This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Work in a Reversible Process-3”.
1. A reversible adiabatic air compressor takes in air at 100 kPa, 25°C and delivers it at 1 MPa. Assuming the specific heat is constant, calculate the specific work.
a) 145.6 kJ/kg
b) 324.6 kJ/kg
c) 178.6 kJ/kg
d) 278.6 kJ/kg
View Answer
Explanation: From data book, for air: cp = 1.004 kJ/kg.K and k = 1.4;
T2 = T1(P2/P1)^[(k-1)/k] = 575.6 K
Specific work (w) = ∆h = cp∆T = 1.004*(575.64 – 298.15)
= 278.6 kJ/kg.
2. The force needed to compress a non-linear spring is given by F = 200x + 30x^2, where F is force in Newton and x is displacement of the spring in meter. Determine the work needed to compress the spring a distance of 0.6 m.
a) 32.46 J
b) 23.43 J
c) 38.16 J
d) 48.87 J
View Answer
Explanation: Work needed to compress the spring (W) = ∫Fdx = ∫(200x + 30x^2)dx
= 100x^2 + 10x^3 = 100*0.6^2 + 10*0.6^3 = 38.16 J.
3. 19 mm thick fibre panels with thermal conductivity of 0.12 W/m.K are used for false ceiling of an AC room. If the floor area of the room is 17.65 m^2 and the temperature difference across the fibre panel is 15°C, calculate the heat transfer rate.
a) 4.672 kW
b) 3.672 kW
c) 2.672 kW
d) 1.672 kW
View Answer
Explanation: Heat transfer rate = kA∆T/x = 0.12*17.65*15/0.019
= 1672.1 Watt = 1.672 kW.
4. Air (an ideal gas) at 227°C and 500 kPa is allowed to expand to a pressure of 100 kPa in an ideal throttling process. What will be the final temperature of the expanded air?
a) 230°C
b) 227°C
c) 300°C
d) 327°C
View Answer
Explanation: In an ideal throttling process, enthalpy is constant;
hence for an ideal gas the temperature is also constant.
Final temperature of the expanded air (Tf) = 227°C.
5. A 250 L rigid tank contains methane at 500 K, 1500 kPa. It is now cooled down to 300 K. Find the heat transfer using ideal gas.
a) –402.4 kJ
b) –502.4 kJ
c) –602.4 kJ
d) –702.4 kJ
View Answer
Explanation: Assume ideal gas, P2 = P1 × (Τ2 / Τ1) = 1500 × 300 / 500 = 900 kPa
m = P1V/RT1 = 1500 × 0.250.5183 × 500 = 1.447 kg
u2 – u1 = Cv(T2–T1) = 1.736(300–500)= –347.2 kJ/kg
1Q2 = m(u2 – u1) = 1.447(-347.2) = –502.4 kJ.
6. A piston cylinder contains 3 kg of air at 20°C and 300 kPa. It is now heated up in a constant pressure process to 600 K. Find the heat transfer.
a) 641 kJ
b) 741 kJ
c) 841 kJ
d) 941 kJ
View Answer
Explanation: Ideal gas: PV = mRT
P2V2 = mRT2 thus V2 = mRT2 / P2 = 3×0.287×600 / 300 = 1.722 m^3
Process: P = constant,
1W2 = ⌠ PdV = P (V2 – V1) = 300 (1.722 – 0.8413) = 264.2 kJ
U2 – U1 = 1Q2 – 1W2 = m(u2 – u1)
1Q2 = U2 – U1 + 1W2 = 3(435.097 – 209.45) + 264.2
= 941 kJ.
7. A steam turbine inlet is at 1200 kPa, 500°C. The actual exit is at 300 kPa, 300°C with an actual work of 407 kJ/kg. What is its reversible work output if ambient temperature is at 25°C?
a) 414.9 kJ/kg
b) 314.9 kJ/kg
c) 214.9 kJ/kg
d) 614.9 kJ/kg
View Answer
Explanation: T0 = 250C = 298.15 K
The turbine is assumed to be adiabatic, so q = 0.
Inlet state: hi = 3476.28 kJ/kg; si = 7.6758 kJ/kg K
he = hi – w(ac) = 3476.28 – 407 = 3069.28 kJ/kg
Actual exit state: Pe = 300 kPa, Te = 300 0C; se = 7.7022 kJ/kg K
w(rev) = (hi – Tosi) – (he – Tose) + q(T0/TH) = (hi – he) + To(se – si)
= (3476.28 – 3069.28) + 298.15(7.7022 – 7.6758) + 0
= 407 + 7.87 = 414.9 kJ/kg.
