Thermodynamics Questions and Answers – Energy Equation-3

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This set of Advanced Thermodynamics Questions & Answers focuses on “Energy Equation-3”.

1. Superheated R-134a at 0.5 MPa, 20°C is cooled in a piston-cylinder at constant temperature to a final two-phase state with quality of 50%. The refrigerant mass is 5 kg, and during the process 500 kJ of heat is removed. Find the necessary work.
a) -67.9 kJ
b) -77.9 kJ
c) -87.9 kJ
d) -97.9 kJ
View Answer

Answer: c
Explanation: Energy Eq.: m(u2 -u1) = 1Q2 – 1W2 = -500 – 1W2
State 1: T1,P1, v1 = 0.04226 m3/kg ; u1 = 390.52 kJ/kg
=> V1 = mv1 = 0.211 m3
State 2: T2 , x2 ⇒ u2 = 227.03 + 0.5 × 162.16 = 308.11 kJ/kg,
v2 = 0.000817 + 0.5 × 0.03524 = 0.018437 m3/kg
=> V2 = m(v2) = 0.0922 m3
work = -500 – m(u2 – u1) = -500 – 5 × (308.11 – 390.52) = -87.9 kJ.
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2. Air at 600 K flows with 3 kg/s into a heat exchanger and out at 100°C. How much (kg/s) water coming in at 100 kPa, 20°C can the air heat to the boiling point?
a) 0.37 kg/s
b) 0.17 kg/s
c) 0.27 kg/s
d) 0.57 kg/s
View Answer

Answer: c
Explanation: C.V. :Heat Exchanger, No external heat transfer and no work.
Writing the Steady State Energy Equation (SSEE) and putting values,
we get the water flow rate at the exit is 0.27 kg/s.

3. Nitrogen gas flows into a convergent nozzle at 200 kPa, 400 K and very low velocity. It flows out of the nozzle at 100 kPa, 330 K. If the nozzle is insulated find the exit velocity.
a) 681.94 m/s
b) 581.94 m/s
c) 481.94 m/s
d) none of the mentioned
View Answer

Answer: d
Explanation: C.V.: Nozzle; steady state; one inlet and exit flow; insulated so it is adiabatic.
SSEE: h1 + 0 = h2 + [(V2)2] / 2
[(V2)2] = 2(h1 – h2) = 2Cp(T1 – T2)
= 2 × 1.042 (400 – 330)
= 145.88 kJ/kg = 145 880 J/kg
V2 = 381.94 m/s.
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4. A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350°C and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Find the specific work.
a) 382.3 kJ/kg
b) 482.3 kJ/kg
c) 582.3 kJ/kg
d) 682.3 kJ/kg
View Answer

Answer: b
Explanation: SSEE is W/m = (h1 – h2) + [(V1)2 – (V2)2]/2 + g(z1 – z2)
here z1=z2 and V2=0 hence w = (h1 – h2) + [(V1)2]/2
h1 = 3157.65 kJ/kg, h2 = 2675.46 kJ/kg
wT = 3157.65 – 2675.46 + ½ (152/1000) = 482.3 kJ/kg.

5. A steam turbine has an inlet of 2 kg/s water at 1000 kPa, 350°C and velocity of 15 m/s. The exit is at 100 kPa, x = 1 and very low velocity. Find the power produced.
a) 664.6 kW
b) 764.6 kW
c) 864.6 kW
d) 964.6 kW
View Answer

Answer: d
Explanation: SSEE is W/m = (h1 – h2) + [(V1)2 – (V2)2]/2 + g(z1 – z2)
here z1=z2 and V2=0 hence w = (h1 – h2) + [(V1)2]/2
h1 = 3157.65 kJ/kg, h2 = 2675.46 kJ/kg
wT = 3157.65 – 2675.46 + ½ (152/1000) = 482.3 kJ/kg
thus power produced = (2 kg/s)(482.3 kJ/kg) = 964.6 kW.
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6. 10kg of water in a piston-cylinder exists as saturated liquid/vapour at 100 kPa, with a quality of 50%. It is now heated till the volume triples. The mass of piston is such that a cylinder pressure of 200 kPa will float it. Find the heat transfer in the process.
a) 23961 kJ
b) 24961 kJ
c) 25961 kJ
d) 26961 kJ
View Answer

