Q2 3 and this becomes, letting ρ ρ ρ1 2 3= = ,

1000 0 02 12 1000 0 0122× × × = + ×π . & . .m ∴ =& . .m 2 5 08 kg / s

4.33 v dA A Vr

10

2 2

1

∫ = . vrr

rdrr

max . .0

2

12

21

1 2 0025 2∫ −

= × ×π π

∴ = × ×2 0054

0025 22

2π πvmax. . . ∴v max = 1 m/s. ∴ =v r( ) 1

005

2

2−

r.

.m / s

56

4.34 & & & . . ( ) . & ..

m m m y y dy min out= + × × × = − + × × ×

+∫ ρ ρ ρ2 2 10 10 20 100 2 1 2 102

0

1

Note: We see that at y = 0.1 m the velocity u(.1) = 10 m/s. Thus we integrate to y = 0.1, and between y = 0.1 and 0.2 the velocity u = 10.

443

2ρ ρ ρ= +

+ & .m ∴ =& .m 0 6667ρ = 0.82 kg/s.

4.35 V h u y dyh

1 10

= ∫ ( ) . 10 05 10 20 100 2

0

× = −∫. ( )y y dyh

= −

10 10100

32 3h h .

∴666.7 h3 − 200 h2 = −1. This can be solved by trial-and-error: h = .06: −.576 ? −1. h = .07: −.751 ? −1. h = .08: −.939 ? −1. h = .083: −.997 ? −1.

h = .084: −1.016 ? −1. ∴h = 0.0832: or 8.32 cm. Note: Fluid does not cross a streamline so all the flow that enters on the left leaves on the right. The streamline simply moves further from the wall.

4.36 ( )& . . ( )/

m VdA y y y dy= = − − ×∫∫ ρ 2 2 1 3545 6 9 2 50

1 32

( )= − − +∫22 6 2 127 9 3 192 2 3

0

1 3

y y y y dy. ./

= 4.528 slug/sec.

V u= = × =23

23

243max .fps (See Prob. 4.31b).

ρ =+2 2 1 942

. . = 2.07 slug/ft3. ∴ = × × ×

ρV A 2 0743

513

. = 4.6 slug/sec.

Thus, ρV A m≠ & since ρ = ρ(y) and V = V(y) so that V Vρ ρ≠ . 4.37 A V A V1 1 2 2= . π π× × = × ×. ( . . ) .01 8 2 2 042

2V cos 30 o ∴V2 = 0.05774 m/s.

4.38 2000 43

0015 9000 533

3

3

× × × ×π .m of H Om of air

m of airs

2 = 1.5 × (1.5h). ∴h = 0.565 m.

4.39 Use Eq. 4.3.3: 0 1 1 1 1= − + ⋅∫∂ρ∂

ρt

d V V n Av

$ . v

V n V1 1 1⋅ = −$ .

∴ = −ρ∂ρ∂1 1 1A Vt

Vtire . ( . ).

37 14 7 1441716 520

196

180 172+

×× ×

× = ×π

∂ρ∂t

57

∴ =∂ρ∂t

3 01 10 5. .×−

− slugft sec3

4.40 & & & .m m min = +2 3 V1 = 20 m/s (see Prob. 4.31c). 20 1000 02 10 1000 022 2

3× × = + ×π π. . .V ∴ V3 = 12.04 m/s.

4.41 0 2 3 1= + = + + −ddt

m m ddt

m m m mc v net c v. . . .& & & &

∴ = − − = × × × − − × ×ddt

m m m mc v. . & & & . .1 2 32 21000 02 20 10 1000 02 10π π

= 2.57 kg/s. 4.42 The control surface is close to the interface at the instant shown. ∴Vi = interface velocity. ρ ρe e e i i iA V A V= .

1 15 3008000

287 67312 2.5 .

..× × × =

××π π V i

∴Vi = 0.244 m/s. 4.43 Assume an incompressible flow: 4 1 2 2Q A V= . 4 1500 60 2 4 2× = ×/ ( ) .V ∴ =V2 12.5 . fps

4.44 For an incompressible flow (low speed air flow)

udA A VA

=∫ 2 2

1

. 20 0 8 0 151 5 22

0

0 2

y dy V/.

. . .× = ×∫ π

20 0 856

0 2 0 156 5 22× = ×. . . ./ π V ∴ =V2 27 3. . m / s

4.45 A V v dA A Ve e1 1 2+ =∫

π π π( . . ).

..

01 0 025 4 200 10 025

2 0 12 22

22

0

0 025

− × + −

= ×∫

rrdr Ve

0 1178 01963 0 0314. . . .+ = Ve ∴ =Ve 10 0. . m / s 4.46 Draw a control volume around the entire set-up:

0 2 2 1 1= + −dm

dtV A V Atissue ρ ρ

= +−

−& & ( tan ) &m

d dh h htissue ρπ ρπ φ2

2 2

2 12

14

Ve nn

V i

58

or

& & & tan .m d d h h htissue =−

+

ρπ φ

222

2 12

12

4

4.47 The width w of the channel is constant throughout the flow. Then

0 2 2 1 1= + −dmdt

A V A Vρ ρ . 0 2 2 1 1= + −ddt

whL A V A V( )ρ ρ ρ

0 100 0 2 8 4 0 2= × + × − ×ρ ρ ρdhdt

w w w. . . ∴ =& . .h 0 008 m / s

4.48 0 2 2 1 1= + −dmdt

A V A Vρ ρ

= + × × − × −& ( . . / ).m 1000 0 003 0 02 10 10 602 6π ∴ = × −& . .m 3 99 10 4 kg / s

4.49 ρ ρ1 1 1 2 2 2A V A V= . &m A V1 2 2 2= ρ . 400 10 900 0 2 0 0510 100 6 2e Ve

− −× × = × ×/ . . .π ∴ =Ve 207 m / s.

4.50 0 3 3 1 1 1 2= + − −dmdt

Q A V mρ ρ & where m Ah= ρ .

a) 0 1000 0 6 1000 0 6 60 1000 0 02 10 102 2= × + × − × × −π π. & . / . .h ∴ =& .h 0 0111 m / s or 11.1 mm / s

b) 0 1000 0 6 1000 0 01 0 202= × + × − −π . & . .h ∴ =& . .h 0 00884 m / s or 8.84 mm / s

c) 2 20 1000 0.6 1000 1.0/60 1000 0.02 5 10.hπ π= × + × − × × −& 0.000339 m/s or 0.339 mm/s.h∴ =& 4.51 A V A V1 1 2 2= where A 2 is an area just under the top surface.

a) π π× × = ×−0 02 10 602 10 2. ( tan )/e hdhdt

t o

∴ = −h dh e dtt2 100 001333. ./ ∴ = − +−h e t3 100 04 0 04. . ./ Finally, h t e t( ) . ( ) ./ /= − −0 342 1 10 1 3

b) 0 04 10 10 60 1010. ( tan ) &/× × = ×−e h ht o

∴ = −hdh e dtt0 2309 10. ./ ∴ = − +−h e t2 104 62 4 62. . ./ Finally, /10 1/2( ) 2.15(1 ) .th t e−= −

59

4.52 &W T pAVdudy

A belt= + +ω µ

= × × + × × × + × × × ×−20 500 2 60 400 0 4 0 10 181 10 100 0 0 85π / . .5 . .5 . = + + =1047 800 0 000724 1847. W 4.53 If the temperature is essentially constant, the internal energy of the c.v. does not change and the flux of internal energy into the pipe is the same as that leaving the pipe. Hence, the two integral terms are zero. The losses are equal to the heat transfer exiting the pipe. 4.54 80% of the power is used to increase the pressure while 20% increases the internal energy ( &Q = 0 because of the insulation). Hence,

& ~ . &m u W∆ = 0 2 1000 0 02 4 18 0 2 500× × = ×. . . .∆T ∴ =∆T 0 836. . Co

4.55 (D) 2 22 1

2PW V VQ gγ

−=

&2 1 1200 200

. .0.040

Pp p Wγ γ γ− −

+ =×

&

4040 kW and energy req'd = 47.1 kW.

0.85PW∴ = =&

4.56 & .WQ H

PP

p

=γη

5 7469800 200 87

× =× ×Q

.. ∴ =Q 0 01656. . m / s3

4.57 − = − ×&

&. .

Wmg

T 40 0 89

a) & . .WT = × × × =40 0 89 200 9 81 69 850 W

b) & . ( / ) .WT = × × × =40 0 89 90 000 60 9 81 523 900 W c) & . ( / ) .WT = × × × × =40 0 89 8 10 3600 9 81 7766 100 W

4.58 10000000. 0.89 50. 1.273 m/s

100 3 60 9.8T

TW

z VAVg V

ηρ

− = ∆ = × ∴ =× × × ×

&

4.59 V

gp

zV

gp

z12

11

22

222 2

+ + = + +γ γ

.

122 32 2

636

64 4

2 2

22 2×

+ = +. .

.h

h

8 23620 1

22 2..

.= +h

h Continuity: 3 × 12 = h2 V2.

3 ft

V1h2

V2

60

This can be solved by trial-and-error. 2 8': 8.24 ? 8.31h = 2 7.9': 8.24 ? 8.22h = Qh2 = 7 93. .'

2 1.8': 8.24 ? 8.00h = 2 1.75': 8.24 ? 8.31h = Qh2 = 1 76. '.

4.60 V

gz

Vg

z hL12

122

22 2+ = + + . ∴

×+ =

×+ +

42 9 81

216

2 9 810 2

2

22 2. .

. .h

h

∴ 2 615 0 815 22

2. . / .= +h h Trial-and-error provides the following:

h2 2 2 615 2 63= =.5: . ? . h2 2 45 615 2= =. : . ? .59. 2 ∴ =h2 2 47. m

h2 0 65 2 615 2= =. : . ? .58 h2 0 64 615 2 63= =. : . ? . . 2 ∴ =h2 0 646. m

4.61 Manometer: Position the datum at the top of the right mercury level.

9810 4 98102

1000 9810 13 6 4 9810 22 222

1× + + + × = × × + × +. ( . ) .z pV

p

Divide by γ = 9810: . . . .42

13 6 4 222 2

21+ + + = × + +z

p Vg

pγ γ

(1)

Energy: V

gp

zV

gp

z12

11

22

222 2

+ + = + +γ γ

. (2)

Subtract (1) from (2): With z1 = 2 m, V

g12

212 6 4= ×. . . ∴V1 = 9.94 m/s

4.62 The manometer equation (see Prob. 4.61) is

0 42

13 6 4 222 2

21. . . .+ + + = × + +z

p Vg

pγ γ

(1)

Energy: V

gp

zV

gp

zV

g12

11

22

22

22

2 20 05

2+ + = + + +

γ γ. . (2)

Subtract (1) from (2): With z1 = 2 m, and with V2 = 4V1 (continuity)

18

212 6 0 41

2.. . .

Vg

= × ∴V1 = 7.41 m/s.

4.63 (A) 220

V=

21 2 1

2

V p p

g

− −+

22

2120

. 0 . 7 200000 Pa.2 9.8 9810

pp

γ−

= + ∴ =×

4.64 Q = 120 × 0.002228 = π × 112

2

1

V . ∴V1 = 12.25 fps.

61

Continuity: π π×

= ×

112

112

2

1

2

2V V.5

. ∴V2 = 5.44 fps.

Energy: V

gp V

gp V

g12

1 22

2 12

2 20 37

2+ = + +

γ γ. .

∴ = × + −

p2

2 2

60 144 62 4 0 6312 25

64 45 4464 4

. ..

..

. = 8702.9 psf or 60.44 psi

4.65 Q = 600 × 10-3/60 = π × .022 V1. ∴V1 = 7.958 m/s.

320.023 3

3 3 20

1 110 1

0.02 6.67 0.02

yV dA wdy

AV wα

= = − ∫ ∫ × ×

VA VA2

1 1

2

2

2

04 7 95806

= =×. .

. = 3.537 m/s.

Energy: V

gp V

gp

hL12

1 22

2

2 2+ = + +

γ γ.

∴ =−

×+

−hL

7 958 32 9 81

690 000 700 0009810

2 2. .537.

= 1.571 m

4.66 V Q A1 1 2

0 0803

= =×

/..π

= 28.29 m/s. ∴V2 = 9V1 = 254.6 m/s.

Energy: Vg

p Vg

p Vg

12

1 22

2 12

2 22

2+ = + +

γ γ. .

∴ =×

−×

p1

2 2

9810254 62 9 81

0 828 292 9 81

..

..

. = 32 1 106. .× Pa

4.67 a) Across the nozzle: 2 2

1 2.07 .025 .V Vπ π× = × ∴V2 = 7.84 V1.

Energy: Vg

p Vg

p12

1 22

2

2 2+ = +

γ γ.

22

1 17.84 1

9810 .2 9.81

p V−

∴ =×

For the contraction: π π× = ×. . .07 0521

23V V ∴V3 = 1.96 V1.

Energy: Vg

p Vg

p12

1 32

3

2 2+ = +

γ γ.

Manometer: γ γ× + = × +. . . .15 136 151 3p p ∴ = × +p p1 312 6 15γ γ

. . .

Subtract the above 2 eqns: Vg

Vg

Vg

12

32

2 12

212 6 15

2196

2+ × = =. . . .

∴ − = × ×( . ) . . .196 1 12 6 15 2212V g ∴V1 = 3.612 m/s. ∴p1 = 394 400 Pa.

62

From the reservoir surface to section 1:

V

gp

zV

gp

z02

00

12

112 2

+ + = + +γ γ

H = +3 61219 62

2..

394 4009810

= 40.0 m.

b) Manometer: γ γ× + = × +. . . .2 136 21 3p p ∴ = × +p p1 312 6 2γ γ

. . .

Energy: Vg

p Vg

p12

1 32

3

2 2+ = +

γ γ. Also, V3 = 1.96 V1.

∴ + × =V

gV

g12 2

12

212 6 2

1962

. ..

. ∴V1 = 4.171 m/s.

The nozzle is the same as in part (a): ∴p1 = 534 700 Pa. From the reservoir surface to the nozzle exit:

V

gp

zV

gp

z02

00

22

222 2

+ + = + +γ γ

. ∴ = =×

HV

g22 2

232 7

2 9 81.

. = 54.5 m.

4.68 a) Energy: V

gp

zV

gp

z02

00

22

222 2

+ + = + +γ γ

. ∴ = = × ×V gz2 02 2 9 81 2 4. . = 6.862 m/s.

Q = AV = .8 × 1 × 6.862 = 5 49. . m / s3 For the second geometry the pressure on the surface is zero but it increases with depth. The elevation of the surface is 0.8 m.

∴ = +zV

gh0

22

2. ∴ = − = × ×V g z h2 02 2 9 81 2( ) . = 6.264 m/s.

∴Q = .8 × 6.264 = 5 01. . m / s3 Note: z0 is measured from the channel bottom in the 2nd geometry.

∴z0 = H + h.

b) V

gp

zV

gp

z02

00

22

222 2

+ + = + +γ γ

. ∴ = = × × +

=V gz2 02 2 32 2 622

21 23. . fps.

∴Q = AV = (2 × 1) × 21.23 = 42.5 cfs. For the second geometry, the bottom is used as the datum:

∴ = + +zV

gh0

22

20 . ∴ = + −

Vg

H h h22

2( ) .

∴ = = × ×V gH2 2 2 32 2 6. = 19.66 fps. ∴Q = 39.3 cfs.

63

4.69 From the reservoir surface to the exit: Continuity:

V

gp

zV

gp

z KV

g02

00

22

22

12

2 2 2+ + = + + +

γ γ. V V1 2

2

2

0308

=..

= .1406 V2.

102

51406

222 2

22

= + ×V

gV

g.

∴V2 = 13.36 m/s. ∴Q = 13.36 × π × .0152 = 0 00944. . m / s3 The velocity in the pipe is V1 = 1.878 m/s.

Energy 0 → A: 1018782 9 81 9810

818782 9 81

32 2

=×

+ +×

+.

..

..

.pA ∴pA = 65 500 Pa.

Energy 0 → B: 1018782 9 81 9810

2 018782 9 81

102 2

=×

+ +×

+.

..

..

.pB ∴pB = −5290 Pa.

Energy 0 → C: 1018782 9 81 9810

12 2 818782 9 81

2 2

=×

+ + +×

..

..

..

pC ∴pC = −26 300 Pa.

Energy 0 → D: 1018782 9 81 9810

0 518782 9 81

2 2

=×

+ + +×

..

..

.pD ∴pD = 87 500 Pa.

4.70 V

gp

zV

gp

z02

00

22

222 2

+ + = + +γ γ

. 80 0009810

+ =×

42 9 81

22V.

. ∴V2 = 19.04 m/s.

a) Q A V= = × ×2 22025 19 04π . . = 0 0374. . m / s3

b) Q A V= = × ×2 2209 19 04π . . = 0 485. . m / s3

c) Q A V= = × ×2 2205 19 04π . . = 0 1495. . m / s3

4.71 a) p

zV

gV

g0

022

12

21

2γ+ = + .54 . 80 000

9810+ = +4

162

12

12

12V

gV

g.54 . ∴V1 = 3.687 m/s.

Q A V= = × ×1 1205 3 687π . . = 0 0290. . m / s3

b) A V A V1 1 2 2= . V V V1

2

2 2 20905

3 24= =..

. .

80 0009810

+ = +42

2 33 24

222 2

22V

gV

g.

.. ∴V2 = 3.08 m/s. ∴ Q A V= 2 2 = 0 0784. . m / s3

c) 80 0009810

+ = +42

12

22

22V

gV

g.5 . ∴V2 = 9.77 m/s. ∴ Q A V= 2 2 = 0 0767. . m / s3

4.72 (C) Manometer: 2

21 22

VH p g p

gγ ρ+ = + or

22

19810 0.02 .2V

p gg

ρ× + =

Energy: 2 1000007.96

. 3.15.2 9.81 9810

K K= ∴ =×

64

Combine the equations: 2

119810 0.02 1.2 . 18.1 m/s.

2V

V× = × ∴ =

4.73 Manometer: .6.13 21 pzHpzH ++=++ γγγγ ∴ = +p

Hp1 212 6

γ γ. .

Energy: p V

gp V

g1 1

22 2

2

2 2γ γ+ = + .

Combine energy and manometer: 12 62

22

12

. .HV V

g=

−

Continuity: Vd

dV2

12

22 1= . ∴ = × −

V H g

dd1

2 14

2412 6 2 1. / .

∴ = =×

−

Q V

d H gd d

d112

14

24

1 2

12

4 412 6 2

1π

π ./

/

= 12 35 12

22

14

24

1 2

./

d dH

d d−

4.74 Use the result of Problem 4.73:

a) Q = 12.35 × . ..

. .

/

16 082

16 082 2

4 4

1 2

×−

= 0 0365. . m / s3

b) Q = 12.35 × . ..

. .

/

24 084

24 082 2

4 4

1 2

×−

= 0 0503. . m / s3

c) Using English units with g = 32.2: Q d dH

d d=

−

22 37 1

222

14

24

1 2

. ./

Q = ×

−

22 3712

14

10 1225

2 2

4 4

1 2

./

.5 .

/

= 1.318 cfs.

d) Q = 22.37 ×113

15 121 3333

22

4 4

1 2

×

−

/.

/

= 2.796 cfs.

4.75 (B) 2

20.040

. 7.96 m/s.2 0.04

LV p Q

h K Vg Aγ π

∆= = = = =

×

2 1000007.96

. 3.15.2 9.81 9810

K K= ∴ =×

4.76 a) Energy from surface to outlet: V

gH2

2

2= . ∴ =V gH2

2 2 .

Energy from constriction to outlet: p V

gp V

g1 1

22 2

2

2 2γ γ+ = + .

65

Continuity: V V1 24= . With p1 = pv = 2450 Pa and p2 = 100 000 Pa,

24509810

162 9 81

2100 000

98101

2 9 812+

×× = +

××

. ..gH gH

∴H = 0.663 m.

b) With p1 = 0.34 psia, p2 = 14.7 psia,

..

..

.34 144

62 4162

214 7 144

62 41

22

×+ =

×+

ggH

ggH ∴H = 2.21 ft.

4.77 Continuity: V V1 24= . Energy surface to exit: V

gH2

2

2= .

Energy constriction to exit: p V

gp V

gv

γ γ+ = +1

22 2

2

2 2.

∴ = +−

= − = − × ×p pV V

gp Hv 2

22

22

2

162

15 100 000 15 65 9810γ γ . = 4350 Pa.

From Table B.1, T = 30°C.

4.78 Energy surface to surface: z z hL0 2= + . ∴ = +30 20 22

22Vg

.

Continuity: V1 = 4V2. ∴V12 = 160 g. ∴V2

2 = 10 g.

Energy surface to constriction: 30160

294 0009810 1= +

−+

gg

z( )

∴z1 = −40.4 m. ∴H = 40.4 + 20 = 60.4 m.

4.79 Continuity: V V2

2

2 1106

= = 2.778 V1.

Energy: V

gp V

gp1

21 2

22

2 2+ = +

γ γ.

Vg

Vg

12 2

12

2200 2 778

224509810

+ = + 000

9810.

.

∴V1 = 7.67 m/s. ∴Q = π × .052 × 7.67 = 0 0602. . m / s3 4.80 Velocity at exit = Ve . Velocity in constriction = V1 . Velocity in pipe = V2 .

Energy — surface to exit: V

gHe

2

2= . ∴ =V gHe

2 2 .

Continuity across nozzle: VDd

Ve2

2

2= . Also, V V1 24= .

Energy — surface to constriction: HV

gpv= +1

2

2 γ.

66

a) 51

216

22 5

974

4= × × ×

+

−g

Dg

..

5509810

∴ =D 0 131. m

b) 151

216

8 122 15

34 14 7 14462 4

4

4= × ×

+

−g

Dg

( / )(. . )

.. ∴ = ′ ′′D 0 446. or 5.35

4.81 Energy — surface to exit: 32

42

117722

22

22= + ∴ =

Vg

Vg

V. . .

Energy — surface to “A”: 31177

2 9 811176 100

3 111772 9 81

2

=×

+−

+ + +×

..

( ) .5.

..

0009810

H

∴ =H 8.57 . m

4.82 & . .m A V= = × ×

× =ρ π1 94

112

120 5 0792

slug / sec.

& . .. .

/ . , .WP = ×−

×+

×

=5 079 32 2 30 120

2 32 2120 144

62 40 85 12 950

2 2 ft - lbsec

or 23.5 Hp

4.83 & . .m A V= = × × × =ρ π1000 02 40 50 272 kg / s.

2040

2 9 81 98100 82

2

000 = 50.27 9.81102

×−

×+

.

/ . .∆p

∴ = ×∆p 1 088 106. . Pa

4.84 (C) 2 22 1

2PW V VQ gγ

−=

&.

pγ∆

+

160.040 400 16 kW. 18.0 kW.

0.89P

PW

W Q pη

= ∆ = × = = =&

&

4.85 − = × ×−×

+−

×& .

..

. .WT 2 1000 9 810 10 22 9 81

6000 87

2 0009810

∴ = ×& . .WT 1304 10 6 W

We used V Q A2 2 2

225

10 2= =×

=/.

.π

m / s.

4.86 V V1 2 2 2

4503

15 94503 75

10 19=×

= =×

=π π

. ..

. . fps fps

−

× = × ×

−×

+−

10 000

1746

550 450 194 32 21019 159

2 32 218 140 144

62 4

2 2

,.

. .. .

.( )

..ηT

∴ =ηT 0 924.

67

4.87 a) & & & ( ) .Q W mgV V

gp p

z zcg

T TSv− =

−+ − + − + −

22

12

2

2

1

12 1 2 12 γ γ

The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13.

γ γ11

12

2 2

85 9 81287 293

9 92600 9 81

28720

= =××

= =×

=p gRT T T

..

. ..

.. N / m

5003

∴ − − ××

+ − + −

(

..5

.( ) .1

2 9 81600

2085 716

9 812932

2 500 000) = 5 9.81200 000

500 000

9.92

2 TT

∴ =T2 572 K or 299 C .o

Be careful of units! p cv2 600 716= =⋅

000 Pa, J

K kg.5

b) −60 000 + 1 500 000 = same as above. ∴ =T2 560 K or 287 C.o

4.88 γ γ11

12

14 7 144 32 21716 520

0 076460 144 32 2

1716 7600 213= =

× ××

= =× ×

×=

p gRT

. .. .

.. .

lbft

lbft3 3

c mg AVg AVv = = = = × ×

× =4296 213 124

600 6972ft - lb

slug - R lb /sec.o . & . .ρ γ π

Use Eq. 4.5.17 with Eqs. 4.5.18 and 1.7.13:

& & & ( ) .Q W mgV V

gp p c

gT T z zc

v+ =−

+ − + − + −

22

12

2

2

1

12 1 2 12 γ γ

− × × + =×

+×

−×

+ −

10 778 697 697

6002 32 2

60 144213

14 7 1440764

429632 2

300 602

. & .. .

.. .

( )W c

∴ =& .Wc 40 600ft - lbsec

or 73.8 Hp

4.89 Energy — surface to exit: − = − +

& & . .W mg

Vg

VgT Tη 2

222

220 4 5

2

V mg Q2 2

156

13 26 15 9810 147=×

= = = × =π

γ.

. & m / s. 150 N / s.

− × =×

− +×

∴ =& .

..

..

. & .W WT T08 1472 981

20 4 513262 9 81

2

15013.26

5390 kW2

4.90 (D) 2 24.58 7.16

36.0 15 3.2 . 416 000 Pa2 9.81 9810 2 9.81

BB

pp+ = + + ∴ =

× ×

In the above energy equation we used

2

20.2

with 4.42 m/s.2 0.2

LV Q

h K Vg A π

= = = =×

68

4.91 Energy — surface to “C”:

& . &.

..

. .W mgP × + × =×

+ +×

8 10

102 9 81

2007 7

102 9 81

770 52 2 000

9810

( & . . .5 & .mg AVg WP= = × × × × = ∴ =ρ π1000 05 10 9 81 770 522 N / s.) 700 W

Energy — surface to “A”: 3010

2 9 81 98101

102 9 81

1692 2

=×

+ +×

∴ =.

.5.

. .p

pAA 300 Pa

Energy — surface to “B”: 2 2 2

2 2B O B O B

P P B OV V p p V

W mg z z Kg g

ηγ

− −= + + − +

& &

522 9 81 9810

30 1510

2 981706

2

700 .8 = 770.510

100 Pa2

××

+ − +×

∴ =

..

.. .

ppB

B

4.92 Manometer: γ γ γ γ ρ× + + = × + + +2012

13 62012 21 1 2 2

22

z p z pV

. .

∴ + + = × + + +2012

13 62012 21

12

2 22

zp

zp V

gγ γ. .

Energy: V

gz

pH z

p VgT

12

11

22 2

2

2 2+ + = + + +

γ γ.

∴ = × + − =×

=2012

13 62012 2

1813

51612

1 2. . .V

gH VT fps.

π

∴ = × +×

= = = × × ×H W Q HT T T T12 62012

5162 32 2

62 3 62 4 18 9 62 32

..

.. ' . & . . . γ η

= 62 115, .980 ft - lbsec

or Hp

4.93 Energy—across the nozzle: 2 2 2

1 1 2 22 1 12

5. 6.25 .

2 2 2

p V p VV V V

g gγ γ+ = + = =

2 2 2

1 11

6.25400 000. 4.58 m/s

9810 2 9.81 2 9.81V V

V∴ + = ∴ =× ×

, 7.16 m/sAV = , 2 28.6 m/s.V =

Energy—surface to exit:

2 2 228.6 4.58 7.16

15 1.5 3.2 .2 9.81 2 9.81 2 9.81PH + = + +

× × × 36.8 m.PH∴ =

2/ 9810 ( .01 ) 28.6 36.8/.85 3820 W.P P PW QHγ η π∴ = = × × × × =&

69

Energy —surface to “A”:

2 27.16 7.16

15 3.2 . 39 400 Pa2 9.81 9810 2 9.81

AA

pp= + + ∴ =

× ×

Energy —surface to “B”:

2 24.58 7.16

36.0 15 3.2 . 416 000 Pa2 9.81 9810 2 9.81

BB

pp+ = + + ∴ =

× ×

4.94 (A) VQA

= =×

=0 1

0419 892

..

.π

m / s.

Energy —surface to entrance: HV

gp

z KV

gP = + + +22

22

22

2 2γ.

∴ =×

+ + +×

=H P19 892 9 81

18050 5 6

19 892 9 81

201 42 2.

..

..

. 000

9810 m.

∴ = = × × =& / . . / . .W QHP P Pγ η 9810 0 1 2014 0 75 263 000 W

4.95 Energy —surface to exit: 102

2 22

22

22

22

= + + +V

gp

zV

gγ. .

∴ = = = × ∴ =V Q d d2 22

27 83 0 02 7 83 4 0 0570. . . / . . m / s. m .π 4.96 Depth on raised section = y 2 . Continuity: 3 3 2 2× = V y .

Energy (see Eq. 4.5.21): 32

32

0 42

22

2gV

gy+ = + +( . ).

∴ = + − + =3 0599

23 059 4 128 0

2

22 2 2

322. , . .

g yy y y

or

Trial-and-error: 22

2

2.0: .11 ? 0. 1.85 m.

1.8: .05 ? 0.

yy

y

= − ∴ == +

22

2

2.1: .1 ? 0. 2.22 m.

2.3: .1 ? 0.

yy

y

= − ∴ == +

The depth that actually occurs depends on the downstream conditions. We cannot select a “correct” answer between the two. 4.97 Mass flux occurs as shown. The velocity of all fluid elements leaving the top and bottom is approximately 32 m/s. The distance where u y= = ±32 m /s is 2 m.

m1

m3

m2

m3

.

.

..

70

To find &m 3 use continuity:

& & & . ( ) & .m m m y dy m1 2 32

30

2

2 4 10 32 2 28 10 2= + × × = + +∫ ρ ρ

∴ = − × +

=& . .m 3 640 10 28 283

53 3ρ ρ ρ

Rate of K.E. loss = & &mV

mV u

dy112

312 3

0

2

22

22

210− − ∫ρ

= − − +∫1280322

53 3 32 10 282

2 2 3

0

2

ρ ρ ρ. ( )y dy

[655360 54579 507320] 115000 .Wρ= − − = 4.98 The average velocity at section 2 is also 8 m/s. The kinetic-energy- correction factor for a parabola is 2 (see Example 4.9). The energy equation is:

V

gp V

gp

hL12

12

22

2

2 2+ = + +

γα

γ.

82 9 81

1502

82 9 81

1102 2

×+ =

×+ +

. ..

0009810

0009810

hL

∴ =hL 0 815. . m

4.99 VA

VdA y dy= = + = × +

=∫∫

1 12

2812

28 223

29 3323

0

2

( ) . m / s

α = =×

+∫ ∫1 1

2 29 33283

33

2 3

0

2

A VV dA y dy

.( )

[ ]=×

× + × × + × × + =1

2 29 3328 2 3 28 2 3 3 28 2 5 2 7 1 0053

3 2 3 5 7

./ / / .

4.100 a) 2 2 40.01

2 2 2 20

1 1 20 0.01 0.0110 1 2 5 m/s

20.01 0.01 0.01 4 0.01

rV VdA rdr

Aπ

π

= = − = − = ∫ ∫ × ×

απ

π= =× ×

−

∫ ∫

1 10 01 5

10 10 01

233

2 33

2

2

3

0

0 01

A VV dA

rrdr

. .

.

=×

−××

+××

−×

=

20000 01 5

0 012

3 0 014 0 01

3 0 016 0 01

0 018 0 01

2 002 3

2 4

2

6

4

8

6.. .

...

..

.

71

b) 2 30.02

2 20

1 1 10 0.0210 1 0.02 6.67 m/s

0.02 0.020.02 3 0.02

yV VdA wdy

A w

= = − = − = ∫ ∫ ×

320.02

3 33 3 2

0

1 110 1

0.02 6.67 0.02

yV dA wdy

AV wα

= = − ∫ ∫ × ×

=×

−××

+××

−×

=

10000 02 6 67

0 023 0 023 0 02

3 0 025 0 02

0 027 0 02

13

3

2

5

4

7

6. ..

.

...

..

.541

4.101 VA

VdAR

urR

rdr un

nn

n

nR

= = −

= −+

−+

∫∫

1 11 2 2

2 1 12

1

0ππmax

/

max

K E VV

dA urR

rdr u Rn

nn

n

nR

. . max

/

max=

= −

= −+

−+

∫∫ρ

ρπ ρπ

23

33 2

02 21 2

3 2 3

a) V u u= − −

=25

1156

0 758max max.

K E R u R u. . .max max= −

=ρπ ρπ2 3 2 358

513

0 24

αρ

ρπ

ρπ= =

×=

K E

AV

R u

R u

. . .

..max

max12

0 2412

0 7581102

3

2 3

2 3 3

b) V u u= − −

=27

1578

0 817max max.

K E u R R u. . .max max= −

=ρπ ρπ3 2 2 3710

717

0 288

αρ

ρπ

ρπ=

=×

=K E

AVV

R u

R uu

. . .

..

.max

maxmax

2

2 3

22 2

2

0 288

0 8170 817

2

1 056

c) V u u= − −

=29

199

100 853max max.

