chapter 4 thermo

5/26/2018 Chapter 4 Thermo

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65

Chapter 4

First Law of Thermodynamics

The first law of thermodynamics is simply the conservation of energy principle

and can be defined as, the energy can be neither created nor destroyed, it can only

change from one form to other with certain process such as, combustion, chemical and

mechanical. The conservation of energy principle may be expressed as follows: the net

change (increased or decreased) in the total energy of the system during a process is

equal to the difference between the total energy entering and the total energy leaving the

system.

=

systemtheofenergy

totalin theChange

systemtheleaving

energyTotal

systemtheentering

energyTotal

The above relation is often referred to as the energy balance. The energy is a property

and the value of a property does not change unless the state of the system changes. Also,

the first law of thermodynamics is called the law of the conservation of heat and work

undergoing a cycle, and then for a change of state of a system. Many measurements

were made during a cycle (control mass) for various amounts of work and heat, and the

results were compared. The results were always proportional and the observations led to

the formulation of the first law of thermodynamics as,

= WQ

The symbol, Q , which is called the cyclic integral of the heat transfer, represents the

net heat transfer during the cycle, and W , the cyclic integral of the work, represents

the net work during the cycle.


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The First Law of Thermodynamics of a Control Mass

We now considered the first law of thermodynamics for a control mass that undergoes a

change of state. We begin by introducing of the energy property, which is given by the

symbolE. Consider a system that undergoes a cycle, in which it changes from state 1 to

state 2 by process A, and returns from state 2 to state 1 by process B as shown in Fig. 4-

1. Considering the two separate processes, we have,

+=+1

2

2

1

1

2

2

1 BABA WWQQ

Now consider another cycle in which the control mass change from state 1 to state 2 by

process C and returns to state 1 by process B, as before. For this cycle we can write,

+=+1

2

2

1

1

2

2

1 BcBC WWQQ

Subtracting the two equations, we get,

=2

1

2

1

2

1

2

1 CACA WWQQ

or, by rearranging,

=2

1

2

1)()( CA WQWQ

Fig. 4-1 Control mass undergoing a cycle

Since A and C represents arbitrary processes between state 1 and 2, the quantity

WQ is the same for all processes between state 1 and state 2. Therefore,

WQ depends only on the initial and final states and not on the path. We conclude


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67

that this is a point function and this property is the energy of the mass and is given by

the symbolE. Thus we can write,

WdEQ

dEWQ

+=

= or

BecauseEis a property, its derivative is written dE. The above equation is integrated

from an initial state 1 to a final state 2, we have,

121212 WEEQ +=

The determination of the energy change of a system during a process involves the

evaluation of the energy of the system at the beginning and at the end of the process as

follows,

12 EEEEE initialfinalsystem ==

The energy can be exist in various forms such as internal (sensible, latent, chemical, and

nuclear), kinetic, potential, electric, and magnetic, and their sum is the total energyEof

a system. In the absence of electric, magnetic and surface tension effects ( i.e., for

simple compressible systems), the change in total energy of a system during a process is

the sum of the changes in its internal, kinetic, and potential energies and can be

expressed as,

PEKEUE ++=

Where,

)(

)vv(2

1

)(

12

2

1

2

2

12

zzmgPE

mKE

uumU

=

=

=

or

++= )()vv(

2

1)( 12

2

1

2

212 zzguumE

Most of engineering systems are stationary and no change of its velocity and elevation

during process, it means that, 0== PEKE , and the total energy change reduces to

UE = . The energy can be transferred to or from the system in three forms: Heat,


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WorkandMass Flow. For a closed system or fixed mass, the only two forms of energy

are the heat and work interaction.

Heat Transfer, Q

The heat transferred to a system (heat gain) increases the internal energy of the

system, and the heat transferred from a system (heat loss) decreases the internal energy

of a system.

Work, W

The energy interaction that is not caused by a temperature difference between a

system and surroundings is called work. A rising piston, rotating shaft, and electrical

wire crossing the system boundaries are work done.

