1

CHAPTER ONE

Basic Concepts of Thermodynamics

Thermodynamics is the science that deals with the conversion of

heat into mechanical energy. It is based upon observations of common

experience, which have been formulated into thermodynamic laws. These

laws govern the principles of energy conversion.

Thermodynamics has very wide applications as basis of thermal

engineering. Almost all process and engineering industries, agriculture,

transport, commercial and domestic activities use thermal engineering.

However, energy technology and power sector are fully dependent on the

laws of thermodynamics.

Types of system

The analysis of thermodynamic processes includes the study of the

transfer of mass and energy across the boundaries of the system. On the

basis, the system may be classified mainly into three parts: (1) Closed

system (2) Open System (3) Isolated system.

(1) Closed System:

Consists of a fixed amount of mass, and (no) mass can enter or

leave a closed system, but energy in the form of heat or work can

cross the boundary, as shown below in figure.

(2) Open system:

Is a selected region in space. It usually encloses a device that

involves mass flow, such as a compressor, turbine, diffuser, nozzle, heat

Closed System

.constm

(Control mass)

No mass to (or)

out off the system Energy (Yes) in a form

of Heat (or) Work

2

Open

System

(Control

Volume) Mass in

Mass out Energy in

Energy out

exchanger, and throttling valve (expansion valve), as shown in figure

below:

(3) Isolated System

In an Isolated system, neither energy nor masses are allowed to

cross the boundary. The system has fixed mass and energy. No such

system physically exists. Universe is the only example, which is perfectly

isolated system.

Other Special System

1. Adiabatic System: A system with adiabatic walls can only exchange

work and not heat with the surrounding. All adiabatic systems are

thermally insulated from their surroundings. Example is Thermos flask

containing a liquid.

2. Homogeneous System: A system, which consists of a single phase, is

termed as homogeneous system. For example, Mixture of air and water

vapor, water plus nitric acid and octane plus heptanes.

3. Heterogeneous System: A system, which consists of two or more

phase, is termed as heterogeneous system. For example, Water and steam,

Ice plus water and water plus oil.

Isolated System Heat ( Q ) and Work (W ) = 0

3

For defining any system, certain parameters are needed. Properties

are those observable characteristics of the system, which can be used for

defining it. For example, pressure, temperature, volume.

Properties further divided into three parts:

* Intensive Properties

Intensive properties are those, which have same value for any part

of the system or the properties that are independent of the mass of the

system. For example; pressure and temperature

* Extensive Properties

Extensive properties are those, which dependent Upon the mass of

the system and do not maintain the same value for any part of the system.

For example; mass, volume, energy and entropy.

* Specific Properties

The extensive properties when estimated on the unit mass basis result in

intensive property, which is also known as specific property. For

example: specific heat, specific volume and specific enthalpy.

a) DENSITY ( ):

Density is defined as mass per unit volume;

Density = mass/ volume = V

m (kg/m 3 )

Hg = 13.6 × 10 3 (kg/m 3 )

OH 2 = 1000 (kg/m 3 )

b) Specific Volume ( v ):

It is defined as volume occupied by the unit mass of the system. Its

unit is (m 3 /kg). Specific volume is reciprocal of density.

v =

1

m

V (m 3 / kg)

4

* Differentiate amongst gauge pressure, atmospheric pressure and

absolute pressure. Also give the value of atmospheric pressure in bar

and mm of Hg.

While working in a system, the thermodynamic medium exerts a

force on boundaries of the vessel in which it is contained. The vessel may

be a container, or an engine cylinder with a piston etc. The exerted force

(F) per unit area (A) on a surface, which is normal to the force, is called

intensity of pressure or simply pressure (P). Thus

P = F/A= .g.h

It is expressed in Pascal (1 Pa = 1 N/m 2 ),

(1 bar = 10 5 Pa),

Standard atmosphere (1 atm =1.0132 bar 1 bar),

Or, technical atmosphere (1 kg/cm 2 or 1 atm).

1 atm means 1 atmospheric absolute.

The pressure is generally represented in following terms:

1. Atmospheric pressure

2. Gauge pressure

3. Vacuum (or vacuum pressure)

4. Absolute pressure

5

* Atmospheric Pressure (P atm ):

It is the pressure exerted by atmospheric air on any surface. It is

measured by a barometer. Its standard values are;

1 Patm = 760 mm of Hg i.e. column or height of mercury

1 Patm = .g.h. = 13.6 × 10 3 × 9.81 × 760/1000

1 Patm = 101.325 (kN/m 2 ) = 101.325 (kPa) 100 (kPa)

1 Patm = 1.01325 (bar) 1 (bar)

When the density of mercury is taken as 13.595 (kg/m 3 ) and acceleration

due to gravity as 9.8066 (m/s 2 )

* Gauge Pressure (P gauge ):

It is the pressure of a fluid contained in a closed vessel. It is always

more than atmospheric pressure. It is measured by an instrument called

pressure gauge (such as Bourden’s pressure gauge).

* Vacuum pressure (P vac ):

It is the pressure of a fluid, which is always less than atmospheric

pressure. Pressure (i.e. vacuum) in a steam condenser is one such

example. It is also measured by a pressure gauge but the gauge reads on

negative side of atmospheric pressure on dial. The vacuum represents a

difference between absolute and atmospheric pressures.

* Absolute Pressure (P abs ):

It is that pressure of a fluid, which is measured with respect to

absolute zero pressure as the reference. Absolute zero pressure can occur

only if the molecular momentum is zero, and this condition arises when

there is a perfect vacuum. Absolute pressure of a fluid may be more or

less than atmospheric depending upon, whether the gauge pressure is

expressed as absolute pressure or the vacuum pressure.

Pabs = Patm + Pgauge

Pabs = Patm – Pvac

6

* Define thermodynamic equilibrium of a system and state its

importance. What are the conditions required for a system to be in

thermodynamic equilibrium? Describe in brief?

Equilibrium is that state of a system in which the state does not

undergo any change in itself with passage of time without the aid of any

external agent. Equilibrium state of a system can be examined by

observing whether the change in state of the system occurs or not. If no

change in state of system occurs then the system can be said in

equilibrium.

Form of Energy

1. Microscopic Forms: are those related to the molecular structure of

a system and the degree of the molecular activity. The sum of all

microscopic forms of energy is called the Internal Energy (U) of a

system.

2. Macroscopic Forms: are those a system possesses as a whole with

respect to some outside reference frame, such as Kinetic Energy

(KE) and Potential Energy (PE). The macroscopic energy is

related to the motion and influence of some external effects such as

gravity, magnetism, electricity and surface tension.

a) Kinetic Energy (KE): is the energy that a system possesses

as a result of its motion.

