Category: Thermodynamics Objective Questions

Thermodynamics Multiple Choice Questions on “Polytropic Process-1”.

1. A polytropic process(n = − 1) starts with P = 0, V = 0 and ends with P= 600 kPa, V = 0.01 m³. Find the boundary work done.
a) 1 kJ
b) 2 kJ
c) 3 kJ
d) 4 kJ
Answer: c
Clarification: W = ⌠ PdV
= (1/2)(P1 + P2)(V2 – V1)
= (1/2)(P2 + 0)( V2 – 0)
= (1/2)(600*0.1)
= 3 kJ.

2. The piston/cylinder contains carbon dioxide at 300 kPa, with volume of 0.2 m³ and at 100°C. Mass is added at such that the gas compresses with PV^(1.2) = constant to a final temperature of 200°C. Determine the work done during the process.
a) -80.4 kJ
b) -40.4 kJ
c) -60.4 kJ
d) -50.4 kJ
Answer: a
Clarification: Work done = (P2V2 – P1V1)/(1-n) and mR = (P1V1)/T1 = 0.1608 kJ/K
Work done = 0.1608(473.2 – 373.2)/(1 – 1.2) = -80.4 kJ.

3. Neon at 400 kPa, 20°C is brought to 100°C in a polytropic process with n = 1.4. Find the work done.
a) -52.39 kJ/kg
b) -62.39 kJ/kg
c) -72.39 kJ/kg
d) -82.39 kJ/kg
Answer: d
Clarification: For Neon, k = γ = 1.667 so n < k, Cv = 0.618, R = 0.412
1w2 = [R/(1-n)](T2 – T1) = -82.39 kJ/kg.

4. A mass of 1kg of air contained in a cylinder at 1000 K, 1.5 MPa, expands in a reversible adiabatic process to 100 kPa. Calculate the work done during the process using Constant specific heat.
a) 286.5 kJ
b) 386.5 kJ
c) 486.5 kJ
d) 586.5 kJ
Answer: b
Clarification: Process: 1Q2 = 0, 1S2 gen = 0 => s2 = s1
T2 = T1(P2/P1)^[(k-1)/k] = 1000(0.1/1.5)^0.286 = 460.9 K
1W2 = -(U2 – U1) = mCv(T1 – T2)
= 1 × 0.717(1000 – 460.9) = 386.5 kJ.

5. A cylinder/piston contains 1kg methane gas at 100 kPa, 20°C. The gas is compressed reversibly to a pressure of 800 kPa. Calculate the work required if the process is isothermal.
a) -216.0 kJ
b) -316.0 kJ
c) -416.0 kJ
d) -516.0 kJ
Answer: b
Clarification: Process: T = constant. For ideal gas then u2 = u1 1W2 = 1Q2 and ∫ dQ/T = 1Q2/T
1W2 = 1Q2 = mT(s2 – s1) = -mRT ln(P2/P1)
= -0.51835× 293.2 ln(800/100) = -316.0 kJ.

6. A cylinder/piston contains 1kg methane gas at 100 kPa, 20°C. The gas is compressed reversibly to a pressure of 800 kPa. Calculate the work required if the process is polytropic, with exponent n = 1.15.
a) -314.5 kJ
b) -414.5 kJ
c) -514.5 kJ
d) -614.5 kJ
Answer: a
Clarification: Process: Pv^(n) = constant with n = 1.15 ;
T2 = T1(P2/P1)^[(n-1)/n] = 293.2(800/100)^0.130 = 384.2 K
1W2 = ∫ mP dv = m(P2v2 – P1v1)/(1 – n) = mR (T2 – T1)/(1 – n)
= 1*0.51835(384.2 – 293.2)/(1 – 1.15) = -314.5 kJ.

7. Helium in a piston/cylinder at 20°C, 100 kPa is brought to 400 K in a reversible polytropic process with exponent n = 1.25. Helium can be assumed to be an ideal gas with constant specific heat. Find the specific work.
a) -587.7 kJ/kg
b) -687.7 kJ/kg
c) -787.7 kJ/kg
d) -887.7 kJ/kg
Answer: d
Clarification: Process: Pv^(n) = C & Pv = RT => Tv^(n-1) = C
Cv = 3.116 kJ/kg K, R = 2.0771 kJ/kg K
v2 / v1 = (T1 / T2 )^[1/(n-1)] = 0.2885
P2 / P1 = (v1 / v2)^(n) = 4.73 => P2 = 473 kPa
W = (P2 v2 – P1 v1)/(1-n) = R(T2-T1)/(1-n) = -887.7 kJ/kg.

