Thermodynamics Questions and Answers – Energy Equation-2

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This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Energy Equation-2”.

1. A 100L rigid tank contains nitrogen at 3 MPa, 900 K. The tank is then cooled to 100 K. What is the heat transfer for this process?
a) −490.7 kJ
b) −590.7 kJ
c) −690.7 kJ
d) −790.7 kJ
View Answer

Answer: c
Explanation: V = constant hence work = 0 ; Energy Eq: m(u2 – u1) = 1Q2 – 1W2
State 1: v1 = 0.0900 m3/kg => m = V/v1 = 1.111 kg, u1 = 691.7 kJ/kg
State 2: 100 K, v2 = v1 = V/m
interpolating for v (between 200kPa and 400kPa) we get
P2 = 200 + 200 (0.09 – 0.1425)/(0.0681 – 0.1425) = 341 kPa
u2 = 71.7 + (69.3 – 71.7)(0.09 – 0.1425)/(0.0681 – 0.1425) = 70.0 kJ/kg,
1Q2 = m(u2 – u1) = 1.111 (70.0 – 691.7) = −690.7 kJ.
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2. A rigid container has 0.75kg water at 1200 kPa, 300°C. Now, the water is cooled to a final pressure of 300 kPa. Find the heat transfer in the process.
a) -2348 kJ
b) -1348 kJ
c) -2148 kJ
d) -1148 kJ
View Answer

Answer: d
Explanation: Energy Eq.: U2 – U1 = 1Q2 – 1W2 ; V = constant hence work = 0
State 1: 300°C, 1200 kPa => superheated vapor,
thus v = 0.21382 m3/kg, u = 2789.22 kJ/kg
State 2: 300 kPa and v2 = v1 and v2 < vg two-phase we get T2 = Tsat = 133.55°C
x2 = (v2 – vf)/v(fg) = (0.21382 – 0.001073)/0.60475 = 0.35179
u2 = uf + (x2)*u(fg) = 561.13 + x2 1982.43 = 1258.5 kJ/kg
1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) = 0.75 (1258.5 – 2789.22) = -1148 kJ.

3. A cylinder fitted with a piston contains 2kg of superheated R-134a vapour at 100°C, 350 kPa. The cylinder is then cooled so that R-134a remains at constant pressure till it reaches a quality of 75%. Calculate the heat transfer in this process.
a) -174.6 kJ
b) -274.6 kJ
c) -374.6 kJ
d) -474.6 kJ
View Answer

Answer: b
Explanation: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: P = constant ⇒ 1W2 =⌡⌠PdV = P∆V = P(V2 – V1) = Pm(v2 – v1)
State 1: h1 = (490.48 + 489.52)/2 = 490 kJ/kg
State 2: h2 = 206.75 + 0.75 ×194.57 = 352.7 kJ/kg (350.9 kPa)
1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1)
1Q2 = 2 × (352.7 – 490) = -274.6 kJ.
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4. Ammonia at 0°C, quality 60% is contained in a 200L tank. The tank and ammonia is now heated to a final pressure of 1 MPa. Determine the heat transfer for the process.
a) 520.75 kJ
b) 620.75 kJ
c) 720.75 kJ
d) 820.75 kJ
View Answer

Answer: c
Explanation: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: Constant volume hence 1W2 = 0
State 1: two-phase state and v1 = 0.001566 + x1 × 0.28783 = 0.17426 m3/kg
u1 = 179.69 + 0.6 × 1138.3 = 862.67 kJ/kg, m = V/v1 = 0.2/0.17426 = 1.148 kg
State 2: P2 , v2 = v1 superheated vapor ⇒ T2 ≅ 100°C, u2 ≅ 1490.5 kJ/kg
1Q2 = m(u2 – u1) = 1.148(1490.5 – 862.67) = 720.75 kJ.