2

. 2YmKE

Where:

s

mvelocityY

kgmassm

7

b) Potential Energy (PE): is the energy that system possesses as

result of its elevation in a gravitational field.

zgmPE ..

Where:

2smonaccelerati nalgravitatio

g

melevationz

kgmassm

c) The total energy of a system consists of the Kinetic (KE), Potential

(PE) and Internal (U) energies and is expressed as:

1000

..

2000

. 2 zgmYmUPEKEUE kJ

Or, on a unit mass basis:

1000

.

2000

2 zgYupekeue

kg

kJ

1000

.

2000.

1000

..

2000

.

12

2

1

2

212

12

2

1

2

212

zzgYYuumPEKEUE

zzgmYYmUUPEKEUE

8

* Phase Change of Pure Substance:

(1) Compressed Liquid

(2) Saturated Liquid

(3) Saturated Liquid-Vapor Mixture

(4) Saturated Vapor

(5) Superheated Vapor

Explaining:

(1) If we improve only one item of the following conditions

Compressed Liquid

:1 At given Temperature T and Pressure P :

givensatgiven

givensatgiven

PTT

TPP

@

@

@:2

f

f

f

f

s

h

u

v

s

h

u

v

givenT or givenP

Then,

PorTf

PorTf

PorTf

PorTf

s

h

u

v

s

h

u

v

)(@

)(@

)(@

)(@

1

2 3

4

5

T

v

Saturated Liquid Line

Saturated Vapor Line

9

(2) For Saturated Liquid, in problem (example) will say (saturated liquid

water, R-12, R-134a, R-22, ammonia, nitrogen…etc) at a define

Temperature (or) Pressure, then the same procedure as Compressed

Liquid could be used to find specific volume, internal energy, enthalpy

and entropy as shown below:

PorTf

PorTf

PorTf

PorTf

s

h

u

v

s

h

u

v

)(@

)(@

)(@

)(@

Example (1): Rigid tank contains water at temperature C90 and pressure

of KPa125 . Find: (1) the specific volume of water, (2) internal energy,

(3) enthalpy of water and (4) entropy.

Solution:

By using table (7-1a):

CKPaKPa

TPP givensatgiven

90@14.70125

@

Compressed liquid water

On the other hand, it could be proved by using table (7-1b):

MPaCC

PTT givensatgiven

125.0@99.10590

@

Compressed liquid water

KkgkJss

kgkJhh

kgkJuu

kgmvv

Cf

Cf

Cf

Cf

.1925.1:)4(

92.376:)3(

85.376:)2(

001036.0:)1(

90@

90@

90@

3

90@

10

Example (2): Frictionless piston-cylinder device contains water at a

pressure of 450 KPa and entropy of 1.532

Kkg

kJ.

. Find: (1) specific

volume of water, (2) enthalpy and (3) temperature of water.

Solution:

At a given pressure 450 (KPa) = 0.45 (MPa) and by using table (7-1b):

8207.1532.1 45.0@ MPafgiven ss Compressed liquid water

CTT

kgkJhh

kgmvv

MPasat

MPaf

MPaf

93.147:)3(

25.623:)2(

001088.0:)1(

45.0@

45.0@

3

45.0@

Example (3): Saturated liquid ammonia at 27.5 C , find: (1) pressure of

ammonia, (2) internal energy, (3) entropy, (4) specific volume and (5)

enthalpy of ammonia.

Solution:

Given temperature in question is between (25 and 30) C and the

ammonia is saturated liquid, then:

kgkJ

hhhh

kgm

vvvv

KkgkJ

ssss

kgkJ

uuuu

KPaPP

PP

CfCf

Cfammonia

CfCf

Cfammonia

CfCf

Cfammonia

CfCf

Cfammonia

CsatCsat

Csatammonia

335.3102

42.32225.298

2:)5(

001669.02

001680.0001658.0

2:)4(

.1607.1

2

2005.11210.1

2:)3(

525.3082

46.32059.296

2:)2(

1.10852

0.11672.1003

2:)1(

30@25@

5.27@

330@25@

5.27@

30@25@

5.27@

30@25@

5.27@

30@25@

5.27@

11

Example (4): 7 liters rigid tank contains 10 kg of refrigerant R-12 at 400

(KPa). Find: (1) enthalpy, (2) entropy and (3) temperature of R-12.

Solution:

12-R liquid Compressed0007299.00007.0

19:

0007.010

007.0

10

007.07

4.0@12

3

12

3

MPafR

R

vv

bTable

kgm

m

Vv

kgm

mLV

CTT

KkgkJss

kgkJhh

MPasat

MPaf

MPaf

15.8:)3(

.1691.0:)2(

64.43:)1(

4.0@

4.0@

4.0@

Example (5): Piston cylinder device contains 5 kg of saturated liquid

refrigerant R-134a at 17 C . Find: (1) enthalpy, (2) volume ( 3m ), and (3)

pressure inside piston-cylinder device.

Solution:

C20

C17

C16

Interpolation

69.71 h 26.77

kg

kJh

0008062.0 v 0008157.0

kg

mv

12

..:)3(

4004.0000808575.0*5*

000808575.0

10*5.9*25.00008062.01620

1617

0008062.00008157.0

0008062.0:)2(

0825.73

3925.157.5*25.069.714

1

57.5

69.71

1620

1617

69.7126.77

69.71:)1(

3

3

6

WH

LmvmV

kg

mv

vv

kg

kJh

hh

h

(3) For Saturated Liquid-Vapor Mixture Region

a)

PorTfg

PorTfg

PorTfg

PorTfg

PorTf

PorTf

PorTf

PorTf

s

h

u

v

s

h

u

v

s

h

u

v

)(@

)(@

)(@

)(@

)(@

)(@

)(@

)(@

with a given quality ( x ),

then:

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

ssxssxss

hhxhhxhh

uuxuuxuu

vvxvvxvv

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

..

..

..

..

b) Find quality (x) if in example (problem) are given masses of liquid and

vapor with certainly Temperature (T) or Pressure (P) of pure substance:

gf

g

total

g

mm

m

m

mx

, where: 10 x , then:

13

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

PorTfPorTgPorTfPorTfgPorTf

ssxssxss

hhxhhxhh

uuxuuxuu

vvxvvxvv

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

)(@)(@)(@)(@)(@

..

..

..

..

Example (6): Rigid tank contains ammonia at 40 C and internal energy

of 1000 (kg

kJ ). Determine: (1) specific volume of ammonia, (2)

enthalpy, and (3) pressure of ammonia.

Solution:

From table (8-1) and at T = 40 C :

1341100074.368

1000

1341&74.368

40@40@

40@40@

CgCf

CgCf

uuu

kg

kJu

kg

kJu

kg

kJu

Saturated Liquid-Vapor Mixture Ammonia

KPaPP

kgkJhxhh

kgmvxvv

Then

xx

uxuu

CTsat

TfgTf

TfgTf

CfgCf

9.1554:)3(

65.1085)8.1098*65.0(43.371.:)2(

05458.0)08141.0*65.0(001725.0.:)1(

:

65.0)2.972.(74.3681000

.