8. Consider air in a cylinder volume of 0.2 L at 7 MPa, 1800K. It now expands in a reversible polytropic process with exponent, n = 1.5, through a volume ratio of 8:1. Calculate the work for the process.
a) 1.61 kJ
b) 1.71 kJ
c) 1.81 kJ
d) 1.91 kJ
Answer: c
Clarification: Process: PV^(1.50) = constant, V2/V1 = 8
State 1: P1 = 7 MPa, T1 = 1800 K, V1 = 0.2 L, m1=P1V1/RT1 = 2.71×10-3 kg
State 2: T2 = T1 (V1/V2)^(n-1) = 1800(1/8)^(0.5) = 636.4 K
1W2 = ⌠ PdV = mR(T2 – T1)/(1 – n)
= 2.71×10^(-3) × 0.287(636.4 – 1800)/(1-1.5) = 1.81 kJ.

9. A cylinder/piston contains carbon dioxide at 300°C, 1 MPa with a volume of 200L. The total external force acting on the piston is proportional to V3. This system is allowed to cool to room temperature, 20°C. Find the work.
a) -24.4 kJ
b) -34.4 kJ
c) -44.4 kJ
d) -54.4 kJ
Answer: a
Clarification: PV^(-3) = constant
State 1: m = P1V1/RT1 = (1000 × 0.2)/(0.18892 × 573.2) = 1.847 kg
P2 = P1(T2/T1)^[n/(n-1)] = 1000(293.2/573.2)^(3/4) = 604.8 kPa
V2 = V1(T1/T2)^[1/(n-1)] = 0.16914 m^3
Work = ⌠ PdV = (P2V2 – P1V1)/(1-n) = [604.8 × 0.16914 – 1000 × 0.2] / [1-(-3)]
= -24.4 kJ.

10. A cylinder/piston contains 100L of air at 25°C, 110 kPa. The air is compressed in a reversible polytropic process to a final state of 200°C, 800 kPa. Assume the heat transfer is with the ambient at 25°C. Find the work done by the air.
a) -11.28 kJ
b) -21.28 kJ
c) -31.28 kJ
d) -41.28 kJ
Answer: b
Clarification: m = P1V1 /(RT1) = 110 × 0.1/(0.287 × 298.15) = 0.1286 kg
T2/T1 = (P2/P1)^[(n-1)/n] => 473.15/298.15 = (800/110)^[(n-1)/n]
⇒ (n-1)/n = 0.2328 hence n = 1.3034
V2 = V1(P1/P2)^(1/n) = 0.1(110/800)^(0.7672) = 0.02182 m^3
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = (800 × 0.02182 – 110 × 0.1)/(1 – 1.3034)
= -21.28 kJ.

11. A mass of 2 kg ethane gas at 100°C, 500 kPa, undergoes a reversible polytropic expansion with n = 1.3, to a final temperature of 20°C. Find the work done.
a) 43.7 kJ/kg
b) 53.7 kJ/kg
c) 63.7 kJ/kg
d) 73.7 kJ/kg
Answer: d
Clarification: P2 = P1(T2/T1)^[n/(n-1)] = 500(293.2/373.2)^(4.333) = 175.8 kPa
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = R(T2-T1)/(1-n)
= 0.2765(293.2-373.2)/(1-1.30) = 73.7 kJ/kg.

12. A piston/cylinder contains air at 100 kPa, 300 K. A reversible polytropic process with n = 1.3 brings the air to 500 K. Any heat transfer if it comes in is from a 325°C reservoir and if it goes out it is to the ambient at 300 K. Find the specific work.
a) -171.3 kJ/kg
b) -181.3 kJ/kg
c) -191.3 kJ/kg
d) -201.3 kJ/kg
Answer: c
Clarification: Process : Pv^(n) = C
Work = ⌠PdV = (P2V2 – P1V1)/(1-n) = R(T2-T1)/(1-n)
= 0.287 (500 – 300)/(1 – 1.3) = -191.3 kJ/kg.

13. A cylinder/piston contains saturated vapour R-22 at 10°C; the volume is 10 L. The R-22 is compressed to 60°C, 2 MPa in a reversible polytropic process. If all the heat transfer during the process is with the ambient at 10°C, calculate the work done.
a) −6.26 kJ
b) −7.26 kJ
c) −8.26 kJ
d) −9.26 kJ
Answer: b
Clarification: State 1: P1 = 0.681 MPa, v1 = 0.03471; m = V1/v1 = 0.01/0.03471 = 0.288 kg
State 2: v2 = 0.01214 m^3/kg; P2/P1 = 2.0/0.681 = (0.03471/0.01214)^(n)
=> n = 1.0255
Work = ⌠PdV = m(P2v2 – P1v1)/(1-n)
= 0.288(2000 × 0.01214 – 681 × 0.03471)/(1 – 1.0255) = −7.26 kJ.

