thermo i chapter 3

Website: http://www.guider55.4t.com Thermodynamics I

40

Chapter 3 Heat and Work

Energy can be transferred to or from a closed system in two forms of heat and

work. It is important to distinguish between these two forms of energy. An energy

transfer to or from a closed system is heat, if it is caused by temperature difference

between the system and its surroundings. Otherwise it is work, and it is caused by force

acting through a distance.

Heat Transfer In thermodynamics, the term of heat simply means heat transfer. Also, the

transfer of heat into a system is referred to as heat addition and the transfer of heat out

of a system as heat rejection. A process during which there is no heat transfer is called

an adiabatic process. The word adiabatic means not to be passed and the system is well

insulated so that only a negligible amount of heat can pass through the boundary as

shown in Fig. 3-1.

Fig. 3-1 An adiabatic process, a system exchange no heat with its surroundings

There are two ways a process can be adiabatic, the system is well insulated so that only

a negligible amount of heat can pass through the boundary, or both the system and

surrounding are at the same temperature. An adiabatic process should not be confused

with an isothermal process (in isothermal process the temperature still constant during

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41

the process). Even though there is no heat transfer during an adiabatic process, the

energy content and the temperature of a system can still be changed by other means

such as work. As a form of energy, heat has energy units, kJ or Btu. The amount of heat

transferred during the process between two stats is denoted by QQ or 12 . Heat transfer per

unit mass of a system is denoted as,

)/( kgkJmQq =

The heat transfer rate Q& has the unit kJ/s, which is equivalent to kW. When the amount

of heat transfer during a process is determined by integrating Q& over the time interval of

the process,

)( 2

1kJdtQQ ∫= &

When Q& remains constant during a process, the relation above reduces to

tQQ ∆= &

Where 12 ttt −=∆ is the time interval during the process.

Energy Transfer by Work Work like heat, is an energy interaction between a system and its surroundings.

As mentioned above, energy can cross the boundary of a closed system in the form of

heat or work. Therefore, if the energy crossing the boundary of a closed system is not

heat, it must be work. More specifically, work is the energy transfer associated with

force acting through a distance, such as rising piston, a rotating shaft and an electric

wire crossing the system boundaries.

The work done during a process between two states is donated by 12W or simply W in kJ.

The work done per unit mass of a system is denoted by w and is expressed as

)/( kgkJmWw =



42

Heat and work are directional quantities, and the complete description of a heat or work

interaction requires the specification of both magnitude and direction. The generally

accepted formal sign convection for heat and work is as the follows: heat transfer to a

system and work done by a system are positive sign and, the heat transfer from a system

and work done on a system are negative sign as Fig. 3-2.

Fig. 3-2 Heat and work are directional quantities

Both heat and work are path function as shown in Fig. 3-3.

Fig. 3-3 Properties are point function, but heat and work are path function

The total work done during process 1-2 as follows,

12

2

1 12 not WWWWw −=∆=∫ δ

The total work is obtained by the following the process path and adding the differential

amount of work done ( Wδ ) along the path. The integral of Wδ is not ( 12 WW − ), which is

meaningless since the work is not a property.



43

Mechanical Forms of Work There are several ways of doing work, each in some way related to a force acting

through a distance. In elementary mechanics, the work done by a constant force on a

body displaced a distance s in the direction of the force is given by

) ( kJsFW =

If the force is not constant, the work done is obtained by adding the differential amounts

of work

)( 2

1kJdsFW ∫=

In many thermodynamics problems, mechanical work is only form of work involved. It

is associated with the movement of the boundary of a system or with the movement of

entire system as a whole. Some common forms of mechanical work are discussed

below.

Moving Boundary Work

Consider the gas enclosed in the piston-cylinder device as shown in Fig. 3-4. The

initial pressure of the gas is P, the total volume is V, and the cross-sectional area of the

piston is A. If the piston is allowed to move a distance ds in a quasi-equilibrium manner,

the differential work done during this process is,

Fig. 3-4 Piston cylinder device to represent boundary work

dVPdsPAdsFWb ===δ

This expression also explains why the moving boundary work is sometimes called the

dVP work. The total boundary work done, during the process as piston moves is



44

obtained by adding all the differential works from the initial state 1 to final state 2 as

shown in Fig. 3-5.

Fig. 3-5 The area under the curve represents the boundary work.