40@

@@

3

@@

40@40@

Example (7): Piston-cylinder device at 550 (KPa) contains 9 (kg) liquid

water and 3 (kg) saturated vapor. Determine: (1) quality (x), (2) enthalpy,

(3) entropy of water and (4) temperature of water.

14

Solution:

CTT

KkgkJsxss

kgkJhxhh

mm

m

m

mx

MPasatwater

MPafgMPaf

MPafgMPaf

gf

g

t

g

48.155:)4(

.1203.3892.4*25.08973.1.:)3(

18.11802097*25.093.655.:)2(

25.04

1

93

3:)1(

55.0@

55.0@55.0@

55.0@55.0@

(4) For Saturated Vapor: in problem (example) will say (saturated vapor

water, R-12, R-134a, R-22, ammonia, nitrogen…etc) at a given

Temperature (or) Pressure. To find specific volume, internal energy,

enthalpy and entropy, the procedure is shown below:

PorTg

PorTg

PorTg

PorTg

s

h

u

v

s

h

u

v

)(@

)(@

)(@

)(@

Example (8): Tank contains 3.5 kg of saturated vapor nitrogen at 120 K.

Determine: (1) shv &, , (2) the volume of tank, and (3) the temperature

of nitrogen.

Solution: Saturated vapor nitrogen and from table (12-1):

KPaMPaPP

Lmkg

mvkgmV

KkgkJss

kgkJhh

kgmvv

Ksat

Kg

Kg

Kg

5.25135135.2

119.28028119.0008034.0*5.3*

.6253.4

075.75

008034.0

120@

33

120@

120@

3

120@

15

(5) If we improve only one item of the following conditions

Superheated Vapor

(1): At given Temperature T and Pressure P :

givensatgiven

givensatgiven

PTT

TPP

@

@

@:)2(

g

g

g

g

s

h

u

v

s

h

u

v

givenT or givenP

Example (9): Rigid tank contains Methane at 175K and 0.4 MPa,

find shv &, ?

Solution:

KKPaKPa

TPP givensatgiven

175@6.2777400

@

Superheated Methane

By using table (13-2):

KkgkJs

kgkJh

kgmv

.728.9

81.351

21799.03

Example (10): Frictionless piston cylinder device contains water at

250 C and 700 KPa. Determine: shuv &,, ?

Solution:

hSMPaCC

PTT

or

hSCMPaMPa

TPP

givensatgiven

givensatgiven

.7.0@97.164250

@

:

.250@973.37.0

@

Superheated water, then from table (7-2):

16

KkgkJ

sss

kgkJ

hhh

kgkJ

uuu

kgm

vvv

MPaMPa

MPaMPa

MPaMPa

MPaMPa

.11.7

2

0384.71816.7

2

6.29532

29502.2957

2

2.27182

5.27159.2720

2

34345.02

2931.03938.0

2

8.0@6.0@

8.0@6.0@

8.0@6.0@

38.0@6.0@

Example (11): Frictionless piston cylinder device contains ammonia at

45 C and 0.35 MPa. Determine: shv &, ?

Solution:

CKPaKPa

TPP givensatgiven

45@1782350

@

Superheated ammonia and from table (8-2):

KkgkJ

sss

kgkJ

hhh

kgm

vvv

CC

CC

CC

.8192.5

2

8557.57828.5

2

9.15572

5.15693.1546

2

43107.02

43865.04235.0

2

50@40@

50@40@

350@40@

Example (12): Rigid tank contains 1.5 kg of refrigerant R-12 at pressure

of 0.4 MPa and enthalpy of 240 (kJ/kg). Find: (1) temperature of R-12,

specific volume (v), internal energy (u), entropy (s), and (2) V )( 3m ,

H )(kJ , U )(kJ and S )(K

kJ

Solution:

We have two items (P and h), then from table (9-1b):

hShh

kgkJh

kgkJhMPaP

ggiven

MPagMPafgiven

.)97.190()240(

97.190&64.434.0@ 4.0@4.0@

17

Superheated R-12, then from table (9-2):

K

kJsmS

kJhmH

kJumU

mvmV

KkgkJs

kgkJu

kgmv

CT

2681.18454.0*5.1*

360240*5.1*

045.32403.216*5.1*

08686.005791.0*5.1*:)2(

.8454.0

03.216

05791.0

80:)1(

3

3

Example (13): A rigid tank with a volume of 2.5 ( 3m ) contains 5 kg of

saturated liquid-vapor mixture of water at 75 C . Now the water is slowly

heated, determine: (1) the temperature at which the water is completely

vaporized (final temperature), (2) the initial and final pressure, and (3)

show the process on vT diagram with respect to saturation lines.

Solution:

(1): 2

3

1 5.05

5.2v

kgm

m

Vv

(Rigid tank)

gvv 2 (given in question), then from table (7-1a) and by searching

through gv until we reach to the value near of 0.5

kg

m3

.

T

v

.1 constP

.2 constP

1

2

21 vv

2

3

21

1

1

..

5

5.2

75

:

VaporSaturated

kgm

mVV

CT

Mixture

waterData

18

By interpolation:

KPaMPaPP

CTT

991.36836899.03613.04154.0

3613.0

5089.04463.0

5089.05.0

71.140140145

140

5089.04463.0

5089.05.0

22

22

CTT final

71.1402

KPaPP

KPaPPP

final

Csatinitial

99.368

58.38:)2(

2

75@1

Example (14):

Frictionless piston-cylinder device contains 1.5 kg of refrigerant R-134a

at 10 C and 600 KPa, now heat is added at a constant pressure until the

enthalpy reaches 150 kgkJ . Determine: (1) the final temperature of R-

134a ( 2T ), (2) SUV &, , (3) work done during this process ( 21W ). If

more heat is added to R-12 until the entropy becomes 1 KkgkJ . , find the

(4) 3333 &,, huvT , (5) work done during this process ( 32W ), (6) plot vT

diagram.