K E R u R u. . .max max= −

=ρπ ρπ2 3 2 3912

921

0 321

αρ

ρπ

ρπ= =

×=

K E

AV

R u

R u

. . .

..max

max12

0 32112

0 8531034

3

2 3

2 3 3

72

4.102 Engine power = F V mV V

u uD × +−

+ −

∞

& ~ ~22

12

2 12

& & ( )m g F V mV V

c T Tf f D v= +−

+ −

∞

22

12

2 12

4.103 &W F VDη = ×

10

5930

1003600

01513401000

1003−

×

×

×

× = ×mkm

kgm

kJkg

kms

0003600

3

3 q f .

∴ =q f 48 030 kJ / kg

4.104 02 2

32222

22

12

11 2= + + − − − +α

γ γνV

gp

zV

gp

zLV

gD

0 22 9 81

0 35 3210 1809 81 0 02

2 6

2=×

− + ×××

−V V.

.. .

.

V V V Q2 514 4 3 434 0 0 235 7 37 10+ − = ∴ = = × −. . . . . m / s and m / s3

4.105 Energy from surface to surface: HV

gp

zV

gp

z K VgP = + + − − − +2

22

212

11

2

2 2 2γ γ.

a) HQ

QP = +× × ×

= +40 50 04 2 9 81

40 50 72

22

π . ..

Try Q H HP P= = =0 25 43 2 58. : . (energy). (curve) Try Q H HP P= = =0 30 44 6 48. : . (energy). (curve) Solution: Q = 0 32. . m / s3

b) HQ

QP = +× × ×

= +4020

0 04 2 9 8140 203

2

22

π . .

Try Q H HP P= = =0 25 52 7 58. : . (energy). (curve)

Solution: Q = 0 27. m / s3 Note: The curve does not allow for significant accuracy. 4.106 Continuity: A V A V A V1 1 2 2 3 3= + π π π× × = × × + × ∴ =0 06 5 0 02 20 0 03 11112 2 2

3 3. . . . .V V m / s Energy: energy in + pump energy = energy out

& & & &mV p

W mV p

mV p

P P112

12

22

23

32

3

2 2 2+

+ × = +

+ +

ρη

ρ ρ

73

1000 0 06 552

1200 85 1000 0 02 20

202

30022

22

π π× × +

+ = × × +

. . & .

0001000

0001000

W P

+ × × +

1000 0 03 11 11

11 112

50022

π . .. 000

1000

∴ =&WP 26 700 W 4.107 (A) After the pressure is found, that pressure is multiplied by the area of the

window. The pressure is relatively constant over the area.

4.108 V

gp V

gp1

21 2

22

2 2+ = +

γ γ.

2

2 1 12 4 .( /2)

dV V V

d= =

a) V V1

212

2 9 81200 16

2 9 81×+ =

×. ..

0009810

∴ =V1 5164. m / s.

( )p A F m V V1 1 2 1− = −& . 200 000 03 1000 03 5 164 4 5 164 51642 2 π π× − = × × × −. . . ( . . ).F ∴ =F 339 N.

b) V V1

212

2 9 81400 16

2 9 81×+ =

×. ..

0009810

∴ =V1 7 303. m / s.

400 000 03 1000 03 7 303 4 7 303 7 3032 2 π π× − = × × × −. . . ( . . ).F ∴ =F 679 N.

c) 2 2

1 116 200 000.

2 9.81 9810 2 9.81V V

+ =× ×

1 5.164 m/s.V∴ =

2 2200 000 .06 1000 .06 5.164(4 5.164 5.164).Fπ π× − = × × × − 1356 N.F∴ =

d) 2 2

1 11

16 30 144. 17.24 fps.

2 32.2 62.4 2 32.2V V

V×

+ = ∴ =× ×

2 2 230 1.5 1.94 (1.5/12) 17.24 (4 1). 127 lb.F Fπ π× × − = × × × − ∴ =

e) 2 2

1 11

16 60 144. 24.38 fps.

2 32.2 62.4 2 32.2V V

V×

+ = ∴ =× ×

2 2 260 1.5 1.94 (1.5/12) 24.38 (4 1). 254 lb.F Fπ π× × − = × × × − ∴ =

f) 2 2

1 11

16 30 144. 17.24 fps.

2 32.2 62.4 2 32.2V V

V×

+ = ∴ =× ×

2 2 230 3 1.94 (3/12) 17.24 (4 1). 509 lb.F Fπ π× × − = × × × − ∴ =

4.109 V

gp V

gp1

21 2

22

2 2+ = +

γ γ. V V V2

2

2 1 193

9= = .

V1

2

2 9 812 81

2 9 81×+ =

×. ..

000 0009810

V12

∴ =V12 50.

74

p A F m V V m V1 1 2 1 18− = − =&( ) & 2 045 1000 045 8 502 2 000 000π π× − = × × ×. .F ∴ =F 10 180 N .

4.110 2 2

1 1 2 2 .2 2V p V p

g gγ γ+ = + 0

2.01 .006 .15. 11.1 m/s.e eV V Vπ × = × × ∴ =

ΣF m V Vx x x= −&( ).2 1

a) V V V2

2

2 1 1108

1= = .562 . V V1

212

2 9 81400 2 441

2 9 81×+ =

×..

..

0009810

∴ =V1 23.56 . m / s

∴ − = −p A F m V V1 1 2 1&( ). 400 05 1000 05 23 232 2 000π π× − = × × ×. . .56(.562 .56).F ∴ =F 692 N.

b) V V V2

2

2 1 1106

2 778= = . . V

gV

g12

12

2400 7 716

2+ =

0009810

.. ∴ =V1 10 91. . m / s

400 05 1000 05 10 91 1778 10 912 2 000π π× − = × × ×. . . ( . . ).F ∴ =F 1479 N.

c) V V V2

2

2 1 1104

6 25= = . . V

gV

g12

12

2400 39 06

2+ =

0009810

.. ∴ =V1 4.585 . m / s

400 05 1000 05 4 5 25 42 2 000π π× − = × × ×. . .585( . .585).F ∴ =F 2275 N.

d) V V V2

2

2 1 1102

25= = . V

gVg

12

12

2400 625

2+ =

0009810

. ∴ =V1 1132. . m / s

400 05 1000 05 1132 24 11322 2 000π π× − = × × ×. . . ( . ).F ∴ =F 2900 N.

4.111 (C) 2 2

1 1 2 2

2 2V p V p

g gγ γ+ = +

2 2

1(6.25 1) 12.73

. 9810 3085000 Pa.2 9.81

p− ×

= × =×

21 1 2 1( ). 3085000 0.05 1000 0.1 12.73(6.25 1)p A F Q V V Fρ π− = − × × − = × × −

17500 N.F∴ =

4.112 V V pV V

g2 1 122

12 2 2

4 1202

62 4120 30

2 32 213= = =

−

=

−×

= fps. 080 psf.γ .

.,

2 2

1 1 2 11.5 1.5

( ) 13,080 1.94 30( 120 30) 1072 lb.12 12x xF p A m V V π π = − − = − × × − − =

&

4.113 V VV

gp V

gp V

gp

2 112

1 22

2 12

142 2

152

= + = + ∴ =

. . .γ γ γ

a) V12 2 9 81

15 9810200 000 26 67=

××

× =.

. . ∴ = =V V1 2516 20 7. . m / s, m / s.

p A F m V Vx x x1 1 2 1− = −&( ). ∴ = × + × × =Fx 200 000 04 1000 04 5 16 11392 2 2π π. . . . N

75

F m V Vy y y= −& ( ).2 1 ∴ = × × =Fy 1000 04 5 16 20 7 5372π . . ( . ) . N

b) V12 2 9 81

15 9810400 000 53 33=

××

× =.

. . ∴ = =V V1 27 30 29 2. . m / s, m / s.

p A F m V Vx x x1 1 2 1− = −&( ). ∴ = × + × × =Fx 400 000 04 1000 04 7 3 22802 2 2π π. . . . N F m V Vy y y= −& ( )2 1 = 1000 04 7 3 29 2 10712π × × × =. . ( . ) . N

c) V12 2 9 81

15 9810800 000 106 7=

××

× =.

. . ∴ = =V V1 210 33 41 3. . m / s, m / s.

2 2 2 21 1 1 1 800 000 .04 1000 .04 10.33 4560 N.xF p A A Vρ π π= + = × + × × =

F m Vy y= & ( )2 = 1000 04 10 33 41 3 21402π × × =. . ( . ) . N

4.114 V V2

2

2 14010

80= = m / s. V

gp V

gp1

21 2

22

2 2+ = +

γ γ

∴ =×

−×

= ×p1

2 269810

802 9 81

52 9 81

3 19 10. .

. Pa.

p A F m V Vx x1 1 2 1− = −& ( ). ∴ = × × − × × − =F 3 19 10 2 1000 2 5 80 5 3536 2 2. . . ( )π π 000 N. 4.115 A V A V1 1 2 2= . π π× × = −. (. . ) .025 4 025 022 2 2

2V

∴ =V2 1111. m / s. p V

gp V

g1 1

22 2

2

2 2γ γ+ = + .

2 2

111.11 4

9810 53700 Pa.2 9.81

p −

= = ×

1 1 2 1( ). p A F m V V− = −&

2 2 53 700 .025 1000 .025 4(11.11 4) 49.6 N.F π π∴ = × − × × − =

4.116 Continuity: . . . .7 1 71 2 2 1 V V V V= ∴ =

Energy: V

gp

zV

gp

z12

11

22

222 2

+ + = + +γ γ

V V

V V12

12

1 22 9 817

492 9 81

1 0 495 3 467×

+ =×

+ ∴ = =.

..

. . . , . m / s.

Momentum: F F R m V Vx1 2 2 1− − = −&( ) 9810 35 7 1 9810 05 0 1 1 1000 1 1 3 467 3 467 495× × − × × − = × × × −. (. .5) . ( . .5) (. .5) . ( . . )R x ∴ =Rx 1986 N.

∴R x acts to the left on the water, and to the right on the obstruction.

F

V2

p1A1

Fp1A1

F2

F1

Rx

76

4.117 Continuity: 6 2 301 2 2 1 V V V V= ∴ =. . . Energy (along bottom streamline):

V

gp

zV

gp

z12

11

22

222 2

+ + = + +γ γ

2 2

2 2/9006 0.2.

2 9.81 2 9.81V V

+ = +× ×

2 1 10.67, .36 m/s.V V∴ = =

Momentum: F F F m V V1 2 2 1− − = −&( ) 9810 3 6 4 9810 1 2 4 1000 2 4 10 67 10 67 36× × − × × − = × × × −( ) . (. ) (. ) . ( . . )F ∴ =F 618 000 N . (F acts to the right on the gate.) 4.118 a) 8 6 2 2× =. .V y F F m V V1 2 2 1− = −& ( ).

γ γ ρ× × − = ××

−

. . .

..3 6

26 8

8 682

22

wy

y w wy

γρ

236 4 8 8

66

4 8 8 29 812

2 2

22 2(. ) .

.. (. )

..

.− = ×−

∴ + =× ×

yy

yy y

y y y22

2 26 7 829 0 2+ − = ∴ =. . . .51 . m (See Example 4.12.)

b) y y yg

y V2 1 12

1 12 2 21

28 1

24 4

89 81

4 12 3 23= − + +

= − + + × ×

=. .

.. . . m

c) y y yg

y V2 1 12

1 12 2 21

28 1

22 2

832 2

2 20 6 12= − + +

= − + + × ×

=

.. . ft

d) y y yg

y V2 1 12

1 12 2 21

28 1

23 3

832 2

3 30 11= − + +

= − + + × ×

=

..54 . ft

4.119 Continuity: V y V y V y y y2 2 1 1 2 1 2 14 4= = ∴ =. .

Use the result of Example 4.12: y y yg

y V2 1 12

1 12

1 212

8= − + +

/

a) y 2 4 8 3 2= × =. . . m

1 / 2

2 21

1 83.2 .8 .8 .8 .

2 9.81V

= − + + × ×

∴ =V1 8 86. m / s.

b) y 2 4 2 8= × = ft.

812

2 28

32 222

12

1 2

= − + + × ×

..

/

V ∴ =V1 25 4. fps.

F2

F1

F

77

4.120 V =×

=9

3 31 m / s. 1

2 9 813

2 9 81

212

1×+ =

×+

. ..

Vy V y1 1 1 3= × .

∴ = +3 0519 62

312

1

..

.V

V Trial-and-error:

V

V

1

1

7 3 05 2 93

7 2 3 05 3 06

= =

= =

: . ? .

. : . ? .

V1 7 19417

==..

m / s.y m.1

1/2

2 22

1 8.417 .417 .417 7.19 1.90 m.

2 9.81y

= − + + × × =

V2 19 7 19 417× = ×. . . . V2 1= .58 . m / s

4.121 Refer to Example 4.12: γ γ ρy

y w w wy

V y11

11 12

3 6 6 10 1060

6 10− × × = × × −

= ⋅. ( ).

∴ − =−

γρ

236 600

612 1

1

( ) .yy

y ∴ + = = ∴ = =( )

.. . . , . .y y y V1 1 1 16

120032 2

37 27 3 8 15 8 ft fps

4.122 Continuity: 20 015 032

22× × = ×π π. . .V

∴ =V2 5 m / s. Momentum: p A p A m V V1 1 2 2 2 1− = −&( ).

60 03 03 1000 015 20 5 2022

2 2 000π π π× − × = × × −. . . ( ).p ∴ =p2 135 kPa.

4.123 2

1 1 2 2 2 2.05

2 . 15 30 m/s.2 .025

V A V A Vπ

π

×= = =

×

p V

gp V

gp1 1

22 2

2

1

2 2

2 29810

30 152 9 81

337γ γ

+ = + ∴ =−

×=.

. 500 Pa.

( )2 1 1 1 1. ( ).x x xF m V V p A F m VΣ = − − = −& &

∴ = + = × + × × =F p A mV1 1 12 2 2337 05 1000 05 15 4420& . . . 500 Nπ π

4.124 & . .m 1

21000 03 12 33 93= × × =π kg / s. & . .m 3

21000 02 8 10 05= × × =π kg / s. ∴ = − = = × ∴ =& & & . . . .m m m V V2 1 3

22 223 88 1000 03 8 446π m / s.

Energy from 1 → 2: V

gp V

gp

p12

1 22

22

2

2 2500

8 4462 9 81

9810+ = + ∴ =−×

×γ γ

...

000+122

= 536 300 Pa.

p1A1 p2A2

p1A1 p2A2

p3A3

RyRx

78

Energy from 1 → 3: V

gp V

gp1

21 3

23

2 2+ = +

γ γ.

∴ = +−

×=p3

2 2

50012 82 9 81

9810 540 000 000 Pa..

p A p A R m V m V m Vx x x x1 1 2 2 2 2 3 3 1 1− − = + −& & & . ∴ = × − × + × − × =R x 500 03 536 03 33 93 12 23 88 8 446 1032 2 000 300 Nπ π. . . . . . p A R m V m V m Vy y y y3 3 3 3 2 2 1 1− = + −& & & .

∴ = × − × − =Ry 540 10 05 8 759 000 .02 N2π . ( ) .

4.125 a) ( )ΣF m V V F mV VmAx x x= − − = − =& . & .&

2 1 1 11

ρ

=×

3001000 052π .

= 38.2 m/s

∴ = × =F 300 38 2 11. . 460 N b) − = − −F m V Vr B& ( )(cos ).1 1α

∴ = × − =F 30028 238 2

38 2 10 6250..

( . ) . N

c) − = − −F m V Vr B& ( )(cos ).1 1α

∴ = × − − =F 30048 238 2

38 2 10 18..

( . ( )) . 250 N

4.126 a) − = −F m V Vx x&( ).2 1 200 1 941 2512

2

12=

×.

..π V ∴ =V1 55 fps.

b) − = − −F m V Vr B& ( )(cos ).1 1α 200 1 9412512

302

12=

−.

.( ) .π V ∴ =V1 85 fps.

c) − = − −F m V Vr B& ( )(cos ).1 1α 200 1 9412512

302

12=

+.

.( ) .π V ∴ =V1 25 fps.

4.127 a) − = −F m V Vx x&( ).2 1 − = × × −700 1000 04 302

1 1 1π . ( cos ).V V Vo ∴ =V1 32 24. m / s. ∴ = = × × =& . . . .m A Vρ π1 1

21000 04 32 24 162 1 kg / s

b) − = − −F m V Vr B& ( )(cos ).1 1α − = × − −700 1000 04 8 866 121

2π . ( ) (. ).V ∴ =V1 40 24. m / s. ∴ = = × × =& . . .m A Vρ π1 1

21000 04 40 24 202 kg / s

c) − = − −F m V Vr B& ( )(cos ).1 1α − = × + −700 1000 04 8 866 121

2π . ( ) (. ).V ∴ =V1 24 24. m / s.

∴ = = × × =& . . . .m A Vρ π1 121000 04 24 24 1218 kg / s

4.128 (D) 2 1( ) 1000 0.01 0.2 50(50cos60 50) 2500 N.x x xF m V V− = − = × × × − = −o&

FV1

V2

79

4.129 a) − = − = ×

× −R m V Vx x x& ( ) . ( cos ).2 1

2

1 94112

120 120 60 120π o ∴ =Rx 305 lb.

R m V Vy y y= − =

× × ×& ( ) . ( . ).2 1

2

1 94112

120 120 866π ∴ =Ry 528 lb.

b) − = − − = ×

× × −R m V Vx r B& ( )(cos ) . (.5 ).1

2

1 1941

1260 60 1α π ∴ =Rx 76 2. . lb

R m V Vy r B= − =

× × ×& ( )sin . ( . ).1

2

1941

1260 60 866α π ∴ =Ry 132 lb.

c) − = − − = ×

× × −R m V Vx r B& ( )(cos ) . (.5 ).1

2

1 1 94112

180 180 1α π ∴ =Rx 686 lb.

R m V Vy r B= − =

× × ×& ( )sin . ( . ).1

2

1941

12180 180 866α π ∴ =Ry 1188 lb.

4.130 V RB = = × =ω 0 30 15.5 m / s. − = − − = × × × −R m V Vx B& ( )(cos ) . (.5 ).1

21 1000 025 40 25 1α π ∴ =Rx 982 N.

∴ = = × × =& .W R Vx B10 10 982 15 147 300 W 4.131 a) − = − = × − −R m V Vx x x& ( ) . ( cos ).2 1

24 02 400 400 60 400π o ∴ =Rx 1206 N.

R m V Vy y y= − = × ×& ( ) . ( sin ).2 124 02 400 400 60π o ∴ =Ry 696 N.

b) − = − − = × − −R m V Vx r B& ( )(cos ) . ( . ).12 2120 1 4 02 300 5 1o π ∴ =Rx 679 N .

R m V Vy r B= − = × × ×& ( )sin . . .12 24 02 300 866α π ∴ =Ry 392 N.

c) − = − − = × − −R m V Vx r B& ( )(cos ) . ( . ).12 2120 1 4 02 500 5 1o π ∴ =Rx 1885 N.

R m V Vy r B= − = × × ×& ( )sin . . .12 24 02 500 866α π ∴ =Ry 1088 N.

4.132 − = − − = × × − − −F m V Vx B& ( )(cos ) . ( ) ( .5 ).1

2 2120 1 4 02 400 180 1o π ∴ =Rx 365 N.

VB = × =12 150 180. m / s. & .W = × × =15 365 180 986 000 W The y-component force does no work. 4.133 (A) 2

2 1( ) 1000 0.02 60 (40cos45 40) 884 N.x r rx xF m V V π− = − = × × × × − =o& Power 884 20 17700 W.x BF V= × = × =

4.134 a) Refer to Fig. 4.16: 1 1 1

21 1

750sin sin45 507 fps.

750cos 300 cos45r r

rr

V VVV

β

β

= =∴

=− =

o

o

Note: V V V V V V Vx x r B r B r2 1 2 2 1 1 1 2 1− = − + − − = − +cos cos (cos cos ).α α α α

∴ = + =

× × + =R mVx r& (cos cos ) .

.5(cos cos ) .1 2 1

2

01512

750 507 30 45 48 9α α π o o lb.

80

∴ = = × × =& . ,W R Vx B15 15 48 9 300 220 000 ft - lbsec

or 400 Hp.

b) 750 60

750 300 605541 1

1 11 2

sin sincos cos

.β

β=

− =

=V

VV Vr

rr r

o

o fps =

∴ = + = ×

× × + =R mVx r& (cos cos ) .

.5(cos cos ) .1 2 1

2

01512

750 554 30 60 46 4α α π o o lb.

∴ = = × × =& . ,W R Vx B15 15 46 4 300 209 000 ft - lbsec

or 380 Hp.

c) 750 90

750 300 906871 1

1 11 2

sin sincos cos

.β

β=

− =

=V

VV Vr

rr r

o

o fps =

∴ = + = ×

× × + =R mVx r& (cos cos ) .

.5(cos ) .51 2 1

2

01512

750 687 30 0 36α α π o lb.

∴ = = × × =& . ,W R Vx B15 15 36 5 300 164 300 ft - lbsec

or 299 Hp.

4.135 a) Refer to Fig. 4.16: 100 30

100 30 2036 9 83 31 1

1 11 1

sin sincos cos

. , .o

oo=

− =

∴ = =V

VVr

rr

αα

α m / s.

2 22 2

2 2

sin60 83.3sin 71.5, 48 .

cos60 83.3cos 20

VV

V

αα

α

= = =

= −

oo

o

− = − = × × − − ∴ =R m V V Rx x x x& ( ) . ( .5 cos cos ).2 121000 015 100 71 60 100 30 8650π o o N .

∴ = = × × = ×& . .W V RB x12 12 20 8650 2 08 106 W

b) 100 30

100 30 4047 68 351 1

1 11 1 2

sin sincos cos

, .o

oo=

− =

∴ = = =V

VV Vr

rr r

αα

α m / s.

2 22 2

2 2

sin60 68.35sin 38.9 m/s, 29.5 .

cos60 68.35cos 40

VV

V

αα

α

= = =

= −

oo

o

− = − = × × − − ∴ =R m V V Rx x x x& ( ) . ( . cos cos ).2 121000 015 100 38 9 60 100 30 7500π o o N .

∴ = = × × = ×& . .W V RB x12 12 40 7500 3 60 106 W

c) 100 30

100 30 5053 8 61961 1

1 11 1 2

sin sincos cos

. , .o

oo=

− =

∴ = = =V

VV Vr

rr r

αα

α m / s.

2 22 2

2 2

sin60 61.76sin 19.32 m/s, 15.66 .

cos60 61.96cos 50

VV

V

αα

α

= = =

= −

oo

o

81

− = − = × × − − ∴ =R m V V Rx x x x& ( ) . ( . cos cos ).2 121000 015 100 19 32 60 100 30 6800π o o N .

∴ = = × × = ×& . .W R Vx B12 12 6800 50 4 08 106 W

4.136 a) Refer to Fig. 4.16: 50 30

50 302500 86 61 1

1 112 2sin sin

cos cos.

o

o

=− =

∴ = − +V

V VV V Vr

B rr B B

αα

2 2 2 2 22 1

2 2

30sin60 sin 900 30 .

30cos60 cosr

r r B Br B

VV V V V

V V

α

α

= ∴ = = + +

− =

o

o

Combine the above: VB = 13 72. m / s. Then, α α1 259 4 42 1= =. , . .o o − = − = × × − − ∴ =R m V V Rx x x x& ( ) . ( cos cos ).2 1

21000 01 50 30 60 50 30 916π o o N .

∴ = = × × =& . .W V RB x15 15 13 72 916 188 500 W

b) 50 30

50 302500 86 61 1

1 112 2sin sin

cos cos.

o

o

=− =

∴ = − +V

V VV V Vr

B rr B B

αα

∴ =VB 14 94. m / s.

2 2 2 22

2 2

30sin70 sin 900 20.52 .

30cos70 cosr

r B Br B

VV V V

V V

α

α

= ∴ = + +

− =

o

o α 1 41 4= . o ,α 2 48 2= . o

22 1( ) 1000 .01 50( 30cos70 50cos30 ). 841 N.x x x xR m V V Rπ− = − = × × − − ∴ =o o&

∴ = = × × =& . .W V RB x15 15 14 94 841 188 500 W

c) 50 30

50 302500 86 61 1

1 112 2sin sin

cos cos.

o

o

=− =

∴ = − +V

V VV V Vr

B rr B B

αα

∴ =VB 16 49. m / s

2 2 2 22

2 2

30sin80 sin 900 10.42 .

30cos80 cosr

r B Br B

VV V V

V V

α

α

= ∴ = + +

− =

o

o α 1 43= o , α 2 53 7= . o

− = − = × × − − ∴ =R m V V Rx x x x& ( ) . ( cos cos ).2 121000 01 50 30 80 50 30 762π o o N .

∴ = = × × =& . .W V RB x15 15 16 49 762 188 500 W

4.137 To find F, sum forces normal to the plate: ( )n out nF m VΣ = & 1 .nV −

a) [ ]∴ = × × × − − =F 1000 02 4 40 40 60 11. . ( sin ) .o 080 N (We have neglected friction)

2 2 3 3 10 ( ) 40sin30 .tF m V m V mΣ = = + − − × o& & & Bernoulli: V V V1 2 3= = .

∴ = − −

= +

∴ = = × =

=

Continuity:

kg / s.

kg / s.0 75 75 320 240

802 3 1

1 2 3

2 1

3

& & .5 && & &

& . & .&

m m mm m m

m m

m

b) 1 201.94 120( 120sin60 ) 3360 lb.

12 12F∴ = − × × × − =o (We have neglected friction)

ΣF m V m V mt = = + − − ×0 120 302 2 3 3 1& & ( ) & sin .o Bernoulli: V V V1 2 3= = .

82

2 12 3 1

1 2 33

20.75 .75 1.94 120 0 0.5

144Continuity: 22.6 slug/sec. and 9.7 slug/sec.

m mm m mm m m m

∴ = = × × ×∴ = − − = + = =

& && & && & & &

4.138 F m Vr r n= = × × × + =& ( ) . . ( ) sin1

21000 02 4 40 20 60 24o 940 N.

F Wx = = ∴ = × =24 30 21 21 20 432 940 600 N. 600 000 Wcos & .o 4.139 F m V V F Vr r n B x B= = × × − = −& ( ) . . ( ) sin . ( ) sin .1

2 2 21000 02 4 40 60 8 40 60o o

& ( ) . ( ).W V F V V V V VB x B B B B B= = − × = − +8 40 75 6 1600 802 2 3

dWdV

V V VB

B B B

&( ) . .= − + = ∴ =6 1600 160 3 0 13 332 m / s.

4.140 (A) Let the vehicle move to the right. The scoop then diverts the water to the

right. Then 2 1( ) 1000 0.05 2 60 [60 ( 60)] 720000 N.x xF m V V= − = × × × × − − =&

4.141 1(rF m V= & 2)(cos 1) 1000 .1 .6 ( )( 2) 120 .B B B BV V V Vα− − = × × − − =

At 2120 1000

0 : 120 133 300 N. 3600

t F× = = × =

2133 300 1.33 m/s

100 000oa = =

−

= =−

∴ − = ∫∫F

mdVdt

V dVV

dtB B B

B

t120100

00122

2033 33

16 67

000 . . .

.

.

∴ −

= ∴ =1

16 671

33 330012 26 6

. .. . . t. sect

4.142 1( )(cos 1) 90 .8 2.5 13.89 ( 13.89)( 1) 34700 N.r BF m V V α= − − = × × × × − − =&

50 100013.89 m/s 34700 13.89 482 000 W or 647 Hp.

3600BV W× = = ∴ = × =

&

4.143 See the figure in Problem 4.141.

F m V V V Vr B B B= − − = × × × − − =& ( )(cos ) . . ( )( )1 1 1000 06 2 2 24α V .B2

− = ∴ − =F mVdVdx

V VdVdxB

BB B

B. . 24 50002

− = − = − ∴ =∫∫245000

245000

27 78 250 458250

27 78

0

m

dx dVV

x n n xB

B

x

. . . ..

l l

V2

V1

= 0

F

83

4.144 − = − − = ×

− −F m V V V Vr B B& ( )(cos ) .

.) ( ).1 1

21 19412512

2α π2

(

∴ = − =F V VdVdtB

B0 1323 2012. ( ) .

At t VdVdt

VBB= = =0 0 0 1323 1

2, . . . Then 20

With dVdt

VB = =6 30 11, . . fps

For tdV

Vdt

VVB

B BB

VB

>−

= =−

− ∴ =∫∫030 1

0 006615 0 013231

30 11

30 18 572

0

2

0

,( . )

. . .. .

. . . fps

4.145 For this steady-state flow, we fix the boat and move the upstream air. This provides us with the steady-state flow of Fig. 4.17. This is the same as observing the flow while standing on the boat.

& . . . ( .W FV F F V=×

∴ = =1 150 1000

36001440 13 89 20 000 = N m / s)

F m V VV

V V= − = ×+

− ∴ =& ( ). ..

( . ). .2 12 2

2 21440 1 23 113 892

13 89 30 6 m / s.π

∴ = = ×+

=Q A V3 321

30 6 13 892

69 9π. .

. m / s.3

η pVV

= = =1

3

13 8922 24

0 625..

. . or 62.5%

4.146 Fix the reference frame to the aircraft so that V1200 1000

360055=

×= .56 m / s.

V m22320 1000

360088 89 1 2 1 1

55 88 892

329=×

= ∴ = × ×+

=. & . ..56 .

.5 m / s. kg / s.π

F = − =329 88 89 55 10.5( . .56) 980 N. = × ∴ =∆ ∆p pπ 1 1 28902. . . Pa & .W F V= × = ×1 10 980 55.56 = 610 000 W or 818 Hp

4.147 Fix the reference frame to the boat so that V1 208860

29 33= × = . fps.

V F m V V2 2 1

2

408860

58 67 1 941012

29 33 58 672

58 67 29 33= × = ∴ = − = ×

+−. & ( ) .

. .( . . ) fps. π

= 5460 lb.

& . , .W F V= × = × =1 5460 29 33 160 000 ft - lbsec

or 291 Hp

V2

F

VB

84

& .. .

. .m = × ×

+=194

1012

29 33 58 672

186 22

π slug / sec

4.148 Fix the reference frame to the boat: V V1 210 20= = m / s, m / s. ∴Thrust = &( ) . ( ) .m V V2 1 1000 0 2 20 10 2000− = × − = N & .W F V= × = × =1 2000 10 20 000 W or 26.8 Hp

4.149 0 2 2 1 0 1 2 20 0 11 1 1 1 1 1. . . . ( ) ( . ).= = × × ∴ = ∴ = ∴ = −V A V V V V y y m / s. m / s. max

flux in = 2 2 1000 20 1 80013

2672 2 23

0

1

0

1

ρV dy y dy= × − = =∫∫ (. )...

000 N .

The slope at section 1 is −20. ∴ = − +V y y A2 20( ) .

Continuity: A V A V V V1 1 2 2 2 12 2= ∴ = =. m / s. V A

V AV A2

22

005 1

1 2( )

(. )/ .

== −

∴ = −

2 1 2 2 2 202= − ∴ = ∴ = −A A V y y/ . .5. ( ) .5 .

flux out = 2 1000 2 20 8001253

8000 001532

3

0

05

0

05

( .5 )( . )

[ . ]..

− =−

=∫ y dy

y 000

0003

= 408 3. N . ∴change = 408 − 267 = 141 N.

4.150 a) β = =−

× ×= =∫ ∫V dA

V A

y dy2

2

2 2

0

1

2

32 20 1

1 2 1 04000 1

343

(. )

. .. .

.

b) See Problem 4.149: V y y y V2 220 0 125 05 0 2( ) ( . ), . .= − ≥ ≥ = m / s.

β = =−

× ×=

−=∫ ∫V dA

V A

y dyy

2

2

2 2

0

05

2

3

005

2 20 125

2 1 1 02000

1253

1 021( . )

. .( . )

. .

.

.

4.151 From the c.v. shown: ( ) .p p r r Lw o1 2 0

2 2− =π τ π

∴ = = ∴ =× ×

× × × −τ µwo

w w

p rL

dudr

dudr

∆

20 03 144 75 122 30 2 36 10 5.

. . /.

= 191 ft / sec

ft.

4.152 Write the equation of the parabola: V r Vrr

( ) .max= −

1

2

02

p1A1 p2A2

τw2πroL

85

Continuity: π π× × = −

∴ =∫.

..max max

.

006 8 1006

2 1622

20

006

Vr

rdr V m / s.

Momentum: p A p A F V dA mV1 1 2 22

1− − = −∫Drag ρ & .

40 006 1000 16 1006

2 1000 006 8 82 22

2

22

0

006

000 Dragπ π π× − = × −

− × × × ×∫.

..

.

Fr

rdr

4 9 651 7 238.524 . . .− = −FDrag ∴ =FDrag N2 11. .

4.153 & ( ) . ( ) .m A V V y dA y dytop = − = × × − +

=∫∫ρ ρ1 1 2

2

0

2

1 23 2 10 32 28 10 65 6 kg / s.