Energy Balance of Closed System

Noting that the energy can be transferred in the forms of heat and work and the

net transfer of a quantity is equal to the difference between the amounts transferred inlet

and outlet, and the energy balance can be defined as,

systemoutinoutin EWWQQ = )()(

The first law of thermodynamics for closed system and fixed boundary (rigid tank) as

shown in Fig. 4-2, and the kinetic and potential energies are negligible, 0== PEKE ,

then the thermodynamics first law relation becomes,

21 muWQmuWQ ooii ++=++

or )( 12 uumWQ =


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Fig. 4-2 Closed system and fixed boundary (stationary).

For no interaction work done and heating process, the first law takes,

)( 12 uumQ =

For cooling process, the first law takes,

)( 12 uumQ =

The first law of thermodynamics for closed system and moving boundary as shown in

Fig. 4-3, and the kinetic and potential energies are negligible, 0== PEKE , then the

thermodynamics first law relation becomes,

)()( 2211

2211

PvumWQPvumWQ

PVmuWQPVmuWQ

ooii

ooii

+++=++++++=+++

Fig. 4-3 Closed system and moving boundary (frictionless piston-cylinder).

or )( 12 hhmWQ =

For no interaction work done and heating process, the first law takes,

)( 12 hhmQ =

For cooling process, the first law takes,


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70

)( 12 hhmQ =

Cooling process with work done from paddle wheelA rigid tank contains a hot fluid that is being stirred by a paddle wheel as shown in Fig.

4-4. The tank is not insulated and heat transferred to surrounding, the kinetic and

potential energies are negligible, 0== PEKE , then the first law becomes,

)( 12 uumWQ =

Applying the sign role or the directions of heat and work in the system, the first law

takes the following form,

)(

)()(

12

12

uumQW

uumWQ

=

=

Or applying the sum of energy at inlet that is equal the sum of energy at outlet,

21 muWQmuWQEE ooiioi ++=++=

There is no heat transfer inter the system, Qi=0, and no work done leave the system,

Wo=0,.

21 muQmuW oi +=+ or the form of, )( 12 uumQW outinpw =

or )( 12 uumQW =

Fig. 4-4 Paddle wheel stirred fluid in closed system


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71

Electric Heating of Fluid in Insulated Rigid Tank

A rigid tank contains fluid that is being heat by an electric heater as shown in Fig.

4-5. The tank is insulated and no heat transferred to surrounding, the kinetic and

potential energies are negligible, 0== PEKE , then the thermodynamics first law

becomes,

)( 12 uumWQ =

Applying the sign role or the directions of heat and work in the system, the first law

takes the following form,

)(

)(

)()(

12

12

12

uumtIV

uumW

uumW

e

e

=

=

=

Where, Vis the volt,Iis the current, and tis the heating time.

Fig. 4-5 Electric heating process

Energy Balance of Steady-flow System

A large number of engineering devices such as turbine, compressor, and nozzles

are operating for long period of time under the same conditions, (steady-flow process).

During the steady-flow process, no intensive or extensive properties within the control

volume change with time. Thus, the volume V, the mass m, and the total energy content

Eof the control volume remain constant as shown in Fig. 4-6.


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72

Fig. 4-6 Steady-flow process and control volume

As a result, the boundary work is zero for steady flow system (because the volume is

constant in control volume Vcv= constant.), and the sum of mass or energy entering the

control volume must be equal to the sum of mass or energy leaving the control volume

(since mcv= constant and Ecv=constant). Then, the rate of the general energy and mass

balance for steady flow process as follows,

and, outinoutin mmEE &&&& ==

or 0and0 == CVCV mE

During the steady flow process, the fluid properties at inlet or exit remain constant.

Also, the rate of heat and work interaction between a steady flow system and its

surrounding do not change with time. For steady flow process, there are some

applications has multi-inlet and multi-exit as shown in Fig. 4-7.

Fig. 4-7 Steady flow process in general

So, the first law of thermodynamics for steady flow process becomes for that case,

)2

v()

2

v(

22

o

o

ooooi

i

iiii gzhmWQgzhmWQ ++++=++++ &&&&&&


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73

For single stream flow, one inlet and one exit as shown in Fig. 4-8, the first low of

thermodynamics becomes,

Fig. 4-8 Steady flow process for one inlet and one exit

)2

v()

2

v(

22

o

o

ooooi

i

iiii gzhmWQgzhmWQ ++++=++++ &&&&&&

And, )2

v()