CT C140

Interpolation

5089.0

gv

4463.0

C145

5.0

3613.0 MPaP 4154.0

19

Solution:

(1):

MPaCC

PTT

or

CKPaKPa

TPP

KPaMPaPP

PCTat

givensatgiven

givensatgiven

CsatCsat

sat

6.0@58.2110

@

:

10@25.415600

@

25.41541525.02

44294.038756.0

210:

12@8@

1

For the first state: R-12 is compressed liquid

KkgkJ

sss

kgkJ

hhh

kgkJ

uuu

kgm

vvv

CfCf

CfCf

CfCf

CfCf

.24495.0

2

2545.02354.0

2

45.632

18.6673.60

2

13.632

83.6543.60

2

00079275.02

0007971.00007884.0

2

12@8@

1

12@8@

1

12@8@

1

312@8@

1

For the first state of R-12

kgkJh

kgkJh

kgkJh

kgkJh

kgkJhMPaPPat

MPaggivenMPaf

MPagMPaf

19.25915048.79

19.259&48.796.0:

06@)(206@

06@06@12

For the second state: R-12 is saturated liquid-vapor mixture

KkgkJs

ssxssxss

kgkJu

uuxuuxuu

kgmv

vvxvvxvv

xx

hhxhhxhh

CTT

MPafMPagMPafMPafgMPaf

MPafMPagMPafMPafgMPaf

MPafMPagMPafMPafgMPaf

MPafMPagMPafMPafgMPaf

MPasat

.53894.02999.09097.0.392.02999.0

..

612.14199.7874.238.392.099.78

..

0138655.00008196.00341.0.392.00008196.0

..

392.071.179.48.79150

..

58.21

2

06@06@06@06@06@2

2

06@06@06@06@06@2

3

2

06@06@06@06@06@2

06@06@06@06@06@2

6.0@2

20

K

kJssmSSH

kJuumUUU

mvvmVVV

44098.024495.053894.0.5.1.

723.11713.63612.141.5.1.

019608.00007927.0013865.0.5.1.:)2(

1212

1212

3

1212

kJvvPmVPW 76.11019608.0*600...:)3( 1221

KkgkJs

KkgkJs

KkgkJsMPaPPPat

MPaggiven .9097.0

.1

.1&6.0::)4(

06@

123

For the third state: R-12 is superheated refrigerant

Now, from table (11-2) at P = 0.6MPa and s = 1 (kJ/kg.K):

kJvvPmVVPW

kgkJh

kgkJu

kgmv

CT

14.23013865.003958.0*600*5.1...

23.288

48.264

03958.0

50

232332

3

3

3

3

3

21

H.W (1):

(1): a) water at 80 C and 500 kPa. Find: ?&,, shuv

b) saturated liquid water at 950 kPa. Find: ?&,,, shuvT

c) water at 325 kPa and x = 0.4. Determine: ?&,,, shuvT

d) water at 160 C and h 1682 kgkJ . Determine: ?&,,, shuvP

e) saturated vapor water at 125 C . Find: ?&,,, shuvP

f) saturated vapor water at 900 kPa. Determine: ?&,,, shuvT

g) water at 350 C and 800 kPa. Find: ?&,, shuv

h) water at 200 kPa and u 2830 kgkJ . Find: ?&,, shvT

(2): a) refrigerant R-134a at -12 C and 600 kPa. Find: ?&, shv

b) refrigerant R-134a at 20 C and v 0.022 kgm3 . Find: ?&, shT

c) saturated vapor R-134a at 320 kPa. Find: ?&,, shvT

d) refrigerant R-134a at 100 C and 600 kPa. Find: ?&, shv

(3): A piston-cylinder device contains initially 50 (L) of liquid water at

25 C and 300 (KPa). Heat is added to the water at a constant pressure

until the entire liquid is vaporized. Determine: (1) mass of water, (2) the

final temperature (2T ), (3) the total enthalpy change ( H ), (4) the work

done during this process (boundary work), and (5) show vT diagram

with respect to saturation lines.

(4): A 0.5 ( 3m ) rigid vessel initially contains saturated liquid-vapor

mixture of water at 100 C . The water is heated until it reaches the critical

state. Determine the volume occupied by the liquid at the initial state.

22

CHAPTER TWO

THE FIRST LAW OF THERMODYNAMICS

"CLOSED SYSTEMS"

1. Conservation of Energy Principle:

One of the fundamental laws of the nature is the first law of

thermodynamics (conservation of energy principle). It simply states that

during an interaction, energy can change from one form to another but the

total amount of energy remains constant.

The conservation of energy principle for closed system or fixed

mass may be expressed as:

system theofenergy total

in the decreases (or) increasesNet

work(or)heat a as system the

from (or) tonsfersenergy traNet

EWQ kJ

Where:

Q net heat transfers across system boundaries kJ

W net work done in all forms kJ

E net change in total energy of a system kJ

The change in total energy E during a process can be expressed

as the sum of the changes in internal U , kinetic KE and potential

PE energies:

1000

.

2000.

1000

..

2000

.

12

2

1

2

212

12

2

1

2

212

zzgYYuumPEKEUE

zzgmYYmUUPEKEUE

Sometimes it is convenient to consider the work term in two parts:

otherb WW & , where: bW represents to (boundary work) and otherW

represents all forms of work except the boundary work. Therefore,

23

1000

.

2000.

1000

..

2000

.

12

2

1

2

212

12

2

1

2

212

zzgYYuumWWQ

zzgmYYmUUWWQ

otherb

otherb

kJ

Or:

1000

.

2000

12

2

1

2

212

zzgYYuuwwq otherb

kg

kJ

For stationary closed system: 0&0 PEKE

For adiabatic process: 0Q

2. Work: is the energy transferred associated with force acting through a

distance.

sFW * kJ

Where:

W work kJ

F force kN

s displacement m

The equation above states that the work done by a constant force

( F ) on a body which is displaced a distance ( s ) in the direction of force.

If the force ( F ) is not constant, the work done is obtained by adding the

differential amount of work.

2

1

*dsFW kJ

The work done per unit mass of a system is denoted w and is defined as:

m

Ww

kg

kJ

24

2.1 Mechanical Forms of Work:

a) Moving Boundary Work bW :

12

1212

2

1

*

:

***

*

***

*

*

vvPw

or

vvPmVVPW

dVPW

dVPdSAPW

APFA

FP

dSFW

.

.

.

constV

constP

constm

Frictionless Piston-

cylinder Device

constP

constV

constm

.

..

Rigid Tank

P

T

V V

21 VV 21 VV

2

1 V

P

T

1 2 outQ

outQ

inQ inQ

V

21 PP

1P

.constP 2P

1V 2V

kJ

kg

kJ

25

For Ideal Gases and during expansion or compression processes,

pressure and volume are related by:

Isothermal Process (1T =

2T = constant):

The ideal gas relation states:

TRmVP ...

Where:

KetemperaturT

KkgkJconstgasR

kgmassm

mvolumeV

kPapressureP

..

3

Then the above equation for ideal gas and for isothermal process can be

written as:

1

222

1

211

1

20

1

2

2

1

2

1

2

1

ln..ln..ln...

ln*

***

***

V

VVP

V

VVP

V

VTRmW

V

VCW

V

dVCdV

V

CdVPW

V

CP

CTRmVP

b

b

b

o

Polytropic Process (nCVP . ):

1

1

1

2

2

1

2

1

.1

...

