MAHALAKSHMI
ENGINEERING COLLEGE
TIRUCHIRAPALLI-621213.
Department: Mechanical
Sem ester: III
Subject Code: ME2202
Subject Name: ENGG. THERMODYNAMICS
UNIT - 1
1. 1 kg of gas at 1.1 bar, 27oC is compressed to 6.6 bar as per the law
pv1.3 = const.
Calculate w ork and heat transfer, if
(1) When the gas is ethane (C2H6) w ith molar mass of 30kg/k mol and c pof2.1 kJ/kg K.
(2) When the gas is argon (Ar) w ith molar mass of 40kg/k mol and c pof 0.52 kJ/kg K.
(AUC DEC ’05)
Given: Polytropic process
Mass (m) = 1kg
Pressure (P1) = 1.1bar
=1.1×100
=110KN/m2
Temperature (T1) = 27°C
=27+273
=300K
Pressure (P2) = 6.6bar
=6.6×100
=660KN/m2
Molar mass of ethane (m) =30Kg/K mol
Cp of ethane = 2.1KJ/K mol
Molar mass of argon = 40Kg/K mol
Cp of argon = 0.52KJ/Kg.K
PV 1.3 = C
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To Find:
1. Work Transfer (W)
2. Heat Transfer (Q)
Solution:
Step:1
Case:1 (Ethane)
Work transfer (W) =
Where,
P1V 1 = mRT1
V 1=
V1 =
V 1 = 680.2363m3
Where,
P1V 1n = P2V 2
V2 =
=
V 2 = 802.2438m3
Work transfer (W) =
=
W = -1515516.892KJ
Heat transfer (Q) = W ×
= -1515516.892
Q = -378879.223KJ
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Case:2 (Argon)
Work transfer (W) =
Where,
P1V 1 = mRT1
V 1=
V1 =
V 1 = 906.9818m3
Where,
P1V 1n = P2V 2
V2 =
=
V 2 = 1166.0760m3
Work transfer (W) =
=
W = -2232807.207KJ
Heat transfer (Q) = W ×
= -2232807.207
Q = 558201.8018KJ
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2.
In an air compressor, air flow s steadily at the rate of 0.5 kg/sec. At entry to the compressor,
air has a pressure of 105kPa and specific volume of 0.86 m3/kg and at exit of the
compressor those corresponding values are 705 kPa and 0.16 m3/kg. Neglect Kinetic and
Potential energy change. The internal energy of air leaking the compressor is 95 kJ/kg
greater than that of air entering. The cooling w ater in the compressor absorbs 60kJ/sec. of
heat from the air. Find pow er required to derive the compressor.
(AUC MAY ’06)
Given:
Mass (m) = 0.5 kg/s
Entering velocity (C1) = 7 m/s
Entering Pressure (P1) = 100 KPa = 100 KN/m2
Entering volume (V 1) = 0.95 m3/Kg
Leaving velocity (C2) = 5 m/s
Leaving Pressure (P2) = 700 KPa = 700 KN/m2
Leaving volume (V 2) = 0.19 m3/Kg
Change in internal energy (U2- U1) = 90 KJ/Kg
Heat absorbs (Q) = 58 KW
To Find:
1. Compute the rate of shaft w ork input to the air in KW (or)
Work input
2. Ratio of inlet pipe to outlet pipe (
)
Solution:
Step:1
m (P1V 1 +
+Z1.g) + Q = m ((U2- U1)+P2V 2 +
+Z2.g) + W
Assume Z1= Z2
0.5 ((100×0.95)+
) + 58 = 0.5 (90+(700×0.19) +
)+ W
47.51225+58 = 111.50625 + W
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W = 105.51225-111.50625
W = 5.994KW
Note;
“-“ sign indicates that the w ork is done on the system
Step:2 To find ratio of inlet to outlet dia (
)
From continuity equation
=
=
=
=3.57
= 3.57
= 3.57
= √3.57
The ratio of inlet to outlet dia of the pipe(
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) = 1.8894
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3.