2

v(

22

o

o

oooi

i

iii gzhmWQgzhmWQ ++++=++++ &&&&&&

In such cases, it is common practice to assume that the heat to be transferred into

the system (heat input) at a rate ofQ& , and work produced by the system (work output) at

a rate of W& and the inlet and outlet are denoted by subscripts 1 and 2. The first law of

thermodynamics for single stream steady flow systems becomes,

)](2

vv)[(

)]2

v()

2

v[(

12

2

1

2

212

1

2

112

2

22

zzghhmWQ

gzhgzhmWQ

+

+=

++++=

&&&

&&&

For specific quantity, the energy balance on a unit mass basis defined as,

)(2

vv)( 12

2

1

2

212 zzghhwq +

+=

Some Steady Flow Applications

Many engineering devices operate under the same conditions for long periods of

time and never change throughout the operating time such as, turbines, compressors,

pumps, boilers, heat exchangers, nozzles and diffusers and these devices can be

analyzed as steady flow processes.


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74

Turbine

A turbine is a rotary machine which used in thermal power station as shown in Fig. 4-9,

whose purpose is the production of shaft work by expansion the fluid. The work done in

the turbine is positive since it done by the fluid on the moving blades to causing

rotational velocity of the turbine shaft.

Fig. 4-9 Adiabatic turbine process

The energy balance ofo this steady-flow system is,

)](

2

vv)[( 12

2

1

2

212 zzghhmWQ +

+= &&&

If the process occurred in the turbine which insulated fro surrounding (adiabatic process,

0=Q& ) we get,

kWzzghhmWPower )](2

vv)[( 12

2

1

2

212 +

+== &

For some cases, by neglecting the kinetic and potential energy, we obtain,

kWhhmWPower )( 12 == &

Compressors and Pumps

The purpose of compressors (gas) or pumps (liquid) is the same, to increase the pressure

of a fluid by adding shaft work as shown in Fig. 4-10.


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Fig. 4-10 Compressor and pump process

For gas compressors, by neglecting the kinetic energy at inlet only and potential energy

at inlet and outlet, and some heat transferred to the surrounding, we obtain,

]v2

1)[(

or]v2

1)[()(

2

212

2

212

kWhhmQWpower

hhmWQ

outin ++==

+=

&&&

&&&

For ideal gas, and the compression is adiabatic process, and neglecting the kinetic

energy, we get

kWTTCmpower

kWhhmWpower

aveP

in

)(

)(

12

12

=

==

&

&&

For pumps, by neglecting the kinetic and potential energy,

)( 12 kWhhmWpower in == &&

For pumps, the increasing pressure is usually occurred at constant temperature, and the

liquid temperature does not change throughout the process. So, the change only in the

pressure and the steady flow energy equation for pumps can be write as,

)( 12 kWPPvmWpower in == &&

Nozzles and Diffusers

Nozzles and diffusers are usually utilized in jet engine, rockets and air craft. The nozzle

is a device that increases the velocity of a fluid. A diffuser is a device that increases the


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76

pressure of the fluid. That is, nozzles and diffuser perform opposite tasks as shown in

Fig. 4-11.

Fig. 4-11 Nozzle and diffuser

The energy balance for this steady flow system which no heat transferred, no work and

potential energy is zero, we get for nozzle,

2

v

0inlet vatvelocityfluid2

vv)(0

2

2

12

1

2

1

2

212

=

+=

hh

hh Q

For diffuser,

2

v

0outlet vatvelocityfluid2

vv)(0

2

112

2

2

1

2

212

+=

+=

hh

hh Q

Heat Exchanger

The heat exchangers are usually used for cooling of a hot fluid inside the coil by using

cooled fluid outside the coil as Fig. 4-12. The energy balance of heat exchanger, the

steady flow system which no heat transferred to the surrounding, no work done, and

kinetic and potential energies are negligible. The heat rejected from hot fluid, 1m& equal

to the heat added to the cooled fluid 2m& , then we get,

)()( 342121 hhmhhm = &&


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77

Fig. 4-12 Heat exchanger process

Throttling ProcessA throttling process occurs when a fluid flowing in a line suddenly encounters a

restriction in the flow passage such as, valves and capillary tubes as shown in Fig. 4-13.

Fig. 4-13 Throttling process in valves and capillary tubes

For throttling process, the steady flow system which no heat transferred, no work done,

and kinetic and potential energies are negligible, we get,

21 hh =

Usually the throttling process is called constant enthalpy process.