− = + − = + + × − × ×∫∫F

V dA m V m V y dytop21 23 28 10 65 6 32 1 23 20 322

1 1 12 2 2

0

2

ρ & & . ( ) . . .

∴ =F 3780 N.

4.154 a) & & & ( ) . . ( ).

m m m A V u y dA y y dytop = − = − = × × − − ×

∫∫1 2 1 1

2

0

1

1 23 1 2 8 20 100 8 2ρ ρ

= = =0 656 0 1 8. . ( ) ). kg / s. (Note: for y u y

Momentum: − = − + × − × × ×∫F y y dyDrag ρ ρ64 20 100 2 656 8 1 2 82 2 2

0

1

( ) . ..

= × + − × ∴ =1 23 6 83 5 25 123 12 8 2 1. . . . . . . NDragF

b) To find h: 8 8 20 100 2

0

1

h y y dy= −∫ ( ) ..

∴ =×

−×

=h20 1

2100 001

30 0667

2. .. m.

Momentum: − = − − × × ×∫F y y dyDrag 1 23 64 20 100 2 123 0667 2 82 2 2

0

1

. ( ) . . ..

= × −1 23 6 83 10. . .50. ∴ =FDrag N2 1. .

4.155 a) Energy: V

gz

Vg

z hL12

122

22 2+ = + + . See Problem 4.118(a).

82 9 81

0 619122 9 81

22 2

×+ =

×+ +

..

..

.51 .hL ∴ =hL 1166. m.

∴ = × × × × =losses = 900 W / m of width.γA V hL1 1 9810 6 1 8 1166 54(. ) .

b) See Problem 4.120: V

gz

Vg

z hL12

122

22 2+ = + + .

86

7 192 9 81

4171

2 9 811 9

2 2..

..58

.. .

×+ =

×+ + hL ∴ =hL 1025. m.

∴ = × × × × =losses = 300 WγA V hL1 1 9810 417 3 7 19 1025 90. . .

c) See Problem 4.121: 5 172 9 81

1163

2 9 812

2 2..

..

.×

+ =×

+ + hL ∴ =hL 0 0636. m.

∴ = × × × =losses = / m of width.γA V hL1 1 9810 116 5 17 0 0636 3740 W. . . 4.156 See Problem 4.122: V V p p1 2 1 220 5 60 135= = = = m / s, m / s, kPa, kPa.

Then, V

gp V

gp

h hL L12

1 22

22 2

2 220

2 9 8160 5

2 9 81135

+ = + +×

+ =×

+ +γ γ

.. .

. 000

9810 000

9810

∴ = =×

∴ =h Kg

K KL 11 472

202 9 81

0 5622

..

. . . m =V

12

4.157 Continuity: V D Vd VdD

V12 2

1

2

2= ∴ =. .

Energy: V

gH t

Vg

V gH t12 2

2 22+ = ∴ =( ) . ( ).

Momentum: ΣF Fddt

V d V m V Vd sdt

ax I x xc v

x xx

x− = − + −

=∫( ) & ( ). .

. .

ρ 2 1

2

2 v

∴ − = = − ∫a m td

V V m t md

V t dtx o

t

( ) ( ). ( ) ( ) .ρπ

ρπ2 2

04 4

But, V dHdt

dHdt

dD

gHdHH

dD

gdt Hgd

Dt H o1

2

2 1 2

2

21 2

2

22 22

2= − ∴ − = ∴ − = ∴ = +. . . ./

/

∴ = +

+

−

∫ad

ggdD

t Hd

ggdD

t H dt mx o

t

o o

ρπρ

π2 2

2

22

0

2

242

22 4

222

4.158 This is a very difficult design problem. There is an optimum initial mass of water

for a maximum height attained by the rocket. It will take a team of students many hours to work this problem. It involves continuity, energy, and momentum.

4.159 VmA

Vee

= =× × ×

=&

.. .

ρ π4

1000 4 00419 892 m / s. Velocity in arm =

v v vM r V d V ri k Vi AdrI

c v

= × × − = × − ×∫ ∫( ) $ ( $ $). .

.

2 4 20

3

Ω Ωρ ρ

87

= − = −∫8 0 360

3

ρ ρAV k rdr AV kΩ Ω$ . $..

Σv v v v v v

Mddt

r V d V r V V n dA i V j V k V Ac v

e e e ec s

= × − = × ⋅ = × +∫ ∫0 0 3 707 707 and ( ) . ( $) . $ (. $ . $) .. . . .

ρ ρ ρ

The z-component of v v vr V V n dA V Ae e

c s

× ⋅ = ×∫ ( $) . . .. .

ρ ρ3 707 2

Finally, − = = × × =( ) . . . ,M AV V A AV A VI z e e e e0 36 4 3 707 2ρ ρΩ . Using 0 36 4 3 707 19 89. . . . .Ω = × × × ∴ =Ω 46 9. . rad / s 4.160 A moment

vM resists the motion thereby

producing power. One of the arms is shown.

v

M ri k Vi Adr AV k rdr AV kI = × − × = − = −∫∫ 4 2 8 2 7780

10 12

0

25$ ( $ $) $ . $.

/.

Ω Ω Ωρ ρ ρ

Σv v v v v v

M Mkddt

r V d V r V V n dA V A kc v

e ec s

= × − = × ⋅ = × ×∫ ∫$, ( ) , ( $) $.. . . .

and 01012

42ρ ρ

Thus, M + ×

× × = × ×

×2 778 1947512

2009

30 2001012

1941 412

42

22

. ..

./

.π π

∴ =M 309 ft - lb. & .W M= = × =Ω 309 30 9270 ft - lb / sec 4.161 & . . .m A V V V= = = × ∴ =10 1000 01 31 82

0 0ρ π m / s.

Continuity: V V V re02 201 01 006 05π π× = × + × −. . . ( . ).

02.01 .006 .15. 11.1 m/s.e eV V Vπ × = × × ∴ =

∴ = − − = −V V r V re0 19 1 05 42 4 212. ( . ) . .

v

M ri k V i Adr ri k r i AdrI = × + × + × + × −∫∫2 2 2 2 42 4 212005

2

0

05$ ( $ $) $ [ $ ( . )$]

.

..

Ω Ωρ ρ

= + −∫∫4 4 42 4 21202

05

2

0

05

Ω ΩV Ak rdr Ak r r drρ ρ$ $ ( . ).

..

= × × × × + × ×4 31 8 1000 01 052

4 1000 0122

2Ω Ω. . . $ .π πk

42 42

2 05212

32 052 2 3 3.

(. . ) (. . ) $− − −

k

x

y r

Ve

V

Ω

88

= + =( . . ) $ . $.0 05 0 3 0 35Ω Ω Ωk k

ri V j V dr rdr k ke e$ ( $) . . . $ . $.

.

.

.

.

× − × = − × × = −∫∫ ρ 006 111 1000 006 13 862

05

2

05

2

∴ − = −0 35 13 86. . .Ω ∴ =Ω 39 6. . rad /s

4.162 1000 1000500

2= ∴ = = ⋅M MΩ. N m.

v

M ri k V r i r drI r r= × − × ×∫ $ ( $ ( )$ ) .2 2 02Ω ρ π

= ∫0 08 2

0

. ( ) $.πΩ r V r drkR

Continuity: V r r V R V r RV rr r( ) . cos . . ( ) . / .2 02 30 2 02 0 866π π× = × ∴ =o

v v vo or V V n dA R R V V R k V V kr r r r

c s

× ⋅ = − + × = − +∫ ( $ ) ( sin ) cos . $ . ( . ) $.. .

ρ ρ πΩ 30 30 2 02 00301 35 5

∴ − − = − + ∴ − − =∫2 16 32 00301 35 52 1 1333 00

152. . ( .5 ). . .

.

V r dr V V V Vr r r r r

∴ = ± + × =V r12

52 1 52 1 4 1333 70 92( . . ) . m / s.

The flow rate is Q A Ve r= = × × × × =cos . . . . . .30 2 15 02 70 9 866 1 16o π m / s3

4.163 See Problem 4.159. V Ve = = × =19 8900802

19 89 3182

2...

. . m / s. m / s.

v

M ri k Viddt

k ri Adr A AI e= × − × + −

×

= × = ×∫4 2 01 004

0

32 2$ ( $ $) $ $ . . , . .

.

ΩΩ

ρ π π

= − − = − −∫ ∫8 4 360 360

32

0

3

ρ ρAV k rdr Addt

k r dr AV k Addt

kΩΩ

ΩΩ$ $ $ $.

. .

( ) ( $) $.. .

v v vr V V n dA V A kz e e

c s

× ⋅ =∫ ρ 212 2

Thus, 360 36 212 31 8 3732AV A ddt

V A ddte eΩ

Ω ΩΩ+ = + = or . .

The solution is Ω = +−Ce t31 8 11 73. . . The initial condition is Ω( ) .0 0= ∴ = −C 1173. . Finally, Ω = − −11 73 1 31 8. ( ) ..e t rad / s 4.164 This design problem would be good for a team of students to do as a project. How large a

horsepower blower could be handled by an average person?

89

CHAPTER 5

The Differential Forms of the

Fundamental Laws

5.1 0 = − + ⋅∫ ∫∂ρ∂

ρt

d V V ndAc v c s. . . .

$ .v

Using Gauss’ theorem:

0 = − + ∇ ⋅ − = + ∇ ⋅

−∫ ∫∫∂ρ∂

ρ∂ρ∂

ρt

d V V d Vt

V d Vc v c vc v. . . .. .

( ) ( ) .v v v v

Since this is true for all arbitrary control volumes (i.e., for all limits of integration), the integrand must be zero:

∂ρ∂

ρt

V+ ∇ ⋅ =v v

( ) .0

This can be written in rectangular coordinates as

− = + +∂ρ∂

∂∂

ρ∂

∂ρ

∂∂

ρt x

uy

vz

w( ) ( ) ( ).

This is Eq. 5.2.2. The other forms of the continuity equation follow.

5.2 & & .m m mtin out

element− =∂

∂

( )ρ θ ρ∂

∂ρ θv rd dz v

rv dr r dr d dzr r r( ) ( )− +

+

+ − +

ρ ρ∂

∂θρ θθ θ θv drdz v v d drdz( )

+ +

− +

+

= +

ρ θ ρ

∂∂

ρ θ∂∂

ρ θv r dr d dr vz

v dz r dr d drt

r dr d drdzz z z2 2 2( ) .

Subtract terms and divide by rd drdzθ :

− −+

− −+

=+ρ ∂

∂ρ

∂∂θ

ρ∂∂

ρ∂∂

ρθvr r

v r drr

vr z

v r drr t

r drr

rr z( ) ( ) ( ) / / .1 2 2

Since dr is an infinitesimal, ( ) / ( / ) / .r dr r r dr r+ = + =1 2 1 and Hence,

∂ρ∂

∂∂

ρ∂

∂θρ

∂∂

ρ ρθt rv

rv

zv

rvr z r+ + + + =( ) ( ) ( ) .1 1 0 This can be put in various forms.

90

5.3 & & .m m mtin out

element− =∂

∂

ρ θ θ φ ρ∂∂

ρ θ θ φv rd r d vr

v dr r dr d r dr dr r r( ) sin ( ) ( ) ( )sin− +

+ +

+ +

− +

+

ρ θ φ ρ

∂∂θ

ρ θ θ φθ θ θv dr rdr

d v v d dr rdr

d2 2

sin ( ) sin

+ +

− +

+

ρ θ ρ∂

∂φρ φ θφ φ φv dr r dr d v v d dr r dr d

2 2( )

= +

∂∂

ρ θ θ φt

rdr

drd d2

2

sin

Because some areas are not rectangular, we used an average length ( / ).r dr+ 2 Now, subtract some terms and divide by rd d drθ φ :

− − −+

−+

ρ θ ρ θ∂∂

ρ θ∂∂θ

ρ θθv vr

v r drr

vr dr

rr r rsin sin ( )sin ( ) ( ) sin2 2

−+

=+

∂

∂φρ

∂ρ∂

θφ( ) sinvr

dr

r t

r dr

r2 2

2

Since dr is infinitesimal ( ) / ( / ) / .r dr r r r dr r+ = + =2 2 1 and Divide by r sinθ and there results

∂ρ∂

∂∂

ρ∂∂θ

ρθ

∂∂φ

ρ ρθ φt rv

rv

rv

rvr r+ + + + =( ) ( )

sin( )1 1 2 0

5.4 For a steady flow ∂ρ∂t

= 0. Then, with v w= = 0 Eq. 5.2.2 yields

∂∂

ρ ρρ

xu du

dxu d

dx( ) .= + =0 0 or

Partial derivatives are not used since there is only one independent variable.

5.5 Since the flow is incompressible DDt

ρ= 0. This gives

3 2 3

1 200 1 200ˆ ˆ ˆ ˆcos2 sin2r rp p

p i i i ir r r r r

θ θ∂ ∂ ρ ρ

θ θ∂ ∂θ

∴∇ = + = − −

v or

ux

wz

∂ρ∂

∂ρ∂

+ = 0.

Also, v v∇ ⋅ =V 0, or ∂

∂∂∂

ux

wz

+ = 0.

91

5.6 Given: ∂∂

∂ρ∂t z

= ≠0 0, . Since water can be considered to be incompressible, we

demand that DDt

ρ= 0. ∴ u

xw

z∂ρ∂

∂ρ∂

+ = 0, assuming the x-direction to be in the

direction of flow. Also, we demand that v v∇ ⋅ =V 0, or ∂

∂∂∂

ux

wz

+ = 0.

5.7 We can use the ideal gas law, ρ =p

RT. Then, the continuity equation

DDt

Vρ

ρ= − ∇ ⋅v v

becomes, assuming RT to be constant, 1RT

DpDt

pRT

V= − ∇ ⋅v v

or

1p

DpDt

V= −∇ ⋅v v

.

5.8 a) Use cylindrical coordinates with v v zθ = = 0:

1 0r r

rvr∂∂

( ) =

Integrate:

rv Cr = . ∴ =vCrr .

b) Use spherical coordinates with v vθ φ= = 0:

1 022

r rr vr

∂∂

( ) =

Integrate:

r v Cr2 = . ∴ =v

Crr 2

.

5.9 DDt

Vux

vy

ρρ ρ

∂∂

∂∂

= − ∇ ⋅ = − +

= − × + × = −

⋅

v v2 3 200 1 400 1 1380. ( ) .

kgm s3

5.10 In a plane flow, u u x y v v x y= =( , ) ( , ). and Continuity demands that ∂∂

∂∂

ux

vy

+ = 0.

If u ux

= =const, then ∂∂

0 and hence ∂∂

vy

= 0. Thus, v = const also.

92

5.11 If u C v C= =1 2 and , the continuity equation provides, for an incompressible flow,

∂∂

∂∂

∂∂

∂∂

ux

vy

wz

wz

w C+ + = ∴ = =0 0 3. . and

The z-component of velocity w is also constant. We also have

DDt t

ux

vy

wz

ρ ∂ρ∂

∂ρ∂

∂ρ∂

∂ρ∂

= = + + +0

The density may vary with x, y, z and t. It is not, necessarily, constant.

5.12 ∂∂

∂∂

ux

vy

+ = 0. ∴ + = A vy

∂∂

0. ∴ = − + v x y Ay f x( , ) ( ).

But, v x o f x( , ) ( ).= =0 ∴ = − v Ay.

5.13 ∂∂

∂∂

ux

vy

+ = 0. ∴ = − = −+ −

+= −

−+

∂∂

∂∂

vy

ux

x y x xx y

x yx y

( )5 ( )( ) ( )

2 2

2 2 2

2 2

2 2 2

5 2 5 5

∴ =−

++ =

++ = ∴ =

+∫v x yy xx y

dy f xy

x yf x f x v

yx y

( , )( )

( ) ( ). ( ) . .5 5 5

052 2

2 2 2 2 2 2 2

5.14 From Table 5.1: 1 1 110

42r r

rvr

vr rr

∂∂

∂∂θ

θθ( ).

sin .= − = − +

∴ = +

+ = −

+∫rvr

dr f rr

fr 104

104

2

.sin ( )

.sin ( ).θ θ θ θ

. (. , ) ...

sin ( ) . ( ) .2 2 10 242

0 0v f fr θ θ θ θ= × −

+ = ∴ =

∴ = −

vrr 100 4

2

.sin .θ

5.15 From Table 5.1: 1 1 201

12r r

rvr

vr rr

∂∂

∂∂θ

θθ( ) cos .= − =−

+

∴ = − +

+ = − −

+∫rvr

dr f rr

fr 20 11

201

2 cos ( ) cos ( ).θ θ θ θ

v f fr ( , ) ( )cos ( ) . ( ) .1 20 1 1 0 0θ θ θ θ= − − + = ∴ =

∴ = − −

vrr 20 112 cos .θ

93

5.16 From Table 5.1, spherical coordinates: 1 12

2

r rr v

rvr

∂∂ θ

∂∂θ

θθ( )sin

( sin ).= −

∴ = +

1 110

4022

23r r

r vr rr

∂∂ θ

θ θ( )sin

sin cos .

∴ = +

+ = −

+∫r v rr

dr f rr

fr2

3210

402 10

80cos ( ) cos ( )θ θ θ θ

4 2 10 2802

0 02v f fr ( , ) cos ( ) . ( ) .θ θ θ θ= × −

+ = ∴ =

∴ = −

vrr 1080

3 cos .θ

5.17 Continuity: ∂∂

ρ ρρ

xu du

dxu d

dx( ) . .= ∴ + =0 0

ρ = =××

= =−

×=

pRT

dudx

18 1441716 500

0 00302526 4532 2 12

2193. ./

slugft

fps / ft.

∴ = − = − × = −ddx u

dudx

ρ ρ .. .

00302486

219 0 00136 slug / ft 4

5.18 [ ]∂∂

∂∂

∂∂

ux

vy x

e ex x+ = − − = −− −0 20 1 20. ( )

Hence, in the vicinity of the x-axis:

∂∂

vy

e v ye Cx x= = +− −20 20 and .

But v y C= = ∴ =0 0 0 if . . v ye ex= = =− −20 20 0 2 02( . ) .541 m / s

5.19 [ ]1 0 20 1 20r r

rv vz z

e erz z z∂

∂∂∂

∂∂

( ) . ( )+ = − − = −− −

Hence, in the vicinity of the z-axis:

1 202

202

r rrv e rv r e Cr

zr

z∂∂

( ) .= = +− − and

But v r Cr = = ∴ =0 0 0 if . . v re er

z= = =− −10 10 0 2 0 2712( . ) . m / s 5.20 The velocity is zero at the stagnation point. Hence,

0 1040

22= − ∴ =R

R. m

The continuity equation for this plane flow is ∂∂

∂∂

ux

vy

+ = 0. Using ∂∂ux

x= −80 3 ,

94

we see that ∂∂

vy

x= − −80 3 near the x-axis. Consequently, for small ∆y ,

∆ ∆v x y= − −80 3 so that v = − − =−80 3 0 1 0 2963( ) ( . ) . . m / s 5.21 The velocity is zero at the stagnation point. Hence

040

10 22= − ∴ =R

R. m

( )1 1 40 10 202

22

2

r rr v

r rr

rr∂

∂∂∂

= − = −( ) .

Near the negative x-axis continuity provides us with

( )1 20r

vrsin

sin .θ

∂∂θ

θθ =

Integrate, letting θ = 0 from the y-axis: v Cθ θ θsin cos= − +20

Since v θ = 0 when θ = =90 0o , . C Then, with α = =−tan.

. ,1 0 13

1 909o

v θθθ

= − = − = − =20 2088 09188 091

200 03330 999

0 667cossin

cos .sin .

..

. m / s

5.22 Continuity: ∂∂

∂∂

ux

vy

vy

ux

+ = ∴ = − = −−

×= −0 13 5 11 3

2 005220. . .

.. m / s

m∆∆

∆∆

∴ = − = − ∴ = − × = −∆ ∆v v y v0 220 220 004 0 88. . . . m / s

b) a u uxx = = × + =

∂∂

12 6 220 2772. ( ) . m /s 2

5.23 ΣF may y= . For the fluid particle occupying the volume of Fig. 5.3:

τ∂τ

∂τ

∂τ

∂τ

∂τ

∂yyyy

zyzy

xyxy

ydy

dxdzz

dzdxdy

xdx

dydz+

+ +

+ +

2 2 2

− −

− −

− −

τ

∂τ

∂τ

∂τ

∂τ

∂τ

∂yyyy

zyzy

xyxy

ydy

dxdzz

dzdxdy

xdx

dydz2 2 2

+ =ρ ρg dx dy dz dx dy dzDvDty

Dividing by dx dy dz , and adding and subtracting terms:

∂τ

∂

∂τ

∂

∂τ

∂ρ ρxy yy zy

yx y zg Dv

Dt+ + + = .

5.24 Check continuity:

∂∂

∂∂

∂∂

ux

vy

wz

x y x xx y

x y y yx y

+ + =+ −

++

+ −+

=( )10 ( )

( )( )10 ( )

( ).

2 2

2 2 2

2 2

2 2 2

10 2 10 20

95

Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7 give, with g gx y= = 0:

. u u p

u vx y x

∂ ∂ ∂ρ ρ

∂ ∂ ∂+ = −

2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 310 10 10 10 20 100( )

( ) ( ) ( )

p x y x y xy x y yx x y x y x y x y x y

∂ρ ρ ρ

∂− − +

∴ = − − =+ + + + +

. v v p

u vx y y

∂ ∂ ∂ρ ρ

∂ ∂ ∂+ = −

2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 310 20 10 10 10 100( )

( ) ( ) ( )

p x xy y x y x y yy x y x y x y x y x y

∂ρ ρ ρ

∂− − +

∴ = − − =+ + + + +

2 2 2 2 2 2 2 2 2100 100 100ˆ ˆ ˆ ˆ ˆ ˆ ( ).

( ) ( ) ( )

p p x yp i j i j xi yj

x y x y x y x y

∂ ∂ ρ ρ ρ∂ ∂

∴∇ = + = + = ++ + +

v

5.25 Check continuity (cylindrical coord from Table 5.1):

1 1 101

1 101

102 2r r

rvr

vr r r rr

∂∂

∂∂θ

θ θθ( ) cos cos .+ = +

+

−+

= ∴ It is a possible

flow. For Euler’s Eqs. (let ν = 0 in the momentum eqns of Table 5.1) in cylindrical coord:

22

2 22 2 3

100 1 1 201 sin 10 1 cosr r

rv vv vp

vr r r r r r r r

θ θ∂ ∂∂ ρρ ρ ρ θ ρ θ

∂ ∂ ∂θ = − − = + − −

− +

−

101

110

102

22

ρθ

r r rsin .

4

1 100 11 sin cosr

rv v v v vp

vr r r r r r

θ θ θ θ∂ ∂∂ ρρ ρ ρ θ θ

∂θ ∂ ∂θ = − − − = −

− −

− +

10 1 1 20 100 1 12 3 2

2

ρ θ θρ

θ θr r r r

cos sin sin cos .

3 2 3

1 200 1 200ˆ ˆ ˆ ˆcos2 sin2r rp p

p i i i ir r r r r

θ θ∂ ∂ ρ ρ

θ θ∂ ∂θ

∴∇ = + = − −

v

5.26 This is an involved problem. Follow the steps of Problem 5.25. Good luck!

( )∂

∂ρ ρ

∂∂

ρ∂∂θ

θ φ θpr

v v

rv v

rvr

vr

r r=+

− −2 2

1r

p v vr

vvr

vr

vrr

∂∂θ

ρ ρ∂∂

ρ∂∂θ

θ θ θ θ= − − −( )

96

5.27 2 2. .

3 3p p V p p V

µ µλ λ ∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅

v v v v

∂∂

$ $ $ $.

ss

ss

nR

nR

≅ = − = −∆∆

∆α∆α

∂∂

∂θ∂

$ $ $$ .

st

st

nt

nt

≅ = =∆∆

∆θ∆

2

ˆ ˆ.DV V V V

V s V nDt t s t R

∂ ∂ ∂θ∂ ∂ ∂

∴ = + + −

v

For steady flow, the normal acc. is −

VR

2

, the tangential acc. is V Vs

∂∂

.

5.28 For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to

vΩ.

Thus, Euler’s equation becomes

2 ( ) .DV d

V r r p gDt dt

ρ ρ Ω

+ Ω × + Ω × Ω × + × = − ∇ −

v vv v v v vv v v

5.29 2xxu

px

∂τ µ

∂= − + Vλ+ ∇ ⋅

v v30 psi.= −

τ τyy zz p= = − = −30 psi.

xyu vy x

∂ ∂τ µ

∂ ∂= + 5 5.1

10 30 1440 18 10 psf.12

− − = − × = ×

τ ττ

τxz yzxx

= = =××

= ×−

−0 18 1030 144

4 17 105

8. . . xy

5.30 2 2 3

9/5 2 13/5 9/5 2 13/516 16 8 16

. ( , ) ( ).3

v u y y y yv x y f x

y x C x C x Cx C x

∂ ∂∂ ∂

= − = − ∴ = − +

v x o f x C C( , ) . ( ) . . . ./= ∴ = = ∴ =0 0 8 1000 0 03184 5 ∴ = −− −u x y yx y x( , ) ./629 98904/ 5 2 8 5 v x y y x y x( , ) ./ /= −− −252 52702 9 5 3 13 5

τ µ∂∂xx p ux

= − + = − + = −2 100 0 100 kPa.

τ τyy zz p= = − = −100 kPa.

5 4/5 52 10 629 1000 5.01 10 Pa.xyu vy x

∂ ∂τ µ

∂ ∂− − − = + = × × = ×

τ τxz yz= = 0.

97

5.31 Du uDt t

∂∂

= ( ) .u v w u V ux y z

∂ ∂ ∂∂ ∂ ∂

+ + + = ⋅∇

v v

Dv vDt t

∂∂

= ( ) .u v w v V vx y z

∂ ∂ ∂∂ ∂ ∂

+ + + = ⋅∇

v v

Dw wDt t

∂∂

= ( )u v w w V wx y z

∂ ∂ ∂∂ ∂ ∂

+ + + = ⋅∇

v v

ˆ ˆˆ ˆ ˆ ˆ( ) ( ) .DV Du Dv Dw

i j k V ui vj wk V VDt Dt Dt Dt

∴ = + + = ⋅∇ + + = ⋅∇v

v v v v v

5.32 Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible effects:

2 ˆˆ ˆ3 3 3

DVp g V Vi Vj Vk

Dt x y zµ ∂ µ ∂ µ ∂

ρ ρ µ∂ ∂ ∂

= − ∇ + + ∇ + ∇ ⋅ + ∇ ⋅ + ∇ ⋅v

v v v v v v v vv

= −∇ + + + + +

∇ ⋅

v v v v vp g V

xi

yj

zk Vρ µ∇

µ ∂∂

∂∂

∂∂

2

3$ $ $

∴ = −∇ + + ∇ + ∇ ∇ ⋅ρ ρ µµDV

Dtp g V V

v v v v v v v2

3( ).

5.33 If u=u(y), then continuity demands that ∂∂

vy

v C= ∴ =0. .

But, at y=0 (the lower plate) v=0. ∴ = =C v x y0 0, ( , ) . and

2 2 2

2 2 2.x

u u uDu u u u u pu v w g

Dt t x y z x x y z

∂ ∂ ∂∂ ∂ ∂ ∂ ∂ρ ρ ρ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∴ = + + + = − + + + +

∴ = − +02

2

∂∂

µ∂p

x

u

ay.

0 .Dv pDt y

∂ρ

∂= = −

ρ∂∂

ρ∂∂

ρDwDt

pz

gpz

g= = − + − ∴ = − −0 0( ). .

5.34 Continuity: ( ) 0. .r rrv rv Cr

∂∂

= ∴ = At 0, . 0.rr v C= ≠ ∞ ∴ =

10 .rDv p

Dt r∂

ρ ∂= = −

10 .

Dv pDt r

θ ∂ρ ∂θ

= = −

98

z zDv vDt t

∂ρ ρ

∂= rv+ z vv

rθ∂

∂+ z z

zv v

vr z

∂ ∂∂θ ∂

+2 2

2 2 21 1z z zv v vp

z r rr r

∂ ∂ ∂∂µ

∂ ∂∂ ∂θ

= − + + +

2

2zv

z

∂

∂+

∴ = − + +

0 12

2

∂∂

µ∂∂

∂∂

pz

vr r

vr

z z .

5.35 Continuity: 1 0 0 022 2

1r rr v r v C r r v Cr r r

∂∂

( ) . . , . .= ∴ = = = ∴ = At

2

22

cot .v vpr r rθ θ∂

ρ µ θ∂

− = − + −

22 2 2

1 10

sin

v vpr

r r rr rθ θ∂∂ ∂

µ∂θ ∂ ∂ θ

= − + −

01

= −r

psin

.θ

∂∂φ

5.36 For an incompressible flow

v v∇ ⋅ =V 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and

5.3.3:

ρ∂∂

µ∂∂

∂∂

µ∂∂

∂∂

∂∂

µ∂∂

∂∂

ρDuDt x

pux y

uy

vx z

uz

wx

gx= − +

+ +

+ +

+2 .

= − + + + + + +

+

∂∂

µ∂∂

µ∂∂

µ∂∂

µ∂∂

∂∂

∂∂

∂∂

ρpx

ux

uy

uz x

ux

vy

wz

g x

2

2

2

2

2

2

∴ = − + + +

+ρ

∂∂

µ∂∂

∂∂

∂∂

ρDuDt

px

ux

uy

uz

g x

2

2

2

2

2

2 .

2 .yDv u v v v w

p gDt x y x y y z z y

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ρ µ µ µ ρ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= + + − + + + +

= − + + + + + +

+

∂∂

µ∂∂

µ∂∂

µ∂∂

µ∂∂

∂∂

∂∂

∂∂

ρpy

vx

vy

vz y

ux

vy

wz

g y

2

2

2

2

2

2

∴ = − + + +

+ρ

∂∂

µ∂∂

∂∂

∂∂

ρDvDt

py

vx

vy

vz

gy

2

2

2

2

2

2 .

2 zDw u w v w w

p gDt x z x y z y z z

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ρ µ µ µ ρ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + + + + − + +

99

= − + + + + + +

+

∂∂

µ∂∂

µ∂∂

µ∂∂

µ∂∂

∂∂

∂∂

∂∂

ρpz

wx

wy

wz z

ux

vy

wz

gz

2

2

2

2

2

2

∴ = − + + +

+ρ

∂∂

µ∂∂

∂∂

∂∂

ρDwDt

pz

wx

wy

wz

gz

2

2

2

2

2

2 .

5.37 If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3., with µ µ= ( , , )x y z we arrive at

2 2 2

2 2 2 2xDu p u u u u u v u w

gDt x x x y y x z z xx y z

∂ ∂ ∂ ∂ ∂µ ∂ ∂µ ∂ ∂ ∂µ ∂ ∂ρ ρ µ

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ ∂

= − + + + + + + + + +

5.38 If plane flow is only parallel to the plate, v w= = 0. Continuity then demands that ∂ ∂u x/ .= 0 The first equation of (5.3.14) simplifies to

u u

ut x

∂ ∂ρ

∂ ∂+ v+

uw

y∂∂

+u pz x

∂ ∂∂ ∂

= −

xgρ+

2

2u

x

∂µ

∂+

2 2

2 2u u

y z

∂ ∂

∂ ∂+ +

2

2u ut y

∂ ∂ρ µ

∂ ∂=

We assumed g to be in the y-direction, and since no forcing occurs other than due to the motion of the plate, we let ∂ ∂p x/ .= 0

5.39 From Eqs. 5.3.10, −+ +

= − + +

− ∇ ⋅

τ τ τ µ ∂∂

∂∂

∂∂

λxx yy zz p

ux

vy

wz

V3

23

v v

.

2 2. .

3 3p p V p p V

µ µλ λ ∴ = − + ∇ ⋅ ∴ − = − + ∇ ⋅

v v v v

5.40 ˆˆ ˆ( ) ( )V V u v w ui vj wkx y z

∂ ∂ ∂⋅∇ = + + + + ∂ ∂ ∂

v v

ˆ( )w w w v v v

V V u v w u v w iy x y z z x y z

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂∇ ⋅∇ = + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

×v v

ˆu u u w w wu v w u v w j

z x y z x x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

ˆv v v u u uu v w u v w k

x x y z y x y z ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

+ + + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Use the definition of vorticity: ˆˆ ˆ( ) ( ) ( )w v u w v u

i j ky z z x x y

ω∂ ∂ ∂ ∂ ∂ ∂

= − + − + −∂ ∂ ∂ ∂ ∂ ∂

v

100

ˆˆ ˆ( ) ( ) ( ) ( ) ( )w v u w v x

V ui vj wky z x z x y x y z

ω ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

⋅∇ = − + − + − + + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

vv

ˆˆ ˆ( ) ( ) ( ) ( )w v u w v u

V u v w i j kx y z y z z x x y

ω ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

⋅∇ = + + − + − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

v v

Expand the above, collect like terms, and compare coefficients of ˆ,i ˆ,j and ˆ.k 5.41 Studying the vorticity components of Eq. 3.2.21, we see that /z u yω ∂ ∂= − is the

only vorticity component of interest. The third equation of Eq. 5.3.24 then simplifies to

DDt

zz

ων ω= ∇ 2

2

2z

y

∂ ων

∂=

since changes normal to the plate are much larger than changes along the plate,

i.e., .z z

y x∂ω ∂ω∂ ∂

>>

5.42 If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces

to

DDt

zω= 0

that is, there is no change in vorticity (along a streamline) between sections 1 and 2. Since (see Eq. 3.2.21), at section 1,

ω∂∂

∂∂z

vx

uy

= − = −10

we conclude that, for the lower half of the flow at section 2, .10=

yu

∂∂

This means the velocity profile at section 2 is a straight line with the same slope of the profile at section 1. Since we are neglecting viscosity, the flow can slip at the wall with a slip velocity u0 ; hence, the velocity distribution at section 2 is u y u y2 0 10( ) .= + Continuity then allows us to calculate the profile: V A V A1 1 2 2=

12

10 0 04 0 04 10 0 02 2 0 02 0 30 0( . )( . ) ( . / )( . ). .× = + × ∴ =w u w u m / s.

