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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 =
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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.
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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
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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.
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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.
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46
2
2
1
1
21222111 0or and 2 State , 1 State
TV
TV
dPPPmRTVPmRTVP
=∴
====
The boundary work as follows,
)( 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
The boundary work as follows,
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 .
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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
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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
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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 ∆=
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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
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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
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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
=
==
=→
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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
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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
===
=
=−=
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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
=−
×−×=
−−
==∫=
=
=
=
==
==
∫
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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
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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
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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
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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&
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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
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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
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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
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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
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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)