)( 2

1kJdVPWb ∫=

The quasi-equilibrium expansion process described above is shown on P-V diagram and

the differential area dA is equal to PdV. The total area A under the process curve 1-2 is

obtained by adding the differential areas.

Area2

1

2

1dVPdAA ∫∫ ===

In actual processes of car engine, the boundary work done Wb is used to overcome

friction between the piston and the cylinder, to push atmospheric air out of the way, and

to rotate the crank shaft.

)(2

1dxFAPFWWWW crankatmfrictioncrankatmfrictionb ∫ ++=++=

Boundary Work During Constant Volume Process

The constant volume heat addition process of air as ideal gas inside a rigid tank is

shown in Fig. 3-6. The pressure and temperature are increased but the volume kept

constant and the following relations obtained.



45

Fig. 3-6 Constant volume process

2

2

1

1

21222111 0or and 2 State , 1 State

TP

TP

dVVVmRTVPmRTVP

=∴

====

The boundary work as follows,

02

112 === ∫ PdVWWb

Boundary Work During a Constant Pressure Process

Fig. 3-7 Constant pressure process

A frictionless piston-cylinder device contains air as ideal gas and heat is transferred

through boundary. The pressure inside the cylinder kept constant and the following

relations are obtained.



46

2

2

1

1

21222111 0or and 2 State , 1 State

TV

TV

dPPPmRTVPmRTVP

=∴

====


)( 12

2

1

2

112 VVPPdVPPdVWWb −==== ∫∫

Boundary Work During a Constant Temperature Process

Fig. 3-7 Constant temperature process

A frictionless piston-cylinder device contains air as ideal gas, and the compression

process is quasi-equilibrium. The temperature inside the cylinder kept constant by

releasing heat to surrounding and the following relations are obtained.

CVPVP

CmRTPVTTmRTVPmRTVP

==∴===

==

2211

21

222111

or and 2 State , 1 State


1

211

1

22

1

2

1

2

1

2

112 lnlnln1VV

VPVV

CVCdVV

CdVVCPdVWWb ======= ∫∫∫

In the above relation, mRTVPVP or by replaced becan 2211 ,

Also, 2112 /by replaced becan / PPVV .



47

Polytropic Process

Fig. 3-8 P-V diagram for polytropic process

During actual expansion and compression processes of gases, pressure and volume are

often related by nn CVPCPV −== or , where n and C are constants. The work done

during a polytropic process as shown in Fig. 3-4 can be represent as,

nVPVPW

nVVPVVP

nVV

CW

nVCdVCVdVPW

b

nnnnnn

b

nn

b

−−

=

+−−

=+−

−=

+−===

+−+−+−+−

+−−∫∫

1

11

1

1122

1111

1222

11

12

2

1

12

1

2

1

Since CVPVP nn == 2211 . For ideal gas mRTPV = , this equation can also be written as

1for 1

)( 12 ≠−

−= n

nTTmRWb

For the special case of 1=n the boundary work becomes

1

22

1

12

1ln

VV

PVdVCVPdVWb ∫∫ === −

For an ideal gas this result is equivalent to the isothermal process.

Shaft Work

Energy transmission with a rotating shaft is very common in engineering practice as

shown in Fig. 3-9. Often the torque T applied to the shaft is constant, which means that



48

the force F is also constant. The work done during n revolutions is determined as

follows,

Fig. 3-9 Shaft work

rTFFrT =→=

This force acts through a distance s, which is related to the radius r by

nrs 2π=

Then the shaft work is determined from

)( 2 2 kJTnrnrTsFWsh ππ ===

The power transmitted through the shaft is the shaft work done per unit time, which can

be expressed as

kWTnWPower sh 2 && π==

Where n& is the number of revolutions per unit time.

Spring Work

Fig. 3-10 Elongation of a spring under the influence of a force



49

When a force is applied on a spring as shown in Fig 3-10, the length of the spring

changes by a differential amount dx under the influence of a force F and the work done

is obtained as,

dxFW spring =δ

For linear elastic springs, the displacement x is proportional to the force applied.

)( kNkxFxF =→∝

Where k is the spring constant and has unit kN/m. The spring work is as follows,

)( )(21

2

21

22

2

1

22

1KJxxkxkkxdxW

kxdxdxFW

spring

spring

−===

==

∫

δ

Electrical Work

Fig. 3-11 Electrical work

The electric power as the result of passing electric current I through resistance R as

shown in Fig. 3-11.