..

nn

b

n

b

nn

VVn

CW

dVVCdVPW

VCPCVP

n

VPVPWb

1

.. 1122 kJ (or)

n

TTRmWb

1

.. 12 kJ

26

b) Gravitational Work gW :

It can be defined as the work done by or against a gravitational

force field.

kJ

zzgmW

JzzgmW

dzgmdzFW

g

g

g

1000

..

..

...

12

12

2

1

2

1

This expression is easily recognized as the change in potential

energy PE .

c) Acceleration Work aW :

s

mYdt

ds

dsdt

dYmW

dt

dYaamF

dsFW

a

a

2

1

2

1

..

&.

.

kJYYm

W

JYYm

dYYmW

a

a

2

1

2

2

2

1

2

1

2

2

.2000

.2

..

This expression is easily recognized as the change in kinetic

energy KE .

27

d) Shaft Work shaftW :

nW

or

nnrr

W

shaft

shaft

...2

:

...2...2.

Where: torque = rF.

r moment arm m

n number of revolutions per unit time (rps)

W the power transmitted through a shaft kW

e) Spring Work ( springW ):

xkF

dxFW

spring

springspring

.

.

2

1

Where: k Spring constant m

kN

x Displacement m

Then, the work done by spring can be expressed as:

2

1

2

2.2

xxk

Wspring kJ

The work done by the system against spring can be expressed as:

1221 .

2VV

PPareaW

kJ

kJ

kW

28

2.2 Electrical Work:

2

1

.. dtIVWe

Where: V Potential difference

I Number of electrons (currant)

t Time interval (second)

If both IV & remains constant during the time interval t , then

the electrical work will be:

1000

.. tIVWe

kJ

2.3 Specific Heats:

The specific heat is defined as the energy required for rising the

temperature of a unit mass of a substance by one degree. In

thermodynamics, we are interested in two kinds of specific heats; the

specific heat at constant volume VC and the specific heat at constant

pressure PC .

a) Internal Energy, Enthalpy and Specific Heats of Liquids

and Solids:

First law of Thermodynamics:

PEKEUWWQ otherb kJ

For "Closed System" (Constant Volume):

0bW and for stationary system: 0& PEKE

The first law of Thermodynamics becomes:

V

V

other

other

T

uC

uwq

mUWQ

:/

29

T

uCV

Kkg

kJ.

For "Closed System" (Constant Pressure):

First law of Thermodynamics:

PEKEUWWQ otherb kJ

P

P

other

other

other

other

b

T

hC

hwq

mHWQ

VPUHVPUH

VPUWQ

UWVPQ

PEKEVPW

:/

.&.

.

.

0&&.

T

hCP

Kkg

kJ.

b) Internal Energy, Enthalpy and Specific Heats of Ideal

Gases:

vPuhTRvP

TChT

hC PP

.&..

.

TRuh

and

TRuh

.

:

.

kg

kJ

TTRTCTC

TRuh

TCuT

uC

VP

VV

:/...

.

.

RCC VP

30

Example (1): A rigid tank contains 5 (kg) of air at 500 (kPa) and 150 C .

As a result of heat transfer to the surroundings, the pressure inside the

tank drops to 400 (kPa). Determine: (1) the final temperature of the air in

the tank, (2) the work done during this process, (3) find the amount of

heat transferred to the surrounding, and (4) plot VP diagram.

Solution:

CKRm

VPTTRmVP

TankmVV

mP

TRmVTRmVP

39.6538.383287.0*5

214.1*400

.

....

214.1

214.1500

273150*287.0*5.....:)1(

121222

3

21

3

1

11111

21:

0:)2(

VVBecause

Wb

kJQ

KkgkJCTable

TTCmQ

TTCTCuT

uC

umQ

PEKEWW

PEKEumWWQ

V

V

VVV

otherb

otherb

67.5442338.383*276.0*5

.276.0:2

**

**

.

0&,,

.:)3(

12

12

kPaP

CTkPaP

kgm

Tank

air

400

150&500

5

2

11

outQ

V

P

1P

2P

21 VV

)1(

)2(

31

Example (2):

A piston cylinder device initially contains 400 L of air at 1 bar

and 80 C . The air is now compressed to 0.1 3m in such a way that the

temperature inside the cylinder remains constant. Determine: (1) the work

done during this process, and (2) plot VP diagram.

Solution:

kPaPkPabar

mLVLm

1001001

4.040010001

:)1(

1

3

1

3

.8021 constCTT Isothermal process

kJmkPaV

VVPW

V

VVP

V

VVP

V

VTRmW

b

b

45.554.0

1.0ln*4.0*100ln..

ln..ln..ln...

3

1

222

1

222

1

211

1

20

Example (3): Determine the power required to accelerate a 900 (kg) car

from rest to velocity of 80 (km/h) in 20 (s) on a level road.

Solution:

kWs

kJ

t

WW

kJYYm

dYYmW

aa

a

1.1120

2.222

2.22203600

1000*80*

1000*2

900.

2.. 2

22

1

2

1

2

2

CTT

mV

barP

LV

air

80

1.0

1

400

:

21

3

2

1

1

)1(

)2(

P

V

1V 2V

1P

2P

32

Example (4): A piston-cylinder device contains 0.05 3m of a gas

initially at 0.2 MPa. At this state, a linear spring that has a spring constant

of 150 mkN is touching the piston but exerting no force on it. Now heat

is transferred to the gas causing the piston to rise and to compress the

spring until the volume inside the cylinder is doubled. If the cross-

sectional area of the piston is 0.25 2m , determine: (1) the final pressure

inside the cylinder 2P , (2) the total work done by the gas, (3) the fraction

of this work done against the spring to compressed it, and (4) plot

VP diagram.

Solution:

springPPP 12:)1( … (1)

cc

spring

springA

xk

A

FP

. … (2)

mA

V

A

VVx

VV

A

VV

A

Vx

cc

cc

2.025.0

05.02

.2

111

12

12

… (3)

Substitute the value of equation (3) into equation (2):

kPakPakPaPPP

kPam

mmkNP

springfinal

spring

320120200

12025.0

2.0*150

2

2

12

1

3

1

2

2

2.0

05.0

25.0

VV

MPaP

mV

mAc

)1(

)2(

P

V

1V 2V

1P

2P

inQ

33

(2): the work done by the gas against the spring (area under the process

curve):

kJareaW

VVPP

areaW

1305.0.2

320200

.2

1221

(3): kJxxk

Wspring 302.0*2

150.

2

222

1

2

2

Example (5): A rigid tank contains a hot fluid, which is cooled while

being stirred by a paddle wheel. Initially the internal energy of the fluid is

800 kJ . During the cooling process, the fluid loses 500 kJ of heat and

the paddle wheel dose 100 kJ of work on the fluid. Determine the final

internal energy of the fluid.