Thermodynamics Test on “Entropy Generation in a Closed and Open System”.

1. 1 kg of air at 300 K is mixed with 1 kg air at 400 K in a constant pressure process at 100 kPa and Q = 0. Find the entropy generation in the process.
a) 0.0207 kJ/K
b) 0.0307 kJ/K
c) 0.0407 kJ/K
d) 0.0507 kJ/K
Answer: a
Clarification: U2 – U1 + W = U2 – U1 + P(V2 – V1) = H2 – H1 = 0
H2 – H1 = mA(h2 – h1)A + mB(h2 – h1)B = mACp(T2 – TA1) + mBCp(T2 – TB1) = 0
T2 =(mATA1 + mBTB1)/(mA + mB) = (TA1/2) + (TB1/2) = 350 K
1S2 gen = mACp ln(T2/TA1) + mBCp ln(T2/TB1)
= 1 × 1.004 ln [350/300] + 1 × 1.004 ln[350/400]
= = 0.15477 – 0.13407 = 0.0207 kJ/K.

2. A window receives 200 W of heat transfer at its inside surface of 20°C and transmits this 200 W from its outside surface at 2°C to ambient air at 5°C. Find the window’s rate of entropy generation. thermodynamics-test-q2″ alt=”thermodynamics-test-q2″ width=”198″ height=”211″ class=”alignnone size-full wp-image-166428″>
a) 0.015 W/K
b) 0.025 W/K
c) 0.035 W/K
d) 0.045 W/K
Answer: d
Clarification: S (inside) = 200/293.15 = 0.682 W/K
S (window) = 200/275.15 = 0.727 W/K
S (ambient) = 200/268.15 = 0.746 W/K
Window only: S(gen win) = S(window) – S(inside)
= 0.727 – 0.682 = 0.045 W/K.

3. An insulated cylinder/piston contains R-134a at 1 MPa, 50°C, volume of 100 L. The R-134a expands, dropping the pressure in the cylinder to 100 kPa. The R-134a does 190 kJ of work against the piston during this process. Is that possible?
a) yes
b) no
c) cannot be determined
d) none of the mentioned
Answer: a
Clarification: v1 = 0.02185 m^3/kg, u1 = 409.39 kJ/kg,
s1 = 1.7494 kJ/kg K, m = V1/v1 = 0.1/0.02185 = 4.577 kg
m(u2 – u1) = 1Q2 – 1W2 = 0 – 190 hence u2 = u1 − 1W2/m = 367.89 kJ/kg
T2 = -19.25°C ; s2 = 1.7689 kJ/kg K
m(s2 – s1) = ⌡⌠dQ/T + 1S2(gen) = 1S2(gen)
1S2(gen) = m(s2 – s1) = 0.0893 kJ/K
This is possible since 1S2(gen) > 0.

4. A hot metal piece is cooled rapidly to 25°C, removing 1000 kJ from the metal. Calculate the change of entropy if saturated liquid R-22 at −20°C absorbs the energy so that it becomes saturated vapor.
a) 1.950 kJ/K
b) 2.950 kJ/K
c) 3.950 kJ/K
d) 4.950 kJ/K
Answer: c
Clarification: R-22 boiling at -20°C; m = 1Q2 /h(fg) = 1000/220.327 = 4.539 kg
∆S(R-22) = ms(fg) = 4.539(0.8703) = 3.950 kJ/K.

5. A hot metal piece is cooled rapidly to 25°C, removing 1000 kJ from the metal. Calculate the change of entropy if energy is absorbed by ice.
a) 2.662 kJ/K
b) 3.662 kJ/K
c) 4.662 kJ/K
d) 5.662 kJ/K
Answer: b
Clarification: Ice melting at 0°C; m = 1Q2 /h(fg) = 1000/333.41 = 2.9993 kg
∆S(H2O) = ms(ig) = 2.9993(1.221) = 3.662 kJ/K.

6. A hot metal piece is cooled rapidly to 25°C, removing 1000 kJ from the metal. Calculate the change of entropy if energy is absorbed by vaporizing liquid nitrogen at 101.3 kPa pressure.
a) 9.929 kJ/K
b) 10.929 kJ/K
c) 11.929 kJ/K
d) 12.929 kJ/K
Answer: d
Clarification: Nitrogen boiling at 101.3 kPa; m = 1Q2 /h(fg) = 1000/198.842 = 5.029 kg
∆S(N2) = ms(fg) = 5.029(2.5708) = 12.929 kJ/K.

