Analisis Hitungan a S U F - Engg

8/2/2019 Analisis Hitungan a S U F - Engg

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Counts Analysis of the Uniform Steady Flow Experiments

The first experiment1. Calculate the flow discharge (Qmeasured)

Given :

hVnotch = 9,15 cm

p = 27 cm

g = 981 cm/s2

h/p = 0,339

= 35

Kv = 0,183 cm (obtained from graph)

Ce = 0,5821 (obtained from graph)

Solution:

Qmeasured = Ce15

82g tan

2

( hVnotch + Kv)

2,5

= 0,5821

15

89812 tan

2

35

0,183+9,15 2,5

= 1153,766 cm3/s

2. Calculate the sectional area of flow (A)Given:

h ( h average) = 0,983 cm

b = 30 cm

Solution:

A = b x h= 30 x 0,983

= 29,5 cm2

3. Calculate the flow velocity (v)Given:

Q = 1153,766 cm3/s

A = 29,5 cm2


2/22

Solution:

v = A

Q

=29,5

1153,766

= 39,111 cm/s

4. Calculate Viscocity at T = Given:

cm

2/s

cm2/s

Solution:

By using Linier Interpolation Method

()

( )

x 310 cm2/s

5. Calculate the number Renold (Re)Given:

v = 39,111 cm/s

h average = 0,983 cm

=3

10446,8 cm2/s

Solution:

Re =

averagehv

=3

10446,8

983,0111,39

= 4553,499

The flow included the type of turbulent flow for Re> 1000


3/22

6. Calculating the circumference of wet (P)Given:

b = 30 cm


Solution:

P = b + 2h

= 30 + (2 x 0,983)

= 31,967 cm

7. Calculate the hydraulic radius (R)Given:

A = 29,5 cm2

P = 31,967 cm

Solution:

R =P

A

=

31,967

29,5

= 0,923 cm

8. Calculating empirical formula based on coefficientGiven:

R = 0,923 cm

Sf = So = 0,0047 cm

v = 39,111 cm/s

Solution:

a. Manning Coefficient

nv

SfR 21

32


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=

39,111

0,0047923,0 21

32

= 0,00166

b. Strickler Coefficient

ks =2

13

2

SfR

v

=2

13

2

0,0047983,0

39,111

= 601,862

c. Chezy Coefficient

C = SfR

v

=0047,0983,0

39,111

= 593,860


5/22

The second experiment1. Calculate the flow rate (Qmeasured)

Given :

hVnotch = 12 cm

p = 27 cm

g = 981 cm/s2

h/p = 0,444

= 35



Solution:

Qmeasured = Ce15

82g tan

2

( hVnotch + Kv)

2,5

= 0,591915

89812 tan

2

35

0,183+12 2,5

= 2283,954 cm3/s



b = 30 cm

Solution:

A = b x h

=30 x 1,333

= 40 cm2


Q = 2283,954 cm3/s

A = 40 cm2

Solution:

v =

A

Q


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=40,0

2283,954

= 57,099 cm/det


cm2/s

cm2/s

Solution:


()

( )

x 310 cm2/s


v = 57,099 cm/s

h aerage = 1,333 cm

=3

10446,8 cm2/s

Solution:

Re =

averagehv

=3

10446,8

333,1099,57

= 9013,946

The flow included the type of turbulent flow for Re> 10006. Calculating the circumference of wet (P)

Given:

b = 30 cm

h ( h ataverage) = 1,333 cm

Solution:

P = b + 2h


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= 30 + (2 x 1,333 )

= 32,667 cm


A = 40 cm2

P = 32,667 cm

Solution:

R =P

A

=32,667

40,0

= 1,223 cm


R = 1,224 cm

Sf = So = 0,0047 cm

v = 57,099 cm/s

Solution:

a. Manning Coefficientn

v

SfR 21

32

=57,099

0,0047224,1 21

32

= 0,00137

b. Strickler Coefficientks

21

32

SfR

v


8/22

=2

1

3

2

0047,0224,1

099,57

= 727,682


C = SfR

v

=0047.0224,1

099,57

= 59,860


9/22

The third experiment1. Calculate the flow rate (Qmeasured)

Given :

hVnotch = 12,30 cm

p = 27 cm

g = 981 cm/s2

h/p = 0,456

= 35



Solution:

Qmeasured = Ce15

82g tan

2

( hVnotch + Kv)

2,5

= 0,593615

89812 tan

2

35

0,183+12,30 2,5

= 2434,044 cm3/s



b = 30 cm

Solution:

A = b x h

= 30 x 1,683

= 50,5 cm2


Q = 2434,044 cm3/s

A = 50,5 cm2

Solution:


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v =A

Q

=50,5

2434,044

= 48,199 cm/s


cm2/s

cm2/s

Solution:


()

( )

x 310 cm2/s


v = 48,199 cm/s

R = 1,513 cm

=3

10446,8 cm2/s

Solution:

Re =

Rv

=3

10446,8

513,148,199

= 9606,299


Given:

b = 30 cm



11/22

Solution:

P = b + 2h

= 30 + (2 x 1,683)

= 33,367 cm


A = 50,5 cm2

P = 33,367 cm

Solution:

R =P

A

=33,367

50,5

= 1,513 cm


R = 1,513 cm

Sf = So = 0,0047 cm

v = 48,199 cm/s

Solution:

a. Manning Coefficientn

v

SfR 21

32

=48,199

0,00471,513 21

32

= 0,00187



12/22

ks =

21

32

SfR

v

=2

1

32

0047,0513,1

48,199

= 533,338


C = SfR

v

=0047.0513,1

48,199

= 571,477


13/22

The fourth experiment1. Calculate the flow rate (Qmeasured)

Given :

hVnotch =13,50 cm

p = 27 cm

g = 981 cm/s2

h/p = 0,500

= 35



Solution:

Qmeasured = Ce15

82g tan

2

( hVnotch + Kv)

2,5

= 0,601515

89812 tan

2

35

0,18313,50 2,5

= 3102,841 cm3/s



b = 30 cm

Solution:

A = b x h

= 30 x 1,700

= 51,0 cm2


Q = 3102,841 cm3/s

A = 51,0 cm2

Solution:


14/22

v =A

Q

=51,0

3102,841

= 60,840 cm/s


cm2/s

cm2/s

Solution:


()

( )

x 310 cm2/s


v = 60,840 cm/s

R = 1,527 cm

=3

10446,8 cm2/s

Solution:

Re =

Rv

=3

10446,8

527,160,840

= 12245,800


Given:

b = 30 cm



15/22

Solution:

P = b + 2h

= 30 + (2 x 1,700)

= 33,400 cm


A = 51,00 cm2

P = 33,400 cm

Solution:

R =P

A

=33,400

51,0

= 1,527 cm

8. Calculating empirical formula based on coefficient

Given:

R = 1,527 cm

Sf = So = 0,0047 cm

v = 60,840 cm/s

Solution:


nv

SfR 21

32

=60,840

0,00471,527 21

32

= 0,00149



16/22

ks =

21

32

SfR

v

=2

13

2

0047,01,527

60,840

= 669,254

c. Chezy CoefficientC

SfR

v

=0047.0527,1

60,840

= 718,172


v = 60,840 cm/s

R = 527,1 cm

=3

10446,8 cm2/s

Solution:

Re =

Rv

=3

10446,8

527,160,840

x

= 12245

The flow included the type of turbulent flow for Re> 1000


17/22

The fifth Experiment1. Calculate the flow rate (Qmeasured)

Given :

hVnotch =14,80 cm

p = 27 cm

g = 981 cm/s2

h/p = 0,548

= 35



Solution:

Qmeasured = Ce15

82g tan

2

( hVnotch + Kv)

2,5

= 0,612315

89812 tan

2

35

0,18314,80 2,5

= 3963,309 cm3/s



b = 30 cm

Solution:

A = b x h

= 30 x 2,000

= 60,0 cm2


Q = 3963,309 cm3/s

A = 60,0 cm2

Solution:

v =

A

Q


18/22

=60,0

3963,309

= 66,055 cm/s


cm2/s

cm2/s

Solution:


()

( )

x 310 cm2/s


v = 66,055 cm/s

R = 1,765 cm

=3

10446,8 cm2/s

Solution:

Re =

Rv

=3

10446,8

765,1055,66

x

= 15641,759


Given:

b = 30 cm


Solution:

P = b + 2h


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= 30 + (2 x 2,000)

= 34,000 cm


A = 60,0 cm2

P = 34,000 cm

Solution:

R =P

A

=34,000

60,0

= 1,765cm


R = 1,765 cm

Sf = So = 0,0047 cm

v = 66,055 cm/s

Solution:


n

v

SfR 21

32

=66,055

0,00471,765 21

32

= 0,00152


ks =2

13

2

SfR

v


20/22

=2

13

2

0047,0765,1

055,66

= 659,254

c. Chezy CoefficientC

SfR

v

=0047.0765,1

055,66

= 725,306


21/22

DISCUSSION

1. Tail gate function is to equalize or control the water level is raised orelevated water level at every turn of the discharge. But from an experiment

conducted each height h is not the same. This is caused because each turn

of the discharge will affect the water level.

2. From the experiments have been conducted Manning coefficient valuesobtained do not approach the price of 0.01 m -1 / 3 sec or 0.00215 cm-1

/ 3 dt. This is because the slope of the basic channels that are too big

which causes a higher flow rate so that the value of Manning coefficient

become large.

3. Renold Numbers 1000, included turbulent flow.

4. Heights h1, h2, h3 on one trial is not the same as the channel bottom slopeis too large and lack of precision when measuring height.


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CONCLUSION

1. From the steady uniform flow experiments can be known that in the uniformflow discharge, muka air,kecepatan dan berat jenis yang tetap sepanjang

waktu dan seragam diberbagai tempat yang di tinjau.

2. Setelah melakukan lima kali percobaan didapatkan nilai rata-rataManning,Strickler, dan Chezy sebesar:

Manning : 0,0030 Strickler : 339,0172 Chezy : 362,254

3. Nilai koefisienManning mempunyai nilai yang berkebalikan dengan nilaikoefisien Strickler ( n =

Ks

1)

Analisis Hitungan a S U F - Engg

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