In an isentropic flow through nozzle, air flow s at the rate of 600 kg/hr. At inlet to
the nozzle, Pressure is 2 MPa and temperature is 127oC. The exit pressure is 0.5 MPa. Initial air
velocity is 300 m/s determines (i) Exit velocity of air (ii) Inlet and exit area of nozzle.
(AUC DEC ’06)
Given:
Mass of flow rate (m1) = 600Kg/hr
=600÷3600
=0.1666Kg/sec
Inlet pressure (P1) = 2MPa×10^6 N/m2
Inlet temp (T1) = 127°C + 273
=400K
Outlet pressure (P2) = 0.5MPa×10^6 N/ m2
Inlet air velocity (C1) = 300m/s
To Find:
1.Exit velocity of air (C2)
2. inlet and exit area (d1,d2)
Solution:
Step:1
=
(
)^
=
= 1.4859
T2= 1.4859× T1
T2= 1.4859×400
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T2= 594.3977 k
Step:2
C2 = √{2×m[Cp(T2-T1) +
]
C2 = √{2×0.1666[1.005(594.3977-400) +
]
C2 =282.4937m/s
Step:3
Inlet mass flow rate (m) =
Where,
P1V 1 = mRT1
V 1=
V 1=
V 1 = 0.0574 m3/Kg
M1 =
A1 =
A1 =
A 1 =3.1876×10^-5
A1 =
×d1^2
3.1876×10^-5 =
×d1^2
d1 = √
d1 = 0.0063m×1000
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d1 = 6.3706mm
Step:4
=
Where,
=
(
=
(
)^
)^
= 2.692
V 2 = 2.692 V 1
V 2 = 2.692×0.0574
V 2 = 0.155m3/Kg
=
A2 =
A2 =
A 2 = 3.1410×10^-5
×d2^2 = 3.1410×10^-5
d2^2 =
D2 = 0.01078m ×1000
D2 = 10.7883mm
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4. A centrifugal pump delivers 2750 kg of w ater per minute from initial pressure of 0.8 bar
absolute to a final pressure of 2.8 bar absolute. The suction is 2 m below and the delivery is5 m
above the centre of pump. If the suction and delivery pipes are of 15 cm and 10 cm diameter
respectively, make calculation for pow er required to run the pump.
(AUC DEC ’06)
Given:
Mass (m) = 2750 Kg/min = 2750÷60
= 45.8333Kg/s
Initial pressure (P1) = 0.8bar ×100
= 80KN/m2
Final pressure (P2) = 2.8bar ×100
= 280 N/m2
Z1 = -2m (below the centre of pump)
Z2 = 5m (above the centre of pump)
Dia , d1 = 15cm÷100
=0.15m
d2 = 10cm÷100
=0.1m
To Find:
Pow er (P) or Work (W)
Solution:
Step:1
The steady flow energy equation is,
m[
+
+P1V 1] = m [
L.VIJAYAKUMAR /A.P-MECH
+
+ P2V 2] + W
Page 9
W= m[
+
+(P1V 1- P2V 2)]
Where,
M=
×d12×C1×ρ
C1 =
=
C1 = 2.5936m/s
C2 =
=
C2 = 5.8356m/s
Step:2
W = 45.83 [
+
+(80×1000 -280×1000)]
W =[-0.06867-0.01366-200×10^3]
W = 9166003.773KJ/Kg
W=
W = 91.6600KJ/Kg
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6.
A blow er handles 1 kg/sec of air at 293 K and consumes a pow er of
15kw .Theinletandoutletvelocitiesofairarel00 m/sec and 150 m/sec respectively. Find the exit air
temperature, assuming adiabatic conditions. Take Cp of air as 1.005 kJ/kg-K. (AUC DEC ’07)
Given:
Mass(m) = 1Kg/s
Temp (T1) = 293K
Pow er (P) or Work (W) = 15 KW
Inlet velocity (C1) = 100m/s
Outlet velocity (C2) = 150m/s
Cp = 1.005KJ/Kg.k
To Find:
Exit air temp (T2)
Solution:
Step:1
m (h1 +
+Z1.g) + Q = m (h2 +
+Z2.g) + W
Note;
Neglect datum head (Z 1) = (Z2) = 0
Q = 0 (Adiapatic process
m (h1 +
) = m (h2 +
1 (h1 +
)+W
) = 1 (h2 +
(h1 – h2) =[
) + 15
] + 15
(h1 – h2) =21.25
Cp(T1 – T2) = 21.25
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1.005(293 – T2) = 21.25
293 – T2 =
293 – T2= 21.1442
293 –21.1442 =T2
T2 = 271.8557K
7. A room for four persons has tw o fans, each consuming 0.18 kW pow er and three 100W
lamps. Ventilation ant at the rate of 0.0222 kg/sec enters w ith an enthalpy of 84 kJ/kg and
leaves w ith an enthalpy of 59 kJ/kg. If each person puts out heat at the rate of0.175 kJ/sec,
determine the rate at w hich heat is to be removed by a room cooler, so that a steady state is
(AUC DEC ’07)
maintained in the room.