Finally, u y y2 0 3 10( ) .= +

101

5.43 No. The first of Eqs. 5.3.24 shows that, neglecting viscous effects,

xx y z

D u u uDt x y zω ∂ ∂ ∂

ω ω ω∂ ∂ ∂

= + +

so that ω y , which is nonzero near the snow surface, creates ω x through the term / ,y u yω ∂ ∂ since there would be a nonzero ∂ ∂u y/ near the tree.

5.44 k T ndAt

Vgz u d V

Vgz u

pV ndA

c vc s c s

v v∇ ⋅ = + +

− + + + +

⋅∫∫ ∫$ ~) ~ $

. .. . . .

∂∂

ρρ

ρ2 2

2 2

v v v v∇⋅ ∇ − = + +

− + ∇ ⋅ + + +

−∫∫ ∫( ) ~ ~

. .. . . .

k T d Vt

Vgz u d V V

Vgz u

pd V

c vc v c v

∂∂

ρ ρρ

2 2

2 2

∴ − ∇ + + +

+ ∇ ⋅ + + +

− =∫ k T

tV

gz u VV

gz up

d Vc v

22 2

2 20

∂∂

ρ ρ ρ ρρ

~ ~ .. .

v v

2 2 2

2 2 2V V p V

V gz Vt t

∂ ∂ρρ ρ ρ

∂ ρ ∂

+ ∇ ⋅ + + = + ∇ ⋅

v v v v V pV V V g z

t∂

ρ∂ ρ

∇+ ⋅ + ⋅∇ + + ∇

v vv v v v

0.=

continuity momentum

∴ − ∇ + + ⋅∇ = ∴ = ∇k Tt

u V u DuDt

k T2 20∂∂

ρ ρ ρ~ ~ .~

.v v

5.45 Divide each side by dxdydz and observe that

2 2 2

2 2 2, ,y dy yx dx x z dz z

T TT T T Ty yx x z zT T T

dx dy dzx x z++ +

∂ ∂∂ ∂ ∂ ∂−− −∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂= = =

∂ ∂ ∂

Eq. 5.4.5 follows.

5.46 ( / )Du D h p Dh Dp p D Dh Dp pV

Dt Dt Dt Dt Dt Dt Dtρ ρ

ρ ρ ρ ρ ρρ ρ

− = = − + = − + − ∇ ⋅

v%

where we used the continuity equation: / .D Dt Vρ ρ= − ∇ ⋅v

Then Eq. 5.4. 9 becomes

2Dh Dp pV K T p V

Dt Dtρ ρ

ρ − + − ∇ ⋅ = ∇ − ∇ ⋅

v v

which is simplified to

2Dh DpK T

Dt Dtρ = ∇ +

102

5.47 See Eq. 5.4.9: ~ . .u cT cTt

uTx

vTy

wTz

k T= ∴ + + +

= ∇ ρ

∂∂

∂∂

∂∂

∂∂

2

Neglect terms with velocity: ρ ∂∂

c Tt

k T= ∇2 .

5.48 The dissipation function Φ involves viscous effects. For flows with extremely large velocity gradients, it becomes quite large. Then

ρcDTDtp = Φ

and DTDt

is large. This leads to very high temperatures on reentry vehicles.

5.49 u r ur

r r y= − ∴ = − ×10 1 10 000 2 105( ). . ) ( takes the place of 2 ∂∂

From Eq. 5.4.17, Φ =

= ×212

4 102

2 10µ∂∂

µuy

r .

At the wall where r = = × × × × = ⋅−0 01 1 8 10 4 01 10 725 2 10. . . . m , N / m s2Φ

At the centerline ∂∂ur

= =0 0 so Φ .

At a point half-way: Φ = × × × × = ⋅−1 8 10 4 005 10 185 2 10. . . N / m s2

5.50 (a) Momentum: ∂∂

ν∂∂

ut

uy

=2

2

Energy: ρ∂∂

∂∂

µ∂∂

c Tt

K Ty

uy

= +

2

2

2

.

(b) Momentum: ρ ∂∂

µ∂∂

∂µ∂

∂∂

ut

uy y

uy

= +2

2

Energy: ρ∂∂

∂∂

µ∂∂

c Tt

K Ty

uy

= +

2

2

2

.

103

CHAPTER 6

Dimensional Analysis and Similitude

6.1

gV

Vg

pg

zg

VV

gpg

z12

12

11

12

22

222 2

+ +

= + +

ρ ρ.

12

12

1

12

1

12

22

12

2

12

2

12+ + = + +

pV

gzV

VV

pV

gzVρ ρ

.

or 12

12

1

12

1

12

2

22

2

22

22

12+ + = + +

pV

gzV

pV

gzV

VVρ ρ

6.2 a) [ ]& . .m FTL

= =⋅⋅

=⋅

∴kgs

N sm s

N sm

2

b) [ ]p FL

= ∴N

m 2 . 2

c) [ ]ρ = =⋅⋅

=⋅

∴kgm

N sm m

N sm

3

2

3

2

4 . .FTL

2

4

d) [ ]µ =⋅

∴N sm

2 . FTL2

e) [ ]W FL= ⋅ ∴N m.

f) [ ]& .W FLT

=⋅

∴N m

s

g) [ ]σ = ∴N / m . FL

6.3 (A) The dimensions on the variables are as follows:

2 2

2 3 2 2/

[ ] [ ] , [ ] , [ ] , [ ]L L ML ML T M L

W F V M d L p VT TT T L LT

= × = × = = ∆ = = =&

First, eliminate T by dividing W& by ∆p. That leaves T in the denominator so divide by V leaving L2 in the numerator. Then divide by d2. That provides

2W

pVdπ =

∆

&

6.4 12 5 2, , , .T e r

fR R RR R

µ

ρω ρω

∴ =

l

104

6.5 (A) ( , , , , ). The units on the variables on the rhs are as follows:V f d l g ω µ=

12

[ ] , [ ] , [ ] , [ ] , [ ]L ML

d L l L g TTT

ω µ−= = = = =

Because mass M occurs in only one term, it cannot enter the relationship.

6.6 [ ] [ ] [ ] [ ]V f VLT

LML

MLT

= = = = =( , , ). , , , .l lρ µ ρ µ 3

∴There is one π − term: πρ

µ1 =Vl

.

∴ = = ∴ = =π π ρµ1 1 2

0fV

C( ) , ReConst. or Const.l

6.7 [ ] [ ] [ ] [ ]V f d VLT

MT

ML

d L= = = = =( , , ). , , , .σ ρ σ ρ 2 3

∴ = ∴ = = ∴ =πσ

ρπ π

σρ1 2 1 1 2

02V d

f CV d

C. ( ) , onst. or We = Const.

6.8 [ ] [ ] [ ] [ ]V f H g m VLT

gL

Tm M H L= = = = =( , , ). , , , . 2

∴ = ∴ = ∴ =π π1

0

2 1gHm

VC V gH C. . / .

6.9 [ ] [ ] [ ] [ ] [ ] [ ]V f H g m V LT

H L g LT

m M ML

MLT

= = = = = = =( , , , , ). , , , , , .ρ µ ρ µ 2 3

Choose repeating variables H g, ,ρ (select ones with simple dimensions-we couldn’t select V, H, and g since M is not contained in any of those terms):

π ρ π ρ π µ ρ1 2 31 1 1 2 2 2 3 3 3= = =VH g mH g H ga b c a b c a b c, , .

∴ = = = = =πρ

πρ

πµ

ρµ

ρ1

0

2 3 3 3 2 3

Vg H

VgH

mH gH gH

. . ./

∴ =

VgH

fmH gH

1 3 3ρµ

ρ, .

Note: The above dimensionless groups are formed by observation: simply combine the dimensions so that the π − term is dimensionless. We could have set up equations similar to those of Eq. 6.2.11 and solved for 1 1 1 2 2 2 3 3 3, , and , , c and , , .a b c a b a b c But the method of observation is usually successful.

6.10 [ ] [ ] [ ] [ ] [ ]F f d V F MLT

d L V LT

MLT

MLD D= = = = = =( , , , , ). , , , , .l µ ρ µ ρ 2 3

105

π ρ π ρ π µ ρ1 2 31 1 1 2 2 2 3 3 3= = =F V dV VD

a b c b c a a b cl l l, , .

∴ = = =πρ

π πµ

ρ1 2 2 2 3F

Vd

VD

l l l, , .

∴ =

FV

fd

VD

ρµ

ρl l l2 2 1 , .

We could write π

π πππ

1

22 2

2

3

2

1=

f , or

Fd V

fd dV

D

ρµ

ρ2 2 2=

l, . This is equivalent

to the above. Either functional form must be determined by experimentation.

6.11 [ ] [ ] [ ] [ ] [ ]F f d V F MLT

d L V LT

MLT

MLD D= = = = = =( , , , , ). , , , , .l µ ρ µ ρ 2 3

π µ π µ π ρ µ1 2 31 1 1 2 2 2 3 3 3= = =F d V d V d VD

a b c a b c a b c, , . l

By observation we have 1 2 3, , .DF VdVd d

ρπ π π

µ µ= = =

l

∴ =

FVd

fd

VdD

µρ

µ1l

, .

Rather than π π π1 1 2 3= f ( , ), we could write

ππ

ππ

1

32 2

3

1=

f , , an acceptable form:

FV d

fd Vd

D

ρµ

ρ2 2 2=

l, .

6.12 [ ] [ ] [ ] [ ] [ ] [ ]h f d g h L MT

d L ML T

g LT

= = = = = = =( , , , , ). , , , , , .σ γ β σ γ β 2 2 2 21

Select d g, ,γ as repeating variables.

π γ π σ γ π β1 21 1 1 2 2 2= = =hd g d ga b c a b c, , . 3

∴ = = =π πσ

γπ β1 2 2 3

hd d

, , .

∴ =

hd

fd1 2

σγ

β, . Note: gravity does not enter the answer.

6.13 [ ] [ ] [ ] [ ]F f m R FMLT

m MT

R LC C= = = = =( , , ). , , , .ω ω 2

1

∴ = = ∴ = ∴ =π ωω ω

ω1 2 22F m R

Fm R

Fm R

C F Cm RCa b c C C

C. .

6.14 [ ] [ ] [ ] [ ]σ σ= = = = =f M y I MLT

M MLT

y L I L( , , ). , , , . 2

2

24 ∴ =π σ1 M y Ia b c .

106

Given that 1b = − , 1 Const. C .I My

yM Iσ

π σ= = ∴ =

6.15 [ ] [ ] [ ]V f ddpdx

VLT

MLT

d Ldpdx

ML T

=

= = =

=( , , . , , , .µ µ 2 2

∴ =

π µ1 V d

dpdx

a bc

. Let’s start with the ratio µ

dp dx/ so that “M” is accounted for.

Then the π 1 − term is µV

dp dx d/.

2 Hence,

πµ

µ1 2

2

= = ∴ =V

dp dx dV

d dp dx/

/.

Const. Const

6.16 [ ] [ ] [ ] [ ]V f H g V LT

H L g LT

ML

= = = = =( , , ). , , , .ρ ρ 2 3

∴ = = = ∴ =π ρρ

1

0

VH g Vg H

V gHa b c Const. Const. .

Density does not enter the expression.

6.17 [ ] [ ] [ ] [ ] [ ] [ ]3 2( , , , , ). , , , , , .

L M M LV f H g d V H L g d L

T LT L Tµ ρ µ ρ= = = = = = =

π ρ π µ ρ π ρ1 21 1 1 2 2 2 3 3 3= = =VH g H g dH ga b c a b c a b c, , . 3

Repeatingvariables

H g, , .ρ

π πµ

ρπ1 2 3 2 3= = =

VgH gH

dH

, , ./

∴ = =

π π π

µ

ρ1 1 2 3 1 3

fVgH

fgH

dH

( , ), , . or

6.18 ∆p f V d L= ( , , , , , ).ν ε ρ

[ ] [ ] [ ] [ ] [ ] [ ] [ ]2

2 3, , , , , , .

M L L Mp V d L L L L

T TLT Lν ε ρ∆ = = = = = = =

Repeating variables: V d, , .ρ

π ρ π ν ρ π ρ π ε ρ1 2 3 41 1 1 2 2 2 3 3 3 4 4 4= = = =∆pV d V d L V d V da b c a b c a b c a b c, , , .

∴ = = = =πρ

πν

π πε

1 2 3

∆pV V d

Ld d

, , , . 2 4

π π π πρ

ν ε1 1 2 3 4 2 1= ∴ =

fp

Vf

VdLd d

( , , ). , , . ∆

107

6.19 F f V c h r wD = ( , , , , , , , , )ρ µ φ α where c is the chord length, h is the maximum thickness, r is the nose radius, φ is the trailing edge angle, and α is the angle of attack. Repeating variables: V c, , .ρ The π − terms are

πρ

πρµ

π π π φ π π α1 2 2 2 3 4 5 7= = = = = = =F

V cV c c

hcr

cw

D , , , , , , . 6

Then,

F

V cf

V c ch

cr

cw

D

ρρµ

φ α2 2 1=

, , , , ,

6.20 [ ] [ ] [ ] [ ] [ ] [ ]3

22

( , , , , ). , , , , 1, .L L

Q f R A e S g Q R L A L e L s gT T

= = = = = = =

There are only two basic dimensions. Choose two repeating variables, R and g. Then, π π π π1 2 3 4

1 1 2 2 3 3 4 4= = = =QR g AR g eR g sR ga b a b a b a b, , , .

∴ = = = =π π π π1 5 2 2 2 3 4Q

gRA

ReR

s/ , , , .

∴ = ∴ =

π π π π1 1 2 3 4 5 1 2f

Q

gRf

AR

eR

s( , , ). , , .

6.21 [ ] [ ] [ ] [ ]2 2 3( , , , ). , , , , .p p

L L M MV f h g V h L g

T T T Lσ ρ σ ρ = = = = = =

Repeating variables: h g V h g h gpa b c a b c, , . , .ρ π ρ π σ ρ ∴ = =1 2

1 1 1 2 2 2

∴ = = ∴ =

π π

σρ

σρ1 2 2 1 2

V

hg ghV

ghf

ghp p, . .

6.22 F f V e I dD = ( , , , , , ).µ ρ Repeating variables: V d, , .ρ

[ ] [ ] [ ] [ ] [ ] [ ] [ ]2 3, , , , , 1, .D

ML L M MF V e L I d L

T LTT Lµ ρ= = = = = = =

π ρ π µ ρ π ρ π ρ1 2 3 41 1 1 2 2 2 3 3 3 4 4 4= = = =F V d V d e V d I V dD

a b c a b c a b c a b c, , , .

∴ = = = =πρ

πµρ

π π1 2 2 2 3 4F

V d V ded

ID , , , .

∴ =

FV d

fV d

ed

ID

ρµρ2 2 1 , , .

6.23 F f V D gD s= ( , , , , , ).ρ ρ µ Repeating variables: V D, , .ρ

[ ] [ ] [ ] [ ] [ ] [ ] [ ]2 3 3 2, , , , , , .D s

ML L M M M LF V D L g

T LTT L L Tρ ρ µ= = = = = = =

108

π ρ π ρ ρ π µ ρ π ρ1 2 3 41 1 1 2 2 2 3 3 3 4 4 4= = = =F V D V D V D gV DD

a b cs

a b c a b c a b c, , , .

∴ = = = =πρ

πρρ

πµ

ρπ1 2 2 2 3 4 2

FV D VD

gDV

D s, , , .

∴ =

FV D

fVD

gDV

D s

ρρρ

µρ2 2 1 2, , .

6.24 F f V d e r cD = ( , , , , , , ).µ ρ Repeating variables: V d, , .ρ

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .FMLT

VLT

MLT

ML

d L e L r L cLD = = = = = = = =2 3 2

1 µ ρ

π ρ π µ ρ π ρ π ρ π ρ1 2 3 4 51 1 1 2 2 2 3 3 3 4 4 4 5 5 5= = = = =F V d V d eV d rV d cV dD

a b c a b c a b c a b c a b c, , , , .

∴ = = = = =πρ

πµ

ρπ π π1 2 2 2 3 4 5

2FV d Vd

ed

rd

cdD , , , , .

∴ =

FV d

fVd

ed

rd

cdD

ρµ

ρ2 2 12, , , .

6.25 f g V d fT

MLT

ML

V LT

d L= = = = = =( , , , ). [ ] , [ ] , [ ] , [ ] , [ ] .µ ρ µ ρ 13

Repeating variables, V d f V d V da b c a b c, , . , ρ π ρ π µ ρ1 21 1 1 2 2 2= =

∴ = = ∴ =

π π

µρ

µρ1 2 1

fdV Vd

fdV

gVd

, . .

6.26 F f V c tL c= ( , , , , ). , ρ αl Repeating variables: V c, , . ρ l

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .FMLT

VLT

cLT

ML

L t LL c= = = = = = =2 3 1 ρ αl

π ρ π ρ π ρ π α ρ1 2 3 41 1 1 2 2 2 3 3 3 4 4 4= = = =F V cV tV VL

a bc

c a bc

c a bc

c a bc

cl l l l, , , .

∴ = = = =πρ

π π π α1 2 2 2 3 4F

VcV

tL

c cl l, , , .

∴ =

FV

fcV

tL

c cρα

2 2 1l l, , .

6.27 2

2 31

( , , , , ). [ ] , [ ]= , [ ]= , [ ] , [ ] , [ ] .ML M M

T f d t T d L t LT LTT L

ω ρ µ ω ρ µ= = = = =

Repeating variables: d Td d t da b c a b c a b c, , . , , . 3ω ρ π ω ρ π µ ω ρ π ω ρ1 21 1 1 2 2 2 3 3 3= = =

109

1 2 32 5 2, , . T t

dd d

µπ π π

ρω ρω∴ = = =

3 51 12 5 2 2 , . , .

T t tf W d f

d dd d d

µ µρω

ρω ρω ρω

∴ = =

&

6.28 F f V d LD c= ( , , , , , , ) ρ µ ρ ω where d is the cable diameter, L the cable length, ρc the cable density, andω the vibration frequency. Repeating variables: V d, , . ρ The π − terms are

πρ

πρ

µπ π

ρρ

πω1 2 2 2 3 4 5= = = = =

FV d

Vd dL

Vd

D

c

, , , ,

We then have

F

V df

Vd dL

Vd

D

cρρ

µρ

ρ ω2 2 1=

, , ,

6.29 ∆p f D h d d= ( , , , , , ). ω ρ 1 0 Repeating variables: D, , . ω ρ

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ]∆pM

LTD L h L

TML

d L d L= = = = = = =2 3 1 01

ω ρ

πρω

π π π1 2 2 2 31

40= = = =

∆pD

hD

dD

dD

, , , .

∴ =

=∆p

Df

hD

dD

dD

Wρω 2 2 1

1 0, , . & force × velocity = ∆pD D2 × ω .

∴ =

& , , .W D fhD

dD

dD

ρω 3 51

1 0

6.30 T g f d H N h= ( , , , , , , ). , ω ρl Repeating variables: ω ρ, , . d

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .TMLT

fT T

d L H L L N h LML

= = = = = = = = =2

2 3

1 11 ω ρl

πρω

πω

π π π π1 2 5 2 3 4 5 6= = = = = =T

df H

d dN

hd

, , , , , . l

∴ =

Td

gf H

dN

hdρω ω2 5 1 , , , , .

d

l

6.31 Q f H w g= ( , , , , , ). µ ρ σ Repeating variables: H g, , . ρ

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .QLT

H L w L gL

TMLT

ML

MT

= = = = = = =3

2 3 2 µ ρ σ

∴ = = = =π π πµ

ρπ

σρ1 5 2 3 3 4 2

Q

gH

wH gH gH

, , , .

110

∴ =

Q

gHf

wH gH gH5 1 3 2, , .

µ

ρ

σρ

6.32 d f V V Dj a= ( , , , , , , ). σ ρ µ ρ Repeating variables: V Dj , , . ρ

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .d L VLT

VLT

D LMT

ML

MLT

MLj a= = = = = = = = σ ρ µ ρ2 3 3

π π πσ

ρπ

µρ

πρρ1 2 3 2 4 5= = = = =

dD

VV V D V Dj j j

a, , , , .

∴ =

d

Df

V

V V D V Dj j j

a1 2

, , , . σ

ρ

µ

ρ

ρ

ρ

6.33 T f H h R t= ( , , , , ).ω µ ρ , , Repeating variables: ω ρ, , . h

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .TMLT T

H L h L R L t LMLT

ML

= = = = = = = =2

2 3

1 ω µ ρ

πρω

π π π πµ

ρω1 2 5 2 3 4 5 2= = = = =T

dHh

Rh

th h

, , , ,

∴ =

Td

f Hh

Rh

th hρω

µρω2 5 1 2, , , .

6.34 µ ρ= f D H g V( , , , , , ) l . D = tube dia., H = head above outlet, l = tube length.

Repeating variables: D VVD

HD D

gDV

, , . , , , ρ πµ

ρπ π π1 2 3 4 2= = = =

l

∴ =

µρVD

fHD D

gDV1 2

, , . l

6.35 T f R e r= ( , , , , ). , , ω ρ µ l Repeating variables: R, , .ω ρ

[ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] , [ ] .TMLT

R LT

ML

e L r LMLT

L= = = = = = = =2

2 3

1 ω ρ µ l

πρω

π π π πµ

ρω1 2 5 2 3 4 5 2= = = = =T

ReR

rR R R

, , , , l

12 5 2, , , .T e r

fR R RR R

µ

ρω ρω

∴ =

l

6.36 y f V y g2 1 1= ( , , , ). ρ Neglect viscous wall shear.

[ ] , [ ] , [ ] , [ ] , [ ] .y L VLT

y LML

gL

T2 1 1 3 2= = = = = ρ Repeating variables: V y1 1, , . ρ

111

π π12

12

1

12= =

yy

gyV

, . ( ρ does not enter the problem).

∴ =

yy

fgyV

2

1

1

12 .

6.37 f g d V fT

d L LML

MLT

VLT

= = = = = = =( , , , , ). [ ] , [ ] , ] , [ ] , [ ] , [ ] . [ l lρ µ ρ µ1

3

Repeating variables: d V, , . ρ (l = length of cylinder).

π π πµ

ρµ

ρ1 2 3 1. , , . , .= = = ∴ =

fdV d V d

fdV

fd Vd

l l

6.38 QQ

VV

pp

VV

F

FVV

m

p

m m

p p

m

p

m m

p p

p m

p p

m m m

p p p

= = =ll

ll

2

2

2

2

2 2

2 2, ,( )

( )

∆∆

ρρ

ρρ

ττ

ρρ

ρρ

ρρ

m

p

m m

p p

m

p

m m m

p p p

m

p

m m m

p p p

VV

TT

VV

QQ

VV

= = =2

2

2 3

2 3

3 2

3 2, ,&&

ll

ll

( &Q has same dimensions as .)W&

6.39 (A) Re Re . m mm p

m

V Lν

= p p

p

V L

ν= . 12 9 108 m/s.p

m pm

LV V

L∴ = = × =

6.40 A) 5

61.51 10

Re Re . . 4 10 461 m/s.1.31 10

p p pm m mm p m p

m p m p

V L LV LV V

Lν

ν ν ν

−

−×

= = ∴ = = × =×

6.41 a) Re Re . . .m pm m

m

p p

p

m

p

p

m

V d V d VV

d

d= = ∴ = =

ν ν7

QQ

VV

Q QVV

m

p

m m

p pm p

m

p

m

p

= ∴ = = × × =ll

ll

2

2

2

2 21 5 71

70 214. . . . m / s3

&& . & .

WW

VV

Wm

p

m m m

p p pm= = × = ∴ = × =

ρρ

3 2

3 23

2717

7 7 200 1400ll

kW

b) Re Re ...

. .m pm

p

p

m

m

p

VV

d

d= ∴ = = × =

νν

79

134 85

32

32

11.5 4.85 0.148 m /s.

71

4.85 200 466 kW7

m

m

Q

W

= × × =

= × × =&

112

6.42 a) Re Re . . .m pm m

m

p p

p

m

p

p

m

V d V d VV

d

d= = ∴ = =

ν ν5

2

2 21 1

5. 800/5 160 kg/s.55

m m m mm p

p p p p

m Vm m

m V

ρ

ρ= = × ∴ = = =

& l& &

& l

∆∆

∆ ∆pp

VV

p pm

p

m m

p pm p= = ∴ = = × =

ρρ

2

225 25 25 600 15. . 000 kPa

b) Re Re ...

. .m pm

p

p

m

m

p

VV

d

d= ∴ = = × =

νν

58

1143 51

221

800 3.51 112 kg/s. 600 3.51 7390 kPa.5

m mm p= × × = ∆ = × =&

6.43 a) Re Re . . .m pm m

m

p p

p

m

p

p

m

V d V d VV

d

d= = ∴ = =

ν ν10

FF

VV

F Fm

p

m m m

p p pm p= = × ∴ = =

ρρ

2 2

2 2210 10

ll

110

= 1. lb2 .

b) Re Re ...41

. .m pm

p

p

m

m

p

VV

d

d= ∴ = = × =

νν

101 061

7 52

p

p mF Fρ

=mρ

2 22

2 2 21

10 10 17.68 lb. 7.52

p p

m m

V L

V L= × × =

6.44 Re Re . . .m pm m

m

p p

p

m

p

p

m

m

p

m

p

V V VV

= = ∴ = = = assuming l l l

lν ννν

νν

10 1

∴ = =V Vm p10 1000 km / hr. This velocity is much too high for a model test; it is in the compressibility region. Thus, small-scale models of autos are not used. Full-scale wind tunnels are common.

6.45 Re Re . . .m pm m

m

p p

p

m

p

p

m

m

p

V V VV

= ∴ = ∴ = l l l

lν ννν

Water: VV

V Vm

p

p

mm p m p= = = ∴ = =

l

l10 10 900 assuming km / hr.ν ν .

Air: V Vm pp

m

m

p

= = ××

×=

−

−

l

lνν

90 1015 101 10

135

6

. 500 km / hr.

Neither a water channel or a wind tunnel is recommended. Full-scale testing in a water channel is suggested.

113

6.46 Re Re . . / / .m pm m

m

p p

pm p p m m p

V VV V= = ∴ = = = if

l ll l

ν νν ν10

∴ = × =Vm 10 50 500 m / s. This is in the compressibility range so is not recommended. Try a water channel for the model study. Then

6

51 10

10 0.662.1.5 10

pm m

p m p

VV

νν

−

−×

= = × =×

l

l 33.1 m/s.mV∴ =

This is a possibility, although 33.1 m/s is still quite large.

2 2

22 2 2

( ) 1000 10.662 3.56.

( ) 1.23 10D m m m m

D p p p p

F VF V

ρ

ρ= = × × =

l

l

6.47 5

31.06 10

Re Re . . 2.5 15.5 10

p p pm m mm p m p

m p m p

V d VV dd d

Vν

ν ν ν

−

−×

= = ∴ = = × ××

= 0.0048 ft.

Find oilν using Fig. B.2. Then

2

22

1.941 1.11.

1.94 0.9m m m

p p p

p Vp V

ρ

ρ

∆= = × =

∆ ×

6.48 Re Re . . . . .m pm m

m

p p

pm p

p

m

m

p

p

m

V VV V= = ∴ = = × × ×− l l l

l

l

lν ννν

0 1 025 10 3

If ll

lpp

mmV≅ = = =5 5

00252000 0 005 cm, then and m / s.

..

We could try l p mV≅ =50 0 05 cm, but m / s.. Each of these Vm ' s is quite small — too small for easy measurements. Let’s try a wind tunnel. Then,

V Vm pp

m

m

p

p

mp= = × × ×

××

= =−−

−

l

l

l

ll

νν

01 025 101 10

18 100 28 53

3

5. ..

. m / s if cm. Or,

if l p mV= =50 2 8 cm, m / s.. This is a much better velocity to work with in the lab. Thus, choose a wind tunnel.

6.49 Re Re . . . . .m pm m

m

p p

pm p

m

m m

p

p p

m

p

V V Vg

Vg

VV

= ∴ = = ∴ = ∴ = Fr Fr l l

l lν ν

2 2 130

VV

m

p

p

m

m

p

m

p

m

pm= = = ∴ = ∴ = × −

l

lνν

νν

νν

ν301

301

1646 1 10 9. . . . m / s Impossible!2

6.50 (C) 2

Fr Fr . mm p

m m

Vl g

=2p

p p

V

l g=

1. 2 0.5 m/s.

4m

m pp

lV V

l∴ = = × =

114

6.51 (A) From Froude’s number mm p

p

lV V

l= . From the dimensionless force we have:

2 2

* * 22 2 2 2 2 2 or . 10 25 25 156000 N.

p p pmm p p m

m m m p p p m m

F V lFF F F F

V l V l V lρ ρ= = ∴ = = × × =

6.52 Fr Fr m /sm pm

m m

p

p pm p

m

p

Vg

Vg

V V= ∴ = ∴ = = =. . . .2 2

10 160

1 29l l

ll

( )( )

. ( ) ( ) . .FF

VV

FV

VFD m

D p

m m m

p p pD p

p

m

p

mD m= ∴ = × = × × = ×

ρρ

2 2

2 2

2

2

2

22 660 60 10 216 10

ll

l

l N

6.53 Fr Fr m pm

m m

p

p p

m

p

m

p

Vg

V

gVV

= = ∴ =. .2 2

l lll

.

a) QQ

VV

Q QVV

m

p

m m

p pm p

m

p

m

p

= ∴ = = × × =ll

ll

2

2

2

2 22110

110

0 00632. . . m / s3

b) FF

VV

F FV

Vm

p

m m m

p p pp m

p

m

p

m

= ∴ = = × × =ρρ

2 2

2 2

2

2

2

2212 10 10

ll

l

l. . 12 000 N

6.54 Neglect viscous effects. Fr Fr fpsm pm

p

m

pp

VV

V= ∴ = = ∴ =. . . .ll

110

63 2

FF

VV

F FV

Vm

p

m m m

p p pp m

p

m

p

m

= ∴ = = × × =ρρ

2 2

2 2

2

2

2

220 8 10 10 800

ll

l

l. . . lb

6.55 Neglect viscous effects, and account for wave (gravity) effects.

Fr Fr m pm

m m

p

p p

m

p

m

p

m

p

m m

p p

Vg

Vg

VV

VV

= ∴ = ∴ = =. . .//

.2 2

l lll

ll

ωω

∴ = = × × =ω ωm pm

p

p

m

VV

l

l600

110

10 1897 rpm.

TT

VV

T TV

Vm

p

m m m

p p pp m

p

m

p

m

= ∴ = = × × = ⋅ρρ

2 3

2 3

2

2

3

331 2 10 10 120

ll

l

l. . . 000 N m

6.56 Fr Fr 6100

m pm

m m

p

p p

m

p

m

p

m

p

p

m

Vg

Vg

VV

= ∴ = ∴ = = ∴ =. . . . .2 2

278l l

ll

ll

l

l

115

6.57 Check the Reynolds number:

Re .pp p

p

V d= =

×= ×−ν

15 210

30 1066

This is a high-Reynolds-number flow.

Re / . .m =×

= ×−

2 2 3010

1 33 1065

This may be sufficiently large for similarity. If so,

&& . .W

WVV

m

p

m m m

p p p

= = × = × −ρρ

3 2

3 2

3

3 262

151

302 63 10l

l

∴ = × × =−& ( . ) / . .Wp 2 2 15 2 63 10 16336 kW 6.58 This is due to the separated flow downwind of the stacks, a viscous effect.

∴ Re is the significant parameter. Re.

. .p =×

×= ×−

10 415 10

26 7 1055 This is a

high-Reynolds-number flow. Let’s assume the flow to be Reynolds number independent above Re = ×5 10 5 (see Fig. 6.4). Then

55

4/20Re 5 10 . 37.5 m/s.

1.5 10m

m mV

V−×

= × = ∴ ≥×

6.59 Re.

. .p =×

×= ×−

20 1015 10

13 3 1056 This is a high-Reynolds-number flow.

Let Re.4

.. .m

mm

VV= =

××

∴ ≥−101 5 10

3 7555 m / s for the wind tunnel.

Re . . .mm

mV V= =

××

∴ ≥−10 11 10

1 056 m / s for the water channel.