)( WVIWe =&

The electric work done during the time interval t∆ is expressed as,

)( 2

1kJdtVIWe ∫=

When both V and I remains constant during the time interval t∆ , we obtain the

following relation

)( kJtVIWe ∆=



50

Flow Work

Fig. 3-12 Flow work

Unlike closed systems, control volumes involved mass flow across their

boundaries, and some work is required to push the mass into or out the control volume.

This work is called flow work or flow energy as shown in Fig. 3-12. To obtain a relation

for flow work, consider a fluid element of volume V and pressure P with cross-sectional

area A, the force applied on the fluid element by the imaginary piston to move the fluid

element distance L is,

)( kJPVPALFLWPAF

flow ====

For unit mass, the flow work is obtained as

)/( kgkJPvWflow =

Total energy of a flowing flow

The total energy in a fluid element is the sum of internal energy, kinetic energy and

potential energies as,

)/( v21 2 kgkJgzupekeue ++=++=

Where, v and z are the velocity and elevation of the fluid element. By adding the flow

energy to the energy of the fluid element, we obtain the total energy of the flowing

element θ as,

)/( v21 2 kgkJgzuPvpekeuPvePv +++=+++=+=θ



51

With combination huPv =+ in the above relation, we get the flow energy per unit mass

as,

)/( v21 2 kgkJgzh ++=θ

When a fluid stream with uniform properties is flowing at a mass flow rate of m& , the

rate of energy flow with that stream is θm& as follows,

)( )v21( 2 kJgzhmmE ++== &&& θ

Examples of Heat, Energy and Work

1. A piston of mass 15 kg is lowered 1.5 m in the standard gravitational field. Find the

required force and work involved in this process.

Solution

mNzFFdzWNmgF

. 725.2205.115.147 15.14781.915

=×=×=∫=

=×==

2. An escalator raises a 200 kg bucket of concert 25 m in 0.5 minute. Determine the

total amount of work done and the instantaneous rate of work during the process.

Solution

kWskJsmNtWWPowermNmgzEPW

635.1/ 635.1/. 1635)605.0/(49050/. 490502581.9200.

===×=∆==

=××===&

3. A gas is compressed from an initial volume of 0.52 m3 to a final volume of 0.1 m3.

During the quasi-equilibrium process, the pressure changes with volume according to

the relation of P = aV + b, where a = -1200 kPa/m3 and b = 600 kPa. Calculate the

work done during this process, a) by plotting the process on a P-V diagram and

finding the area under the process curve, and b) by performing the necessary

integrations.

Solution



52

kJVVPPWocesstheUnderAreaW

kPabaVPkPabaVP

76.95)52.01.0(2/)48024()(2/)(21 Pr

4806001.01200 2460052.01200

2 1 process under the area thefindingby Work

1221

22

11

−=−×+−=−×+=→==+×−=+=

−=+×−=+=→

or the work by integration as follows,

kJmkNW

VVbVVadVbaVpdVW

76.95. 76.95)52.01.0(600)52.01.0()2/1200(

)()(2/)(22

122

12

2

−=−=−+−×−=

−+−=+∫=∫=

4. A cylinder fitted with a frictionless piston contains 8 kg of superheated refrigerant R-

134a at 400 kPa and 100 oC. The setup is cooled at constant pressure until the

refrigerant reaches a quality of 25 %. Calculate the work done in the process.

Solution

kgmCo

/ 07327.0vVapor dSuperheate , 100 T kPa, 400 Pat 1, State

134a-R Table FromP P Pressure,Constant 2, 1 Process

31

11

21

=

==

=→



53

kJmPPdVW

kgmv

x fgf

904.191)07327.00133.0(4008)vv(

/ 0133178.0

)0007904.00509.0(25.00007904.0)vv(vvVapor Wet 0.25, xkPa, 400 Pat 2, State

1221

32

2

22

−=−××=−=∫=

=

−×+=−+===

−

5. A mass of 5 kg of air at 150 kPa and 15 oC is contained in a gas-tight, frictionless

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

During the process, heat is transferred from the air such that the temperature inside

the cylinder remains constant. Calculate the work input during this process.