Solution:

kJUU

UUUWQ

PEKEW

PEKEUWWQ

other

b

otherb

400800100500

0&,

22

12

Notice that the (heat) is negative (-) since it is (from the system); so is

the (work-other) since it is (done on the system).

Example (6): A piston-cylinder device contains 25 (g) of saturated water

vapor which is maintained at a constant pressure of 300 kPa. A resistance

heater within the cylinder is turned and passes a current of 0.2 (A) for 5

(minutes) from 120-V source. At the same time, a heat lose of 3 kJ

occurs.

(a) Show that for a closed system, the boundary work bW and the change

in internal energy U in the first law of thermodynamics can be

combined into one term H for a constant pressure process.

34

(b) Determine the final temperature of the steam 2T ,

(c) Find the boundary work bW and

(d) Show the process on vPvT & diagrams.

Solution:

HWQ

VPUH

VPUWQ

UWVPQ

VPW

UWWQ

PEKE

PEKEUWWQa

other

other

other

b

otherb

otherb

.

.

.

.

0&

:)(

12

12

.1000

..

..

0&

:)(

hhmtIV

Q

hhmhmHWQ

UWWQ

PEKE

PEKEUWWQb

other

otherb

otherb

12.1000

..hhm

tIVQ

… (1)

For the first state and at: vaporsatMPaP .&3.01 , then from

table (7-1b):

kg

kJhh MPag 3.27253.0@1 … (2)

Substitute the value of equation (2) into (1):

kgkJh

h

3.2865

3.2725.1000

25

1000

60*5*2.0*1203

2

2

kg

kJhkg

kJh MPag 3.27253.2865 3.0@2 Superheated steam

35

From table (7-2) and at:

kg

kJhMPaP 3.2865&3.0 22

` CT 2002

kJvvPmW

kgmvtable

kgmvvbtable

vvPmWc

b

MPag

b

828.06058.07163.0*300*025.0..

7163.0:)27(

6058.0:)17(

..:

12

3

2

3

3.0@1

12

(d): vPvT & diagrams [H.W]

Example (7): 1 (kg) of a perfect gas occupied a volume of 0.85 3m at

15 C and at constant pressure of 1 bar. The gas is first heated at constant

volume and then at constant pressure, determine the VP CC & of the gas

if the heat capacity ratio is 4.1k

Solution:

VP

V

P CCC

Ck *4.14.1 . … (1)

RCC VP (For ideal gas)

V

VV

P

VVP

C

Rk

C

R

C

C

CRCC

1

1

:/

1

k

RCV … (2)

KkgkJ

Tm

VPR

TRmVP

.3.027315*1

85.0*100

.

.

...

1

11

111

Then, from equations (2) and (1) by substituting the value of (R):

KkgkJCKkgkJC PV .05.1&.75.0 Respectively

36

Example (8): A piston-cylinder device contains gas, which initially has

pressure, volume and temperature as 275 (kPa), 90 (L) and 185 C

respectively. Now the gas is compressed at a constant pressure until the

temperature becomes 15 C . Calculate: (1) the amount of heat transferred,

and (2) the work done during this process.

Take: KkgkJCKkgkJR P .005.1&.29.0

Solution:

VPUH

VPUQ

UWVPQ

VPW

UWWQ

PEKEW

PEKEUWWQ

other

b

otherb

other

otherb

.

.

.

.

0&,

:)1(

12..... TTCmTCmhmHQ PP … (1)

kg

TR

VPmTRmVP 186.0

273185*29.0

09.0*275

.

....

1

11111

… (2)

Substitute (2) into equation (1):

kJQ

Q

TTCmTCmhmHQ PP

778.31

18515*005.1*186.0

..... 12

12..:)2( VVPVPWb

kg

mP

TRmVTRmVP

3

2

22222 0565.0

275

27315*29.0*186.0.....

kJVVPVPWb 185.909.00565.0*275.. 12

37

Example (9): A rigid tank has a volume of 20 (L) and is filled with a

superheated steam at 300 kPa and 250 C . Determine: (1) the amount of

heat transferred to the surroundings when the steam pressure in the tank

drops to 100 kPa, and (2) show the process on vT diagram.

Solution:

12..

0&,,

.:)1(

uumumQ

pekeWW

pekeumWWQ

otherb

otherb

… (1)

From table (7-2): kgkJukgmv 7.2728&7964.0 1

3

1 … (2)

kgv

Vm

vv

025.07964.0

02.0

1

1

21

… (3)

For the 2nd state and at MPakPaP 1.01002

kgmvkgmvkgmv

kgkJukgkJu

kgmvkgmv

MPagMPaf

MPagMPaf

MPagMPaf

3

1.0@2

3

2

3

1.0@2

1.0@21.0@2

3

1.0@2

3

1.0@2

694.17964.0001043.0

1.2506&36.417

694.1&001043.0

Second state is saturated liquid-vapor mixture water

kgkJu

uxuu

xx

vvxvvxvv

MPafgMPaf

MPafMPagMPafMPafgMPaf

05.1399

7.2088*47.036.417.

47.0001043.0694.1.001043.07964.0

..

2

1.0@1.0@22

1.0@21.0@1.0@21.0@1.0@22

… (4)

Now, substitute values from equations (2), (3) and (4) into equation (1):

kJQ

uumumQ

39.33

7.272805.1399*025.0.. 12

kPaP

dSuperheateCTkPaP

mLVVV

Water

100

250&300

02.020

2

11

3

21

outQ

38

Example (10): A piston-cylinder device contains 5 kg of refrigerant R-12

at 800 kPa and 60 C . The refrigerant is now cooled at a constant pressure

until it exists as saturated liquid at 20 C . Determine: (1) the amount of

heat transfer to the surroundings and (2) show the process on vT

diagram.

Solution:

hvPu

vPumQ

vPmW

uumumWQ

pekeW

pekeumWWQ

b

b

other

otherb

.

..

..

..

0&,

.:)1(

12

12.. hhmhmQ … (1)

v

)1(

)2(

7964.021 vv

kPaP 1002

kPaP 3001 T

CT

CTkPaPP

kgm

R

20

60&800

5

12

2

121

39

For the 1st state and at CTMPakPaP 60&8.0800 11

kgkJh

tableThen

hS

CMPaMPa

TPP

R

givensatgiven

72.220

29:

.

60@52.18.0

@

1

12

… (2)

For the 2nd state, we have a saturated liquid R-12, then at CT 601 :

kgkJhh

aTable

Cf 87.54

19:

20@2

… (3)

Substitute values from (2) and (3) into equation (1):

kJQ

kg

kJkgQ

hhmhmQ

25.829

72.22087.54*5

.. 12

Or it can be solved by using specific volumes and internal energies as

below:

1212 ....