Either could be selected. The better facility would be chosen.

F

F

V

V FFm

m

m m m

m m m mm

1

2

1 1 1

2 2 2 2

2

2 2

2 2

2

2

2

2

3 23 2

10001 23

215

1416= = ∴ = × =

ρ

ρ

l

l.

. ..

.4 ..4

. . N

&&

.. & ( . )

..

WW

VV

Wm

p

m m m

p p pp= =

××

∴ = × × =ρρ

3 2

3 2

3 2

3 2

3

3

2

2

15 420 10

15 3 22015

104

71ll

100 W

6.60 Re is the significant parameter. This is undoubtedly a high-Reynolds-

number flow. If the model is 4' high then l

lp

m

= 250, and the model’s diameter is

45/250 = 0.18'. For Re ,m = ×3 10 5 we have

Re.

..m

mm

VV= × =

××

∴ ≥−3 1018

1 5 102505

4 fps, and a study is possible.

116

6.61 Mach No. is the significant parameter. M Mm p= .

a) M M . . 200 m/s.pmm p m p

m p

VVV V

c c= ∴ = ∴ = =

FF

VV

Fm

p

m m m

p p pp= ∴ = × × =

ρρ

2 2

2 22 210 1 20 4000

ll

. . N

b) 255.7

200 186 m/s.296

p pp m m

m m

c TV V V

c T= = = =

F FVVp m

m p p

m m m

= = × × × =ρ

ρ

2 2

2 2

2

2210 601 186

20020 2080

l

l. . N

c) 223.3

200 174 m/s.296

p pp m m

m m

c TV V V

c T= = = =

F FVVp m

p

m

p

m

p

m

= = × × × =ρ

ρ

2

2

2

2

2

2210 338 174

20020 1023

l

l. . N

6.62 273

M M . . 250 276 m/s.223.3

pmm p m

m p

VVV

c c= ∴ = ∴ = =

223.3

. 290 262 m/s.273

m m mp

p p p

V c TV

V c T= = ∴ = =

pp

VV

p pV

Vm

p

m m

p pp m

p

m

p

m

o

o

= ∴ = = =ρρ

ρ

ρρ

ρ

2

2

2

2

2

2803388

262290

34 6...

. . kPa, abs

α p = 5o for similarity. (Note: we use ρ m at 2700 m where T = 0°C.)

6.63 a) Fr Fr m pm

m m

p

p p

m

p

m

p

Vg

Vg

VV

= = ∴ =. . .2 2

l lll

ωω

ωm

p

m

p

p

mm

VV

= = × ∴ = × =l

l110

10 20001010

6320. . rpm

b) Re Re . . .m pm m

m

p p

p

m

p

p

m

V V VV

= = ∴ = = l l l

lν ν10

ωω

ωm

p

m

p

p

mm

VV

= = × = ∴ =l

l10

110

1 2000. . rpm

6.64 There are no gravity effects nor compressibility effects. It is a high-Re

flow. TT

VV

T TV

Vm

p

m m m

p p mp m

p

m

p

m

= ∴ = = × × = ⋅ρρ

2 3

2 3

2

2

3

3

2

2312

1560

10 750ll

l

l. . N m

117

ωω

ω ωm

p

m

p

p

mp m

p

m

m

p

VV

V

V= ∴ = = × × =

l

lll

. . . rpm5001560

110

12 5

6.65 Re Re . .m pm m

m

p p

pm p

p

m

V VV V= ∴ = ∴ = = × = m / s.

l l l

lν ν10 10 100

This is too large for a water channel. Undoubtedly this is a high-Re flow. Select a speed of 5 m/s. For this speed,

Re.

,m =×

×= ×−

5 011 10

5 1065 where we used 0.1 ( 1 m,m p= =l l i.e., the

dia. of the porpoise). ω ωm pm

p

p

m

VV

= = × × =l

l1

510

10 5 motions / second.

6.66 ρρ

ρ* * * * * *, , , , , .= = = = = =

o

t tf u uV

v vV

x x yy

l l

Substitute in:

* * * * *

o * * *( ) ( )

0.o oV u V v

ft x y

∂ρ ∂ ρ ∂ ρρ ρ ρ

∂ ∂ ∂+ + =

l l

Divide by ρoV / :l

∴ + + =fV t x

uy

vfV

l l∂ρ∂

∂∂

ρ∂

∂ρ

*

* ** *

** *( ) ( ) . .0 parameter =

6.67 * * * * * * * * *2, , , , , , , , .

V u v w x y z pV u v w x y z p t tf

U U U U Uρ= = = = = = = = =

vv

l l l

Substitute into Euler’s equation and obtain:

Uf Vt

U u Vx

U v Vy

U w Vz

U p∂∂

∂∂

∂∂

∂∂

ρρ

v

l

v

l

v

l

v

l

v*

**

*

**

*

**

*

*

* *

.+ + + = −∇2 2 2 2

Divide by U2 / :l

fU

Vt

u Vx

v Vy

w Vz

pl

v v v v∂∂

∂∂

∂∂

∂∂

*

**

*

**

*

**

*

** * .+ + + = −∇ Parameter =

fUl

6.68 v

v

l

vlv

lV

VU

ttU

ppU

hh* * * * *, , , , .= = ∇ = ∇ = =

ρ 2 Euler’s equation is then

ρ ρ ρU DV

DtU

p g h2 2

l

v

l

v ll

v*

** * * * .= − ∇ − ∇

Divide by ρU 2 / :l

DVDt

pgU

hv v l v*

** * * * .= −∇ − ∇2 Parameter =

gU

l2 .

6.69 There is no y- or z-component velocity so continuity requires that∂ ∂u x/ .= 0 There

is no initial pressure distribution tending to cause motion so∂ ∂p x/ .= 0 The

118

x-component Navier-Stokes equation is then

u u

ut x

∂ ∂∂ ∂

+ v+u

wy

∂∂

+1u p

z x∂ ∂∂ ρ ∂

= −2

2xu

gx

∂ν

∂+ +

2 2

2 2u u

y z

∂ ∂

∂ ∂+ +

(wide plates) This simplifies to

∂∂

ν∂∂

ut

uy

=2

2 .

a) Let u u U y y h t tU h* * */ , / / .= = = and Then

Uh

ut

Uh

uy

2

2

2∂∂

ν ∂∂

*

*

*

*2=

The normalized equation is

∂∂

∂∂

ut

uy

*

*

*

*2Re=

1 2

where Re =Uhν

b) Let u u U y y h t t h* * */ , / / .= = = and Thenν 2

ν ∂

∂ν

∂∂

Uh

ut

Uh

uy2 2

2*

*

*

*2=

The normalized equation is

∂∂

∂∂

ut

uy

*

*

*

*2=2

6.70 The only velocity component is u. Continuity then requires that ∂ ∂u x/ = 0 (replace z with x and vz with u in the equations written using cylindrical coordinates). The x-component Navier-Stokes equation is

ru

vt

∂∂

+vu

rθ∂

∂+

u uu

r x∂ ∂∂θ ∂

+1

xp

gx

∂ρ ∂

= − +2 2

2 2 21 1u u ur rr r

∂ ∂ ∂ν

∂∂ ∂θ+ + +

2

2u

x

∂

∂+

This simplifies to

∂∂ ρ

∂∂

ν∂∂

∂∂

ut

px

ur r

ur

= − + +

1 12

2

a) Let * * * * 2 */ , / , / , = / and / :u u V x x d t tV d p p V r r dρ= = = =

Vd

ut

Vd

px

Vd

ur r

ur

2 2

2

2 1∂∂

ρρ

∂∂

ν ∂∂

∂∂

*

*

*

*

*

*2 *

*

*= − + +

The normalized equation is

∂∂

∂∂

∂∂

∂∂

ut

px

ur r

ur

*

*

*

*

*

*2 *

*

*Re= − + +

1 12

where Re =Vdν

b) Let * * * 2 * 2 */ , / , / , = / and / :u u V x x d t t d p p V r r dν ρ= = = =

ν ∂

∂ρρ

∂∂

ν ∂∂

∂∂

Vd

ut

Vd

px

Vd

ur r

ur2

2

2

2 1*

*

*

*

*

*2 *

*

*= − + +

The normalized equation is

119

∂∂

∂∂

∂∂

∂∂

ut

px

ur r

ur

*

*

*

*

*

*2 *

*

*Re= − + +2 1

where Re =Vdν

6.71 Assume wz

= =0 0 and ∂∂

. The x-component Navier-Stokes equation is then

ut

∂∂

uu v

x∂∂

+ +u u

wy z

∂ ∂∂ ∂

+1 p

x∂

ρ ∂= −

2 2 2

2 2 2xu u u

gx y z

∂ ∂ ∂ν

∂ ∂ ∂+ + + +

With g gx = the simplified equation is

uux

gu

xu

y∂∂

ν∂∂

∂∂

= + +

2

2

2

2

Let u u V x x h y y h* * */ , / / .= = = and Then

Vh

uux

gVh

ux

uy

2

2

2 2*

*

*

*

*2

*

*2

∂∂

ν∂∂

∂∂

= + +

The normalized equation is

uux

ux

uy

**

*

*

*

*

*Re∂∂

∂∂

∂∂

= + +

1 12

2

2

2

2Fr where Fr and = =

Vhg

VhReν

6.72 * * * * * *2 2 2, , , , , .o

u v T x yu v T x y

U U T= = = = = ∇ = ∇l

l l

ρ∂∂

∂∂

cUT T

xUT T

yK

T Tpo o

ol l l

*

*

*

** * .+

= ∇2

2

Divide by ρc UTp o / :l

∂∂

∂∂ ρ

Tx

Ty

Kc U

Tp

*

*

*

** * .+ = ∇

l2 Parameter =

Kc Upµ

µρ l

=1 1Pr Re

.

6.73 ρρρ

* * * * * * *, , , , , , .= = = ∇ = ∇ ∇ = ∇ = =o o o

V VU

t tU ppp

T TT

v

v

l

v

l

v

l1 12

22

momentum: ρ ρµ µ

ooU DV

Dtp

p U V U V**

** * * * * * *( ).

2

22

23l

v

l

v

l

v

l

v v v= − ∇ + ∇ + ∇ ∇ ⋅

Divide by ρoU2 / :l

[ ]ρρ

µρ

**

** * * * * * *( ) .DV

DtpU

pU

V Vo

o o

v v

l

v v v v= − ∇ + ∇ + ∇ ∇ ⋅2

2

energy: ρ ρ**

** * * * * .c T U DT

DtK T T p U p Vv o o o ol l l

v v= ∇ − ∇ ⋅2

2

Divide by ρ o v oc T U / :l

ρρ ρ

**

** * * * * .DT

DtKc U

Tpc T

p Vo v

o

o v o

= ∇ − ∇ ⋅l

v v2

120

The parameters are: pU

RTU

kRTkU

ckU k

o

o

o o

ρ 2 2 2

2

2 2

1= = = =

M.

µρ oUl

=1

Re.

Kc U

Kc

c

c UK

o v p

p

v oρ µµ

ρl l= =

Pr Re.

pc T

RTc T

c c

cKo

o v o

o

v o

p v

vρ= =

−= − 1.

The significant parameters are K , , Re, Pr. M

162

CHAPTER 8

External Flows 8.1

8.2 Re . . . .= = ∴ =× ×

= ×−

−5 5 1 51 1020

3 78 105

5VD Dν

m

8.3

8.4

8.5 (C) 8.6 (C)

AB

C

A-B: favorableB-C: unfavorableA-C: favorable

separated region

inviscid flow

boundary layer near surface

inviscid flow

viscous flow near sphere

no separation

separation

wake

separation

separated region

building

inviscid flow

boundary layer

wake

A-B: favorableB-C: unfavorableA-D: favorableC-D: undefined

AB C

D

separated flow

163

8.7 ( B) 6

0.8 0.008Re 4880.

1.31 10

VDν −

×= = =

×

8.8 5 5 5 1 22 10 812

0009155= ∴ = = × × =−VD V D Vν

ν a) fps./ . . / . .

b) V = × × =−5 388 10 812

0002915. / . . fps. c) V = × × =−5 1 6 10 812

0 0124. / . . fps.

8.9 Re.

. .= =××

= ×−

VD D Dν

201 51 10

13 25 1055

a) Re . . .= × × = ×13 25 10 6 7 9 105 6 ∴Separated flow. b) Re . . . .= × × = ×13 25 10 06 7 9 105 4 ∴Separated flow. c) Re . . .= × × =13 25 10 006 79505 ∴Separated flow.

8.10 F pdA p A p r rdr p pDA

= − = − = −

=∫∫ back back

front

02

0 00

1

1 2 2 12

14

12

( ) π π π

Bernoulli: p V p p∞ ∞+ = ∴ = × × =12

12

1 21 20 24220 0

2ρ . . Pa.

∴ = =FD12

242 380π ( ) N

C F

V AD

D= =×

× × ×=

1

22

2 2

2 3801 21 20 1

0 5ρ π.

.

8.11 F F Ftotal bottom top 000 .3 .3 + 10 000 N.= + = × × × × =20 3 3 2700. .

Flift cos 10 N= =2700 2659o Fdrag 10 N= =2700 469sin o

C F

V AL

L= =×× × ×

=1

2

22

2 26591000 5 3 3

2 36ρ . .

.

C F

V AD

D= =×

× × ×=

1

2

22

2 4691000 5 3 3

0 417ρ . .

.

8.12 F p A Lw F p A Lw Lwu u ul l l o= = × = = × =26 80002 5

4015 000 .cos

F F F LwL u= − =lo ocos cos5 10 21 950

F F F LwD u= − =lo osin sin5 10 1569

164

C F

V A

LwLwL

L= =×

×=

1

2

22

2 213119 750

0 25ρ

950 .

.

C F

V A

LwLwD

D= =×

×=

1

2

22

2 15693119 750

0 0179ρ .

.

8.13 If CD = 1 0. for a sphere, Re = 100 (see Fig. 8.8). ∴ ×=

V . ,1 100ν

ν V = 1000 .

a) V FD= × × = ∴ = × × × ×−1000 1 46 10 0146 12

1 22 0146 05 1 05 2 2. . . . . . m / s. π

= × −3 25 10 7. . N

b) V FD= ×××

= ∴ = × × × × ×−

1000 1 46 10015 1 22

0 798 12

015 1 22 798 05 1 05

2 2.. .

. (. . ) . . . m / s. π

= × −4 58 10 5. . N

c) V FD= × × = ∴ = × × × ×−1000 1 31 10 00131 12

1000 00131 05 1 06 2 2. . . . . m /s. π

= × −6 74 10 6. . N

8.14 a) Re ..

. .= =×

×= × ∴ =−

VD CDν6 5

1 5 102 10 0 455

5 from Fig. 8.8.

∴ = = × × × × × =F V ACD D12

12

1 22 6 25 45 1 942 2 2ρ π N.. . . .

b) Re ..

. .=×

×= × ∴ =−

15 51 5 10

5 10 0 255 CD from Fig. 8.8.

∴ = = × × × × × =F V ACD D12

12

1 22 15 25 2 5 42 2 2ρ π N.. . . .

8.15 (B) Assume a large Reynolds number so that CD = 0.2. Then

2

2 21 1 80 10001.23 5 0.2 4770 N.

2 2 3600DF V ACρ π× = = × × × × × =

8.16 (D) Assume a Reynolds number of 105. Then CD = 1.2.

2 21 1. 60 1.23 40 4 1.2. 0.0041 m.

2 2DF V AC D Dρ= ∴ = × × × × × ∴ =

56

40 0.0041Re 1.64 10 . 1.2. The assumption was OK.

10D

VDC

ν −×

= = = × ∴ =

165

8.17 The velocities associated with the two Re's are

VD1

15 53 10 1 5 10

0445101= =

× × ×=

−Re ..

ν m / s,

VD2

24 56 10 1 5 10

044520= =

× × ×=

−Re ..

ν m /s.

The drag, between these two velocities, is reduced by a factor of 2.5

( ) ( )[ ]C CD Dhigh low and = =0 5 0 2. . . Thus, between 20 m/s and 100 m/s the drag is

reduced by a factor of 2.5. This would significantly lengthen the flight of the ball.

8.18 a) F V AC V C V CD D D D= ∴ = × ×

∴ =12

0 5 12

00238 212

48102 22

2ρ π . . . . .

Re /.

. . : .= =×

×= = = ×−

VD V V C VDν4 12

1 6 102080 5 984 Try fps, Re = 2 105

Try fps, Re = 2.3 105C VD = = ×. : .4 110

b) C V VD = = × ×

× ∴ =0 2 12

00238 212

2 15522

. : . . . 0.5 fps.π

8.19 4 2 12

1000 1 0 267 210

2 102 2 26

5. . . . . Re . .= × × ∴ = =×

= ×−V C V C V VD Dπ

Try C VD = ∴ = = × ∴0 5 0 73 1 46 105. : . Re . . m /s. OK.

8.20 Re.

. . .= =×

×= × ∴ =−

VDCDν

40 215 10

5 3 10 0 756 . (This is extrapolated from

Fig. 8.8.) ∴ = × × × × × =FD

12

1 22 40 2 60 7 812. ( ) . . 900 N

M = 81 900 × 30 = 2 46 106. .× ⋅ N m

8.21 a) Re ..

. . Re . . Re . .1 55

25

3525 05

1 08 101 2 10 1 8 10 2 4 10=

××

= × = × = ×− Assume a

rough cylinder (the air is highly turbulent). ( ) ( ) ( )∴ = = =C C CD D D1 2 30 7 0 8 0 9. , . , . .

∴ = × × × × + × × + × × =FD12

1 45 25 05 10 7 075 15 8 1 20 9 13802. (. . . . . . ) . N

M = × × × × × + × × × + × × × = ⋅12

1 45 25 05 10 7 40 075 15 8 27 5 1 20 9 10 25 7002. (. . . . . . . ) . N m

b) Re ..

. . Re . , Re . .1 54

25

3525 05

1 65 107 6 10 1 14 10 1 5 10=

××

= × = × = ×−

166

( ) ( ) ( )∴ = = = =×

=C C CD D D1 2 38 7 8 101

287 3081 17. , . , . .

.. . kg / m 3ρ

∴ = × × × × + × × + × × =FD12

1 17 25 05 10 8 075 15 7 1 20 8 10202. (. . . . . . ) . N

M = × × × × × + × × × + × × × = ⋅12

1 17 25 05 10 8 40 075 15 7 27 5 1 20 8 10 192. (. . . . . . . ) . 600 N m

8.22 Atmospheric air is turbulent. ∴Use the "rough" curve. ∴ =CD 0 7. .

F V D V D V VD = = × × × ∴ =

×× −10 1

200238 6 7 2000 10 2000

1 6 102 5

2

4. . . . /.

. = 2

2 2 2 2min o

0.0024= 30 104 11.8 psf.

2 2p U v

ρ∞ ∴ = − − = −

∴ = ∴ = =V D V D2 2370 148 0 108. . . '. fps

8.23 Since the air cannot flow around the bottom, we imagine the structure to be mirrored as shown. Then L D C CD D/ / . . .= = ∴ =

∞40 5 8 0 66

Re.

. . . . .minmin = =

××

= × ∴ = × =−

VDCDν

30 21 5 10

4 10 1 0 66 6656

∴ = × × ×+

×

× =FD

12

1 22 302 8

220 66 362. . . 000 N

8.24 . B D WF F F+ =

3 2 2 34 1 4 9810 1000 9810 7.82 .

3 2 3Dr V r C rπ π π× + × = × ×

Re .=×

= ×−

V r Vr210

2 1066 ∴ =V C rD

2 178

a) r V V CD= ∴ =. , . .05 892 m. Re = 10 5 Assume a smooth sphere. Try C VD = ∴ = ×. : . .5 4 22 m / s. Re = 4.22 105 This is too large for Re. Try C VD = ∴ = ×. : . .2 667 m / s. Re = 6.67 105 OK.

b) r V V CD= × =. , . .025 4 452 m. Re = 5 10 4 Try C VD = = ×. : . .2 4 72 m / s. Re = 2.4 105 OK. c) r V V CD= =. , . .005 0892 m. Re = 10 4 Try C VD = = ×. : . .5 133 m / s. Re = 1.33 104 OK. d) r V V CD= × =. , . .001 01782 m. Re = 2 10 3 Try C VD = = ×. : . .4 067 m / s. Re = 1.33 103 OK.

W

FB

FD

167

8.25 3 2 3

24 10 1 10 4 10. .077 .00238 62.4 .

3 12 2 12 3 12B D W DF F F V C Sπ π π + = × + × =

3 24

10/12Re 5.2 10 . 1 .0139 810

1.6 10D

VV V C S

−×

= = × ∴ + =×

a) S V CD= =. . .005 2192 Assume atmospheric turbulence, i.e., rough. Try C V C VD D= = = × ∴ = =. : . . Re . . . . .4 23 4 1 2 10 3 275 fps fps

b) S V C C VD D= = = = = × ∴. . . . : . Re . .02 1090 4 52 2 7 102 5 Try fps OK.

c) S V C C VD D= = = =1 0 58 4 3812. . . . : . 200 Try fps

8.26 Assume a 180 lb, 6' sky diver, with components as shown. If V is quite large, then Re > 2 × 105. F FD W= .

12

00238 2 3 12

1 0 7 2 5 12

1 0 7 1812

2 5 1 0 412

4 1802× × × × × × × × × + × × + ×

×

=. . . . . . . . . .V + 2 π

We used data from Table 8.1. ∴ =V 140 fps.

8.27 From Table 8.2 C F V VD D= = × × × =0 35 12

1 22 3 2 0 35 6832 2. . . . . . .

a) F WD = ××

= ∴ =×

=. & .683 80 10003600

337 337 80 10003600

75002

N. W or 10 Hp

b) V F WD= = × = ∴ = × =25 683 25 427 427 25 102 m / s. N. 700 W or 14.3 Hp. & .

c)V F WD= = × = ∴ = × =27 8 683 27 8 527 527 27 8 142. . . & . . m / s. N. 700 W or 19.6 Hp

8.28 1 2 1 1 400 12

1 12. . . . .F F V AC CD D D D= × = = ρ

1 2 12

1 22 2 3 1 1 1 1 4002. . ( ) . . .× × × × × = ×V

∴ =V 9 5. . m / s

8.29 Re ( . . .= =×

= × ∴ =VD CDν

40 4 42 10 0 355 000 / 3600)0.61.51 10

-5 from Fig. 8.8.

a) F V ACD D= = × × × × × =12

12

1 204 40 0 6 6 0 35 93 62ρ . ( . . . 000 / 3600) N2

b) F L DD = × = = =93 6 0 68 63 7 6 0 6 10. . . / / . . N where c) F L DD = × = =93 6 0 76 71 1 20. . . / N where we can use since only one end is free. The ground acts like the mid-section of a 12-m-long cylinder.

3 ft

6 in 8 in. dia.

2.5 ft

6 in

2.5 ft

18 in

1.1 m

1.2 m

FW

FD

Fy

Fx

168

8.30 a) Curled up, she makes an approximate sphere of about 1.2 m in diameter (just a guess!). Assume a rough sphere at large Re. From Fig. 8.8, CD = 0 4. :

F V ACD D=12

2ρ

80 9 8 12

1 21 0 6 0 4 53 72 2× = × × × × ∴ =. . . . . . .V Vπ m /s

Check Re: Re . ..

. .=×

×= × ∴−

53 7 1 21 51 10

4 27 1056 OK.

b) F V ACD D=12

2ρ . From Table 8.2, CD = 1.4:

80 9 812

121 4 1 4 4 292 2× = × × × × ∴ =. . . . . .V Vπ m / s

Check Re: Re.

.. .=

××

= ×−

4 29 8151 10

2 27 1056 Should be larger but the velocity

should be close.

c) 212D DF V ACρ=

2 2180 9.8 1.21 1 1.4. 17.2 m/s.

2V Vπ× = × × × × ∴ =

Check Re: Re.

.. .=

××

= ×−

17 2 1151 10

114 1056 This should be greater than 107 for

CD to be acceptable. Hence, the velocity is approximate. 8.31 With the deflector the drag coefficient is 0.76 rather than 0.96. The required power (directly related to fuel consumed) is reduced by the ratio of 0.76/0.96. The cost per year without the deflector is Cost = (200 000/1.2) × 0.25 = $41,667. With the deflector it is Cost = 41,667 × 0.76/0.96 = $32,986. The savings is $41.667 − 32,986 = $8,800.

8.32 2 21 1.00238 88 (6 2) 1.1 122 lb.

2 2D DF V ACρ= = × × × × × =

& , .W F VD= × = × =122 88 10 700 ft - lbsec

or 19.5 Hp

8.33 F V ACD D= = × × × × × × =12

12

1 22 27 8 1 6 05 11 10 432 2 2ρ π. ( . . ) . . . N.

& . ( . . ) .W F VD= × × = × × × =2 1043 27 8 16 2 226 W or 1.24 Hp

169

8.34 The projected area is ( . ) . .2 0 32

4 4 6+× = m2

F V ACD D= = × × × × =12

12

1 18 20 4 6 0 4 4342 2ρ . . . N.

Since there are two free ends, we use Table 8.1 with L D/ / . . ,= =4 1 15 3 47 and approximate the force as FD = × =434 0 62 269. . N 8.35 The net force acting up is (use absolute pressure)

3 3up

4 4 1200.4 1.21 9.8 0.5 0.4 9.8 2.16 N

3 3 2.077 293F π π= × × × − − × × =

×

From a force triangle (2.16 N up and FD to the right), we see that tan / .α = F FDup

a) FD = =2 16 80 0 381. /tan . .o

0 381 12

1 21 0 4 0 2 2 502 2. . . . . .= × × × ∴ =V Vπ m / s.

Check Re: Re . ..

. .=××

= ×−

2 5 0 81 51 10

1 33 1055 Too low. Use CD = 0 5. :

0 381 12

1 21 0 4 0 5 1 582 2. . . . . .= × × × ∴ =V Vπ m / s

b) FD = =2 16 70 0 786. /tan . .o

0 786 12

1 21 0 4 0 2 3 602 2. . . . . .= × × × ∴ =V Vπ m / s.

Check Re: Re . ..

. .=××

= ×−

3 6 0 81 51 10

1 9 1055 Too low. Use CD = 0 5. :

0 786 12

1 21 0 4 0 5 2 272 2. . . . . .= × × × ∴ =V Vπ m / s

c) FD = =2 16 60 1 25. / tan . .o

1 25 12

1 21 0 4 0 52 2. . . . . .= × × × ∴ =V Vπ 2.86 m / s

Check Re: Re . ..

. .=×

×= ×−

2 86 0 81 51 10

1 5 1055 ∴OK.

d) FD = =2 16 50 1 81. / tan . .o

1 81 12

1 21 0 4 0 52 2. . . . . .= × × × ∴ =V Vπ 3.45 m / s

Check Re: 55

3.45 0.8Re 1.8 10 .

1.51 10−×

= = ××

Close, but OK.

8.36 Assume each section of the tree is a cylinder. The average diameter of the tree is 1 m. The top doesn't have a blunt end around which the air flows, however,

170

the bottom does; so assume L D/ ( / ) .= × =5 2 2 5 So, use a factor of 0.62 from Table 8.1 to multiply the drag coefficient. The force acts near the centroid of the triangular area, one-third the way up. Finally, F d× = 5000

12

1 21 5 0 4 0 6253

0 6 50002× × ×

× +

= =. ( ) . . . . .V V 54.2 m / s

8.37 Power to move the sign:

F V V AC VD D= ×12

2ρ

= × × × × × =12

1 21 11 11 0 72 1 1 11 11 6572. . . . . J /s.

This power comes from the engine: 657 12 0 3 1 825 10 4= × × ∴ = × −( & . . & . 000 1000) kg /s.m m Assuming the density of gas to be 900 kg/m3,

1825 10 10 3600 6 521000900

0 304. . $683× × × × × × × =−

8.38 The power expended is F V VD × = × =. ( / ) / . . m / s25 88 60 3 281 11 18

12

121 1118 0 5612

121 0 4 0 83 3× × × × = × × × × ×. . . . . .C V CD D

∴ =V 1347. . m / s or 30.1 mph

8.39 & .W F V V AC V AC VD D D= × = × = × =40 74612

12

2 3η ρ ρ

∴ × × = × × × ∴ =40 746 912

122 3 0 35 34 73. . . . . .V V m / s or 125 km / hr

8.40 (C) 5

4 0.02 0.02Re 5000. St 0.21 .

41.6 10

VD fD fVν −

× ×= = = ∴ = = =

×

4 m/s

42 Hz (cycles/second). distance = 0.095 m/cycle.42 cycles/s

Vf

f∴ = = =

8.41 5

.00340 Re 10 000. 40< 10 000. 0.2< 50 m/s.

1.5 10

VV

−×

< < < ∴ <×

low.003

St = 0.12 = . 8 Hz..2

ff

×∴ =

St =.21 =.003

50 Hz.lhigh

ff

×∴ =. 3500

The vortices could be heard over most of the range.

171

8.42 55

640 . 8.13 10 ft.

1.22 10

VD DD

ν−

−> = ∴ < ×

×

106

122 100 0205 000 < ft or 0.24"

VD DD

ν=

×∴ >−.

. . .

8.43 From Fig. 8.9, Re is related to St. St =f D

V V×

=×0 2 1. .

.

Re.

.. .= =

××

= ∴−

VD VV

ν1

15 100 0955 Try St =.21: m / s. Re = 630.

This is acceptable. ∴ =V 0 095. . m / s

8.44 St = Re = Use Fig. 8.9.fDV V

VD V=

×=

×−

.. .

002 2 210 6ν

Try St = 0.21: m / s Re = 38 10 OK.3V = × ∴. . .0191 8.45 Let St = 0.21 for the wind imposed vorticies. When this frequency equals the natural frequency, or one of its odd harmonics, resonance occurs:

f T L d= / ρ π2 2

2 20.21 1030 000/7850 0.016 . 0.525 m

0.016L Lπ

×= × × ∴ =

Consider the third and fifth harmonics:

f T L d= 3 2 2/ .ρ π ∴ =L 1.56 m.

f T L d= 5 2 2/ .ρ π ∴ =L 2.62 m. 8.46 (C) By reducing the separated flow area, the pressure in that area increases thereby reducing that part of the drag due to pressure. Fig. 8.8 Table 8.1

8.47 Re/

.. . . . .=

××

= × = × × × × × ×

=−

88 6 121 6 10

2 8 1012

00238 88 1 0 8 66

1245 2 22 lb.FD

The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1. & .W F VD= × = × =22 88 1946 ft - lb / sec or 3.5 Hp

( )C F WD Dstreamlined lb. ft - lb

sec or 0.12 Hp= ∴ = =0 035 0 77 67 8. . . & . .

8.48 Re ..

.= =××

=−

VDν

3 081 5 10

165 000 ∴ = × × × × × × =FD

12

1 22 3 0 08 2 1 2 78 08222. ( . ) . . . N

The coefficient 1.2 comes from Fig. 8.8 and 0.78 from Table 8.1.

( )C FD Dstreamlined N. % reduction = = ∴ = ∴

−× =. . .

. ..

.35 0 240 822 0 24

0 822100 70 8%

172

8.49 Re . . .= =×

= × ∴ =−

VD CDν2 0 810

1 6 10 0 4566 . from Fig. 8.8.

LD

CD= = ∴ = × =4

0 85 0 62 0 45 0 28

.. . . . .

Because only one end is free, we double the length.

F V ACD D= = × × × × × =12

12

1000 2 0 8 2 0 282 2ρ . . .900 N

If streamlined, CD = × =0 03 0 62 0 0186. . . .

∴ = × × × × × =FD12

1000 2 0 8 2 0 01862 . . .60 N

8.50 V = × =50 1000 3600 13 9/ . m /s. Assume the ends to not be free. ∴Use CD from Fig. 8.8.

( )Re . ..

. . . . .=×

×= × ∴ = =−

13 9 0 021 5 10

1 85 10 1 2 0 354

streamlinedC CD D

& . . . . .W F V V ACD D= × = = × × × × × =12

12

1 2 13 9 0 02 20 1 2 7733 3ρ W or 1.04 Hp

& . . . .Wstreamlined W or 0.26 Hp= × × × × × =12

1 2 13 9 0 02 20 0 3 1933

8.51 V = × =50 1000 3600 13 9/ . m / s. Re . ..

. . .=×

×= × ∴ =−

13 9 0 31 5 10

2 8 10 0 455 CD

We assumed a head diameter of 0.3 m and used the rough sphere curve.

FD = × × × × =12

12 13 9 0 3 4 0 42 2. . ( . / ) . .π 3.3 N

FD = × × × × =12

12 13 9 0 3 4 0 0352 2. . ( . / ) . .π 0.29 N

8.52 σρ

γ=−

=−

×= + =∞

∞

p p

V Vp h pv

12

0 7150

1000150

2 2. .

000 167012

where 000 Pa.atm

∴ =V 20.6 m / s.

8.53 CF

V AL

L= =× × × ×

= ∴ ≅12

200

1000 12 4 100 69 3

2 2ρα

00012

.

. . .o

CF

FDD

D= =× × × ×

∴ = N..

. .0165 12

1000 12 4 104800

2

173

σ crit 000) 1670

12

= >× + −

× ×=. ? ( .