5. Water in a 150L closed, rigid tank is at 100°C, 90% quality. The tank is cooled to −10°C. Calculate the heat transfer during this process.
a) -163.3 kJ
b) -263.3 kJ
c) -363.3 kJ
d) -463.3 kJ
View Answer

Answer: b
Explanation: Energy Eq: m(u2 – u1) = 1Q2 – 1W2; Process: V = constant, 1W2 = 0
State 1: Two-phase thus v1 = 0.001044 + 0.9×1.6719 = 1.5057 m3/kg
and u1 = 418.94+0.9×2087.6 = 2297.8 kJ/kg
State 2: T2, v2 = v1 ⇒ mix of saturated solid + vapour
v2 = 1.5057 = 0.0010891 + x2 × 466.7 => x2 = 0.003224
u2 = -354.09 + 0.003224 × 2715.5 = -345.34 kJ/kg; m = V/v1 = 0.15/1.5057
= 0.09962 kg
1Q2 = m(u2 – u1) = 0.09962(-345.34 – 2297.8) = -263.3 kJ.
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6. A cylinder with constant volume of 0.1 L contains water at critical point. It then cools down to room temperature of 20°C. Calculate the heat transfer from the water.
a) -61.7 kJ
b) -71.7 kJ
c) -81.7 kJ
d) -91.7 kJ
View Answer

Answer: a
Explanation: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
Process: Constant volume ⇒ v2 = v1 hence work done is zero.
State 1: v1 = vc = 0.003155 m3/kg, u1 = 2029.6 kJ/kg and m = V/v1 = 0.0317 kg
State 2: T2, v2 = v1 = 0.001002 + x2 × 57.79
x2 = 3.7×10^(-5), u2 = 83.95 + x2 × 2319 = 84.04 kJ/kg
1Q2 = m(u2 – u1) = 0.0317(84.04 – 2029.6) = -61.7 kJ.

7. A constant pressure piston-cylinder contains 0.2 kg water as saturated vapour at 400 kPa. It is now cooled so that the water occupies half the original volume. Find the heat transfer in the process.
a) –203.9 kJ
b) –233.9 kJ
c) –223.9 kJ
d) –213.9 kJ
View Answer

Answer: d
Explanation: Energy Eq: m(u2 – u1) = 1Q2 – 1W2
and P = constant => 1W2 = Pm(v2 – v1)
thus 1Q2 = m(u2 – u1) + 1W2 = m(u2 – u1) + Pm(v2 – v1) = m(h2 – h1)
State 1: v1 = 0.46246 m3/kg; h1 = 2738.53 kJ/kg
State 2: v2 = v1 / 2 = 0.23123 = vf + x v(fg)
x2 = (v2 – vf) / v(fg) = (0.23123 – 0.001084) / 0.46138 = 0.4988
h2 = hf + (x2)*h(fg) = 604.73 + 0.4988 × 2133.81 = 1669.07 kJ/kg
1Q2 = 0.2 (1669.07 – 2738.53) = –213.9 kJ.
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8. 2kg water at 120°C with a quality of 25% has its temperature raised 20°C in a constant volume process. What is the heat transfer in the process?
a) 877.8 kJ
b) 887.8 kJ
c) 897.8 kJ
d) 907.8 kJ
View Answer

Answer: a
Explanation: Energy Eq.: m (u2 − u1 ) = 1Q2 − 1W2 and V = constant thus work is zero
State 1: T, x1 and v1 = vf + (x1)*v(fg) = 0.00106 + 0.25 × 0.8908 = 0.22376 m3/kg
u1 = uf + (x1)*u(fg) = 503.48 + 0.25 × 2025.76 = 1009.92 kJ/kg
State 2: T2, v2 = v1 < vg2 = 0.50885 m3/kg so two-phase
x2 = (v2 – vf2)/v(fg2) = (0.22376 – 0.00108)/0.50777 = 0.43855
u2 = u(f2) + (x2)*u(fg2) = 588.72 + x2 ×1961.3 = 1448.84 kJ/kg
From the energy equation, 1Q2 = m(u2 − u1) = 2 ( 1448.84 – 1009.92 ) = 877.8 kJ.