7. A piston cylinder has 2.5 kg ammonia at -20°C, 50 kPa. It is heated to 50°C at constant pressure from external hot gas at 200°C. Find the total entropy generation.
a) 0.511 kJ/K
b) 0.611 kJ/K
c) 0.711 kJ/K
d) 0.811 kJ/K
Answer: a
Clarification: v1 = 2.4463 m^3/kg, h1 = 1434.6 kJ/kg, s1 = 6.3187 kJ/kg K
v2 = 3.1435 m^3/kg, h2 = 1583.5 kJ/kg, s2 = 6.8379 kJ/kg K
1Q2 = m(h2 – h1) = 2.5 (1583.5 – 1434.6) = 372.25 kJ
1S2(gen) = m(s2 – s1) – 1Q2/T(gas)
= 2.5 (6.8379 – 6.3187) – 372.25/473.15 = 0.511 kJ/K.

8. A piston/cylinder contains 1 kg water at 20°C, 150 kPa. The pressure is linear in volume. Heat is added from 600°C source until the water is at 1 MPa, 500°C. Find the total change in entropy.
a) 1.751 kJ/K
b) 2.751 kJ/K
c) 3.751 kJ/K
d) 4.751 kJ/K
Answer: c
Clarification: v1 = 0.001002 m^3/kg; u1 = 83.94 kJ/kg; s1 = 0.2966 kJ/kg K
v2 = 0.35411 m^3/kg; u2 = 3124.3 kJ/kg; s2 = 7.7621 kJ/kg K
1W2 = ½ (1000 + 150) 1 (0.35411 – 0.001002) = 203 kJ
1Q2 = 1(3124.3 – 83.94) + 203 = 3243.4 kJ
m(s2 – s1) = 1(7.7621 – 0.2968) = 7.4655 kJ/K; 1Q2/T(source) = 3.7146 kJ/K
1S2 gen = m(s2 – s1) − 1Q2/T(SOURCE) = ∆Stotal
= ∆S(H2O) + ∆S(source) = 7.4655 – 3.7146 = 3.751 kJ/K.

9. 1kg of ammonia is contained in a piston/cylinder, as saturated liquid at −20°C. Heat is added at 100°C until a final condition of 70°C, 800 kPa is reached. Assuming the process is reversible, find the entropy generation.
a) 1.007 kJ/K
b) 1.107 kJ/K
c) 1.207 kJ/K
d) 1.307 kJ/K
Answer: d
Clarification: P1 = 190.08 kPa, v1 = 0.001504 m^3/kg, u1 = 88.76 kJ/kg, s1 = 0.3657 kJ/kg K
v2 = 0.199 m^3/kg, u2 = 1438.3 kJ/kg, s2 = 5.5513 kJ/kg K
1W2 =(1/2)(190.08 + 800)1(0.1990 – 0.001504) = 97.768 kJ
1Q2 = m(u2 – u1) + 1W2 = 1(1438.3 – 88.76) + 97.768 = 1447.3 kJ
1S2(gen) = m(s2 – s1) – 1Q2/T(res) = 1(5.5513 – 0.3657) – (1447.3/373.15)
= 1.307 kJ/K.

10. A piston/cylinder device keeping a constant pressure has 1 kg water at 20°C and 1 kg water at 100°C both at 500 kPa separated by a membrane. The membrane is broken and the water comes to a uniform state with no external heat transfer. Find the entropy generation for the process. thermodynamics-test-q10″ alt=”thermodynamics-test-q10″ width=”269″ height=”126″ class=”alignnone size-full wp-image-166429″>
a) 0.0507 kJ/K
b) 0.0607 kJ/K
c) 0.0707 kJ/K
d) 0.0807 kJ/K
Answer: b
Clarification: m2u2 + P2V2 = m2h2 = mAuA + mBuB+ PV1 = mAhA + mBhB
hA= 84.41 kJ/kg, sA= 0.2965 kJ/kg K; hB = 419.32 kJ/kg, sB= 1.3065 kJ/kg K
h2 =(mA/m2)hA + (mB/m2)hB = (84.41/2) + (419.32/2) = 251.865 kJ/kg
h2 = 251.865 kJ/kg & P2 = 500 kPa; T2 = 60.085°C, s2 = 0.83184 kJ/kg K
1S2(gen) = m2s2 − mAsA – mBsB = 2 × 0.83184 – 1 × 0.2965 – 1 × 1.3065
= 0.0607 kJ/K.