. .759810 4 101

1000 12143

2 ∴No cavitation.

8.54 CF

V AL

L= =× × × ×

= ∴ = 000

12

12

50

1 94 351612

30105 7 3

2 2ρα

.. . . .o

CF

FDD

D= =× × × ×

∴ = lb..

. .027 12

194 351612

301280

2

σ crit 12

= >× + − ×

× ×=16

62 4 16 12 2117 25 144

194 35182

2. ? . / .

.. . ∴No cavitation.

8.55 p pv∞ = × + =×

= ×9810 5 101 1670 16 106 000 = 150 000 Pa. Pa. Re = 20 .810 -6 .

σ σ=−

× ×= ∴ = + = + =

150

1000 200 74 0 1 3 1 74 52

2

000 167012

. . ( )( ) . ( . ) .C CD D

∴ = = × × × × × =F V ACD D12

12

1000 20 4 522 2 2ρ π . . .52 000 N

Note: We retain 2 sig. figures since CD is known to only 2 sig. nos. 8.56 For a 6° angle of attack we find from Table 8.4 CL = 0 95. .

F V AC LL L= = × × × × × = ×12

12

1000 15 4 0 4 95 12 9 82 2ρ . . . . 000

∴ =L 0 69. . m

8.57 ΣF ma a a= − × × = ∴ =. ..

. . a) 1.75 m /s2400 9810 43

2 4009 81

3π

b) 400 981043

24009 81

12

100043

23 3− × × = + × × ×

∴ =π π.

.. . .a a 1.24 m /s 2

8.58 F ma V a a FV

m Va= = × × − ∴ =−

= × −1 1 11000 1 21200

0 2 1000. . . . .

F m m a a FV V

FV

aa= + ∴ =− + −

=−

( ) . .2 2 21200 200 1400 is true acceleration.

∴−

× = −−

−

−

× =% . . error = a a

a

FV

FV

FV

2 1

2

100 1400 1200

1400

100 16 7%

174

8.59 (B) 12

2From Fig. 8.12a 1.1. .LL L

FC C

V cLρ= =

2 2 2 1200 9.811088. 33.0 m/s.

1.23 16 1.1L

WV V

cLCρ× ×

∴ = = = ∴ =× ×

8.60 C F

V ACL

LD= =

×

× × ×= ∴ = =1

2

1000 9 81

412 80 150 496 3 2 0065

2 2ρα

.

.. . . . . .1

2

o

& . . .W F VD= = × × × ×

× =

12

412 80 15 0065 80 102 300 W or 13.8 Hp

8.61 a) CV

VL = =× +

× × ×∴ =122

1500 9 81 300012

1 22 202.

.

.. . 34.5 m / s

b) ( )CV

VL max 50 m / s= =

× +

× × ×∴ =1 72 1500 9 81 3000

12

412 202. .

.. . (at 10 000 m)

c) & . .W F VD= = × × × ×

× =

12

412 80 20 0065 80 132 700 W or 18.4 Hp

where we found CD as follows:

( )C CL Dcruise =

× +

× × ×= ∴ =

1500 9 81 300012

412 80 2067 0065

2

.

.. . . , from Fig. 8.12.

∴Power = 18 40 45

40 9.

.. .= Hp

8.62 CV

VL = =× +

× × ×∴ =122

1500 9 81 300012

1007 202.

.

.. . 38.0 m / s

8.63 ( )C CL Dcruise =

× +

× × ×= ∴ = =

1500 9 81 300012

1 007 80 200 275 0 275

480 0057

2

.

.. . . . .

∴ = = × × × × =& . .W F VD12

1 007 80 20 0 0057 293 400 W or 39.4 Hp

% change = 39 4 18 418 4

100 114%. ..

−× = increase

The increased power is due to the increase in air density.

175

8.64 CV

VL = =× +

× × ×∴ =122

1500 9 81 900012

1 22 202.

.

.. . 39.9 m / s

8.65 CV

VL = =×

× × × ×∴ =172

250 9 8112

122 60 82.

.

.. .

000 69.8 m / s

8.66 a) CV

VL = =×

× × × ×∴ =172

250 9 8112

105 60 875 2

2.

.

.. . .

000 m / s

% change = 75 2 69 869 8

100 7 77%. ..

.−× = increase

b) CV

VL = =×

× × ×∴ = =

×=

172

250 9 8112

1515 60 862 6

1013287 233

15152

..

.. .

..

. 000

m / s kg / m 3ρ

% change = 62 6 69 869 8

100 10 3%. ..

.−× = −

c) CV

VL = =×

× × ×∴ = =

×=

172

250 9 8112

1093 60 8737

1013287 323

10932

..

.. .

..

. 000

m / s kg / m3ρ

% change = 73 7 69 869 8

100 5 63%. ..

.−× = increase

8.67 For a conventional airfoil assume C C CL D L/ . . .= =47 6 0 3 at

0 3 9 8112

0 526 222 200 302 38 10

2

6. .

.. .=

×

× × × ×∴ = ×

m m kg

& . ..

W F VD= = × × × × × =12

0 526 222 200 30 0 347 6

4903 000 W or 657 Hp

8.68 v

vv v v

vv

∇ × + ⋅ ∇ +∇

− ∇

=

∂∂ ρ

νVt

V Vp

V( ) .2 0

v

vv v v v

vv v

∇ × = ∇ × = ∇ ×∇

= ∇ × ∇ =∂∂

∂∂

∂ω∂ ρ ρ

Vt t

Vt

pp( ) . . 1 0

v v v v v∇ × ∇ = ∇ ∇ × = ∇( ) ( )2 2 2V V ω (we have interchanged derivatives)

2 21 1( ) ( ) ( )

2 2V V V V V V ∇ × ⋅ ∇ = ∇ × ∇ − × ∇× = ∇ × ∇

v v v v v v v v v v v( )V ω− ∇ × ×

v v v

( )V ω= ∇ ⋅v v v

( )Vω− ∇ ⋅v vv

( ) ( )V Vω ω+ ⋅ ∇ − ⋅∇v v v vv v

176

, . .x L u U Uy

∂ψ∂

= = ∴ = = ⋅ ∇ − ⋅ ∇ ∇ ⋅ = ∇ ⋅ ∇ × =( ) ( ) ( ) .v v v v v v v v v v v

V V Vω ω ω where 0

There results: ∂ω∂

ω ω ν ωv v v v v v v vt

V V+ ⋅ ∇ − ⋅ ∇ − ∇ =( ) ( ) .2 0

This is written as DDt

Vv

v v v vωω ν ω= ⋅ ∇ + ∇( ) .2

8.69 x-comp: ∂ω∂

∂ω∂

∂ω∂

∂ω∂

ω∂∂

ω∂∂

ω∂∂

ν ωx x x xx y z xt

ux

vy

wz

ux

uy

uz

+ + + = + + + ∇ 2

y-comp:

∂ω

∂

∂ω

∂

∂ω

∂

∂ω

∂ω

∂∂

ω∂∂

ω∂∂

ν∂ ω

∂

∂ ω

∂

∂ ω

∂y y y y

x y zy y y

tu

xv

yw

zvx

vy

vz x y z

+ + + = + + + + +

2

2

2

2

2

2

z-comp:

∂ω∂

∂ω∂

∂ω∂

∂ω∂

ω∂∂

ω∂∂

ω∂∂

ν∂ ω∂

∂ ω∂

∂ ω∂

z z z zx y z

z z z

tu

xv

yw

zwx

wy

wz x y z

+ + + = + + + + +

2

2

2

2

2

2

8.70 xwy

∂ω

∂=

vz

∂∂

− 0. yuz

∂ω

∂= =

wx

∂∂

− 0. 0.zv ux y

∂ ∂ω

∂ ∂= = − ≠

2 2z( ) ; .zz z

D Dw

DT Dtω ω

ω ν ω ν ω= ⋅ ∇ + ∇ ∴ = ∇vv

If viscous effects are negligible, then DDt

zω= 0.

Thus, for a planer flow, ω z = const if viscous effects are negligible.

8.71 a) v v∇ × = −

+ −

+ −

=V

wy

vi

uz

wx

jvx

uy

k∂∂

∂∂

∂∂

∂∂

∂∂

∂∂z

$ $ $ .0 ∴irrotational

∂φ∂

φx

x x f y= ∴ = +10 5 2. ( )

∂φ∂

∂∂yfy

y f y C C= = ∴ = + =20 10 02. . . Let ∴ = +φ 5 102 2x y

b) v v∇ × = + + − =V i j k0 0 8 8 0$ $ ( )$ . ∴irrotational

∂φ∂

φx

y xy f y z= ∴ = +8 8. ( , ) . ∂φ∂

∂∂

∂∂y

xfy

xfy

f f z= + = ∴ = =8 8 0. ( ). and

∂φ∂z

dfdz

z f z C C= = − ∴ = − + =6 3 02. . . Let

177

∴ = −φ 8 3 2xy z

c) v v∇ × = + +

− +

+−

− +

+

=

− −

V i jy x y x

x y

x x y y

x yk0 0

12

2 12

20

2 2 1 2

2 2

2 2 1 2

2 2$ $

( ) ( )$ .

/ /

∴irrotational

∂φ∂

φx

x

x yx y f y=

+∴ = + +

2 2

2 2. ( )

∂φ∂

∂∂

∂∂y

x y yfy

y

x y

fy

f C C= + + =+

∴ = ∴ = =−12

2 0 02 2 1 2

2 2( ) . . . ./ Let

∴ = +φ x y2 2

d) v v∇ × = + +

−+

−−

+

=V i j

y xx y

x yx y

k0 02 2

02 2 2 2 2 2$ $ ( )

( )( )

( )$ . ∴irrotational

∂φ∂

φx

xx y

n x y f y=+

∴ = + +2 2

2 212

. ( ) ( ) l

∂φ∂

∂∂

∂∂y

yx y

yx y

fy

fy

f C C=+

=+

+ ∴ = ∴ = =2 2 2 212

20 0. . . . Let

∴ = +φ ln x y2 2

8.72 ∂ ψ∂

∂ ψ∂

2

2

2

2 0x y

+ = . This requires two conditions on x and two on y.

At , . .x L u U Uy

∂ψ∂

= − = ∴ =

At , . .x L u U Uy

∂ψ∂

= = ∴ =

At y h= − , . = 0 ψ

At , = h. y h Uψ= × (See Example 8.9). The boundary conditions are stated as:

∂ψ∂

∂ψ∂

ψ ψy

L y Uy

L y U x h x h Uh( , ) , ( , ) , ( , ) , ( , ) .− = = − = = 0 2

8.73 uy

y f x vx

dfdx

f x C= = ∴ = + = − = − = ∴ = − +∂ψ∂

ψ∂ψ∂

100 100 50 50. ( ). . .

∴ = −ψ ( , ) .x y y x100 50 (We usually let C = 0.)

ux

x f y vy

dfdy

f y C= = ∴ = + = = = ∴ = +∂φ∂

φ∂φ∂

100 100 50 50. ( ). . .

∴ = +φ( , ) .x y x y100 50

y = h

y = 0x = −L

y

xU

178

8.74 a) ψ θ= 40 .

b) 1 1 1

401

0 02

r r r r r r r∂∂

∂ψ∂θ

∂ ψ∂θ∂

∂∂

∂∂θ

+ −

= + −

=( ) ( ) .

∴It is incompressible since the above continuity equation is satisfied. Note: The continuity equation is found in Table 5.1.

c) ∂φ∂

∂ψ∂θ

φ θr r r

nr f= = ∴ = +1 40

40. ( ) l

∂φ∂θ

∂∂θ

∂ψ∂

= = − = ∴ = =f

rr

f C C0 0. . . Let

∴ =φ 40ln r

d) vr

v a vvr r rr r r

r= = = = −

= −40

040 40

102, . . θ∂∂

∴ =r 5 43. m

8.75 uy

yx y x

yx

f y= =+

= ∴ = − +−∂ψ∂

∂φ∂

φ202

402 21. tan ( ).

vy

xy x

fy

xx y

fy

xx y

f C C= = −+

+ = −+

+ = −+

∴ = =∂φ∂

∂∂

∂∂

401

40 20 2 02 2 2 2 2 2

//

. . . Let

φ = − −40 1tan .yx

8.76 a) ∂ ψ∂

∂ ψ∂

∂ψ∂

2

2

2

22 2 20 10 2

x y xy x y x+ = = + −. ( ) ( ).

∂ ψ∂

2

22 2 220

xy x y= − −( ) − + −80 2 2 2 3x y x y( )

∂ψ∂y

x y y x y y= − + + +− −10 10 10 22 2 1 2 2 2( ) ( ) ( ).

∂ ψ∂

2

22 2 220

yy x y= + −( ) + + − +− −40 802 2 2 3 2 2 3y x y y x y( ) ( ) .

∴ + =+

−+

++

−+

∂ ψ∂

∂ ψ∂

2

2

2

2 2 2 2

2

2 2 3 2 2 2

3

2 2 3

20 80 60 80x y

yx y

x yx y

yx y

yx y( ) ( ) ( ) ( )

=+

+−

+−

+=

+ − −+

=80 80 80 80 80 80 80

02 2

2 2 3

2

2 2 3

3

2 2 3

2 3 2 3

2 2 3

y x yx y

x yx y

yx y

x y y x y yx y

( )( ) ( ) ( ) ( )

.

b) In polar coord: ψ θ θθ

θ θ( , ) sin sin sin sin .r r rr

rr

= − = −10 10 10 102

110

1010

102r r r r

f∂ψ∂θ

θ∂φ∂

φ θ θ= −

= ∴ = +

+cos . cos ( ).

179

2 2

1 1 10 1010 sin 10sin sin . 0.

df dff C

r r d r dr r

∂φ ∂ψθ θ θ

∂θ θ ∂ θ = − + = − = − − = =

.

∴ = +

= ++

φ θ φ10 1 10 102 2r

rx y x x

x ycos ( , ) , or

where we let r x r x ycos .θ = = + and 2 2 2

c) Along the x-axis, vx

= − =∂ψ∂

0 where we let y = 0 in part (a) and

uy x y

yx y x

y= = −+

++

= − =∂ψ∂

10 10 2010 10 02 2

2

2 2 2 2( ). with

Euler’s Eq: ρ∂∂

∂∂

ρ∂∂

uux

px x x

px

= − ∴ −

= −. . 1010 20

2 3

∴ = −

= − +

+ =∫px x

dxx x

C Cρ ρ200 200 50 100

505 3 4 2 . 000.

= −

+1000100 50

502 4x x 000 Pa. (Could have used Bernoulli!)

d) Let ux

x= = − ∴ = ± ∴0 0 10 10 12: . . Stag pts: (1, 0), (−1, 0)

8.77 a) ∂ φ∂

∂ φ∂

∂∂

∂∂

2

2

2

2 2 2 2 2

2 2

2 2 21010 10 10 2

x y xx

x y yy

x yx y x x

x y+ = +

+

+

+

=

+ −+

( )10 (( )

++ −

+=

+ − + + −+

=( )10 ( )

( ) ( ).

x y y yx y

x y x x y yx y

2 2

2 2 2

2 2 2 2 2 2

2 2 2

10 2 10 10 20 10 10 200

b) Polar coord: φ θ= +10 5 2r n rcos .l (See Eq. 8.5.14.)

∂φ∂

θ∂ψ∂θ

ψ θ θr

rr r

r f r= + = ∴ = + +10 10 1 10 102cos . sin ( )

110sin 10sin .

dff C

r r dr∂φ ∂ψ

θ θ∂θ ∂

= − = − = − − ∴ = . 10 sin 10 .rψ θ θ∴ = +

∴ = + −ψ ( , ) tan .x y yyx

10 10 1

c) vy

yx y

= =+

∂φ∂

102 2 . Along x-axis (y = 0) v = 0.

ux

xx y

= = ++

∂φ∂

10 102 2

. Along x-axis ux

= +10 10 .

Bernoulli: 2

2V p

gzρ

+ +2

2V p

gzρ

∞ ∞∞= + + (assume )z z∞=

( / ) . .10 102

102

100 50 2 12 2

2

++ = + ∴ = − +

x pp

x xρ ρ100 000 kPa

180

d) ux

x= = + ∴ = − ∴0 0 10 10 1: . . Stag pt: (−1, 0)

e) ya v=v v

uy x

∂ ∂∂ ∂

+ 0 on axis. xu

x a u vx

∂∂

= − = +2

10 1010 .

uy x x

∂∂

= + −

210( 2,0) (10 5) 12.5 m/s .

4xa ∴ − = − − = −

8.78 2 3

2 5( , ) 5 .

2 3y y

u x y y y Cy

∂ψψ

∂= − = ∴ = − + 2 31

. (3 10 ).6

y yψ∴ = −

q udy y y dy= = − = − = × −∫∫ ( ). .

. ...

50 2

25

0 23

6 667 1022 3

3

0

2

0

2

m / s2

ψ ψ2 12 3 31

63 0 2 10 0 2 0 6 667 10− = × − × − = × −( . . ) . . m / s 2

ω∂∂

= − = − + ≠uy

y1 10 0. ∴φ doesn’t exist.

8.79 ψππ

θ θ θ= + = +30 52

30 52

y r sin .

a) vr rr = = + =1 30 5

20∂ψ

∂θθcos .

At θ π= = ∴ =, . . ' .52

30 0 0833r

rs

s

Stag. pt: ( " , ).−1 0

b) At θ π ψπ π π

= = = +, sin . =.0833, r rs52

302

52 2

∴ = =r y inter .0119 ft.

c) q U H H H= × = ∴ = ∴∆ψπ π. . . = 5

6030 5

2 Thickness = 2 5

30H =

π ft or 1.257".

d) v ur ( , ) cos . . . . .1 30 52

30 2 5 27 5 27 5π π= + = − + = − ∴ = fps

8.80 [ ] [ ] [ ]φππ

ππ

= + + − − + + = + +2

12

1 10 14

12 1 2 2 2 1 2 2 2l l ln x y n x y x n x y( ) ( ) ( )/ /

[ ]− − + +14

1 102 2ln x y x( ) .

ux

x

x

x

x x xv y

y

= =+

+−

−

−+ =

+−

−+ = =

=

∂φ∂ 0

2 2

14

2 1

1

14

2 1

110 1

2 11

2 110 0 0

[ ( )]

( )

[ ( )]

( ) ( ) ( ). . if

30 fpsy

x= 0

= 5π/2

181

At the stagnation point, ux x x

= ∴+

−−

+ = ∴−

=0 12 1

12 1

10 0 21

202

.( ) ( )

. .

∴ = ∴ = ± ∴ ×x x2 1 1 1 049. . . . m. oval length = 2 1.049 = 2.098 m

All the flow from the source goes to the sink, i.e., ππm /s, or m / s for 2 2

20y > .

u yx y y yx

( )( ) ( )

.= =+

−−

++ =

++

=

∂φ∂ 0

2 2 2

14

2

1

14

2

110 1

110

12

0

110 . tan 10 .

2 21

hq dy h h

y

π π− = + = ∴ + = ∫ +

h = 0.143 m so that thickness = 2h = 0.286 m. The minimum pressure occurs on the oval surface at (0,h).

There u =+

+ =1

1 14310 10 982.

. m /s.

Bernoulli: Vp V p p2 2 2 2

2 210 98

2 1000102

+ = + + = +∞ ∞

ρ ρ. . . 10 000

1000

Pamin∴ = −p 280 .

8.81 [ ] [ ] [ ]φππ

ππ

= + + +−

− + + = + +22

1 22

1 2 12

12 2 1 2 2 2 1 2 2 2l l ln x y n x y x n x y( ) ( ) ( )/ /

[ ]− − + +12

1 22 2ln x y x( ) .

ux

x

x y

x

x yv

yx y

yx y

= =+

+ +−

−

− ++ =

− +−

− +∂φ∂

12

2 1

1

12

2 1

12

1 12 2 2 3 2 2 2 2

( )

( )

( )

( ).

( ) ( )

Along the x-axis (y = 0), v = 0 and ux x

=+

−−

+1

11

12.

Set ux x

x x=−

−+

= = ∴ = ±0 11

11

2 2 22: , . . or

Stag. pts.: ( , ), ( , ).2 0 2 0 −

u v( , ) . ( , ) .4 0 14 1

14 1

2 4 0 0=− +

−− −

+ = − =1.867 m / s

u v( , ) . ( , ) .0 4 11 4

11 4

2 0 4 41 4

41 4

02 2 2 2=+

−−+

+ = =+

−+

=2.118 m / s

8.82 φππ

ππ

= + − + + +22

122

12 2 1 2 2 2 1 2l ln x y n x y[ ( ) ] [ ( ) ]/ /

= + − + + +12

112

12 2 2 2l ln x y n x y[ ( ) ] [ ( ) ].

(0, h)y

x

182

ux

xx y

xx y

= =+ −

++ +

∂φ∂ 2 2 2 21 1( ) ( )

.

vy

yx y

yx y

= =−

+ −+

++ −

∂φ∂

11

112 2 2 2( ) ( )

.

At (0, 0) u = 0 and v = 0. At (1, 1)

2 2 2 2 2

2 2 1 10 0.4 m/s. 1.2 m/s.

52 1 1 2 1v u= + = = = + =

+ +

∴ = +

vV i j12 04. $ . $ . m / s

8.83 φππ

ππ

= − + + + + + ∞

22

122

12 2 1 2 2 2 1 2l ln y x n y x U x[( ) ] [( ) ] ./ /

= − + + + + + ∞12

112

12 2 2 2l ln y x n y x U x[( ) ] [( ) ] .

a) Stag. pts. May occur on x-axis, y =0.

ux

xx

xxy

= =+

++

+=

∂φ∂ 0

2 21 110.

2 0.2 1 0. x x∴ + + = ∴no stagnation points exist on the x-axis.

(They do exist away from the x-axis.)

Along the y-axis: u y q udyh

( ) . ( )= = = =∫1012

20

m / s.2π π

∴ = = ∴ =∫π 10 10 0 3140

dy h hh

. m. .

b) uxx

x x x=+

+ ∴ + + = ∴ = −2

11 2 1 0 12

2. . m.

Stag. pt.: (−1, 0) Along the y-axis: u h h= ∴ = × ∴ =1 0 1 3 14. . . . . mπ

c) uxx

x x x=+

+ ∴ + + = ∴ = − −2

10 2 10 1 0 9 902

2. . . . , 0.10 m.

Stag. pts.: (−9.9, 0) , (−0.1, 0). Along the y-axis: u h h= ∴ = ∴ =0 2 0 2 15 71. . . . . . mπ

8.84 φ θ θ= +60 8r

rcos cos .

a) vr r rr = = − + = −

∂φ∂

θ θ θ60

8 860

2 2cos cos cos .

At the cylinder surface v r = 0 for all θ. Hence,

x

y

x

y

x

y

183

60 8 2 7392rr

cc= ∴ =. . m

b) Bernoulli: ∆pU

= = =∞ρ2 2

21000 8

232 000 Pa or 32 kPa

c) vr rθ

∂φ∂θ

θ θ= = − −1 60 82 sin sin .

At r r vc= = − − = −, sin sin sin θ θ θ θ8 8 16

d) ∆pv

= = =ρ 902 2

21000 16

2128

o

000 Pa or 128 kPa

8.85 ψππ

θπ

πθ= + = +

42

202

2 10l ln r n r

At ( , ) ( , ), ( , ) ( , / ).x y r= =0 1 1 2 θ π

vrr ( , / ) ( ) .1 2 1 1

12 2π

∂ψ∂θ

= = =

vrθ π

∂ψ∂

( , / ) .1 2 101

10= − = − = −

vr ( . , / ).

.1 7 4 21 7

1 18π = = , vθ π( . , / ).

.1 7 4 101 7

5 88=−

= −

vr( . , ).

.3 2 0 23 2

0 625= = , vθ ( . , ).

.3 2 0 103 2

3 125=−

=

vr ( , / ) .6 4 26

0 333− = =π , vθ π( , / ) .6 4 106

1 67− =−

= − , etc.

Note: We scaled the radius at each 45° increment to find r.

b) vr

vrr = = −

2 10 and θ . From Table 5.1 (use the l.h.s. of

momentum)

aDvDt

vr

vvr

vrr

rr

r= + = −θ θ∂∂

2 2

= −

− = −

2 2 100 10412 3r r r

= −104 m / s2

aDvDt

v vr

vvr

v vr

rr

rθ

θ θ θ θ∂∂

= + = + =

+

−=

2 10 2 1002 3r r r

( ).

∴ = −va( , ) ( , )0 1 104 0 m / s2

x

y

184

c) v vr ( . , / ) / . . , ( . , / ).

.14 14 4 2 14 14 0 1414 14 14 4 1014 14

0 707π πθ= = = − = −

v vr ( . , / ) / . , ( . , / ) / . .0 1 2 2 0 1 20 0 1 2 10 0 1 100π πθ= = = − = −

Bernoulli: 20 0001.2

13 760 Pa++

= ++

∴ =0 1414 0 707

2 1 220 100

2

2 2 2 2. ..

.p

p

We used ρair3 kg / m at standard conditions.= 1 2.

8.86 Along the y-axis v vrr = = − −0 10 40

2 and θ .

We have set θ π=

2 in Eq. 8.5.27. rc = =

4010

2.

b) vrr = − − ⇒10 40 4 3 5 126 92cos cos . ( , ) ( , . ).θ θ o

vr

v vrθ θθ θ= − − ∴ = − = −10 40 6 96 9 282sin sin . . , . . m /s m / s

c) Use Eq. 8.5.28: p p U= − ∞02 22ρ αsin

Drag = p r d L p r L p p Uc ccos . ./

/

α α ρπ

π

− × = − ∞−∫ 90 90 0

2

2

2

2 2

= − − ×∞∫2 2 202 2

900

2

( sin )cos/

p U r Ld p r Lc cρ α α απ

[ ]= −

− − =∞ ∞ ∞2 2

32 2 8

302

3

0

2

02 2r L p U p U r L r L Uc c csin sin .

/

α ρα

ρ ρπ

CU A

r L U

U r LD

c

c

= = = =∞

∞

∞

Drag12

8 312

2

83

2 6672

2

2ρ

ρ

ρ

( / ). .

8.87 v Ur

U r Ur c= − = = = × =∞ ∞ ∞cos cos . , .θµ

θ µ22 24 1 4 4 Let

For θ π= = − +, . vrr 4 4

2

b) vr

Urc

θ

∂ψ∂

θµ θ

θ θ= − = − − = − −

= −∞ sinsin

sin sin .2 24 41

8

c) p pV v

c = + − = + × −∞∞ρ ρ

θθ2 2 2 2 2

2 21000 4

21000 8

250 000 sin .

∴ = −pc 58 32 2sin θ kPa.

10 m/s

20 m/s

dα

p(α)

α

p90

x = −1

−x

−vr

185

d) Drag = 2 58 32 1 1 26 2 12

0

2

( sin ) cos/

− × × − × ×∫ α α απ

d

= −

− =2 58 32 1

352 42.7 kN. (See the figure in Problem 8.86c.)

8.88 On the cylinder 10002 sin 60sin ,

2 2 3.651cv U

rθ θ θπ π∞Γ

= − − = − −×

where we have

Used 400

3.651 ft.30cr U

µ

∞= = =

If

2 2 2 2 2 2 2 26 6 6 6

( , ) .0318( 6) ( 2) ( 6) ( 2) ( 6) ( 2) ( 6) ( 2)

x x x xu x y

x y x y x y x y

− − + += − + + +

− + − − + + + + − + + +

227 , 313 .θ∴ = o o Stag. pts.: (3.651 ft, 227°) , (3.651 ft, 313°). Max. pressure occurs on the cylinder at a stagnation pt.:

2 2 2 2max o

0.0024= 30 0 1.08 psf.

2 2p U v

ρ∞ ∴ = − − =

Min. pressure occurs at the top of the cylinder where θ = 90 o and the velocity is:

90o

10002 sin 2 30 104 fps

2 2 3.651v U

rθ

π π∞Γ

= − − = − × − =×

2 2 2 2min o

0.0024= 30 104 11.8 psf.

2 2p U v

ρ∞ ∴ = − − = −

8.89 vθ θπ

= − × −×

2 202 4

sin.

.Γ For one stag. pt.: vθ θ= =0 270 at o :

0 2 20 2702 4

2 20 2 4 100 5= − × −×

∴ = × × × =sin.

. . .o ΓΓ

ππ m / s.2

ΓΓ

= ∴ = =×

=22

100 52 4

22 2π ω ω

π πr

rcc

. ..

. 100 rad /s (See Example 8.12.)

Min. pressure occurs where vθ is max, i.e., θ π= / .2 There

vθ π= − × × −

×=2 20 1 100 5

2 480.

. m /s.

∴ = + − = + × − × = −∞∞p p

V vmin Pa

2 2 2 2

2 20 20

21 22 80

21 22 3660ρ ρθ . . .

186

8.90 Γ = = × × × = = = × =2 2 6 120 2 60 28 42 6 3 1 082 2 2 2π ω π π µr r Uc c. / . . . m / s. m / s.2 3

∴ = − × −×

∴ = −vθ θπ

θ2 328 422 6

1256sin.

.sin . .. Impossible. ∴Stag. pt. is off the

cylinder at θ = >270o , . but r rc From Eq. 8.5.29,

vr

Ur r r rθ

∂ψ∂

θµ

θπ π

= − = − − − = − − − − − =∞ sin sin ( ) . ( ) . .2 22

3 1 1 08 1 28 422

0Γ

∴ + = ∴ − + = ∴ =3 1 08 4 523 1 508 0 36 0 1 2122. . . . . . .

r rr r r m.

Stag. pt.: (1.21, 270°). ( ).

..vθ π90

2 328 422 6

1354o = − × −×

= − m / s.

Min. pressure occurs at θ = = = −

= −90

32

13 542

1 22 1062 2

o , :.

. . at Paminr r pc

Max. pressure occurs at θ = = = −

= −270

32

1542

122 4 042 2

o , :.

. . . at Pamaxr r pc

8.91 At 15,000 ft, ρ =. .0015 slug / ft 3 Lift = ρU L∞ = × × × =Γ . , .0015 350 15 000 60 472,000 lb 8.92 Place four sources as shown. Then, with q = 2π for each:

u x y xx y

xx y

xx y

( , )( ) ( ) ( ) ( ) ( ) ( )

=−

− + −+

++ + −

+−

− + +2

2 22

2 22

2 22 2 2 2 2 2

++

+ + +x

x y2

2 22 2( ) ( )

v x yy

x yy

x yy

x y( , )

( ) ( ) ( ) ( ) ( ) ( )=

−− + −

++

− + ++

−+ + −

22 2

22 2

22 22 2 2 2 2 2 +

++ + +

yx y

22 22 2( ) ( )

8.93 Place four sources with q = 0 2. m / s, as shown.2

u x yx

x yx

x yx

x yx

x y( , ) .

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )= −

−− + −

+−

− + ++

++ + −

++

+ + +

0318

66 2

66 2

66 2

66 22 2 2 2 2 2 2 2

v x yy

x yy

x yy

x yy

x y( , ) .

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )= −

−− + −

+−

+ + −+

+− + +

++

+ + +

0318

26 2

26 2

26 2

26 22 2 2 2 2 2 2 2

where q

22

20318

π π=

−= −

. . .

x

y

x

y(6, 2)

187

At (4,3) u( , ) . . .4 3 03182

4 12

4 2510

100 110

100 250 00922= −

−+

+−+

++

++

= m / s

v( , ) . . .4 3 03181

4 11

100 15

4 255

100 250 01343= −

++

++

++

+

= − m /s

8.94 Re . / .crit = ∴ = × =∞U xxT

Tνν ν6 10 300 20005

a) ν = × ∴ = × × =− −1 56 10 2000 1 56 10 0 3124 4. . . ' ft / sec. or 3.74"2 xT

b) ν µρ

= = × ∴ = × × =− −2 1 10 2000 2 1 10 0 424 4. . . ' ft / sec. or 5.04"2 xT

c) ν = × ∴ = × × =− −3 47 10 2000 3 47 10 0 6944 4. . . ' ft /sec. or 8.33"2 xT 8.95 a) Use Re / . .crit = × = × −3 10 10 1 51 105 5xT ∴ =xT 0 453. . m b) Use Re / . .crit = = × −10 10 1 51 106 5xT ∴ =xT 1 51. . m c) Use Re / . .crit = × = × −3 10 10 1 51 105 5xT ∴ =xT 0 453. . m d) Use Re / . .crit = × = × −3 10 10 1 51 105 5xT ∴ =xT 0 453. . m

e) 4 5growth growthRe 6 10 10 /1.51 10 . 0.091 m or 9.1 cmx x−= × = × ∴ = .

Note: A rough plate, high free-stream disturbances, or a vibrated smooth plate all experience transition at the lower Re .crit 8.96 a) Use Re / .crit = × = −3 10 10 105 6xT ∴ =xT 0 03 3. . m or cm b) Use Re / .crit = = −10 10 106 6xT ∴ =xT 0 1 10. . m or cm c) Use Re / .crit = × = −3 10 10 105 6xT ∴ =xT 0 03 3. . m or cm d) Use Re / .crit = × = −3 10 10 105 6xT ∴ =xT 0 03 3. . m or cm

e) 2 2( ) 20 000 2 1000 10 sin ( /2)p x x= − × ×

8.97 Re ..

.crit For a wind tunnel: = × =×

× =×

×∞ ∞

−6 10 2 6 10 21 5 10

5 55

U Uν

m / s.∴ =∞U 4 5.