9. A 25 kg mass moving with 25 m/s is brought to a complete stop with a constant deceleration over a period of 5 seconds by a brake system. The brake energy is absorbed by 0.5kg water initially at 100 kPa, 20°C. Assume that the mass is at constant P and T. Find the energy the brake removes from the mass assuming P = C.
a) 7.6125 kJ
b) 7.7125 kJ
c) 7.8125 kJ
d) 7.9125 kJ
View Answer

Answer: c
Explanation: E2 – E1= ∆E = 0.5 mV2 = 0.5 × 25 × 25^(2)/1000 = 7.8125 kJ.
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10. An insulated cylinder fitted with a piston contains R-12 at 25°C with a quality of 90% and V=45 L. The piston moves and the R-12 expands until it exists as saturated vapour. During this, R-12 does 7kJ of work against the piston. Determine the final temperature, assuming that the process is adiabatic.
a) -5°C
b) -15°C
c) -25°C
d) -35°C
View Answer

Answer: b
Explanation: Energy Eq.: m(u2 − u1) = 1Q2 – 1W2
State 1: (T, x) => v1 = 0.000763 + 0.9 × 0.02609 = 0.024244 m3/kg
m = V1/v1 = 0.045/0.024244 = 1.856 kg and u1 = 59.21 + 0.9 × 121.03 = 168.137 kJ/kg
Q = 0 = m(u2 – u1) + 1W2 = 1.856 × (u2 – 168.137) + 7.0
=> u2 = 164.365 kJ/kg = ug at T2 and T2 comes out to be -15°C.

11. A reactor filled with water having volume 1 m3 is at 360°C, 20 MPa and placed inside a containment room which is well insulated and initially evacuated. Due to a failure, the reactor ruptures and water fills the room. Find the minimum room volume so that the final pressure does not exceed 200 kPa.
a) 257.7 m3
b) 267.7 m3
c) 277.7 m3
d) 287.7 m3
View Answer

Answer: d
Explanation: Mass: m2 = m1 = V(reactor)/v1 = 1/0.001823 = 548.5 kg
Energy: m(u2 – u1) = 1Q2 – 1W2 = 0 – 0 = 0 hence u2 = u1
State 1: v1 = 0.001823 m3/kg; u1 = 1702.8 kJ/kg which is also equal to u2
State 2: P2 = 200 kPa, u2 < ug hence => Two-phase
x2 = (u2 – uf)/u(fg) = (1702.8 – 504.47)/2025.02 = 0.59176
v2 = 0.001061 + 0.59176 × 0.88467 = 0.52457 m3/kg
V2 = m2 v2 = 548.5 ×0.52457 = 287.7 m3.

12. A piston-cylinder has the piston loaded with outside atmospheric pressure and piston mass to a pressure of 150 kPa. It contains water at −2°C, which is heated until the water becomes saturated vapour. Find the specific work for the process.
a) 163.7 kJ/kg
b) 173.7 kJ/kg
c) 183.7 kJ/kg
d) 193.7 kJ/kg
View Answer

Answer: b
Explanation: Energy Eq. per unit mass: u2 – u1 = 1q2 – 1w2
Process: P = constant = P1, => work = P1(v2 – v1)
State 1: T1 , P1 => saturated solid; v1 = 1.09×10^(-3) m3/kg, u1 = -337.62 kJ/kg
State 2: x = 1, P2 = P1 = 150 kPa; v2 = vg(P2) = 1.1593 m3/kg,
T2 = 111.4°C ; u2 = 2519.7 kJ/kg
work = P1(v2 -v1) = 150[1.1593 -1.09×10^(-3)] = 173.7 kJ/kg.

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