Mixing Process

Fig. 4-14 Mixing chamber process

In engineering applications, mixing two streams of fluid in a certain system are occurred

without heat transferred to surrounding, no work done, and kinetic and potential


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78

energies are negligible as shown in Fig. 4-14. The energy balance of steady flow

process as,

321332211 balancemassforand mmmhmhmhm &&&&&& =+=+

Examples of First Law of Thermodynamics

1. Is it possible to compress an ideal gas isothermally in an adiabatic piston-cylinderdevice? Explain.

Impossible to compress an ideal gas isothermally in an adiabatic piston-cylinder

device because during the compression process which to keep the temperature

constant, it is necessary to remove heat from the system boundary, heat is interaction

out the system.

2. Consider two identical rooms, one with a refrigerator in it and the refrigerator door isopen, and the other without one. If all the doors and windows are closed, will the

room that contains the refrigerator be cooler or warmer than the other room? Why?

The room which contain refrigerator is becoming warmer than the other, because

electric work done for the compressor is added to room, work is interaction into the

system.

3. A 2.5 m3rigid tank contains air at 300 kPa and 150 oC. The air is now cooled until itstemperature drops to 40

oC. Determine (a) the final pressure in the tank and (b) the

amount of heat transfer.

Solution


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79

kPaT

TPP

T

T

P

P

P

mRT

P

mRT

CV

kgRTVPmmRTVP

99.221423

313300

or,olume,constant vis21Process

18.6)423287.0/(5.2300/,1,State

1

212

2

1

2

1

2

2

1

1

111111

===

====

====

kJuumQ

kgkJTCukgkJTCu

uumUQ

W,

V

V

096.488)714.303734.224(18.6)(

/734.224313718.0/714.303423718.0

airgas,-IdealFor

)(

0,0P.E0K.E

boundary,fixedandvolumeclosedformicsThermodynaofLawFirst

12

22

11

12

===

======

==

===

4. A mass of 20 kg of air in a frictionless piston-cylinder device is heated from 27 to 80oC by passing current through a resistance heater inside the cylinder. The pressure

inside the cylinder is held constant at 400 kPa during the process, and a heat loss of

30 kJ occurs. Determine the electric energy supplied in kWh.

Solution

)(

0,

0P.E0,K.EAssume,

boundary,movingandsystemclosedformicsThermodynaofLawFirst

12

201

hhmQQ

WWhmWQhmWQ

oi

oioii

+=

==++=++

==

kgkJTCh P /5.301300005.1

airgas,IdealFor

11 ===

kJQ

kgkJTCh

i

P

3.1095)5.301765.354(2030

/765.354353005.122

=+=

===


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80

kWhskWQW ie 3043.03600

3.1095.3.1095Work,Electric ====

5. Water is being heated in a closed vessel on top of a range while being stirred by apaddle wheel. During the process, 3000 kJ of heat transferred to the water, and 150

kJ of heat is lost to the surrounding air. The paddle-wheel work amounts to 250 N.m,

determine the final energy of the system if its initial energy is 200 kJ.

Solution

21

201

Energy,Final,Energy,Initial0,,

0P.E0,K.Eboundary,Fixedsystem,closedformicsThermodynaofLawFirst

umEumEWWWumWQumWQ

fi

opwioii

====++=++

==

Q

kJE

QumWQumE

f

oiif

25.30501502001000

2503000

Energy,Final 12

=++=

++==

6. A classroom that normally contains 60 peoples which one person at rest may beassumed to dissipate heat at a rate of 432 kJ/h. There are 15 light bulbs in the room,

each with a rating of 150 W. The rate of heat transfer to the classroom through the

walls and the windows is estimated to be 15000 kJ/h. If the room air is to be

maintained at a constant temperature of 22oC and the supply fresh air at 35

oC with

volume flow rate of 10 L/s for one person is provided .Determine the number of

window air conditioning units required if the unit cooling capacity is about of 8 kW.