For a water channel: 6 102

100 35

6× =

×∴ =∞

− ∞

UU. . m / s .

8.98 The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19. The pressure distribution (see solution 8.86) on the surface is p p U r xc= − =∞0

2 22ρ α α αsin ( where is zero at the stagnation point). Then

2 2( ) 20 000 2 1000 10 sin ( /2)p x x= − × × = −20 200 22sin ( / )x kPa The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:

188

v rθ θ θ π α α( ) sin sin sin( ) sin= = − − = − − =2 10 10 20 20 ∴ =U x x( ) sin( / )20 2 8.99 U x v r v U x x x rc c( ) . sin . ( ) sin .= = = ∴ = =θ θ α α at since 1 8 8 p x x( ) sin sin= − = −58 32 58 322 2α kPa 8.100 The height h above the plate is h x mx m m( ) . . . . .= + = × + ∴ = −4 1 2 4 15

∴ = − × = ∴ =−

h x x U x h U xx

( ) . . . . ( ) . ( ) .. .

0 4 15 6 4 2 40 4 15

Continuity: or

U xx

( ).

.=−

162 67

Euler’s Eqn: ρ∂∂

∂∂

ρuux

px

dpdx x x

= − ∴ =− −

.. ( . )

16

2 6716

2 67 2

=−

2562 67 3( . )

.x

8.101 a) top out in0 0 0 0

m m m udy udydx udy udydxx x

δ δ δ δρ ρ ρ ρ

∂ ∂= − = + − =∫ ∫ ∫ ∫

∂ ∂& & &

b) 0 ( ) ( )( )2x

dpF p dx p d p dp dδ τ δ δ δΣ = − + + − + +

0 higher order termsdx dpτ δ= − − +

2 2 2out in top

0 0 0 0

2

0 0

( )

( )

mom mom mom u dy u dydx u dy U x udydxx x

u dydx U x udydxx x

δ δ δ δ

δ δ

ρ ρ ρ ρ

ρ ρ

∂ ∂− − = + − −∫ ∫ ∫ ∫ ∂ ∂

∂ ∂= −∫ ∫ ∂ ∂

& & &

2 3

2 5 4 5 23 4 3 4

4 .22 4.65 1.5 10 3 4.65 (1.5 10 3)

y dy

dxδ

− −

×= − +

× × × × × × ×

8.102 τ δ ρ ρδδ

02

00

= − + − ∫∫dpdx

U xddx

udyddx

u dy( )

= − + −

−∫ ∫∫δ ρ ρ ρ

δ δδdpdx

ddx

uUdydUdx

udyd

dxu dy

0

2

00

where we have used gdfdx

dfgdx

fdgdx

U g f udy= − = =

∫. , . Here ρ

δ

0

∴ = − + − − =∫∫τ δ ρ ρ ρδδ

000

dpdx

ddx

u U u dydUdx

udy( ) . ( const. )

189

8.103 dpdx

ddx

U UdUdx

dUdx

Udy U Udy= − = − = −

=∫ ∫

ρρ ρ

δ δ

δ δ

21 12

0 0

where .

∴ = − −

+ −∫ ∫τ δ ρδ

ρ θ ρδ δ

00

2

0

1dUdx

Udy ddx

U dUdx

udy( )

= + − = +∫ρ θ ρ ρ θ ρ δδd

dxU

dUdx

U u dyddx

UdUdx

U d( ) ( ) ( ) .2 2

0

8.104 If dp dxdUdx

ddx

u U u dy/ ( ) .= = = −∞∫0 0 00

then and τ ρδ

τ ρπδ

πδ

ρδπ

πδ

ρδπ

δδ δ

02

0

2

0

2

21

22

2 22

2= −

= − −

= −

∞ ∞ ∞∫

ddx

Uy y

dy Uddx

y yU

ddx

sin sin cos

τ µ∂∂

µπδ0

02

0= ==

∞

uy

Uy

cos .

∴ = ∴ = ∴ =∞ ∞∞ ∞

µπδ

ρδ

δ δν

δνU U d

dxd

Udx x

U2137 11 5 4 792. . . . . .

b) τ µπ

νµ

ν0 21

4 790 328= =∞

∞∞

∞UU

xU

Ux.

. .

c) ∂∂

∂∂

πν

∂∂

∂∂

ux

Ux

y Ux

Ux

ax

Uax a

xvy

=×

=

= −

= −∞

∞∞ ∞

−

sin.

sin cos ./

2 4 79 2

3 2

∴ =

=

∞∞ ∞

∞∞ ∞∫∫v U

yx

Uy

Ux

dy UU U

y dy.

cos . . cos . ./

164328 0316 1893 2

00 ν ν ν ν

δδ

8.105 u Uy d

dxU

y ydy= = −

∞ ∞∫δ

τ ρδ δ

δ

. 02

0

1

= −

=∞ ∞ρδ δ

ρδd

dxU U

ddx

2 2

2 316

.

τ µ∂∂

µδ

µδ

ρδ

δ δν

021

66= = ∴ = ∴ =∞ ∞

∞∞

uy

U UU d

dxd

Udx. .

∴ = = =∞ ∞

∞∞δ

νδ

ντ µ

ν2

012 3 46 0 289U

x xx

UU

Ux

. ( ) . . . .

%error in δ ( ) . .x =−

× =5 3 46

5100 30 8% low.

U

δ u = Uy/δ

y

190

%error in lowτ 0332 289

332100 13%( ) . .

..x =

−× =

8.106 τ ρδ δ

ρδ δδ

δδ

02 2

6

2

0

6

3 1 3 13

1 13

= −

+ +

− −

∞ ∞∫∫

ddx

Uy y

dy Uy y

dy/

//

+ +

− −

∞∫ ρ

δ δδ

δ

Uy y

dy2

2 323

13

23/

= = ∴ =∞∞

∞d

dxU

U ddx

Uρ δ µδ

δδ

µ ρ2 0 13583

22 08( . ) . . / .

Thus, δ τ ρ ρ( ) . / , ( ) ..

. Re ./x vx U x Uv

U xU x= =

=∞ ∞

∞∞

−6 65 0 13586 65

20 4510

2 2 1 2

%error for δ =−

× =6 65 5

5100 33%. . %error for τ 0

0 451 0 3320 332

100 36%=−

× =. .

.

8.107 Continuity from entrance to x: U H u y dy U x H00

2 2= + −∫ ( ) ( )( ).δδ

Write U x U x dy U x dy( ) ( ) ( ) .δδδ

= = ∫∫00

Then, continuity provides

U H u U dy UH00

2= − +∫ ( )δ

= − −∫UH U u dy20

( )δ

= − ∴ =−

UH U U xU H

Hdd

22

0δδ

. ( ) .

If we were to move the walls out a distance δ d x( ), then U x( ) would be constant

since ( )[ ]H d d− +2 2δ δ would be constant; then U x U( ) .= 0 For a square wind tunnel, displace one wall outward 4 0δ d dp dx for / .= 8.108 The given velocity profile is that used in Example 8.13. There we found δ ν= = = = =∞

−5 48 5 48 10 10 0 00173 0 00173 3 0 0036. / . / . . .x U x x m. Assume the streamline is outside the b.l. Continuity is then

10 0 02 102003 003

003 102

20

003

× = −

+ −∫.

. .( . )

. y ydy h

= + − ∴ =0 02 10 0 03 0 021. . . .h h m or 2.1 cm

[ ]δ d

y ydy= − +

= − + =∫

110

1020003

10003

110

03 03 01 0 0012

20

003

. .. . . .

.

m

h − = − =2 2 1 2 0 1. . cm or 0.001 m. The streamline moves away from the wall a distance δ d .

191

8.109 From Prob. 8.107 we found that we should displace the one wall outward 4δ d . From the definition of δ d :

h xy y

dyd( ) = = − +

= − +

=∫4410

1020 10

43

43

2

20

δδ δ

δ δδ

δδ

=×

×

=

−43

5 48 1 86 10 10160 287 303

0 007355

. . //(. )

.x x m

We used δ ( )x found in Example 8.13, ρ ν µ ρ= =p RT/ , / . and

8.110 a) u Uy y

UU

y ydyd= −

= − +

= − + =∞

∞∞∫

32

12

11

32

12

34

18

3753

3

3

30δ δ

δδ δ

δ δ δ δδ

. . .

From Eq. 8.6.16, δ ν νd

xU

xU

= × =∞ ∞

. . . .375 4 65 1 74 %error = 1.2%.

θδ δ δ δ

δδ

= −

− +

=

∞∞∫

1 32

12

132

12

0 13922

3

3

3

30UU

y y y ydy . .

∴ = × =−

× =∞ ∞

θν ν. . . . . .

..139 4 65 0 648 648 644

644100 0 62%x

Ux

U %error =

b) u Uy y y y

dyd= −

= − +

= − + =∞ ∫2 1

23

32

2

2

20δ δ

δδ δ

δ δδ

δδ

. / . See Example 8.13.

∴ = =−

× =∞ ∞

δν ν

dx

Ux

U5 48

31 83 1 83 1 72

1 72100 6 4%. . . . .

.. %error = .

θδ δ δ δ

δ δ δ δ δ δ δδ

= −

− +

= − − + + − =∫ 2 1 2

13

43

24

24

15

13332

2

2

20

y y y ydy . .

∴ = × =−

× =∞ ∞

θν ν. . . . . .

..1333 5 48 0 731 731 644

644100 13 5%x

Ux

U %error = .

c) 0

21 sin 0.363 . See Problem 8.104. 4.79 .

2dy x

dyU

δ π δ νδ δ δ δ

δ π ∞

= − = − = =∫

∴ = × =−

× =∞ ∞

δν ν

dx

Ux

U0 363 4 79 1 74 1 74 1 72

1 72100 1 2%. . . . . .

.. %error =

θπδ

πδ

δπ

πδ

δ δπ

δδδ

= −

= − − + −

= − + =∫ sin sin cos sin . .y y

dyy y

21

22

2 2 22

0 13700

term

∴ = × =−

× =∞ ∞

θν ν. . . . . .

.. .137 4 79 0 654 654 644

644100 1 6%x

Ux

U %error =

192

8.111 a) δ ν= =

× ×

=

∞

−

4 65 4 65 1 6 10 2012

0 07594 1 2

. . . . ./

xU

ft

b) τ ρν

02 2

4 1 2

5323 323 0024 12 1 6 1020 12

9 11 10= = × ×××

= ×∞

∞

−−. . . . . .

/

UxU

psf

c) Drag = 2120 15 1.29

2U

LUν

ρ ∞∞

× × ×

1 / 24

21 1.6 10.0024 12 300 1.29 0.0546 lb.

2 20 12

− ×= × × × × = ×

d) δ∂∂ δ δ

δx

ux

Uy y d

dx=

−

∞=× ×

= = − +

10

4

2

3

44 651 6 10 10

120 0416

32

32

..

. . ft.

∴ = −×

+×

××

= − +−∂

∂ux

y yy y12

32 0416

32 0416

4 652

1 6 1010 12

27 9 161402

3

4

43

. . . .. .

. .

∴ = − = × − × =∫vux

dy∂∂

δ 27 92

041616140

40416 0 01212 4

0

.. . . . fps

8.112 a) δν

= =× ×

=

∞

−

4 65 4 6515 10 6

400221

5 1 2

. ..

. ./

xU

m

b) τ ρν

02 2

5 1 2

0323 323 122 415 10

6 40 00498= = × ×

××

=∞

∞

−

. . ..

. ./

UxU

Pa

c) Drag = 12

12912

122 4 6 5 12915 10

6 402992 2

5 1 2

ρU Lwv

LU∞∞

−

× = × × × × ×××

=. . .

.. .

/

N

d) 3

2 43 3

22

u y y dU

x dx∂ δ∂ δ δ

∞

= − +

2 3

2 5 4 5 23 4 3 4

4 .22 4.65 1.5 10 3 4.65 (1.5 10 3)

y dy

dxδ

− −

×= − +

× × × × × × ×

[ ]∴ = − + ××

= − + ×−∂

∂ux

y y y y4 6166 2 53 104 65

215 10

413

641 2 63 107 35

5 3.. .

. . .

∴ = − = × −×

× =∫vux

dy∂∂

δ 6412

01562 63 10

40156 0 003912

54

0

..

.. . , m / s

where δx=

−

=× ×

=3

5

46515 10 3

401560.

.. m.

193

8.113 a) δ ν= =

× ×

=

∞

−

5 5 1 5 10 210

0 008665 1 2

xU

. . ./

m Use τ ρν

02332= ∞

∞

. .UxU

Drag = τ 02

5

0

332 1 22 101 5 10

102

1 24 0 561wdx

L

= × ××

× =−

∫ . ..

/. . N

b) δ = ××

×

=

−

. . . ..

38 2 1 5 1010 2

0 04535 2

m

Drag = 12

074 12

1 22 10 2 4 074 1 5 1010 2

2 152

2

25 2

ρνU Lw

U L∞∞

−

×

= × × × × ×

××

=. . . . . .

. .

N

8.114 a) δ τ= ××

×

= = × × ×

××

− −

. . . . . . .. .

38 6 1 5 1020 6

0 0949 12

1 22 20 059 1 5 1020 6

5 2

02

5 2

m

= . .6 Pa

b) δ τ= ××

= = × × ×

×

− −

. . . .. .

38 6 1020 6

0 0552 12

1000 20 059 1020 6

6 2

02

6 2

m

= 286 Pa.

8.115 u y Uuy

U yuy

Uy

( ) . . / ./ /= = = =∞ ∞− −

=

∞δ∂∂

δ∂∂

δδ

17

17

6 7 1 7

∂∂

δ

uy

y =

should be zero. Thus, this condition is not satisfied.

τ µ∂∂

µ δ00

1 717

10

= = = ∞=

∞−u

yU

y

/ . Thus, this is unacceptable and

∂∂

uy

at, and near, the wall is not valid.

u Uy y

= −

∞

32

12

3

3δ δ.

u Uy

=

∞ δ

1 7/

.

8.116 a) Drag =

12

0024 20 12 15 074 1 58 1020 12

1060 1 58 1020 12

0 3124 2 4

× × × ×××

−

××

=− −

. ( ) . . . . ..

lb

cubic

turb (power-law)

U

y

uu

194

b) Drag =

12

0024 20 12 15 074 1 58 1020 12

1700 1 58 1020 12

0 2724 2 4

× × × ×××

−

××

=− −

. ( ) . . . . ..

lb

c) Drag =

12

0024 20 12 15 074 1 58 1020 12

2080 1 58 1020 12

0 2524 2 4

× × × ×××

−

××

=− −

. ( ) . . . . ..

lb

8.117 a) Drag = 12

1000 1 2 1 2 074 101 2 1

1060 101 2 1

5 2126 2 6

× × × ××

−

×

=− −

. ( ) .. .

. ..

N

b) Drag = 12

1000 1 2 1 2 074 101 2 1

1700 101 2 1

4 4426 2 6

× × × ××

−

×

=− −

. ( ) .. .

. ..

N

c) Drag = 12

1000 1 2 1 2 074 101 2 1

2080 101 2 1

3 9926 2 6

× × × ××

−

×

=− −

. ( ) .. .

. ..

N

8.118 U∞ = = = ××

×

=60 1000

360016 67 38 100

16 67 102355

2

. ..

..

m / s. 000 1.5 10 m-5

δ

τ ρ02 2

5

5

212

12

1 22 16 67 059 1 5 1016 67 10

0 0618= = × × ××

×

=∞

−

U c f . . . ..

. ..

Pa

b) τ ρ02 2

5

5

2

12

12

1 22 16 67 455

0616 67 101 5 10

0 151= = × ××

×

=∞

−

U c

nf . . .

...

. .

l

Pa

∴ = = ∴ =×

+ ∴ =−u nτδ

δ.

.. .

.. .

.. . .151

1 22351 16 67

3512 44 351

1 5 107 4 5855 m / s. ml

Both (a) and (b) are in error, however, (b) is more accurate. 0.px

∂∂

<

8.119 a) 5 =uτ νδ

ν (See Fig. 8.24 b). ∴ =

× ×= ×

−−δ ν

5 1 5 10351

2 14 105

4..

. . m

b) δµ

δ δτ τ

δ

δ

δ

δδ

ν

d UU u dy

Un

ydy

uU

ny

dy= − = −

+ −∞

∞∞ ∞

∫∫∫1 2 5 2 44 3 74

15

15

0

( ) . . ..

.

l l

( )= − − −

− −

∞

= =uU

y ny

y y ny

yv

τν

δ

δ δ

δ δδ δ

2 5 15 2 44 3 7415 87 8

87 8

585

. . . .. .

.

l l

= + − + − =.

.[ . ] . .351

16 67219 620 008 2188 951 43 7 m

Note: We cannot use zero as a lower limit since the ln-profile does not go to the

195

wall. Hence, we use δ ν ; the lower limit provides a negligible contribution to the integral.

8.120 a) Use Eq. 8.6.40: c

nf =

××

=

−

.

..

. .455

06 300 201 58 10

0 00212

4

2

l

b) τ ρ02 21

212

0024 300 00212 0 229= = × × × =∞U c f . . . . psf uτ = =.

..229

00249 77 fps.

c) δ νν

τ

= = × × = ×− −5 5 1 58 10 9 77 8 09 104 5

u. / . . . ft

d) 3009 77

2 44 9 771 58 10

7 4 0 2284.. .

.. . . .=

×+ ∴ =−ln δ

δ ft

8.121 a) τ ρ02 2

6

2

12

12

1000 10 455

06 10 310

110= = × ××

=∞

−

U c

nf

.

.

.

l

Pa

∴ = = ∴ = =×

= ×−

−uuτ ν

τ

δν110

1000332 5 5 10

3321 51 10

65.

.. . m / s. m

b) u u= = × =5 5 332 1 66τ . . . m / s c) y =. .15δ — Do part (d) first! ∴ = × =y . . . .15 0333 0 005 m

d) 10332

2 44 33210

7 4 0 03336.

. . . . . .= + ∴ =−

ln δδ m

8.122 Assume flat plates with dp dx

nf/ .

.

.

. .= =×

=

−

0523

0610 100

10

00163

6

2 C

l

∴Drag = 2 12

1000 10 10 100 001632× × × × × × =. .163 000 N

To find δ τmax we need u .

τ τ02

6

2

12

1000 10 455

0610 100

10

70 9 70 91000

0 266= × ××

= ∴ = =

−

.

.

. . .

ln

u Pa. m / s.

10266

2 44 26610

7 4 0 896.

. . . . . .= + ∴ =−

ln δδ mmax

8.123 a) Assume a flat plate of width πD. 85

15 600Re 6 10 .

1.5 10

ULν −

×= = = ×

×

196

2 8 1/5 21 1drag 0.073(6 10 ) 1.2 15 600 100 32600 N

2 2fC U L Dρ π π−= = × × × × × × × =

power 32600 15 489000 W or 655 hp or 164 hp/engineDF U= × = × = .

b) 3helium

1000.167 kg/m .

2.077 288p

RTρ = = =

×

air heliumBF W W Vρ= − = ∆ × 2 7(1.2 0.167) 9.8 50 600/2 2.38 10π= − × × × × = ×

payload = 6 6 623.8 10 9.8 1.2 10 12 10 NBF W− = × − × × = ×

8.124 uy

ux x y

vx

uy y

uy y

= = = − = =∂ψ∂

∂∂

∂ ψ∂ ∂

∂ψ∂

∂∂

∂ ψ∂

∂∂

∂ ψ∂

, , , , . 2 2

2

2

2

3

3

Substitute into Eq. 8.6.45 (with dp dx/ ):= 0

∂ψ∂

∂ ψ∂ ∂

∂ψ∂

∂ ψ∂

ν∂ ψ∂y x y x y y

2 2

2

3

3− = .

8.125 We also have

∂ψ∂

∂ψ∂φ

∂φ∂

∂ψ∂η

∂η∂

∂ ψ∂ ∂

∂ ∂ψ ∂∂φ

∂φ∂

∂ ∂ψ ∂∂η

∂η∂x x x x y

yx

yx

= + = +,( / ) ( / )

2

Recognizing that ∂φ ∂ ∂φ ∂ ∂η ∂ ν/ , / , / / ,x y xy

U x= = = − ∞1 02

3 and

∂η ∂ ν/ / ,y U x= ∞ ∂ψ∂ νφ

∂ψ∂η

∂ψ∂

∂ψ∂φ ν

∂ψ∂ηy

Ux

y Ux

= = −∞ ∞, 2 3

∂ ψ∂ ∂ νφ

∂ ψ∂φ∂η νφ

∂ψ∂η νφ

∂ ψ∂η νφ

2 2

3

2

2 32x yU U U y U

= + −

∞ ∞ ∞ ∞-

12

∂ ψ∂ νφ

∂ ψ∂η νφ

∂ ψ∂ νφ

∂ ψ∂η νφ

2

2

2

2

3

3

3

3yU U

yU U

=

=∞ ∞ ∞ ∞,

Equation 8.6.47 then becomes, using U y∞ =/ / ,νφ η

η ∂ψ

∂ηη ∂ ψ

∂φ∂ηη ∂ψ

∂ηη ∂ ψ

∂η∂ψ∂φ

η ∂ψ∂η

η ∂ ψ∂ηy y yx yx x y

2 2 2

2

2

2

2

22 2 2− −

− −

= ∞νν

η ∂ ψ∂η

Ux y

3

3

Multiply by y2 2/η and Eq. 8.6.49 results:

−

+ − = ∞1

2

2 2 2

2

3

3φ∂ψ∂η

∂ ψ∂φ∂η

∂ψ∂η

∂ψ∂φ

∂ ψ∂η

ν∂ ψ∂η νφ

U

197

8.126 uy

U x dFd y

U x FU

xU F= = = =∞ ∞

∞∞

∂ψ∂

νη

∂η∂

ν ην

η ' ( ) ' ( ).

We used Eq. 8.6.50 and Eqs. 8.6.48.

( )vx x

U x F Ux

F U x Fx

= − = − = − −∞∞

∞∂ψ∂

∂∂

νν

ν∂∂η

∂η∂

12

= − − −

∞

∞∞ −1

212

3 2Ux

F U x F y U xνν

ν ' /

= − + = −∞ ∞ ∞ ∞12 2

12

Ux

Fy U

xU

xF

Ux

F Fν

νν ν

η ' ( ' ).

8.127 The results are shown in Table 8.5.

8.128 a) τ 02

5

0 332 1 22 51 5 10

2 50 0124= × ×

××

=−

. ..

. . Pa

b) δ =× ×

=−

515 10 2

50 0122

5.. . m

c) v Ux

F Fmaxmax

m / s= −

=× ×

× =∞−ν

η12

1 5 10 52

8605 0 005275

( ' ) . . . .

d) Q udy UdFd

dy UdFd

dvx

U= = =∞ ∞

∞∫ ∫∫0 00

δ δδ

η ηη

= − =× ×

× =∞∞

−

Ux

UF F

νδ[ ( ) ( )]

.. . /0 5

1 5 10 25

3 28 0 045

m s / m2

8.129 a) τ 02

44332 0024 15

1 6 106 15

2 39 10= × ×××

= ×−

−. ..

. . psf

b) δ =× ×

=−

51 6 10 6

150 04

4.. ft.

c) v Ux

F Fmaxmax

fps= −

=× ×

× =∞−ν

η12

1 6 10 156

8605 0 01724

( ' ) . . . .

d) Q udy Ux

UF= = =

× ×× =∞

∞

−

∫ν

δδ

( ).

. . / .151 6 10 6

153 28 0 394

4

0

ft sec / ft2

8.130 At x = 2m, Re = 5 × 2/10-6 = 107. ∴Assume turbulent from the leading edge.

a) τ ρ02

2

12

0 4550 06

= ∞Un x

.( . Re )l

198

= × ××

=12

1000 5 0 4550 06 10

32 127 2

.( . )

.ln

Pa

b) uτ τ ρ= = =0 32 1 1000 0 1792/ . / . m / s

50 1792

2 44 0 179210

7 4 0 02486.

. . . . .= + ∴ =−

ln m or 24.8 mmδδ

c) Use the 1/7 the power-law equation:

Q y dy= =∫ 5 0 0248 0 1091 7

0

0 0248

( / . ) ./.

m / s / m3

8.131 From Table 8.5 we would select η = 6:

a) 51.5 10 2

6 6 0.0147 m5

xUν

δ−

∞

× ×= = =

b) 515.8 10 6

6 6 0.047 ft or 0.57 in.15

xUν

δ−

∞

× ×= = =

8.132 From Table 8.5 we interpolate for F' .= 0 5 to be

η =−

−− + =

0 5 0 32980 6298 0 3298

2 1 1 1 57. .. .

( ) .

=× ×

∴ =−y y51 5 10 2

0 003855.. . m or 3.85 mm

( )v Ux

F F=

−∞ν

η12

' =× ×

=−1 5 10 5

20 207 0 00127

5.( . ) . m / s

u vy x

∂ ∂τ µ

∂ ∂= + 2"F U

xUν

ρ ∞∞

=

52 1.5 10

0.291(1.2)5 0.011 Pa2 5

−×= =

×

8.133

If v y v y v y= = > = <0 10 0 0 at and at then δ δ ∂ ∂, / and continuity demands that / 0. The component, for must then be greater than ,u x u y U∂ ∂ δ> > as shown in (b); there should be a slight “overshoot”. Also, consider the control volume of (c) where the lower boundary is just above y v y= =δ . , If at large say 0 y = 10δ , then continuity demands that u out the right area be greater than :U an “overshoot”. It is not reasonable to assume that v = const as in (a);

y

y = δ

v

v

y

y = δ

v = 0

v

(a) (b) (c)

Uv

v = 0

u > U

199

reality would demand a profile such as that sketched in (b). The overshoot would be quite small and is neglected in boundary layer theory.

8.134 u Uy y

= −

∞

32

12

3

3δ δ

For the Blasius profile: see Table 8.5. (This is only a sketch. The student is encouraged to draw the profiles to scale.)

8.135

8.136 A: 0. (favorable)px

∂∂

<

B: ∂∂px

≅ 0.

C: 0. (unfavorable)px

∂∂

>

D: 0.px

∂∂

>

E: 0.px

∂∂

<

y U

cubic

Blasius

A B C D

y y y y

2U∞

zero velocitygradient

separationstreamline

backflow

inviscid profile

low velocityoutside b.l.

y

AB

C

D

Eδ

DδΒ

δC

δΑδE

200

CHAPTER 9

Compressible Flow

9.1 Btu ft-lb lbm ft-lb0.24 778 32.2 6012

Btu sluglbm- R slug- Rpc = =

o o

c c Rv p= − = − = =6012 1716 4296 42961

7781

32 2

ft - lbslug - R

ft - lb

slug - RBtuft - lb

sluglbmo o .

Btu

0.171 lbm- R

= o

9.2 c c R c kc cc

kR c

kRp v p v p

pp= + = ∴ = + −

=. . . or 1

1

∴ = −c Rk kp / ( ).1

9.3 If ∆s = 0, Eq. 9.1.9 can be written as

c nTT

R npp

nTT

nppp

c Rp

l l l l2

1

2

1

2

1

2

1

=

=

or

It follows that, using c c R c c kp v p v= + = and / ,

TT

pp

pp

R ckp

2

1

2

1

2

1

11

=

=

−/

.

Using Eq. 9.1.7,

TT

pp

pp

pp

kk

2

1

2 1

2 1

2

1

11

1

2

2

1

1

= =

=

− −ρ

ρρρ

or /

.

Finally, this can be written as

pp

k

2

1

2

1

=

ρρ

.

9.4 Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change:

& &

&~ ~ .

Q Wm

V V p pu uS−

=−

+ − + −22

12

2

2

1

12 12 ρ ρ

201

Enthalpy is defined in Thermodynamics as h u pv u p= + = +~ ~ / .ρ Therefore,

& &

&.

Q Wm

V Vh hS−

=−

+ −22

12

2 12

Assume the fluid is an ideal gas with constant specific heat so that ∆ ∆h c Tp= . Then

( )& &

&.

Q Wm

V Vc T TS

p

−=

−+ −2

212

2 12

Next, let c c R k c c c R k kp v p v p= + = = − and so that / / ( ).1 Then, with the ideal gas law T p R= / ,ρ the first law takes the form

& &

&.

Q Wm

V V kk

p pS−=

−+

−−

2

212

2

2

1

12 1 ρ ρ

9.5 Differentiate p c d xy ydx xdykρ − = = + using ( ) : ρ ρ ρ− − −− =k kdp pk d1 0. Rewrite:

dpd

kp

ρ ρ= .

9.6 The speed of sound is given by c dp d= / .ρ For an isothermal process TR p K K= =/ ,ρ where is a constant. This can be differentiated: dp Kd RTd= =ρ ρ. Hence, the speed of sound is c RT= .

9.7 Eq. 9.1.4 with & &Q W V c TS p= = + =02

2

is: cons't.

V c T V V c T T V V V V c T c Tp p p p

2 2 2 2

2 22

2+ =

++ + =

+ ++ +

( ) ( ) ( ) .∆∆

∆ ∆∆

22 ( )

02 2

V V V∆ ∆∴ = + . .p pc T V V c T h+ ∆ ∴ − ∆ = ∆ = ∆

We neglected (∆V)2. The velocity of a small wave is V c h c V= ∴ = −. . ∆ ∆ 9.8 For water

ρρ

dpd

= ×2110 10 6 Pa

Since ρ = 1000 kg / m we see that3 ,

202

c dp d= / ρ

= × =2110 10 1000 14536 / m / s

9.9 For water cp dp

d= ≅ =

×=

∆∆ρ ρ

2110 101000

14536

m / s.

L = × ×velocity time = 1453 0.6 = 872 m. 9.10 Since c = 1450 m/s for the small wave, the time increment is

∆t dc

= = =10

14500 0069. seconds

9.11 a) M = =× ×

=Vc

2001 4 287 288

0 588.

. .

b) M 600/ 1.4 1716 466 0.567.= × × = c) M = × × =200 1 4 287 223 0 668/ . . .

d) M 600/ 1.4 1716 392 0.618.= × × = e) M = × × =200 1 4 287 238 0 647/ . . . 9.12 c kRT d ct= = × × = ∴ = = × =1 4 287 263 256 256 1 21 309. . . m / s. m 9.13 a) Assume T = 20°C: c kRT= = × × =1 4 287 293 343. m /s. d c t= = × =∆ 343 2 686 m b) Assume T = 70°F: c kRT= = × × =1 4 1716 530 1130. fps. d c t= = × =∆ 1130 2 2260 ft. For every second that passes, the lightning flashed about 1000 ft away. Count 5 seconds and it is approximately one mile away.

9.14 cM

cV

= × × = = =1 4 287 263 256 1. sin . m / s. α

1000sin 0.256. tan 0.2648 . 3776 mL

Lα α= ∴ = = ∴ =

∆t = =37761000

3 776. . s

1000 m

V

L

203

9.15 Use Eq. 9.2.13:

a) cV

V= =× ×

=sin .sin

α or m / s1 4 287 28822

908o

b) cV

V= =× ×

=sin .sin

α or fps1 4 1716 51922

2980o

9.16 Eq. 9.2.4: ∆∆ ∆

Vpc

p

kRT= − = − = −

× ×= −

ρ ρ0 3

00237 1 4 1716 5190 113.

. .. . fps

Energy Eq: V c T V V c T T V V c Tp p p

2 2

2 20+ =

=+ + ∴ = +

( ) ( ). .∆∆ ∆ ∆

∴ = − = − × × − × × =∆ ∆T c V c p/ . ( . ) /( . . ) . .1 4 1716 519 113 0 24 778 32 2 0 021o F

Note: c p = × × = × ×. . . . .24 778 32 2 24 778 32 2Btulbm - F

ft - lbBtu

lbmslug

ft - lbslug - Fo o

Then ftft - lb / (slug - F)

ft lb - sec Fsec ft - lb - ft

F.2 2 2

2

/ sec 2

o

oo=

− −−

= (units can be a pain!)

9.17 a) ρ ρ ρ ρ ρ ρ ρ ρ ρAV AV AdV AVd Ad dV VdA dAdV Vd dA d dAdV= + + + + + + + Keep only the first order terms (the higher order terms—those with more than one differential quantity—will be negligible): 0 = + +ρ ρ ρAdV AVd VdA Divide by ρAV:

dVV

d dAA

+ + =ρ

ρ0

b) Expand the r.h.s. of Eq. 9.3.5 (keep only first order terms):

V kk

p V VdV kk

p dpd

2 2

2 122 1

+−

=+

+−

++ρ ρ ρ

.