11. A 4 L jug of milk at 25°C is placed in refrigerator where it is cooled down to a temperature of 5°C. Assuming the milk has the property of liquid water, find the entropy generated in the cooling process.
a) 0.0215 kJ/K
b) 0.0315 kJ/K
c) 0.0415 kJ/K
d) 0.0515 kJ/K
Answer: c
Clarification: v1 = vf = 0.001003 m3/kg, h = hf = 104.87 kJ/kg; sf = 0.3673 kJ/kg K
h = hf = 20.98 kJ/kg, s = sf = 0.0761 kJ/kg K
P = constant = 101 kPa => 1W2 = mP(v2 – v1);
m = V/v1 = 0.004 / 0.001003 = 3.988 kg
1Q2 = m(h2 − h1) = 3.988 (20.98 – 104.87) = -3.988 × 83.89 = -334.55 kJ
1S2(gen) = m(s2 − s1) − 1Q2/T(refrig)
= 3.988 (0.0761 − 0.3673) − (−334.55 / 278.15) = − 1.1613 + 1.2028
= 0.0415 kJ/K.

12. A pan contains 5 L of engine oil at 20°C, 100 kPa. Now 2 L of hot 100°C oil is mixed into the pan. Find the entropy generation.
a) 0.0728 kJ/K
b) 0.0828 kJ/K
c) 0.0928 kJ/K
d) 0.1028 kJ/K
Answer: a
Clarification: ρ = 885 kg/m3; From energy equation,
T2 = (mA/m2)TA + (mB/m2)TB = (5/7)20 + (2/7)100 = 42.868°C = 316.02 K
S2 – S1 = m2s2 − mAsA – mBsB = mA(s2 – sA) + mB(s2 – sB)
= 0.005 × 885 × 1.9 ln (316.02/293.15) + 0.002 × 885 × 1.9 ln (316.02/373.15)
= 0.6316 – 0.5588 = + 0.0728 kJ/K.

13. Argon in a light bulb is at 90 kPa and heated from 20°C to 60°C with electrical power. Find the total entropy generation per unit mass of argon.
a) 0.01 kJ/kg K
b) 0.02 kJ/kg K
c) 0.03 kJ/kg K
d) 0.04 kJ/kg K
Answer: d
Clarification: 1s2(gen) = s2 – s1 = Cp ln (T2/T1) – R ln (P2/ P1)
= Cp ln (T2/T1) – R ln (T2/ T1) = Cv ln(T2/T1)
= 0.312 ln [ (60 + 273)/(20 + 273) ] = 0.04 kJ/kg K.

14. Oxygen gas in a piston cylinder at 300 K, 100 kPa with volume 0.1m^3 is compressed in a reversible adiabatic process to a final temperature of 700 K. Find the final pressure and volume.
a) 2015 kPa, 0.0116 m3
b) 3015 kPa, 0.0216 m3
c) 1015 kPa, 0.0416 m3
d) 4015 kPa, 0.0216 m3
Answer: a
Clarification: Process: Adiabatic 1q2 = 0, Reversible 1s2 gen = 0
Entropy Eq.: s2 – s1 = ∫ dq/T + 1s2 gen = 0
∴s2 = s1 (isentropic compression process)
P2 = P1( T2 / T1)^(k/k-1) = 2015 kPa
V2 = V1( T2 / T1)^(1/1-k) = 0.1 × (700/300)^(1/1−1.393)
= 0.0116 m^3.

15. Argon in a light bulb is at 90 kPa and heated from 20°C to 60°C with electrical power. Find the total entropy generation per unit mass of argon.
a) 0.02 kJ/kg K
b) 0.03 kJ/kg K
c) 0.04 kJ/kg K
d) 0.05 kJ/kg K
Answer: c
Clarification: Energy Eq. : m(u2 – u1) = 1W2 electrical
Entropy Eq.: s2 – s1 = ∫ dq/T + 1s2 gen = 1s2 gen
Process: v = c & ideal gas ∴ P2/ P1 = T2/T1
1s2 gen = s2 – s1 = Cp ln(T2/T1) – R ln (P2/P1)
= Cpln(T2/T1) – R ln(T2/T1) = Cv ln(T2/T1)
= 0.312 ln{(60+273)/(20+273)} = 0.04 kJ/kg K.

Thermodynamics for various tests,

Thermodynamics Questions & Answers for campus interviews on “Second Law Efficiency-2”.

1. As Ta approaches Tr, second law efficiency
a) half
b) first law efficiency
c) zero
d) unity
Answer: d
Clarification: The lower the Ta, lower will be second law efficiency, here Tr=source temperature and Ta=use temperature.