Solution


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81

3/13.1)27335(287.0

100Denisty,Air

)(capacity,Cooling

rate,flowmassSteady

0k,output wororinputNo,

0P.E0,K.EAssume

volume.controlandflowsteadymics,ThermodynaoflawFirst

mkgRT

P

hhmQQ

mmm

WWhmWQhmWQ

a

oiaio

aoi

oiooooiiii

=+

==

+===

==++=++

==

&

&&&

&&

Units381.2

8

475.22

CapacityUnit

UnitsofNumber

475.22/475.223600

052.80909

/052.809093600)475.29654.309(678.049020

/475.296295005.1

/54.309308005.1

airgas,IdealFor

/49020

4326036001000

1501515000

/678.010601013.1

22

11

3

===

===

=+=

===

===

=

++=++=

===

o

o

o

P

P

i

personslightwalli

aa

Q

kWskJQ

hkJQ

kgkJTCh

kgkJTCh

hkJQ

QQQQ

skgVm &&

7. The steam radiator for a heating system has a volume of 20 L and is filled withsuperheated vapor at 300 kPa and 250

oC. At this moment both inlet and exit valves

to the radiator are closed. Determine the amount of heat that will be transferred to the

room when the steam pressure drops to 100 kPa. Also, show the process on a P-v

diagram with respect to saturation line.

Solution


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82

wet vapor2,statevvv

7964.0vv,100at,table,steamfrom2,State

olume,constant vatoccuredprocessCooling

0251.00.9764

1020

vmass,Steam

07964v72728

250,300at,vapordsuperheatetable,steamfrom1,State

2

211

21

3-

1

3

11

11


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83

kgmx

CVV

kgkJ

xkgm

x

xandkPaP

/30747.0vor,water vapsaturated,%100

olume,constant vatoccuredprocessHeating

/415.7957.2088181.036.417u

uuu/30747.0)001043.0694.1(181.0001043.0v

)vv(vv

181.0,100at,tablesteamfrom1,State

3

22

21

2

fg1f1

3

1

fg1f1

11

==

==

=+=

+==+=

+=

==

CT

kgkJu

kgkJkgmCT

kgm

o

o

2

2

g

3

g

2

3

g2

160

/4.2568

2568.40.30747601

)/(u)/(v)(

%100x,/30747.0vvat,table,steamsaturatedfrom2,State

=

=

===

hrWWt

tWW

skJkWWVIPowerW

kJuumW

uumUWQ

WQ

ee

ee

e

e

e

e

48.4606088.0

88.14183/time,Heating

/

/88.088.08804220

88.14183)415.7954.2568(8)(

)(

work,electric,0boundary,fixedsystem,closedforamicsthermodynoflawFirst

12

12

=

==

=

======

===

==

=

&

&

&

9. A piston-cylinder device contains 1.2 kg of N2 initially at 100 kPa and 27 oC. TheNitrogen is now compressed slowly in a polytropic process during which PV

1.3= C

until the volume is reduced by one-half. Determine the work done and the heat

transfer for this process.


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84

Solution

kPaV

V

V

VPP

VPVPCPV

mVVkgkJTCu

mV

VmRTVP

KTkPaP

n

nnn

V

23.2465.0

100

,Polytropic2,1Process

53424.00685.15.05.02,State/9.222300743.0

Nofgasidealfor,0685.1

3002968.02.1100,equation,gasidealFrom

300,1001,State

3.1

1

1

2

112

2211

3

12

11

2

3

1

1111

11

=

=

=

==

======

=

==

==

KT

mR

VPTmRTVP

35.369

2968.02.1

53424.023.246,equation,gasidealFrom

2

222222

=

===

kJn

VPVPW

kgkJTCu V

32.823.11

0685.110053424.023.246

1

/427.27435.369743.0

112221

22

=

=

=

===

( ) kJuumWQ

umQumW

umWQumWQ

WQWWW

io

oi

ooii

iib

15.144424.2749.222(2.132.82

00

0,0,ork,boundary w,

0P.E0,K.Eboundary,movingvolume,closedamics,thermodynoflawFirst

21

21

21

021

=+=+=

++=++

++=++

====

==

10.Air enters an adiabatic nozzle steadily at 300 kPa, 200 oC, and 30 m/s and leaves at100 kPa and 180 m/s. The inlet area of the nozzle is 80 cm

2. Determine (a) the mass

flow rate through the nozzle, (b) the exit temperature of the air, and (c) the exit area

of the nozzle.