Given:
No of person(np) = 4
No of fan (nf) = 2
(Wf) = 0.18KW (each)
(Wl ) = 0.1KW (each)
Mass of air (m) = 80Kg/hr
=80÷3600
=0.022Kg/s
Enthalpy of air entering (h1) = 84KJ/Kg
Enthalpy of air leaving (h2) = 59KJ/Kg
Heat (Qp) = 630KJ/hr
=630÷3600
=0.175KJ/s
To Find:
Rate of heat is to be removed
Solution:
Step:1
Rate of energy increase = Rate of energy in flow - Rate of energy out flow
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Q1 = -Ƞp Qp
Q1 = -(4×0.175)
Q1 = - 0.7KJ/s or KW
Step:2
=M(h1- h2)
=0.022(84- 59)
=0.55KJ/s
Step:3
W = electrical energy input
W = Ƞf Wf + Ƞl Wl
W = (2×0.18)+( 3×0.1)
W = 0.66KW
Step:4
Rate of heat is to be removed (Q)
Q = Q1 – 0.55 – W
Q = - 0.7 – 0.55 – 0.66
Q = - 1.916KW
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8. One liter of hydrogen at 273 K is adiabatically compressed to one half of its initial volume. in
the change in temperature of the gas, if the ratio of tw o specific heats for hydrogen is 1.4.
(AUC DEC ’07)
Given: adiabatic process
Initial volume (V 1) = 1lit = 1÷1000
= 0.001m3
Temp (T1) = 273K
Initial volume (V 2) = one half of it’s Initial volume
(V 2) = 0.001÷2 = 0.0005 m3
To Find:
Change in temp of gas (T2 – T1)
Solution:
Step:1
=
(
)^(
)
=
(
)^(
=
1.3195
)
T2 = 1.3195×T1
T2 = 1.3195×273
T2 = 360.2256K
Step:2 To find change in temp (T2 – T1)
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= T2 – T1
= 360.2256 – 273
T2 – T1= 87.2256K
9.
The velocity and enthalpy of fluid at the inlet of a certain nozzle are50 m/sec and2800
kJ/kg respectively. The enthalpy at the exit of Nozzle is 2600 kJ/kg. The nozzle is
horizontal and insulated so that no heat transfer takes place from it' Find
(1)
Velocity of the fluid at exit of the nozzle
(2)
Mass flow rate, if the area at inlet of nozzle is 0.09 m2
(3)
Exit area of the nozzle, if the specific volume at the exit of the Nozzle is 0.495 m3/kg.