Solution

kJVVmRTW

VVVP

VVCVCdV

VCPdVW

mPVPVCVPVP

mPmRT

778.572755.2689.0288287.05ln

lnlnln

689.0600/755.2150/

C PVor C T Process, Isothermal 755.2150/288287.05/V 1, State

1

21

1

211

1

22

1

32112

2211

3111

−=×××==

====∫=

=×==

====

=××==

∫

6. A piston-cylinder arrangement shown in Fig. p6 initially contains air at 150 kPa and

400 oC. The setup is allowed to cool to the ambient temperature of 25 oC. a) Is the

piston resting on the stops in the final state? What is the final pressure in the

cylinder? b) What is the specific work done by the air during this process?

Solution



54

CKR

PT

kgmPP

kgmP

RTCTkPaP

o

o

58.63 59.3360.287

0.644150v

/ 644.02877.15.0v21 v, stopsAt

/ 288.1150

673287.0v

400 , 150 1, State

222

31221

3

1

11

11

==×

==

=×===

=×

==

==

kgkJPWWWW

kPaRT

P

CT

CTo

o

/ -96.61.288)-(0.644150)v-v(

8.132644.0

298287.0v

v0.5 v v, 25 state, final 3, state

continou. cooling theand state final befor the stops thereachespiston the, 25

12213221

3

33

1233

2

=×===+=

=×

==

===

>

−−−

7. The refrigerant R-134a is contained in a piston-cylinder as shown in Fig. p7 where

the volume is 11 L when the piston hits the stops. The initial state is -10 oC and 180

kPa with a volume of 8 L. This system is brought indoor and warms up to 20 oC. a)

Is the piston at the stops in the final state? b) Find the work done by the R-134a

during this process.

Solution

kgVm

kgm

kPaPCT o

0718.011135.0/008.0v

/ 11135.0 v vapor,dsuperheate is 134a-R

180 , 10at 1 state 134a,-R of tableFrom

1

31

11

===

=

=−=



55

/ 1532.00718.0011.0v

011.0 , 180 stops,At

3

3

kgmmV

mVkPaP

===

==

205.0)11135.012723.0(1800718.0)v-v(

/ 12723.0 v134a,-R of tableFrom 20 , 180 , 2, State

stops, reach thenot doesPiston the, 25 perature,indoor temAt 75 table,From

12

32

o2221

kJWmPPdVW

kgmCTkPaPPP

CTCT

of

o

=−××==∫=

=

===

=

=

8. During some actual expansion process in piston-cylinder device, the gas have been

observed to satisfy the relationship PVn=C, where n and C are constant. Calculate the

work done when the gas expands from 150 kPa and 0.025 m3 to a final volume of 0.3

m3 for the case of n=1.3.

Solution

kJn

VPVPdV

VCPdVW

kPaVV

PP

VPVPPV

mVkPaP

n

n

nnn

57.63.11

025.01503.093.51

93.53.0

025.0150

C, ,Ploytropic is Process

025.0 , 150 1, State

1122

3.1

2

112

2211

311

=−

×−×=

−−

==∫=

=

=

=

==

==

∫



56

9. A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa as Fig. p9. At

this state, a linear spring that has a spring constant of 150 kN/m is touching the

piston but exert no force on it. Now heat is transferred to the gas, causing the piston

to rise and compress the spring until the volume inside the cylinder double. If the

cross-section area of the piston is 0.25 m2. Determine a) the final pressure inside the

cylinder, b) the total work done by the gas, and c) the fraction of this work done

against the spring to compress it.

Solution

kJW

kxkxdxfdxW

kJVVPPPdVW

kPaAdxk

AFPP

mA

VVAVdx

mVV

mVkPaP

s

s

pp

pp

32/2.0150

2/ work,Spring

13)05.01.0()320200(5.0)()(5.0

curve under the area thefrom calculated becan work theand processliner a is This

32025.0

2.0150200.200

2.025.0

05.01.0 1.005.022

05.0 , 200 1, State

2

2

1221

12

12

312

311

=×=

=∫=∫=

=−×+×=−×+×=∫=

=×

+=+=+=

=−

=−

=∆

=

=×==

==

10. A piston-cylinder device contains 50 kg of water at 150 kPa and 25 oC as Fig. p10.

The cross-section area of the piston is 0.1 m2. Heat is now transferred to the water,

causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the



57

piston reaches a linear spring whose spring constant is 100 kN/m. More heat is

transferred to the water until the piston rises 20 cm more. Determine a) the final

pressure and temperature, and b) the work done during this process. Also, show the

process on a P-V diagram.