2. Which of the following statement explains the concept of energy cascading?
a) the fuel should first be used for high temperature applications
b) the heat rejected from these applications can then be cascaded to applications at lower temperatures
c) it ensures more efficient energy utilization
d) all of the mentioned
Answer: d
Clarification: These are the main features of energy cascading.

3. Second law efficiency of different components can be expressed in different forms.
a) true
b) false
Answer: a
Clarification: It is derived using the exergy balance rate.

4. A steam turbine inlet is at 1200 kPa and 500°C. The actual exit is at 300 kPa having an actual work of 407 kJ/kg. Find its second law efficiency?
a) 0.88
b) 0.98
c) 0.78
d) 0.68
Answer: b
Clarification: To = 25°C = 298.15 K, hi = 3476.28 kJ/kg; si = 7.6758 kJ/kg K
he = hi – w(ac) = 3476.28 – 407 = 3069.28 kJ/kg
Te = 300°C; se = 7.7022 kJ/kg K
wrev = (hi – Tosi) – (he – Tose) = (hi – he) + To(se – si)
= (3476.28 – 3069.28) + 298.15(7.7022 – 7.6758)
= 407 + 7.87 = 414.9 kJ/kg
second law efficiency = w(ac)/w(rev) = 407 / 414.9 = 0.98.

5. A heat exchanger increases the availability of 3 kg/s water by 1650 kJ/kg by using 10 kg/s air which comes in at 1400 K and leaves with 600 kJ/kg less availability. What is the second law efficiency?
a) 0.625
b) 0.825
c) 0.925
d) 0.725
Answer: b
Clarification: I = Φ(destruction) = Φ(in) – Φ(out) = 10 × 600 – 3 × 1650 = 1050 kW
second law efficiency = Φ(out)/Φ(in) = (3 × 1650)/(10 × 600)
= 0.825.

6. A heat engine receives 1 kW heat at 1000 K and gives out 600 W as work with the rest as heat to the ambient. Find the second law efficiency.
a) 0.655
b) 0.755
c) 0.855
d) 0.955
Answer: c
Clarification: First law efficiency = 0.6/1 = 0.6
Φ(H) = [1-(To/Th)]Q = [1-(298.15/100)](1) = 0.702 kW
second law efficiency = 0.6/ 0.702 = 0.855.

7. A heat pump has a COP of 2 using a power input of 2 kW. Its low temperature is To and high temperature is 80°C, with an ambient at To. Find the second law efficiency.
a) 0.11
b) 0.41
c) 0.51
d) 0.31
Answer: d
Clarification: Φ(H) = [1-(To/Th)]Q, Q = β*W = 2*2 = 4 kW
second law efficiency = Φ(H)/W = [1-(298.15/353.15)](4/2)
= 0.31.

8. The condenser in a refrigerator receives R-134a at 50°C, 700 kPa and it exits as saturated liquid at 25°C. The flow-rate is 0.1 kg/s and air flows in the condenser at ambient 15°C and leaving at 35°C. Find the heat exchanger second-law efficiency. thermodynamics-questions-answers-campus-interviews-q8″ alt=”thermodynamics-questions-answers-campus-interviews-q8″ width=”203″ height=”85″ class=”alignnone size-full wp-image-166463″ srcset=”/2017/06/thermodynamics-questions-answers-campus-interviews-q8 203w, /2017/06/thermodynamics-questions-answers-campus-interviews-q8-200×85 200w” sizes=”(max-width: 203px) 100vw, 203px”>
a) 0.77
b) 0.87
c) 0.47
d) 0.67
Answer: a
Clarification: m1h1 + mah3 = m1h2 + mah4 (here m is the mass flow-rate)
ma = m1 × (h1 – h2)/(h4 – h3) = 0.1 × (436.89 – 234.59)/(1.004(35 – 15))
= 1.007 kg/s
ψ1 – ψ2 = h1 – h2 – T0(s1 – s2) = 436.89 – 234.59 – 288.15(1.7919 – 1.1201)
= 8.7208 kJ/kg
ψ4 – ψ3 = h4 – h3 – T0(s4 – s3) = 1.004(35 – 15) – 288.15 × 1.004 × ln (308.15/288.15)
= +0.666 kJ/kg
η(II) = ma(ψ4 – ψ3)/m1(ψ1 – ψ2) = 1.007(0.666)/[0.1(8.7208)] = 0.77.

