L.VIJAYAKUMAR /A.P-MECH Page 1 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, 27 o C is compressed to 6.6 bar as per the law pv 1.3 = const. Calculate w ork and heat transfer, if (1) When the gas is ethane (C 2 H 6 ) w ith molar mass of 30kg/k mol and c p of2.1 kJ/kg K. (2) When the gas is argon (Ar) w ith molar mass of 40kg/k mol and c p of 0.52 kJ/kg K. (AUC DEC ’05) Given: Polytropic process Mass (m) = 1kg Pressure (P 1 ) = 1.1bar =1.1×100 =110KN/m 2 Temperature (T 1 ) = 27°C =27+273 =300K Pressure (P 2 ) = 6.6bar =6.6×100 =660KN/m 2 Molar mass of ethane (m) =30Kg/K mol C p of ethane = 2.1KJ/K mol Molar mass of argon = 40Kg/K mol C p of argon = 0.52KJ/Kg.K PV 1.3 = C
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MAHALAKSHMImahalakshmiengineeringcollege.com/pdf/mec/IIIsem/ME2202...L.VIJAYAKUMAR /A.P-MECH Page 9 4. A centrifugal pump delivers 2750 kg of water per minute from initial pressure
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“-“ sign indicates that the w ork is done on the system
Step:2 To f ind 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( ) = 1.8894
L.VIJAYAKUMAR /A.P-MECH Page 6
3. In an isentropic f low through nozzle, air f low 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 f low 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
L.VIJAYAKUMAR /A.P-MECH Page 7
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 f low rate (m) =
Where,
P1V1 = mRT1
V1=
V1=
V1 = 0.0574 m3/Kg
M1 =
A1 =
A1 =
A1 =3.1876×10^-5
A1 = ×d1^2
3.1876×10^-5 = ×d1^2
d1 = √
d1 = 0.0063m×1000
L.VIJAYAKUMAR /A.P-MECH Page 8
d1 = 6.3706mm
Step:4
=
Where,
= ( )^
= ( )^
= 2.692
V2 = 2.692 V1
V2 = 2.692×0.0574
V2 = 0.155m3/Kg
=
A2 =
A2 =
A2 = 3.1410×10^-5
×d2^2 = 3.1410×10^-5
d2^2 =
D2 = 0.01078m ×1000
D2 = 10.7883mm
L.VIJAYAKUMAR /A.P-MECH Page 9
4. A centrifugal pump delivers 2750 kg of w ater per minute from initial pressure of 0.8 bar
absolute to a f inal 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 f low energy equation is,
m [ + +P1V1] = m [ + + P2V2] + W
L.VIJAYAKUMAR /A.P-MECH Page 10
W = m [ + +(P1V1- P2V2)]
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
L.VIJAYAKUMAR /A.P-MECH Page 11
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 (Z1) = (Z2) = 0
Q = 0 (Adiapatic process
m (h1 + ) = m (h2 + ) + W
1 (h1 + ) = 1 (h2 + ) + 15
(h1 – h2) =[ ] + 15
(h1 – h2) =21.25
Cp(T1 – T2) = 21.25
L.VIJAYAKUMAR /A.P-MECH Page 12
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
maintained in the room. (AUC DEC ’07)
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 f low - Rate of energy out f low
L.VIJAYAKUMAR /A.P-MECH Page 13
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
L.VIJAYAKUMAR /A.P-MECH Page 14
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 specif ic heats for hydrogen is 1.4.
(AUC DEC ’07)
Given: adiabatic process
Initial volume (V1) = 1lit = 1÷1000
= 0.001m3
Temp (T1) = 273K
Initial volume (V2) = one half of it’s Initial volume
(V2) = 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 f ind change in temp (T2 – T1)
L.VIJAYAKUMAR /A.P-MECH Page 15
= T2 – T1
= 360.2256 – 273
T2 – T1= 87.2256K
9. The velocity and enthalpy of f luid 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 f luid at exit of the nozzle
(2) Mass f low rate, if the area at inlet of nozzle is 0.09 m2
(3) Exit area of the nozzle, if the specif ic 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. A2
Solution:
Step:1 To f ind velocity of the f luid at exit of the nozzle (C2)
C2 =√[2 (h1 - h2) + C1^2]
C2 =√[2 (2800 - 2600)×1000 + 50^2]
L.VIJAYAKUMAR /A.P-MECH Page 16
C2 = 634.4288m/s
Step:2 Tofind mass flow rate (m)
m =
Where,
V1 = Specif ic volume is not given so it is assumed to be
V1 = 1m3/Kg
m = =
m = 4.5Kg/s
Step:3 To f ind exit area of nozzle (A2)
=
A2 =
=
A2 = 0.0035m2
L.VIJAYAKUMAR /A.P-MECH Page 17
10. A w ork done by substance in a reversible non-flow manner is in accordance w ith V = (15/p)
m3, 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,
L.VIJAYAKUMAR /A.P-MECH Page 18
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
L.VIJAYAKUMAR /A.P-MECH Page 19
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
L.VIJAYAKUMAR /A.P-MECH Page 20
To Find:
Work done (W)
Dia of the inlet pipe (d1)
Solution:
Step: 1
m (h1 + +Z1.g) + Q = m (h2 + +Z2.g) + W
Where,
Z2 = Z1
m (h1 + ) + Q = m (h2 + ) + W
5 (900 + ) - 25 = 5 (400 + ) + 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 =?
P1V1 = mRT1
V1 =
V1 =
V1 = 4.2875 m2/s
L.VIJAYAKUMAR /A.P-MECH Page 21
m =
A1 =
A1 =
A1 = 0.4275 m2
A1 = ×d12
d12
=
d1 = √
d1 = √
0.7377m = d1
d1 = 0.7377m ×1000
d1 = 737.7736mm
L.VIJAYAKUMAR /A.P-MECH Page 22
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 f inal
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
L.VIJAYAKUMAR /A.P-MECH Page 23
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.