Hence,

02

2 1= +

−++

−

VdV kk

p dpd

pρ ρ ρ

= +−

+ − −+

VdV

kk

p dp p pdd1 2

ρ ρ ρ ρρ ρ ρ

= +−

−

VdV

kk

dp pd1 2

ρ ρρ

where we neglected ρ ρ ρd compared to 2 . For an isentropic process Eq. 9.2.8 gives ρ ρdp kpd= , so the above becomes

204

01 2= +

−−

VdV kk

kpd pdρ ρ

ρ

= +−

−VdV k

kk pd

11

2

( ) ρ

ρ= +VdV k

pd

ρρ2

But d dV V dA Aρ ρ/ / /= − − so that the above equation is

0 = + − −

VdV kp dV

VdAAρ

which can be written as

dAA

Vkp

dVV

= −

2

1ρ

.

Since c kp2 = / ,ρ and M = V/c, this is put in the form

dAA

Vc

dVV

= −

2

21 or ( )dA

AdVV

= −M2 1

c) Substituting in V c c kRT R c k kp= = = −M and we find, , / ( ) / ,2 1

TT

Vc T

cc T

kRTc Tp p p

02 2 2 2

21

21

21= + = + = +

M M

=−

+ = +−M M

221

21 1 1

2k k

kk( ) .

d) &

/( )

/m pk

TRA

p k

T kkR

A

k k

= =+

−

+−

−

−MM

MM

02

1

02

1 2

1 12

1 12

= +−

+−

pk

RTA

kk

k

00

2

12 1

11

2M M

( )

At the critical area A * *, . M = 1 Hence,

& .* ( )m p

kRT

Ak

kk

=+

+−

00

12 11

2

e) Since &m is constant throughout the nozzle, we can equate Eq. 9.3.17 to Eq. 9.3.18:

pk

RTA

kp

kRT

Ak

kk

kk

00

2

12 1

00

12 1

11

21

2M M+

−

=+

+−

+−( ) * ( )

or

AA

kk

kk

*

( )( )=

+ −+

+−1 2 1

1

21

2 1

MM

205

9.18 a) atm 10 69.9 10 79.9 kPa abs.sp p= + = + = 1 69.9 kPa abs.p =

From 1 → s : V p p p

ps

ss

s

k

12

1

11

1

1 1 1 4

2906 79 9

69 90 997+ = =

=

=ρ ρ

ρ ρ. . ..

. ./ / .

kg / m3

∴ + = ∴ =V

V12

1269 79 77 3 900.906

900.997

m / s. . .

b) p ps = + = =26 4 10 36 4 26 41. . . kPa abs. kPa abs.

From 1 → s : V p p p

ps

ss

s

k

12

1

11

1

1 1 1 4

20 412 36 4

26 40 518+ = =

=

=ρ ρ

ρ ρ. . ..

. ./ / .

kg / m 3

V

V12

1226 36 111+ = ∴ =

400.412

400.518

m /s. .

9.19 a) 1/ 1/1.42

31 11

1 1

105. 1.22 1.254 kg/m .

2 101

ks s

ss

p pV pp

ρ ρρ ρ

+ = = = =

2

11

1.4101 000 105 0001.4. 81.3 m/s.

2 .4 1.22 1.254 .4V

V+ = ∴ =

b) 2

11

4000 81.3-81. 81.0 m/s. % error = 100 0.42%.

2 1.22 81.3V

V= ∴ = × =

9.20 Is p pr < × = kPa.. ? . .5283 0 5283 200 105 70 a) p p V kRT pr e e e e< ∴ ∴ = ∴ = = choked flow. M kPa.. . . . .5283 1 105 70

2

1000 2981 4 287

21000 248 1 315 8× =

×+ ∴ = =

.. . .

TT T Ve

e e e K, m / s.

ρ πe m=×

= ∴ = × × × =105 7

287 248 11 484 1 484 01 315 8 0 14732.

. .. . & . . . . . kg / m kg / s3

b) p pV

r ee> ∴ < × + =

M 1000 298 = 000 e2

e

. . . ..

..

.

5283 12

1 44

130 130200 2 3380

1 4

ρρ

ρ ρ0200

287 2982 338 1 7187 257 9=

×= ∴ = ∴ =

.. . . . . kg / m m / s.3

e eV

∴ = × × × =& . . . . .m 1 7187 01 257 9 0 13932π kg / s

9.21 Is p pr < × = psia.. ? . .5283 0 5283 30 15 850

a) p p V kRTr e e e< ∴ = = = 15 choked flow and M psia. . . , . .85 1 15 85 2

0 24 5301 4 17162 778 32 2

0 24 441 7 1030..( . )

. . .× =× ×

×+ ∴ = =

TT T Ve

e e e R, fps.o

Vs1

Vs=0

206

315.85 1440.003011 slug/ft .

1716 441.7eρ×

= =×

2.5.003011 1030 0.01692 slug/sec.

12m π ∴ = × × =

&

b) 15.85. M 1, and 20 psia.r e ep p> ∴ < =

30

30 144 .00475 slug/ft .

1716 530ρ

×= =

×

1/1.4320

.00475 .003556 slug/ft . 30eρ ∴ = =

2 1.4 20 144 0.24 530(778 32.2) .

2 .4 .003556eV ×

× × = +

∴ = ∴ = ×

× =V me 838 9 003556 5

12838 9 0 01627

2

. & . . . . . fps. slug / secπ

(Note: ft-lb ft-lb

0.24 Btu/lbm- R=0.24 778 0.24 778 32.2 .)lbm- R slug- R

c p = × = × ×oo o

9.22 a) p p p Tr e e e< ∴ = ∴ = × = = × = M kPa. K.. . . . . . .5283 1 5283 200 105 7 8333 298 248 30

ρ e eV=×

= = × × =105 7

287 248 31 483 1 4 287 248 3 315 9.

. .. . . . . kg / m m / s.3

∴ = × × × =& . . . . .m 1 483 01 315 9 0 14722π kg / s

b) p p ppp

T Tr ee

e e> ∴ = = ∴ = = kPa, M . . . . . , .5283 130 0 65 81 88400

0

ρ e eV=×

= = × × =130

287 263 41 719 81 1 4 287 263 4 263 5

. .. , . . . . kg / m m / s.3

∴ = × × × =& . . . . .m 1 719 01 263 5 0 14232π kg / s

9.23 a) p p pr e e< ∴ = ∴ = × = M psia. . . . . .5283 1 5283 30 15 850 Te = × =. .8333 530 441 6o R.

∴ =×

×= = × × =ρ e eV15 85 144

1716 441 6003012 1 4 1716 441 6 1030.

.. . . .

slugft

fps.3

& . . . .m = ×

× =003012 5

121030 0 01692

2

π slug / sec

b) p p ppp

T Tr ee

e e> ∴ = = = ∴ = = psia. M . . . . . . . .5283 20 2030

6667 785 0 89000

0

∴ =×

×= = × × =ρ0

20 1441716 472

00356 785 1 4 1716 472 836. . . . fps.Ve

∴ =

× =& . . . .m 00356 5

12836 0 01664

2

π slug /sec

207

9.24 p Te e= × = = × =. . . . .5283 400 211 3 8333 303 252 5 kPa abs K.

V me = × × = ∴ =×

× × =1 4 287 252 5 318 5 211 3287 252 5

05 318 5 7 292. . . & .. .

. . . . m / s. kg /sπ

9.25 p p p Te e= = ∴ = = × =. . . . . .5283 101 191 2 8333 283 235 80 0 kPa kPa abs K.

V me = × × = ∴ =×

× × =1 4 287 235 8 307 8 101287 235 8

03 307 8 1 302. . . &. .

. . . . m / s. kg /sπ

p p p Te e0 02 191 2 382 4 5283 202 0 235 8= × = = = =. . . . . kPa abs. kPa abs. K.

V me e= = ∴ =×

× × =307 8 1 202287 235 8

03 307 8 2 602. . &. .

. . . . m / s since M kg / sπ

9.26 p p p Te e= = ∴ = = × =. . . . . . .5283 14 7 27 83 8333 500 416 60 0 psia psia R.o

1.4 1716 416.6 1000 fps.eV = × × = 30.3203 kg/m and 199.4 kPa abs.e epρ∴ = =

0 0 2 27.83. 0.5283 29.4 psia, 416.6 R, 1000 fps.e e ep p p T V= × = = = =o ∴ =& . .m 0 202 slug / sec 9.27 Treat the pipeline as a reservoir. Then, p pe = =. .5283 264 50 kPa abs.

M and m /s.e eV= = × × =1 1 4 287 8333 283 307 8. (. ) .

& .. (. )

. .m =× ×

× × × =−264 5287 8333 283

30 10 307 8 3 614 kg / s.

∆∆

− = =× ×× ×

=Vm t& .

. / (. . ).

ρ3 61 6 60

2645 287 8333 283333 m3

9.28 5193 3001 667 2077

25193 225 200

225

300

1 667

667× =

×+ ∴ = ∴ =

..

.

. K.

TT T pe

e e e

=97.45 kPa. Next, T p Vt t t= = ∴ = × × =225 97 45 1 667 2077 225 882 6 K, kPa; m / s.. . .

3 2 297.45 = 0.2085 kg/m . 0.2085 × × .03 × 882.6 = 0.0752.077 225t e eVρ π ρ π= ×

×

5193 3002

1 667667

200200 2 077 300

13302 1 667

1 667× = + =×

=

V ppe e

ee

ee

..

./ .

..

ρρ

ρ kPa.

= + × × −VVe

e

23 667

23324 10 9 54. ..

or 3 116 10 63 420 10 91 86 2 3 667. . ..× = + × =−V V Ve e e Trial - and - error: m / s.

30.3203 kg/m and 199.4 kPa abs.e epρ∴ = =

208

9.29 ρ ρ11

12

1 1 4300 100287 293

4 757 4 757 340400

4 236= =+×

= =

=p

RT .. . . . .

/ .

kg / m kg / m3 3

V V V V12

22

2 14 757 10 4 236 5 4 492× × = × × ∴ =. . . . .

V k

kp V k

kp V V1

21

1

22

2

2

12 2

12

2 1 2 1 21 44

400 4 4922

1 44

340+

−= +

−+ = +

ρ ρ. .

.. .

.. 000

4.757 000

4.236

∴ =V1 37 35. . ms

∴ = = × × × =& . . . . .m A Vρ π1 1 124 757 05 37 35 1 395 kg / s

9.30 ρ11

1

45 14 7 1441716 520

0 009634= =+

×=

pRT

( . ) . . slugft

3

slug / ft 3ρ2

1 1 4

009634 50 759 7

008573=

=. .

.. .

/ .

V V V V12

22

2 1009634 4 008573 2 4 495× × = × × ∴ =. . . . .

V V

V12 2

12

121 44

59 7 144 4 4952

1 44

50 7 144 121 9+×

= +×

∴ =..

. . ..

. . ..009634 .008573

fps.

∴ = × × =& . ( / ) . . .m 009634 2 12 121 9 0 10252π slug / sec

9.31 Energy 0 → 2: 1000 3032

1000 322

2 2 2× = + =V

T V kRT .

2 1.627 20 32.5 kPa.p∴ = × =l

1.4.4 3

2 2107.9 5.39

200 5.390 kPa. 0.1740 kg/m .303 .287 107.9

p ρ ∴ = = = = ×

Energy 0 → 1: 1000 3032

10001 4 287

318 4 252 312

1 1× = +×

∴ = =V

V T V

m / s, K.12

.. . .

p1

1 44

1200 252 3303

105 4 105 4287 252 3

1 455=

= =×

=. . .

. .. .

..

kPa. kg / m3ρ

Continuity: 1 455 05 318 4 1744

3 1 4 287 107 9 0 20652 22

2. . . . . . . . .π π× × = × × × ∴ =d

d m

9.32 V kRTT

T T Vt tt

t t t2 1000 293

1 4 2872

1000 244 0 313 1= × =×

+ ∴ = =..

. . .

K. m / s.

∴ =

= ∴ =×

=pt t500 244293

263 5 263 5287 244

3 7631 4

4..

. ..

. . kPa abs. kg / m 3ρ

0 1 2

209

2

2 21.4 1.4

1.4 263 5001000 293 . 3.763 .025 313.1 .075 .

2 .4 3.763e e e

e ee e

V p pVπ ρ π

ρ ρ× = + × × × = × =

2

6 .4293 000= 1.014 10 . Trial-and-error: 22.2 m/s, 659 m/s.2e

e eV

V V−∴ + × =

∴ = ∴ =ρ e ep5 897 0 1987 494 2 4 29. , . . . , . . kg / m kPa kPa abs3

9.33 *

09. 0.997 from Table D.1. 500 .997 498.5 kPa.e e

eA p

ppA

= ∴ = ∴ = × =

and 0

0.00855 from Table D.1. 4.28 kPa abs.ee

pp

p= ∴ =

9.34 M psia, R. t = ∴ = × = = × =1 5283 120 63 4 8333 520 433 3. . . . .p Tt t

o

∴ =ρ t . .01228 slugft 3 & . . . . . .m

ddt

t= = × × ∴ =1 012284

1 4 1716 433 3 0 3192π

ft

pp

T Ve e0

15120

125 2 014 552 520 287 2 014 1 4 1716 287= = ∴ = = × = = × ×. . . , . . . M R, eo

= 684 fps.

AA

dde

e* . . . . . . .= ∴ =×

∴ =1 7084

1 708 3194

0 4172 2

ftπ π

9.35 *

M 4. 10.72, .006586 2000 13.17 kPa, .2381 293 69.76 K.e e eA

p TA

= = = × = = × =

For AA

p pe* . , . . . . . .= = ∴ = = × =10 72 0584 9976 9976 2000 1995 20 M kPa abse

9.36 Let 1150

M 1. Neglect viscous effects. M 0.430.1.4 287 303

t = = =× ×

22

1*

.051.5007. . 0.0816 m or 8.16 cm.

1.5007 1.5007 4t

t tdAA

A dA

ππ ×∴ = ∴ = = = ∴ =

9.37 p Te es= × = = × =. . . . .5283 400 211 3 8333 303 252 5 kPa abs.

303

.96 . 254.5 K. 1.4 287 254.5 319.8 m/s.303 252.5

ee e

TT V

−= ∴ = ∴ = × × =

−

2211.3.05 319.8 7.27 kg/s.

.287 254.5m π∴ = × × =

×&

210

9.38 Isentropic flow. Since k = 1.4 for nitrogen, the isentropic table may be used.

M = =3 4 235: . .*

AA

31003 1.4 297 373 1181 m/s. .9027 kg/m .

.297 373i iV ρ= × × = = =×

∴ = =×

= ∴ = =A mV

Aii i

t

&.

. . ..

. .ρ

109027 1181

0 00938 009384 235

0 00221 m m2 2

At M = = =3 3571 027220 0, . , . .T T p p

∴ = = = = = =T T p pe e0 03733571

1044 10002722

3670.

..

. K or 772 C kPao

9.39 Isentropic flow. Since k = 1.4 for nitrogen, the isentropic table may be used.

M = =3 4 235: . .*

AA

Vi i= × × = =××

=3 1 4 1776 660 3840 15 1441776 660

001843. . . fps. slugft 3ρ

A Ai t=×

= ∴ = =.

.. . .

.. .2

001843 38400283 0283

4 2350 00667 ft ft2 2

At M = = =3 3571 027220 0, . , . .T T p p

∴ = = = = = =T T p pe e0 06603571

1848 1502722

551.

..

.o oR or 1388 F psia

9.40 Assume pe e= =×

=101 101189 1273

4198 kPa. Then kg / m3ρ.

. .

Momentum: 2 2 280 000 9.81. .4198 .25 .

6F mV AV Vρ π

×= = = ×&

1260 m/s.V∴ =

9.41 F mV AV= = =×

=& ..

. .ρ ρ2 101287 873

403 kg / m (Assume gases are air. )3

4 2100 9.81 .403 200 10 . 349 m/s.V V−× = × × ∴ =

9.42 M M t e= = ∴ = =1 4 2 94 02980 0. ; . , . .*

AA

p pee

0.3665 .3665 300 109.95 K,eT T= = × = 0 0 100 .0298 . 3356 kPa abs.ep p p= = ∴ =

i te

Ve = 0

Mt = 1M > 1

M < 1

~

i te

Ve = 0

Mt = 1M > 1

M < 1

~

Ve

FB

p0A

0

211

∴ = × × =Ve 2 94 1 4 287 109 95 618. . . m / s.

∴ =−×

× × + × =FB100

287 109 9505 618 3 22 2 2

. .. . .π π 356 000 412 000 N

9.43 Assume an isentropic flow; Eq. 9.3.13 provides

103

11

22

11..

pp

k k= +

−

−M

Using k = 1.4 this gives M or M2 00424 0 206= =. . . For standard conditions V c= = × × =M m / s0 206 14 287 288 70. . . 9.44 a) 2 2 2 20.9850 1000 . 80 000 0.985 1000( 1000)V p Vρ× = − = × −

V p22 2

2

21

10002

1 44

287 283 080

287 2839850

−+ − ×

= =

×=

..

..

. .ρ

ρ kg / m3

2 22 2

21000 1.4

( 985 1 065 000) 284 300 = 02 2 .4 985

V VV− + − + −

∴ − + ∴ = =3 3784 784 261 3 77422

2 2 2V V V 300 = 0. m / s. kg / m 3ρ . .

Substitute in and find p2 808= kPa.

M K or 473 C1 21000

1 4 287 2832 966 808

287 3 774746=

× ×= =

×=

.. .

. ..T o

M2261

1 4 287 7460 477=

× ×=

.. .

b) M M 0.477. kPa1 2 2 11000 1 4 287 283 2 97 10 12 809 6= × × = ∴ = = =/ . . . . . .p p

T2 22 644 283 748 809 6287 748

3 771= × = ∴ =×

=. . ..

. . K or 475 C kg / m3o ρ

9.45 a) ρ ρ1 2 212 144

1716 500002014 002014 3000=

××

= × =. . . . slugft

3 V

Momentum: 2 212 144 .002014 3000( 3000).p V× − = × − 2 2

2 2

2

3000 1.4 1716 500 0.

2 .4V p

ρ −

+ − × =

VV

V22 2 2

263000 7

6 04219 854 6 042 6 006 10− +

− − ×.

( , . ) . = 0.

∴ − + × ∴ = =6 23 15 10 833 0 0072522

26

2 2V V V, . .000 = 0. fps. slugft

3ρ

p2 102 9= . . psia

M R or 731 F1 23000

1 4 1716 5002 74 102 9 144

1716 007251191=

× ×= =

××

=.

. . ..

.T o o

212

M2833

1 4 1716 11910 492=

× ×=

.. .

b) 1 2 2M 3000/ 1.4 1716 500 2.74. M 0.493. 8.592 12 103.1 psia.p= × × = ∴ = = × =

T2 22 386 500 1193 103 1 1441716 1193

0 00725= × = ∴ =×

×=. . . . .o oR or 733 F slug / ft 3ρ

9.46 [ ]

ρρ

2

1

2

1

1

2

12 2

12

12

12

12

12

2 11

1

1 12

4 2 2

12 1

= =− ++

+

+−

− +=

++ −

pp

TT

k kk

kk k k

kk

M M

M M

MM

( ) ( )( )

.

M12 2

1

12

12

=+

+−k

kpp

kk

. (This is Eq. 9.4.12). Substitute into above:

ρρ

2

1

2

1

2

1

2

1

2 2

1

1 1 1

4 1 1 1

1 1 1

1 1 1=

+ + + −

+ − + + −

=

+ + + −

+ + − +

( ) ( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( )( ).

k kpp

k

k k kpp

k

k kpp

k

k k kpp

=− + ++ + −

k k p pk k p p

1 11 1

2 1

2 1

( ) /( ) /

.

For a strong schock in which pp

kk

2

1

2

1

111

>> =+−

, . ρρ

9.47 Assume standard conditions: T1 115 101= =o C, kPa.ρ ∴ = × × =V1 2 1 4 287 288 680. m / s. 1 2 2M 2. M .5774. 1.688 288 486 K. T= ∴ = = × =

2 4.5 101 454 kPa.p = × = 2 .5774 1.4 287 486 255 m/s.V∴ = × × =

induced 1 2 680 255 425 m/s.V V V∴ = − = − =

The high pressure and high induced velocity cause extreme damage. 9.48 If M then M m /s2 1 15 2 645 2 645 1 4 287 293 908= = ∴ = × × =. , . . . . .V

p2 28 00 200 1600 1600287 2 285 293

8 33= × = =× ×

=. .. ( . )

. . kPa abs kg / m 3ρ

9.49 If M then M fps2 1 15 2 645 2 645 1 4 1716 520 1118= = ∴ = × × =. , . . . . .V

p2 28 00 30 240 240 1441716 2 285 520

0 01695= × = =×

× ×=. .

( . ). . psia slug / ft 3ρ

V1V

2

stationary shock

213

9.50 p T1 1 12615 101 26 4 223 3 1000 1 4 287 223 3 3 34= × = = = × × =. . . / . . . . kPa. K. M ∴ = = × = = × =M kPa. K.2 2 24578 12 85 26 4 339 3 101 223 3 692 5. . . . . . .p T For isentropic flow from → : For M = .458, p = .866 p0 and T T p T= ∴ = = = =. . /. . . /. .960 339 866 391 692 5 960 7210 0 0 kPa abs K or 448 Co 9.51 After the shock M kPa abs.2 24752 10 33 800 8264= = × =. , .p For isentropic flow from → : For M = .475, p = .857 p0 . ∴ = =p0 8264 857 9640/. . kPa abs

9.52 AA

p p pe* . . . . /. . .= ∴ = = ∴ = =4 147 985 101 985 102 50 0 M kPa abse

M kPa. K.t = = × = = × =1 5283 102 5 54 15 8333 298 248 3. . . . . .p Tt t

∴ =×

= = × × =ρt tV54 15

287 248 37599 1 4 287 248 3 315 9

.. .

. . . . . kg / m m / s.3

∴ = × × × =& . . . . .m 7599 025 315 9 0 4712π kg / s If throat area is reduced, M t

remains at 1, ρ πt m= = × × × =. & . . . . .7599 7599 02 315 9 0 3022 kg / m and kg / s3

9.53 p p AA

p pe = = ∴ = =101 4 2 94 9 9182 1 2 1 kPa = M and . . . , / . .*

∴ = = = =p p p1 1 0101 9 918 10 18 2 94 0298/ . . . , / . . kPa. At M ∴ = =p0 10 18 0298 342. /. . kPa abs M kPa abs K.t = = × = = × =1 5283 342 181 8333 293 244 1, . . . .p Tt t

∴ = × × =Vt 1 4 287 244 1 313. . . m / s M kPa abs K.1 1 12 94 10 18 3665 293 107 4= = = × =. , . . . .p T

∴ = × × =V1 2 94 1 4 287 107 4 611. . . . m / s M kPa K.2 24788 101 2 609 107 4 280 2= = = = × =. , . . . .p T Te e

∴ = × × =V2 4788 1 4 287 280 2 161. . . . m / s

9.54 p p AA

p pe = = ∴ = =14 7 4 2 94 9 9182 1 2 1. . . . , / . .* psia = M and

∴ = = = =p p p1 1 014 7 9 918 1 482 2 94 0298. / . . . , / . . psia. At M ∴ = =p0 1 482 0298 49 7. /. . . psia

M psia R.t = = × = = × =1 5283 49 7 26 3 8333 520 433 3, . . . . . .p Tt to

∴ = × × =Vt 1 4 1716 433 3 1020. . . fps

M psia R.1 1 12 94 1 482 3665 520 190 6= = = × =. , . . . .p T o

∴ = × × =V1 2 94 1 4 1716 190 6 1989. . . . fps

214

M psia R.2 24788 14 7 2 609 190 6 497 3= = = = × =. , . . . . .p T Te eo

∴ = × × =V2 4788 1 4 1716 497 3 523. . . . fps

9.55 M kPa K.t t tp T= = × = = × =1 5283 500 264 8333 298 248 3. . . . .

AA

p12

2 1 185

2 56 2 47 0613 500 30 65* . . . , . . .= = ∴ = = × = M

T V1 1451 298 134 4 2 47 1 4 287 134 4 574= × = ∴ = × × =. . . . . . K. m / s M kPa K.2 2 2516 6 95 30 65 213 2 108 134 4 283 3= = × = = × =. , . . . . . .p T

After the shock it’s isentropic flow. At M = =. , . .*516 1 314AA

p A02

2

511 500 255 5 041 314

003825= × = =×

=. . ..

. .* kPa. m2π

AA

p pee r*

..

. . . . . . . .=×

= ∴ = × = =π 05003825

2 05 940 255 5 240 2982

kPa abs = Me

T Ve e=

= ∴ = × × =283 3 213

240273 8 298 1 4 287 273 8 99

2857

. . . . . ..

K. m / s

9.56 p p Tt t= = × = =

=. .

. / .

546 546 1200 655 673 6551200

5850

3 1 3

kPa. K.

∴ =×

= = × × = =ρ t tV655462 585

2 42 1 3 462 585 593 1.

. . . ( . ) kg / m m /s. M3t

& . . . . .m A Vd

dt t tt

t= ∴ = ××

× ∴ =ρπ

m or 6 cm4 2 424

593 0 0602

Te e=

= ∴ =

×=673 101

1200380 2 101

462 380 2575

3 1 3. / .

.. .

. . K kg / m3ρ

Ve

2

21872 380 2 1872 673+ × = ×. . (Energy from → e .) (cp = ⋅1872 J / kg K)

∴ = ∴ = × ∴ =Vd

dee

e1050 4 5754

1050 0 0922

m / s. m or 9.2 cm. . . .π

9.57 M kPa.e = = = × =1 546 546 1000 5460. . .p pe

Te e=

= ∴ =×

=623 5461000

542 546462 542

2 183

1 3..

.. . K.

kgm 3

ρ

Vd

dee

e= × × = = × ∴ =1 3 462 542 571 15 2 184

571 0 1242

. . . . . m / s. m or 12.4 cmπ

215

9.58 M psia. R.e = = × = =

=1 546 150 81 9 1160 81 9150

10093

1 3. . . .

..

p Te eo

∴ =××

= = × × =ρ e eV81 9 1442762 1009

00423 1 3 2760 1009 1903. . . . slugft

fps.3

. . . . .25 004234

1903 0 1992

= × ∴ =πd

dee ft. or 2.39"

9.59 M kPa. K.t = = × = =

=1 546 1200 655 673 655

1200585

3 1 3

. .. / .

p Tt t

∴ = × × = =×

=Vt t1 3 462 585 593 655462 585

2 42..

. . m / s. kg / m3ρ

∴ = × × × =& . . .m 2 42 0075 593 0 2542π kg / s per nozzle

Te =

=673 120

1200396

3 1 3. / .

. K

9.60 M1800

1 4 287 3032 29=

× ×=

.. .

From Fig. 9.15, β = 46 79o o, .

a) 1n46 . M 2.29sin46 1.65. β = ∴ = =o o

2n 2 M .654 M sin(46 20 ). ∴ = = −o o ∴ =M2 1 49. .

p T2 23 01 40 120 4 1 423 303 431= × = = × =. . . . kPa abs K.

V2 1 4 287 431 1 49 620= × × × =. . . m / s

b) β = ∴ = = ∴ = = −79 2 29 79 2 25 541 79 202o o o o. . sin . . . sin( ). M M M 1n 2n

∴ =M2 0 631. .

p T2 25 74 40 230 1 90 303 576= × = = × =. . . kPa abs K.

V2 1 4 287 576 631 303= × × × =. . . m / s c)

V1

V2

= 20o

V1

= 35o

a detached shock

216

9.61 β θ1 40 10= ∴ =o o. . M M M M = 1.58.1n 2n= = ∴ = = − ∴2 40 1 29 791 40 102 2sin . . . sin( ).o o o If θ β2 210 1 58 51 1 58 51 1 23= = = = ∴ =o o o then, with M M M2n 2n. , . . sin . . . ∴ = = − ∴ = = − = − =M M M 3n . sin( ). . . .824 51 10 1 26 10 51 10 413 3 2

o o oβ β

9.62 M M K.1n 2n= = ∴ = = × =3 5 35 2 01 576 1 696 303 5142. sin . . . . .o T

M 2 1 2 2576

35 202 26 20 47=

−= = = ∴ =

.sin( )

. . . .o o

o oθ θ β

M M M M2n 3n 3= = ∴ = = − ∴ =2 26 47 1 65 654 47 20 1 443. sin . . . sin( ). . .o o o

T V kRT3 3 3 31 423 514 731 1 44 1 4 287 731 780= × = = = × × =. . . . K. M m /s 9.63 M M R.1n 2n= = ∴ = = × =3 5 35 2 01 576 1 696 490 8312. sin . . . . .o oT

M 2 1 2 2576

35 202 26 20 47=

−= = = ∴ =

.sin( )

. . . .o o

o oθ θ β

M M M M2n 3n= = ∴ = = − ∴ =2 26 47 1 65 654 47 20 1 443 3. sin . . . sin( ). . .o o o T V kRT3 3 3 31 423 831 1180 1 44 1 4 1716 1180 2420= × = = = × × =. . . .o R. M fps

9.64 M M M 1n 2n1 13 10 28 3 28 1 41 736= = ∴ = = = ∴ =, . . sin . . . .θ βo o o ∴ = × =p2 2 153 40 86 1. . kPa.

M kPa 2 3736

28 102 38 6 442 86 1 555=

−= ∴ = × =

.sin( )

. . . . .o o

p

( ) . .p3 10 33 40 413normal kPa= × = 9.65 At M 1 1 13 49 8 19 47= = =, . , . .θ µo o (See Fig. 9.18.) θ θ1 2 249 8 25 74 8 4 78+ = + = ∴ =. . . . .o M

From isentropic flow table: p ppp

pp2 1

0

1

2

0

20 102722

002452 1 80= = × × =.

. . . kPa

0 22 1 2

1 0

1253 .1795 127K or 146 C. 12.08 .

.3571T T

T TT T

µ= = × × = − =o o

V2 4 78 1 4 287 127 1080 90 25 70 53 12 08 32 4= × × = = + − − =. . . . . . . m / s α o 9.66 θ θ1 26 4 4 65 8= = =. . , . .o o For M (See Fig. 9.18.) ∴ = − =θ 65 8 26 4 39 4. . . .o

T TTT

TT

V2 10

1

2

02273 1

55562381 117 4 1 4 287 117 867= = × = ∴ = × × =

.. . . K. m / s

T2 156= − o C.

217

9.67 θ θ= = =26 4 4 65 8. . , . .o o For M ∴ = − =θ 65 8 26 4 39 4. . . .o

0 22 1

1 0

1490 .2381 210 R or 250 F.

.5556T T

T TT T

= = × = −o o

V2 4 1 4 1716 210 2840= × × =. . fps

9.68 a) θ θ1 2 239 1 39 1 5 44 1 2 72 20 10585

04165= = + = ∴ = = ×. . . . . . ..

.o o M u up

= 14.24 kPa. For θ β= = = = = ∴ =5 2 5 27 2 5 27 1 13 889o o o and M M M1n 2n. , . . sin . . . . ∴ = × =p2 1 32 20 26 4l . . . kPa

M Ml o o= =

−=2

88927 5

2 37.sin( )

. .

b) M M M1n 2n= = ∴ = = = =2 72 5 25 2 72 25 1 15 875. , . . . sin . , . .θ βo o o

M2u =−

=.

sin( ). .875

25 52 56

o o

For M = 36.0 For M2= = + = =2 37 36 5 41 2 58. , . , . .θ θo ol

c) Force on plate = ( . . ) .26 4 14 24 1000− × × =A F

C F

V A

A

AL = =

× ×

× × ×=

cos . .

. .. .5

12

12 2 996 100012

1 4 2 5 20 0000 139

1 12 2

o

ρ

d) C F

V A

A

AD = =

× ×

× × ×=

sin . .

. .. .5

12

12 2 1000 087212

1 4 2 5 20 0000 0122

1 12 2

o

ρ

9.69 β = = = ∴ = × = =19 4 19 1 30 1 805 20 36 1 7862

o o. sin . . . . . . M kPa. M1n 2np

M M2 1 2 3786

19 53 25 54 36 59 36 3 55=

−= = = ∴ =

.sin( )

. . . . . . . .o o

θ θ

p ppp

pp3 2

0

2

3

0

36 1 10188

0122 23 4= = × × =..

. . kPa.

C

A A

V AD =

−

=×

× × ×=

36 12

23 42

5

12

6 35 087212

1 4 4 200 0025

12 2

. . sin. .

.. .

o

ρ

Lift

Drag

FAirfoil surface

M1M

2M

3p

2p

3

shock

218

9.70 If θ β= = →5 41o o with M then Fig. 9.15 =18, .

1n 2nM 4sin18 1.24. M .818. = = ∴ =o

2 1.627 20 32.5 kPa.p∴ = × =l

∴ =−

=M2 l o o

.sin( )

. .81818 5

3 64

At M At M 2u u1 1 2 10 2

0

4 65 8 75 8 4 88 20 002177006586

= = = = =, . . . . . ..

θ o o p ppp

pp

= 6.61 kPa.

CV A

A A A

AL = =

− × − × ×

× × ×=

Lift12

325 5 20 2 6 61 2 1012

14 4 200 0854

12 2ρ

. cos / . / cos

.. .

o o

CV A

A A

AD = =

− × ×

× × ×=

Drag12

32 5 5 6 61 2 1012

1 4 4 200 010

12 2ρ

. sin . / sin

.. .

o o

M1 M2u

M2lshock

shock