9. Steam enters a turbine at 550°C, 25 MPa and exits at 5 MPa, 325°C at a flow rate of 70 kg/s. Determine the second law efficiency.
a) 0.68
b) 0.78
c) 0.88
d) 0.98
Answer: c
Clarification: hi = 3335.6 kJ/kg, si = 6.1765 kJ/kg K,
he = 2996.5 kJ/kg, se = 6.3289 kJ/kg K
Actual turbine: w,ac = hi – he = 339.1 kJ/kg
Rev. turbine: w,rev = w,ac + T0(se – si) = 339.1 + 45.44 = 384.54 kJ/kg
η(II) = w,ac/w,rev = 339.1/384.54 = 0.88.

10. A compressor is used to bring saturated water vapour initially at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the second-law efficiency.
a) 0.651
b) 0.751
c) 0.851
d) 0.951
Answer: d
Clarification: hi = 2778.1 kJ/kg, si = 6.5864 kJ/kg K
Actual compressor: h(e,ac) = 3693.9 kJ/kg, s(e,ac) = 6.7356 kJ/kg K
-w(c,ac) = h(e,ac) – hi = 915.8 kJ/kg
i = T0[s(e,ac) – si] = 298.15 (6.7356 – 6.5864) = 44.48 kJ/kg
w(rev) = i + w(c,ac) = -915.8 + 44.48 = -871.32 kJ/kg
η(II) = -w(rev)/w(c,ac) = 871.32/915.8 = 0.951.

Thermodynamics for campus interviews,

Thermodynamics Problems on “Maxwell’s Equations and TDS Equations”.

1. If a relation exists among variables x,y,z then z may be expressed as a function of x and y as, dz=Mdx+Ndy .
a) true
b) false
Answer: a
Clarification: Here, M,N and z are functions of x and y.

2. A pure substance which exists in a single phase has ____ independent variables.
a) zero
b) one
c) two
d) three
Answer: c
Clarification: Of all the quantities, any one can be expressed as a function of any two others.

3. Which of the following relation is correct?
a) dU=TdS-pdV
b) dH=TdS+Vdp
c) dG=Vdp-SdT
d) all of the mentioned
Answer: d
Clarification: These relations are true for a pure substance which undergoes an infinitesimal reversible process.

4. Maxwell’s equations consists of ____ equations.
a) four
b) three
c) two
d) one
Answer: a
Clarification: Maxwell’s equations consists of four equations.

5. Which of the following is not a Maxwell equation?
a) (∂T/∂V) = -(∂p/∂S)
b) (∂T/∂p) = -(∂V/∂S)
c) (∂p/∂T) = (∂S/∂V)
d) (∂V/∂T) = -(∂S/∂p)
Answer: b
Clarification: The correct equation is (∂T/∂p) = (∂V/∂S).

6. The condition for exact differential is
a) (∂N/∂y) = (∂M/∂x)
b) (∂M/∂y) = (∂N/∂x)
c) (∂M/∂y) = -(∂N/∂x)
d) all of the mentioned
Answer: b
Clarification: This is the condition for perfect or exact differential and here M and N are the functions of x and y.

7. The first TdS equation is
a) TdS=Cv*dT + T(∂T/∂p)dV
b) TdS=Cv*dT – T(∂p/∂T)dV
c) TdS=Cv*dT + T(∂p/∂T)dV
d) TdS=Cv*dT – T(∂T/∂p)dV
Answer: c
Clarification: This equation comes when entropy is defined as a function of T and V and using Cv and Maxwell’s third equation.

8. The second TdS equation is
a) TdS=Cp*dT + T(∂V/∂T)dp
b) TdS=Cp*dT – T(∂V/∂T)dp
c) TdS=Cp*dT + T(∂T/∂V)dp
d) TdS=Cp*dT – T(∂T/∂V)dp
Answer: b
Clarification: This equation comes when entropy is defined as a function of T and p and using Cp and Maxwell’s fourth equation.

9. Which of the following is true?
a) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= infinity
b) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= 0
c) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= 1
d) (∂p/∂V)*(∂V/∂T)*(∂T/∂p)= -1
Answer: d
Clarification: This is the relation between the thermodynamic variables, p,V and T.

10. For getting TdS equations, we assume entropy to be a function of T and V and also of T and p.
a) true
b) false
Answer: a
Clarification: For first TdS equation, we assume entropy as a function of T and V and for second TdS equation, we assume entropy as a function of T and p .

To practice problems on all areas of Thermodynamics,

Thermodynamics Multiple Choice Questions on “Binary Vapor Cycles”.

1. Which of the following fluid can be used in place of water?
a) diphenyl ether
b) aluminium bromide
c) mercury
d) all of the mentioned
Answer: d
Clarification: These fluids are better than water in high temperature range.