Solution

kgmmV

kPaAdxk

AFPP

mdxAVVmVkPaPP

mmVkgmCT

CTkPaP

pp

p

o

o

/ 0044.05022.0v

3501.0

2.0100150.150

22.02.01.02.0 3, State 2.0 , 150 2, State

pressureconstant is 21 Process 05015.0001003.050v

/ 001003.0v v, 25at tableOH fromliquid subcooled is water , 25 , 150 1, State

333

23

323

3212

311

3f112

11

===

=×

+=+=+=

=×+=×+=

===

→=×=×=

===

==

CT

kgm

kgmkP

o 88.138 , wet vaporis 3 State , v v v

/ 5243.0v

,/ 001079.0 v,Pa 350at tableOH From

3g3f

3g

3f32

=∴∴<<

=

==

Q



58

kJWkxkxdxfdxW

kJWVVPPVVPPdVW

s

s

22/2.01002/ work,Spring

98.28)2.022.0()350150(5.0)05015.02.0(150)()(5.0)(

3 2 1 curve, under the area thefrom calculated becan work The

2

2

2332121

=×=

=∫=∫=

=−×+×+−=−×+×+−=∫=

→→

11. A balloon behaves such that the pressure inside is proportional to the diameter

squared. It contains 2 kg of R-134a at 0 oC and 60 % quality. The balloon and R-

134aq are now heated so that a final pressure of 600 kPa is reached. Considering the

R-134a as a control mass, find the amount of work done in the process. Note that, P

∝ d2, V ∝ d3, or P ∝ V2/3.

Solution

kJW

nVPVPdV

VCPdVW

mPPVVkPaP

VPVPCPVCPVVdPdVdP

mmV

kgmxkPaPxCT

n

n

nnn

fgf

o

32.73)3/2(1

0833.082.29224433.06001

24433.0600

82.2920833.0 , 6002 2, State

, Process, PolytropicFor , , ,

0833.004165.02v

/ 0.041650.0007721)-0.0689 (6.00007721.0)vv(vv 82.292 6.0 , 0at 1, State 134a,-R

21

112221

33/211

2

112

2211

3/23/2232

311

311

111

=−−

×−×=

−−

=∫=∫=

=

=

=∴=

=∴=

=∴∝∝∴∝∝

=×==

=×+=−+=

=∴==

−

−

−

−Q

12. A piston-cylinder device contains a gas at pressure of 200 kPa and volume of 0.04

m3. A heat transferred from a Bunsen burner which be placed under the cylinder until

the volume increased to 0.1 m3. Determine the work done by the system during this

process in the following cases, a) the pressure inside the cylinder remains constant,

b) the temperature inside the cylinder remains constant by removing some weights,

and c) during the heat transfer let the weights be removed at such a rate that the



59

expansion PV1.3 = C. Show the three process on P-V diagram and which process

gives lowest work done.

Solution

kJn

VPVPdV

VCW

kPaVV

PPVPVP

CPV

kJVV

VPdVVCW

CPVCT

kJVVPPdVW

CPmV

mVkPaP

n

nnn

n

41.633.11

04.02001.077.601

77.601.0

04.0200 ,

Process, Polytropic c)

33.704.01.0ln04.0200ln

or , Process, eTemperaturConstant b)

12)04.01.0(200)(

Process, PressureConstant a) 1.0 2, State

04.0 , 200 1, State

112221

3.1

2

1122211

1

21121

1221

32

311

=−

×−×=

−−

==

=

=

==

=

=×===

==

=−×=−==

==

==

∫

∫

∫

−

−

−

13. Determine the torque applied to the shaft of a car that transmits 450 hp and rotates at

a rate of 3000 rpm.

Solution

mkNkJTT

nTWPower sh

. 948.1 948.160/3000236.1450

2

==××=×

==π

π&



60

14. A battery is well insulated while being charged by 12.3 V at a current of 6 A. Take

the battery as a control mass and find the instantaneous rate of work and the total

work done over 4 hours.

Solution

kJtWW

smNsJWVIW

ee

e

72.10621000

1360048.73

/. 8.73/ 8.73 8.733.126.

=×××=∆×=

===×==

&

&

15. Air enters a nozzle steadily at a density of 2.21 kg/m3 and velocity of 30 m/s and

leaves at 0.762 kg/m3 and 180 m/s. If the inlet area of the nozzle is 80 cm2, determine

a) the mass flow rate through the nozzle and b) the exit area of the nozzle.