(AUC DEC ’07)
Given:
Inlet velocity of nozzle (C1) = 50m/s
Inlet enthalpy of nozzle (h1) = 2800KJ/Kg
h1 = 2800×10^3J/Kg
no of heat transfer take place (Q) = 0
exit enthalpy (h2) = 2600KJ/Kg
h2
= 2600×10^3J/Kg
To Find:
1. C2
2. m
3. A 2
Solution:
Step:1 To find velocity of the fluid at exit of the nozzle (C2)
C2 =√[2 (h1 - h2) + C1^2]
C2 =√[2 (2800 - 2600)×1000 + 50^2]
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C2 = 634.4288m/s
Step:2 Tofind mass flow rate (m)
m=
Where,
V 1 = Specific volume is not given so it is assumed to be
V 1 = 1m3/Kg
m=
=
m = 4.5Kg/s
Step:3 To find exit area of nozzle (A 2)
=
A2 =
=
A 2 = 0.0035m2
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10. A w ork done by substance in a reversible non-flow manner is in accordance w ith V = (15/p)
3
m , w here p is in bar. Evaluate the w ork done on or by the system as pressure increases from
10 to 100 bar. Indicate w hether it is a compression process or expansion process. If the change
in internal energy is 500kJ, calculate the direction and magnitude of heat transfer. (AUC MAY
’08)
Given:
reversible non-flow manner
V = (15/p) m3
Pressure (P1) = 10bar
Pressure (P2) = 100bar
Change in internal energy (Δu) = 500KJ
To Find:
The direction and magnitude of heat transfer (Q)
Solution:
Step:1
Heat transfer (Q) = W+ Δu
Where,
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Work done = ʃ V dp
= ʃ 15/p dp
=15 ʃ 1/p dp
= 15 [log p ]
= 15 [log 100- log 10]
= 15 [2-1]
=15 [1]
W = 15KJ
Direction:
W = 15
Positive sign indicates the expansion process
Step:2
By first law of thermodynemics,
Q = W+ Δu
Q = 15+ 500
Q = 515KJ
Q = 515 KJ
Positive sign indicates the expansion process
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11. In a Gas turbine installation, the gases enter the turbine at the rate of 5 kg/sec w ith a
velocity of 500 m/sec and enthalpy of 900 kJ/kg and leave the turbine w ith 150 m/sec, and
enthalpy of 400 kJ/kg. The loss of heat from the gases to the surroundings is 25kJ/kg. Assume
R = 0.285 kJ/kg.K, Cp = 1.004 kJ/kg.K and inlet conditions to be at 100 Kpa and
27oC.Determine the diameter of the inlet pipe.
(AUC MAY ’08)
Given:
Mass flow rate (m) = 5Kg/s
Inlet velocity (C1 ) = 50m/s
Inlet enthalpy (h1 ) = 900KJ/Kg
Outlet enthalpy (h1 ) =400KJ/Kg
Outlet velocity (C1 ) = 150m/s
Heat loss (Q) = -25KJ/Kg
R = 0.285KJ/Kg
Cp = 1.004KJ/Kg
Inlet pressure (P1 ) = 100KPa ×1000 = 100000KN/m2
Inlet temperature (T1 ) = 27°C + 273 = 300K
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To Find:
Work done (W)
Dia of the inlet pipe (d1 )
Solution:
Step: 1
+Z 1.g) + Q = m (h2 +
m (h1 +
+Z 2.g) + W
Where,

Z2 = Z1
 m (h1 +
 5 (900 +
) + Q = m (h2 +
) - 25 = 5 (400 +
)+W
)+W
 4481.25 = 2056.25 + W
 4481.25 – 2056.25 = W
W = 2425KW
Step: 2 To find dia of the inlet pipe (d1 )
m=
Where,
V1 =?
P1 V1 = mRT1
V1 =
V1 =
2
V1 = 4.2875 m /s
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m=
A1 =
A1 =
A1 = 0.4275 m2
A1 =
×d12
2
d1 =
d1 = √
d1 = √
0.7377m = d1
d1 = 0.7377m ×1000
d1 = 737.7736mm
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12. A frictionless piston-cylinder device contains 2 kg of nitrogen at 100 Kpa and 300 K.
Nitrogen is now compressed slow ly according to the relation PV 1.4 = C until it reaches a final
temperature of 360 K. Calculate the w ork input during this process.
(AUC DEC ’09)
Given: Polytropic process
Molecular weight of nitrogen (m) = 28
Ɣ = 1.4
Inlet pressure (P1 ) = 100KN/m2
Inlet temperature (T1 ) = 300K
Outlet temperature (T2 ) = 360 K
PV1.4 = C
To Find:
Work input (W)
Solution:
Step: 1
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W=
Where,
Gas constant (R) =
=
= 0.297KJ/ Kg.K
W=
W=
W = -89.1KJ
Direction:
W = -89.1KJ
Negative sign indicates that the work is done on the system.
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