2. Which of the following statement is true?
a) only mercury has been used in place of water
b) diphenyl ether decomposes at high temperatures
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Also aluminium bromide is a possibility which can be considered.

3. Mercury is a better fluid in high temperature range.
a) true
b) false
Answer: a
Clarification: This is because its vaporization pressure is relatively low even at higher temperatures.

4. Why is mercury unsuitable at low temperatures?
a) its saturation pressure becomes very high
b) its specific volume is very low at such a high pressure
c) both of the mentioned
d) none of the mentioned
Answer: d
Clarification: Its saturation pressure becomes very low and specific volume is very large.

5. In a binary cycle, ____ cycles with ____ working fluid are coupled.
a) two, same
b) two, different
c) three, same
d) three, different
Answer: b
Clarification: In a binary cycle, heat rejected by one can be utilized by other.

6. To vaporize one kg of water, ____ kg of mercury must condense.
a) 5-6
b) 6-7
c) 7-8
d) 8-9
Answer: c
Clarification: This depends on the properties of mercury and water.

7. When mercury cycle is added to the steam cycle,
a) the mean temperature of heat addition increases
b) efficiency decreases
c) maximum pressure is high
d) all of the mentioned
Answer: a
Clarification: The increase in the mean temperature of heat addition increases the efficiency and the maximum pressure is also low.

8. Initially in a reciprocating steam engine,
a) a binary cycle was used
b) steam was used in the high temperature
c) ammonia or sulphur dioxide was used in the low temperature range
d) all of the mentioned
Answer: d
Clarification: Steam from engine at higher temperature and pressure was used to evaporate sulphur dioxide or ammonia which operated in another cycle.

9. In a mercury-steam cycle, mercury cycle is called ____ and steam cycle is called ____
a) bottoming cycle, topping cycle
b) topping cycle, bottoming cycle
c) both are called bottoming cycle
d) both are called topping cycle
Answer: b
Clarification: The mercury-steam cycle represents a two-fluid cycle.

10. We can get a tertiary cycle if a sulphur dioxide cycle is added to mercury-steam cycle.
a) true
b) false
Answer: a
Clarification: In this three-fluid cycle, sulphur dioxide cycle is added in the low temperature range.

Thermodynamics Multiple Choice Questions on “Psychrometric Chart and Process”.

1. Which of the following statement is true?
a) the chart is plotted for pressure equal to 760mm Hg
b) the constant wbt line represents adiabatic saturation process
c) the constant wbt line coincides with constant enthalpy line
d) all of the mentioned
Answer: d
Clarification: All these come from the psychrometric chart.

2. In sensible heating or cooling,
a) work done remains constant
b) dry bulb temperature or air remains constant
c) both of the mentioned
d) none of the mentioned
Answer: a
Clarification: The dry bulb temperature of air changes.

3. When humidity ratio of air ____ air is said to be dehumidified.
a) increases
b) decreases
c) remains constant
d) none of the mentioned
Answer: b
Clarification: when it increases, air is said to be humidified.

4. Air can be cooled and dehumidified by
a) circulating chilled water in tube across air flow
b) placing evaporator coil across air flow
c) spraying chilled water to air
d) all of the mentioned
Answer: d
Clarification: These are the ways of cooling and dehumidifying air.

5. Cooling and dehumidification of air is done in summer air conditioning.
a) true
b) false
Answer: a
Clarification: This is a common process in summer air conditioning.

6. Heating and humidification is done in
a) summer air conditioning
b) winter air conditioning
c) both of the mentioned
d) none of the mentioned
Answer: b
Clarification: This is opposite to summer air conditioning.

7. Which of the following is an absorbent?
a) silica gel
b) activated alumina
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: Both of these are examples of absorbents.

8. When air passes through silica gel,
a) it absorbs water vapour molecules
b) latent heat of condensation is released
c) dbt of air increases
d) all of the mentioned
Answer: d
Clarification: This process is called chemical dehumidification.

9. In adiabatic evaporative cooling, heat transfer between chamber and surroundings is ____
a) zero
b) high
c) low
d) none of the mentioned
Answer: a
Clarification: No heat transfer takes place between chamber and surroundings in adiabatic evaporative cooling.

10. The cooling tower uses the phenomenon of evaporative cooling to cool warm water above the dbt of air.
a) true
b) false
Answer: b
Clarification: The cooling tower uses the phenomenon of evaporative cooling to cool warm water below the dbt of air.

11. Cooling towers are rated in terms of
a) approach
b) range
c) both of the mentioned
d) none of the mentioned
Answer: c
Clarification: These are the two factors considered.