8. Find the specific reversible work for a R-134a compressor with inlet state of –20°C, 100 kPa and an exit state of 600 kPa, 50°C. Use a 25°C ambient temperature.
a) -18.878 kJ/kg
b) -28.878 kJ/kg
c) -38.878 kJ/kg
d) -48.878 kJ/kg
View Answer
Explanation: The compressor is assumed to be adiabatic so q = 0.
wrev = T0(se – si) – (he – hi)
hi = 387.22 kJ/kg; si = 1.7665 kJ/kg K
he = 438.59 kJ/kg; se = 1.8084 kJ/kg K
∴ wrev = 298.15(1.8084 – 1.7665) – (438.59 – 387.22)
= -38.878 kJ/kg.
9. A steam turbine receives steam at 6 MPa, 800°C. It has a heat loss of 49.7 kJ/kg and an isentropic efficiency of 90%. For an exit pressure of 15 kPa and surroundings at 20°C, find the actual work.
a) 2233.79 kJ/kg
b) 2423.95 kJ/kg
c) 2483.95 kJ/kg
d) 1648.79 kJ/kg
View Answer
Explanation: Reversible adiabatic turbine
hi = 4132.74 kJ/kg, s = si = 7.6566 kJ/kg K
x = (7.6566 – 0.7548)/7.2536 = 0.9515
h = 225.91 + 0.9515×2373.14 = 2483.95 kJ/kg
w = hi – h = 4132.74 – 2483.95 = 1648.79 kJ/kg.
10. The refrigerant R-22 is contained in a piston/cylinder where the volume is 11 L when the piston hits the stops. The initial state is −30°C, 150 kPa with a volume of 10 L. This system is brought indoors and warms up to 15°C. Find the work done by the R-22 during this process.
a) -0.15 kJ
b) 0.15 kJ
c) -0.35 kJ
d) 0.35 kJ
View Answer
Explanation: Work done at constant Pext = P1.
1W2 = ∫ Pext dV = Pext(V2 – V1)
= 150(0.011 – 0.010) = 0.15 kJ.
11. A piston cylinder contains 1 kg of liquid water at 20°C and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 MPa with a volume of 0.1m^3. Find the final temperature.
a) 400°C
b) 404°C
c) 408°C
d) none of the mentioned
View Answer
Explanation: State 1: Compressed liquid, take saturated liquid at same temperature.
v1 = vf @20°C = 0.001002 m^3/kg
State 2: v2 = V2/m = 0.1/1 = 0.1 m^3/kg and P = 3000 kPa
Superheated vapor close to T = 400°C
Interpolate: T2 = 404°C.
12. A piston cylinder contains 1 kg of liquid water at 20°C and 300 kPa. There is a linear spring mounted on the piston such that when the water is heated the pressure reaches 3 MPa with a volume of 0.1m^3. Find the work in the process.
a) 163.35 kJ
b) 263.35 kJ
c) 363.35 kJ
d) 463.35 kJ
View Answer
Explanation: Work is done while piston moves at linearly varying pressure, so we get:
1W2 = ∫ P dV = area = Pavg (V2 − V1) = (P1 + P2)*(V2 – V1)/2
= 0.5 (300 + 3000)(0.1 − 0.001) = 163.35 kJ.
13. A Carnot heat engine receives heat at 750 K and rejects the waste heat to the environment at 300 K. The net work output of the heat engine is used to drive a Carnot refrigerator, whose COP is 6.14. If the heat removal rate from the refrigerated space is 6.6 kW, determine the rate of heat supply to the heat engine?
a) 2.79 kW
b) 1.79 kW
c) 4.79 kW
d) 7.79 kW
View Answer
Explanation: W(HE) = (1 – T(L)/T(H))*Q(H) = 0.6*Q(H); W(R) = W(HE);
Q(L)/COP(R) = 0.6*Q(H); ∴ Q(H) = Q(L)/(0.6*COP(R))
Rate of heat supply to the heat engine (Q(H)) = 1.79 kW.
14. During the winter season, a room is heated by central heating furnace which delivers the 750 W of heat energy by burning wood pellets. How much power can be saved if heat pump with a COP of 3 is used instead of furnace?
a) 100 W
b) 500 W
c) 1000 W
d) 1500 W
View Answer
Explanation: COP(HP) = QH/W ⇒ W = QH/COP(HP) = 250 W
Power saved by replacing the furnace with heat pump (QL)
= QH – W = 750 – 250 = 500 W.
15. A Korean refrigerator is bought to maintain the cool space at 8°C and a Japanese refrigerator is bought to main the cool space at -20°C in a laboratory room having constant temperature of 25°C. If the both refrigerators run on a reversed Carnot cycle, state the refrigerator that delivers maximum COP.
a) Korean refrigerator
b) Japanese refrigerator
c) both have same COP
d) none of the mentioned
View Answer
Explanation: COP(Korean) = TL/(TH – TL) = (273.15 + 8)/(25 – 8) = 16.54;
COP(Japanese) = TL/(TH – TL) = (273.15 – 20)/(25 – (–20)) = 5.63
Refrigerator that delivers maximum COP = Korean refrigerator.
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