Solution


21/30

85

222222

22

2

2

2

22

22

2

11

11

4

111

3

1

11

111

/,rate,flowMass

99.460,/5.459atairofpropertiesgasidealFrom

/5.459

10002

180

10002

3025.475,

22

nozzle,throughflowsteadyforamicsthermodynoflawfirstFrom

/24.475,473atairofpropertiesgasidealFrom

/5304.0301080209.2rate,flowMass

/209.2473287.0

300

equation,gasidealfrom1,State

VmAVAm

KTkkJh

kgkJh

hV

hV

h

kgkJhKT

skgVAm

mkgRT

P

mRTVP

&&

&

==

==

=

+=

++=+

==

===

=

==

=

223

2

3

2

22

98.3810898.3)1807558.0/(5304.0

/7558.099.460287.0

100

cmmA

mkg

RT

P

===

=

==

11.Steam at 5 MPa and 500 oC enters a nozzle steadily with a velocity of 80 m/s, and itleaves at 2 MPa and 400

oC. The inlet area of the nozzle is 50 cm

2, and heat is being

lost at a rate of 90 kJ/s. Determine (a) the mass flow rate of the steam, (b) the exit

velocity of the steam, and the exit area of the nozzle.

Solution


22/30

86

223

2

222222

2

2

2

22

22

2

11

4

1

11111

3

22

22

3

11

11

953.14104953.1868.589/1512.08335.5

/v,rate,flowMass/868.589

8335.5

90

100026.3247

10002

808.3433,

22

nozzle,throughflowsteadyforamicsthermodynoflawFirst

/8335.506857.0

801050

vrate,flowMass

/1512.0v,/6.3247

400,2at,dsuperheatetable,steamfrom2,State

/06857.0v,/8.3433


cmmA

VmAVAmsmV

VQ

Vh

Vh

skgVA

VAm

kgmkgkJh

CTMPaP

kgmkgkJh

CTMPaP

o

o

o

===

===

+

+=

+++=+

=

===

==

==

==

==

&&

&

12.Air at 80 kPa and 127 oC enters an adiabatic diffuser steadily at a rate of 6000 kg/hand leaves at 100 kPa. The velocity of the air stream is decreased from 230 to 30 m/s

as it passes through the diffuser. Find (a) the exit temperature of the air and (b) the

exit area of the diffuser.

Solution

kgkJh

hV

hV

h

kgkJhKT

/98.426

10002

30

10002

23098.400,

22

nozzle,throughflowsteadyforamicsthermodynoflawfirstFrom

/98.400,400atairofpropertiesgasidealfrom1,State

2

2

2

22

22

2

11

11

=

+=

++=+

==


23/30

87

2

222

3

2

22

222111

22

0679.0)308185.0/()3600/6000(/

/8185.07.425287.0

100

rate,flowMass

7.425,/98.426atairofpropertiesgasidealFrom

mVmA

mkgRT

P

VAVAm

KTkgkJh

===

=

==

==

==

&

&

13.Steam flows steadily through an adiabatic turbine. The inlet conditions of the steamare 10 MPa, 450

oC, and 80 m/s, and the exit conditions are 10 kPa, 90 percent

quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the

change in kinetic energy, (b) the power output, and (c) the turbine inlet area.Solution

skJVV

mEK

kgkJhhxhh

xkPaP

kgmkgkJh

CTMPaP

fgf

o

/4.2310002

80

10002

5012

22.

/35.23458.23929.083.191)(

%90,10atwet vaportable,steamfrom2,State

/02975.0v,/9.3240


222

1

2

2

22

22

3

11

11

=

=

=

=+=+=

==

==

==

&

( ) ( )

( )

223

1

111111

22

12

21

22

12

63.44104625.480/02975.012

/v,v/rate,flowMass

77.1010770)200

80509.324035.2345(12

0.,0,2

cmmA

VmAVAm

MWkWW

EPQzzgVV

hhmWQ

===

==

==

+=

==

++=

&&

&

&&&&

14.Refrigerant, R-134a enters an adiabatic compressor as saturated vapor at -20 oC andleaves at 0.7 MPa and 70 oC. The mass flow rate of the refrigerant is 1.5 kg/s.


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88

Determine (a) the power input to the compressor and (b) the volume flow rate of the

refrigerant at the compressor inlet.