Solution

224-22222

4111

67.38 10 67.38 ,1800.7620.5304 ,

/ 5304.030108021.2Equation, Continuity From

cmmAAVAm

skgVAm

=×=∴××=∴=

=×××== −

ρ

ρ

&

&

16. Refrigerant 134a enters the compressor as saturated vapor at 0.14 MPa and 5 m/s,

and leaves as superheated vapor at 0.8 MP and 50 oC at a rate of 0.04 kg/s and 80

m/s. Determine the rates of energy transfers by the mass into and out of the

compressor. Assume the potential energy to be negligible.

Solution



61

skJhmE

skJhmE

kJ/kghTMPaP

kJ/kghMPaP

out

in

o

/ 504.111000

12

8039.28404.02v

/ 442.91000

12

504.23604.02

v

39.284 134a,-R of tableFrom vapor dsuperheate C, 50 , 8.0 2, State

04.236 134a,-R of tableFrom vapor saturated , 1 x, 14.0 1, State

1

222

2

1

221

1

2

22

1

11

=

×+=

+=

=

×+=

+=

===

===

&

&

Problems 1- The volume of air changes from 100 liters to 250 liters when it expands under a

constant pressure of 10 bar. Draw the pressure- volume diagram for the process and

determine the work done during the process.

2- Determine the work done when the pressure of a gas in a rigid cylinder of volume

0.15 m3 is increased from 1.5 bar to 5 bar, draw the P-V diagram for the process.

3- Air at a pressure of 1 bar and volume 0.6 m3 is compressed to a volume of 0.2 m3 is

such a way that PV = constant. Find the work done on the system during the process

and the final pressure.

4- A gas expands in a cylinder from a volume of 0.18 m3 and a pressure of 5 kgf/cm2 to

a volume of 0.36 m3 according to the law PV1.2= C. Calculate the final pressure and

the work done.

5- A gas expands in a closed system when the pressure changes according to the

equation P= 20V + 15 bar; V is in m3. During the process the volume changes from

0.2 m3 to 0.6 m3. Calculate the work done during the process.

6- A spherical balloon of radius 10 cm contains air at a pressure of 1.5 kgf/cm2. When

the balloon is left exposed in the sun for some time, its radius is found to increase by



62

1 cm, assuming that the pressure in the balloon is directly proportional to its radius,

determine:

a- The index of expansion,

b- The final pressure of the balloon, and

c- The work done by the system during the process.

7- In a piston-cylinder arrangement 1 kg of air at a pressure of 1 kgf/cm2 and volume of

0.2 m3 expands under constant pressure to a volume of 0.8 m3; it then under goes a

constant volume process is such a manner that a change in the air pressure has been

occurred after which the air is compressed according to the law PV = constant until

the air returns to its initial state. Represent the cycle on the diagram and determine

the net work done by the system.

8- A piston cylinder device contains 0.05 m3 of a gas initially at 200 kPa at this state, a

linear spring that has a spring constant of 150 kN/m 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 doubles. If the cross

sectional area of the cylinder is 0.25 m2 ,determine:

a- The final pressure inside the cylinder,

b- The total work done by the gas and

c- The fraction of this work done against the sprig to compress it also find the work

exerted against the piston weight.

9- A balloon behaves such that the pressure inside is proportional to the diameter

squared. It contains 2 kg of R-134a at 0 ºC and 60 % quality. The balloon and R-

134aq are now heated so that a final pressure of 600 kPa is reached. Considering

the R-134a as a control mass, find the amount of work done in the process. Note

that, P ∝ d2, V ∝ d3.

10- The refrigerant R-134a is contained in a piston-cylinder as shown in Fig. (1). Where

the volume is 11 L when the piston hits the stops. The initial state is -10 ºC and 180



63

kPa with a volume of 8 L. This system is brought indoor and warms up to 20 ºC. a)

Is the piston at the stops in the final state? b) Find the work done by the R-134a

during this process.

11- A piston-cylinder device contains 50 kg of water at 150 kPa and 25 ºC as Fig. (2).

The cross-section area of the piston is 0.1 m2. Heat is now transferred to the water,

causing part of it to evaporate and expand. When the volume reaches 0.2 m3, the

piston reaches a linear spring whose spring constant is 100 kN/m. More heat is

transferred to the water until the piston rises 20 cm more. Determine a) the final

pressure and temperature, and b) the work done during this process. Also, show the

process on a P-V diagram.

Fig.(1) Fig. (2)


thermo i chapter 3

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