.

.

0&,

vvPuumvPumQ

UVPQ

VPW

UWQ

PEKEW

PEKEUWWQ

b

b

other

otherb

)1( )2(

T

v 1v 2v

40

Example (11): A piston-cylinder device contains air at 150 kPa and

27 C . At this state the piston is resting on a pair of stops, as shown in

figure below, and the enclosed volume is 400 L. The mass of the piston is

such that a 350 kPa pressure is required to move it. The air is heated now

until its volume has doubled. Determine: (1) the final temperature, (2) the

work done by the air, (3) the total heat added and (4) show P-V diagram.

Solution:

kJQ

kgkJuu

kgkJuu

Table

uumWQ

kgTR

VPm

kJVVPWW

KVP

VPTT

T

VP

T

VP

KK

area

9.76607.21452.1113.697.0140

52.1113&07.214

:6

.

697.027327*287.0

4.0*150

.

.:)3(

1404.08.0*350.:)2(

14004.0*150

8.0*350*27327

.

..

..:)1(

31

1400@3300@1

133131

1

11

13331

11

33

13

3

33

1

11

inQ

13 2VV

kPaP

CTkPaP

LVV

Air

350

27&150

400

:

2

11

21

)1(

)2( )3(

21 VV

P

V

41

Example (12):

Find: (1) 21 & xx , (2)

2T , (3) work done by the system, and (4) plot P-V

diagram.

Solution:

3

2

2

2

2121

55.05

1.0*275

*275

**

*

275*

275125400

mV

k

APV

A

Vk

A

xkP

A

VxAxV

A

xkP

kPaPPPPPP

cspring

cc

spring

c

c

c

spring

springspring

For the first state, at 111 .700&400 hSCTkPaP

First state is superheated steam No quality at this state 11 x

2222

125.0@

125.0@

2

22

3

2

1221

3

11

3

1

3749.1

001048.0125:

8465.02

693.1

2

693.155.0243.2

243.21215.1*2*1215.1

Mixturevvv

v

vkPaPAt

kgmV

v

mV

VVVVVV

mvmVkgmv

gf

MPag

MPaf

)1(

)2(

P

V

1V 2V

1P

2P

outQ

kPaP

CT

kPaP

kgm

water

125

700

400

2

:

2

1

1

mkNk

mAc

5

1.0 2

1V

2V

42

kJVPP

W

CTT

vv

vvx

vvxvvxvv

MPasat

fg

f

fgffgf

87.14455.0*2

125400*

2

99.105

615.0001048.03749.1

001048.08465.0

..

21

125.0@2

22

22

2

22222222

H.W (2):

1- A piston-cylinder device whose piston is resting on top of stops

initially contains 0.5 kg of helium gas at 100 kPa and 25 C . The mass of

the piston is such that 500 kPa of pressure is required to raise it. (1) How

much heat must be transferred to the helium before the piston starts rising

and (2) show the process on P-V diagram

2- A mass of 5 kg of saturated liquid water at 200 kPa is heated at a

constant pressure until the temperature reaches 300 C . Find:

(1) vhu &, , (2) the work done during this process, (3) the amount of

heat added to the system, and (4) show the process on vT diagram

3- A frictionless piston cylinder device initially contains 200 L of

saturated vapor refrigerant R-12 at 800 kPa. The refrigerant R-12 is now

heated until its temperature becomes 50 C . Find: (1) the mass of R-12,

(2) vhu &, , (3) the work done during this process, (4) the amount of

heat added to the system, and (5) show the process on vT diagram

4- Frictionless piston cylinder device contains 8 kg of steam at 500 kPa

and 300 C . The steam now is cooled at a constant pressure until 70% of

it condenses. Determine: (1) quality (x), (2) vhu &, , (3) the work

done during this process, (4) the amount of heat transferred to the

surroundings , and (5) show the process on vT diagram

5- During some actual expansion and compression processes in piston

cylinder device, the gases have been observed to satisfy the

43

relationship .. constVP n . Calculate: (1) the work done when a gas

expands from a state 150 kPa and 0.03 3m to a final volume of 0.2 3m for

(n=1.3), and (2) show the process on P-V diagram

6- A mass of 1.2 kg of air at 150 kPa and 12 C is contained in a piston-

cylinder device. The air now is compressed to a final pressure of 600 kPa.

During this process, heat is transferred from the air such that the

temperature inside cylinder remains constant. Calculate: (1) the work

done during this process, (2) the amount of heat, and (3) show this

process on P-V diagram

7- A piston cylinder device contains 5 kg of water at 150 kPa and 25 C .

The cross-sectional area of the piston is 0.1 2m . Heat is transferred to the

water causing part of it to evaporate and expand. When the volume

reaches 0.2 3m , the piston reaches a linear spring whose spring constant is

100 ( mkN / ). More heat is added to the water until the piston rises 20 cm

more. Determine: (1) the final pressure and temperature, (2) the work

done during this process by spring and by system against spring, (3) the

amount of heat transferred to water, and (4) show this process on P-V

diagram

44

CHAPTER THREE

THE FIRST LAW OF THERMODYNAMICS

"OPEN SYSTEMS – STEADY STATE"

An open system (control volume)is a selected region in space. It

usually encloses a device that involves mass flow, such as a compressor,

diffuser, turbine, nozzle, throttling valve (capillary tube), heat exchanger,

duct, and many other devises.

3.1 Conservation of Mass Principle:

ei mm

unit timeper

(CV) volumecontrol

leaving mass Total

unit timeper

(CV) volumecontrol

entering mass Total

Where: i = subscript stands for inlet

e = subscript stands for exit

Most entering devices such as nozzles, diffusers, turbines, compressors

and pumps involve a single stream (only one inlet and one exit). For this

case, we denoted the inlet state by the subscript (1) and the exit state by

the subscript (2). Then, the previous equation (above) can be written for

single stream steady flow as:

skg

mv

AY

v

AY

v

AYAY

skg

mm

cc

cc

2

22

1

11

222111

21

**

1

****

45

Where:

Density 3mkg

v Specific volume kgm3

Y Average flow velocity sm

cA Cross-sectional area 2m

3.2 Conservation of Volume Principle:

vmV * s

m3

3.3 Conservation of Energy Principle:

The first law of thermodynamics (conservation of energy principle)

for a general steady-flow system with Multi Inlets and Exits can be

expressed as:

1000.

2000.

1000.

2000.

22

ii

ii

ee

ee

zg

Yhm

zg

YhmWQ

kW

For Single Stream systems (One Inlet and One Exit), the conservation

of energy principle can be written as:

1000.

2000.

.