Solution

kgmkgkJh

CTx o

/1464.0v,/31.235

20%100at,tablesteamfrom1,State

3

11

11

==

==

,CompressoradiabaticthroughflowsteadyforamicsthermodynoflawFirst

/01.307

70,7.0atwet vaportable,steamfrom2,State

2

22

kgkJh

CTMPaP o

=

==

( ) ( )

( )

kWW

kWhhmW

EPK.EQzzgVVhhmWQ

55.107PowerInput

55.107)01.30731.235(5.1

0.0,0,2

12

12

2

1

2

212

==

===

===

++=

&

&&

&&&&

15.Air enters the compressor of a gas-turbine plant at ambient conditions of 100 kPaand 27

oC with a low velocity and exit at 1 MPa and 350

oC with a velocity of 80

m/s. The compressor is cooled at a rate of 1200 kJ/min, and the power input to the

compressor is 250 kW. Determine the mass flow rate of the air through the

compressor.

Solution


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89

,CompressorthroughflowsteadyforamicsthermodynoflawFirst

/11.631

623,1at,airofpropertiesgasidealfrom2,State

/19.300

300,100at,airofpropertiesgasidealfrom1,State

2

22

1

11

kgkJh

KTMPaP

kgkJh

KTkPaP

===

=

==

( ) ( )

( )

+=

=

+

+=

2

0.,2

2

212

12

2

1

2

212

VhhmWQ

EPzzgVV

hhmWQ

&&&

&&&

skgm

m

/6884.0

10002

80)19.30011.631()250(

60

1200 2

=

+=

&

&

16.A superheated water vapor is throttled from 8 MPa and 500 oC to 6 MPa. Determinethe final temperature of steam.

Solution

KT

kgkJhMPaP

MPaPkgkJh

hh

kgkJh

CTMPaP o

08.490

/3.3398,6at,levapor tabdsuperheateFrom

6,/3.3398at2,State

enthalpy,constantatoccured21processThrottling

/3.3398

500,8at,vapordsuperheate1,State

2

22

22

21

1

11

=

==

==

=

=

==


26/30

90

17.Liquid water at 300 kPa and 20 oC is heated in a chamber by mixing withsuperheated steam at 300 kPa and 300

oC. Cold water enters the chamber at a rate of

1.8 kg/s. If the mixture leaves the mixing chamber at 60oC, determine the mass flow

rate of the superheated steam required.

Solution

CTkPaP

kgkJhh

CT

hmmhmhmhm

mmm

o

f

o

300,300atvapordsuperheate2,State

/96.83le,water tabsaturatedFrom

20at,watersubcooled1,State

)(balance,Heat

balance,Mass

process.mixingstatesteadyFor

22

1

1

321332211

321

==

==

=

+==+

=+

&&&&&

&&&

skgm

mm

kgkJhh

CT

kgkJh

f

o

/1073.0

93.251)8.1(3.306996.838.1

/93.251le,water tabsaturatedFrom

60at,watersubcooled3,State

/3.3069le,vapor tabdsuperheateFrom

2

22

3

3

2

=

+=+

==

=

=

&

&&

18.Refrigerant, R-134a at 1 MPa and 80 oC is to be cooled to 1 MPA and 30 oC in acondenser by air. The air enters at 100 kPa and 27 oC with a volume flow rate of 1.5

m3/s and leaves at 95 kPa and 60

oC. Determine the mass flow rate of the refrigerant.

Solution


27/30

91

kJ/kgTCh

kgkJCThh

MPaTCT

CTMPaP

kgkJh

CTMPa,P

P

o

f

sat

o

o

sat

o

5.301300005.1

air,ofgasidealfor3,State

side,Air

/49.9130at

liquidsubcooledis2,state,1at30

39.39,1at,tablesaturatedfrom2,ateSt

/2.313

801at,levapor tabdsuperheatefrom1,State

,side134a-R

33

22

2

2

1

11

===

===


28/30

92

Problems

1- Consider two identical rooms, one with a refrigerator in it and the other without one.

If all the doors and windows are closed, will the room that contains the refrigerator

be cooler or warmer than the other room? Why?

2- A fixed quantity of fluid is cooled and simultaneously compressed. If the heat

rejected is 50 kJ and the work done in compression is 34 kJ, determine the change in

internal energy.