12

2

1

2

212

zzg

YYhhmWQ

pekehmWQ

kW

Or:

1000

.2000

12

2

1

2

212

zzg

YYhhwq

kg

kJ

46

Example (1): Air at 100 C and 800 kPa enters the diffuser of a jet engine

steadily with a velocity of 200 m/s and inlet area 0.4 m 2 . The air leaves

the diffuser with a very small value of velocity that can be neglected.

Determine: (1) the mass flow rate of air and (2) the temperature of air at

the exit of diffuser. Take

Kkg

kJCP .005.1

Solution:

(1) One inlet – One exit Device:

skg

kg

m

ms

m

v

AYm

kgm

P

TRvTRvP

TRmVP

v

AYm

mmm

c

c

8.78

015.1

4.0*200*

015.1800

)273100(*287.0***

***

*

3

2

1

11

3

1

11111

111

1

11

21

(2)

KCTT

YYTTC

keTC

TCh

keh

peWQ

pekehmWQ

P

p

p

90.3929.11902000

2000100*005.1

02000

*

0*

*

0

0&,

.

2

22

2

2

1

2

212

Diffuser

(Air) (1) (2)

smY

mA

kPaP

CT

c

200

4.0

800

100

1

2

1

1

1

47

Or it can be solved as below:

02000

0

0&,

.

2

1

2

212

YYhh

keh

peWQ

pekehmWQ

From table (6) and at

kg

kJhKCT 7.373373100 11

kgkJh

h

7.393

02000

20007.373

2

22

2

Now, by returning back to table (4) and from the value of enthalpy we

can find the temperature of exit air (by interpolation):

CKT 79.1193922

Example (2):

Refrigerant-12 (R-12) enters and adiabatic nozzle at 700 (kPa), 100 C

and 20 m/s. The parameters at the exit of the nozzle are 300 (kPa) and

30 C . Determine: (1) the final velocity of R-12, and (2) the ratio ?2

1 A

A

Solution:

(1) For the first state at: CTkPaP 100&700 11 Superheated R-12

kg

kJh 33.2501 (Table 9-2)

Adiabatic

Nozzle

(R-12)

(1) (2)

smY

CT

kPaP

/20

100

700

1

1

1

CT

kPaP

30

300

2

2

48

For the second state at: CTkPaP 30&300 22 Superheated R-12

kg

kJh 12.2072 (Table 9-2)

s

mY

Y

YYhh

peWQ

peYY

hhmWQ

65.294

02000

2033.25012.207

02000

0&,

2000.

2

22

2

2

1

2

212

2

1

2

212

(2)

kg

mvkg

mv3

2

3

1 06573.0&03419.0 (Table 9-2)

6.7

06573.0*20

03419.0*65.294

*

*

**

2

1

21

12

2

1

2

22

1

11

21

c

c

c

c

cc

A

A

vY

vY

A

A

v

AY

v

AY

mm

49

Example (3): Air at 100 kPa and 280K is compressed steadily to 600 kPa

and 400K. The mass flow rate of the air is 0.02 kg/s and the heat loss of

16 (kJ/kg) occurs during the process. Assuming the changes in kinetic

and potential energies are neglected. Determine the necessary power

input to the compressor. Take

Kkg

kJCP .005.1

Solution:

kWW

wmW

kgkJqhhw

kgkJh

kgkJhtable

hhhwq

mhmWQ

pekehmWQ

or

kWW

wmW

kgkJw

w

TChwq

mhmWQ

pekehmWQ

KK

P

74.2

)85.136(*02.0*

85.136

98.400&13.280:)4(

:/.

.

:

732.2

)6.136(*02.0*

6.136

280400*005.116

.

:/.

.

12

400@2280@1

12

Compressor

skg

m 02.0

(1)

(2)

KT

kPaP

280

100

1

1

KT

kPaP

400

600

2

2

?W (Air)

50

Adiabatic

Turbine

(1)

(2)

mz

smY

MPaP

CT

Steam

10

/50

2

400

:

1

1

1

1

MWWout 5

mz

smY

x

kPaP

6

/180

9.0

15

2

2

2

2

Example (4): The power output of an adiabatic turbine is 5 MW and the

inlet conditions of the steam are 2MPa, 400 C , 50 m/s and elevation of

10 m. The exit conditions of the steam are 15 kPa, x2 = 90%, 180 m/s and

elevation of 6 m. Determine: (1) the magnitudes of pekeh &, , (2) the

work done per unit mass of the steam flowing through the turbine, and (3)

the mass flow rate of the steam.

Solution:

(1) At the inlet of turbine is superheated steam, then from table (7-2):

kg

kJh 6.32471

At the exit of turbine we have a saturated liquid-vapor mixture, then from

table (7-1b) and at

kgkJhhkPaP fgf 1.2373&94.22515 222

kgkJhxhh fgf 73.23611.2373*9.094.225. 2222

kgkJ

zzgpe

YYke

hhh

04.01000

106*8.9

1000.

95.142000

50180

2000

8.8856.324773.2361

12

222

1

2

2

12

51

kgkJw

w

adiabaticq

pekehwq

9.870

04.095.148.885

0

(3)

s

kg

kg

kJw

s

kJW

m 74.59.870

1000*5

Example (5): Refrigerant-12 (R-12) enters capillary tube (expansion

valve) of a refrigerator as a saturated liquid at 0.8MPa and is throttled to a

pressure of 0.12MPa. Determine the quality of refrigerant at the final

state and the temperature drop during this process.

Solution:

:NOTICE: Always and for capillary tube (expansion valve) مهم جدا

21 hh

hh exitinlet

At the inlet of capillary tube:

kg

kJhhCTT MPafMPasat 3.67&74.32 8.0@18.0@1

(Table 9-1b)

At the exit of capillary tube:

CTTT

h

hhx

kgkJhh

kgkJhhCTT

fg

f

MPafgMPaf

MPasat

48.5874.3274.25 :Drop eTemperatur

334.048.163

66.123.67

48.163&66.12

3.67&74.25

12

2

22

2

12.0@212.0@2

1212.0@2

52

Example (5): Consider a 15- kW electric heating system. Air enters the

heating section at 100 kPa and 17 C with a volume rate of 150 min

3m . If

heat is lost from the air in the duct to the surroundings at a rate of 200 W.

Determine the exit temperature of air. Take

Kkg

kJCP .005.1

Solution:

raturesexit tempe andinlet at enthapies find order toin (4) tableusingby Or

9.21

17*005.1*3151000

200

..

3832.0

5.2

832.0100

27317*287.0.

.....

.....

0&

.

2

2

12

3

3

3

1

11

111111

12

CT

T

TTCmWQ

skg

kg

mv

s

mV

m

kgm

P

TRv

TRvPTRmVP

TTCmTCmhmWQ

peke

pekehmWQ

P

PP

kW15

(1) (2)

WQ 200

s

mmV

33

5.2min

150

kPaP

CT

100

17

1

1

?2 T