3- A mass of 20 kg of air in a frictionless piston-cylinder device is heated from 27 to 80

C by passing current through a resistance heater inside the cylinder. The pressure

inside the cylinder is held constant at 400 kPa during the process, and a heat loss of

30 kJ occurs. Determine the electric energy supplied in kWh.

4- A piston-cylinder device contains 1.2 kg of N2 initially at 100 kPa and 27 C. The

Nitrogen is now compressed slowly in a polytropic process during which PV1.3

= C

until the volume is reduced by one-half. Determine the work done and the heat

transfer for this process.

5- 1 kg of air can be expanded between two states as it does 20 kJ of work and receives

16 kJ of heat. A second kind of expansion can be found between the same initial and

final states which require a heat input of only 9 kJ. What is the change of internal

energy in the first expansion and what is the work done by the air in the second

expansion?

6- The steam radiator for a heating system has a volume of 20 L and is filled withsuperheated vapor at 300 kPa and 250 C. At this moment both inlet and exit valves

to the radiator are closed. Determine the amount of heat that will be transferred to the

room when the steam pressure drops to 100 kPa. Also, show the process on a P-v

diagram with respect to saturation line.

7- A well-insulated rigid tank initially contains 8 kg of a saturated liquid-vapor mixture

of water at 100 kPa. Initially, three-quarters of the mass in the liquid phase. An


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93

electric resistor placed in the tank is connected to a source of 220 V and a current of

4 A flows through the resistor when the switch is turned on. Determine how long it

will take to vaporized all the liquid in the tank. Also, show the process on a t-v

diagram with respect to saturation lines.

8- A piston-cylinder device initially contains 0.5 m3of saturated water vapor at 200 kPa.

At this state, the piston is resting on a set of stops, and the mass of the piston is such

that a pressure of 300 kPa is required to move it. Heat is now slowly transferred to

the steam until the volume doubles. Show the process on a P-v diagram with respect

to saturation lines and determine (a) the final temperature, (b) the work done during

this process, and the total heat transfer.

9- A system receives 180 kJ by heat transfer at constant volume. Next it rejects 200 kJ

by heat transfer as it receives 50 kJ of work at constant pressure.

a- If an adiabatic process can be found which will restore it to the initial state, how

much work transfer is done during the process.

b-Take the internal energy in the initial state as zero find the corresponding

internal energy at the other two states.

10- 1 kg of a certain fluid is contained in a cylinder at an initial pressure of 20 bar. The

fluid expands according to the law PV2= constant until the volume is doubled. The

fluid is then cooled at constant pressure until the piston regains its original position;

heat is added at constant volume until the pressure rises to the original value of 20

bar. Calculate the network done, for an initial volume of 0.05 m3.

11- In the turbine of a gas turbine unit the gases flow through the turbine is 17 kg/s and

the power developed by the turbine is 14 MW. The enthalpies of the gases at inlet

and outlet are 1200 kJ/kg and 360 kJ/kg respectively, and the velocities of the gases

at inlet and outlet are 60 m/s and 150 m/s, respectively. Calculate the rate of heat

transfer.


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94

12- Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam

are 10 MPa, 450 C, and 80 m/s, and the exit conditions are 10 kPa, 90 percent

quality, and 50 m/s. The mass flow rate of the steam is 12 kg/s. Determine (a) the

change in kinetic energy, (b) the power output, and (c) the turbine inlet area.

13- Air enters the compressor of a gas-turbine plant at ambient conditions of 100 kPa

and 27 C with a low velocity and exit at 1 MPa and 350 C with a velocity of 80

m/s. The compressor is cooled at a rate of 1200 kJ/min, and the power input to the

compressor is 250 kW. Determine the mass flow rate of the air through the

compressor.

14- Steam at 5 MPa and 500 C enters a nozzle steadily with a velocity of 80 m/s, and it

leaves at 2 MPa and 400 C. The inlet area of the nozzle is 50 cm2, and heat is

being lost at a rate of 90 kJ/s. Determine (a) the mass flow rate of the steam, (b) the

exit velocity of the steam, and the exit area of the nozzle.

15- Air at 80 kPa and 127 C enters an adiabatic diffuser steadily at a rate of 6000 kg/h

and leaves at 100 kPa. The velocity of the air stream is decreased from 230 to 30

m/s as it passes through the diffuser. Find

a- The exit temperature of the air and

b- The exit area of the diffuser.

chapter 4 thermo

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