Answer: c
Explanation: m(u2 − u1) = 1Q2 − 1W2
Process: v = constant until P = Plift , then P is constant.
State 1: Two-phase; u1 = 417.33 + 0.5 × 2088.72 = 1461.7 kJ/kg
and v1 = 0.001043 + 0.5 × 1.69296 = 0.8475 m3/kg
State 2: v2, P2 ≤ Plift => v2 = 3 × 0.8475 = 2.5425 m3/kg ;
Interpolate: T2 = 829°C, u2 = 3718.76 kJ/kg
=> V2 = mv2 = 25.425 m3
1W2 = P(lift)(V2 −V1) = 200 × 10 (2.5425 − 0.8475) = 3390 kJ
1Q2 = m(u2 − u1) + 1W2 = 10×(3718.76 − 1461.7) + 3390 = 25961 kJ.
7. A 1L capsule of water at 150°C, 700 kPa is placed in a larger insulated (otherwise evacuated) vessel. The capsule breaks resulting which its contents fill the entire volume. If the final pressure is not to exceed 125 kPa, find the vessel volume?
a) 115 L
b) 125 L
c) 135 L
d) 145 L
View Answer
Answer: a
Explanation: m2 = m1 = m = V/v1 = 0.916 kg
Process: expansion with 1Q2 = 0, 1W2 = 0
Energy: m(u2 – u1) = 1Q2 – 1W2 = 0 ⇒ u2 = u1
State 1: v1 = vf = 0.001091 m3/kg; u1 = uf = 631.66 kJ/kg
State 2: P2 , u2 ⇒ x2 =(631.66 – 444.16)/2069.3 = 0.09061
v2 = 0.001048 + 0.09061 × 1.37385 = 0.1255 m3/kg
V2 = m(v2) = 0.916 × 0.1255 = 0.115 m3 = 115 L.

8. A vertical cylinder fitted with a piston contains 5 kg of R-22 at 10°C. Heat is transferred causing the piston to rise until the volume has doubled. Additional heat is transferred until the temperature inside reaches 50°C, at which point the pressure inside the cylinder is 1.3 MPa. Find the work done.
a) 34.1 kJ
b) 44.1 kJ
c) 54.1 kJ
d) 64.1 kJ
View Answer

Answer: a
Explanation: Process: 1 -> 2 -> 3
As piston floats, pressure is constant (1 -> 2) and the volume is constant for the second part (2 -> 3). So we have: v3 = v2 = 2 × v1
State 3: (P,T) v3 = 0.02015 m3/kg, u3 = 248.4 kJ/kg
v1 = 0.010075 = 0.0008 + x1 × 0.03391 => x1 = 0.2735
u1 = 55.92 + 0.2735 × 173.87 = 103.5 kJ/kg
State 2: v2 = 0.02015 m3/kg, P2 = P1 = 681 kPa this is still 2-phase
Work = P1(V2 – V1) = 681 × 5 (0.02 – 0.01) = 34.1 kJ.
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9. A 250L rigid tank contains methane at 1500 kPa, 500 K. It is now cooled down to 300K. Find the heat transfer.
a) –402.4 kJ
b) –502.4 kJ
c) –602.4 kJ
d) –702.4 kJ
View Answer

Answer: b
Explanation: Assume ideal gas, P2 = P1 × (Τ2 / Τ1) = 1500 × 300 / 500 = 900 kPa
m = P1V/RT1 =(1500 × 0.25)/(0.5183 × 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.

10. A rigid container has 2kg of carbon dioxide gas at 1200 K, 100 kPa that is heated to 1400 K. Find the heat transfer using heat capacity.
a) 231.2 kJ
b) 241.2 kJ
c) 251.2 kJ
d) 261.2 kJ
View Answer

Answer: d
Explanation: Energy Eq.: U2 – U1 = m (u2- u1) = 1Q2 − 1W2
Process: ∆V = 0 ⇒ 1W2 = 0
For constant heat capacity we have: u2- u1 = Cv (T2- T1)
1Q2 = mCv (T2- T1) = 2 × 0.653 × (1400 –1200) = 261.2 kJ.
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11. A piston cylinder contains 3kg 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) 941 kJ
b) 951 kJ
c) 961 kJ
d) 971 kJ
View Answer

Answer: a
Explanation: Ideal gas PV = mRT
P2V2 = mRT2; V2 = mR T2 / P2 = 3×0.287×600 / 300 = 1.722 m3
Process: P = constant, work = ⌠ PdV = P (V2 – V1) = 300 (1.722 – 0.8413) = 264.2 kJ
Energy equation: U2 – U1 = 1Q2 – 1W2 = m(u2 – u1)
Q2 = U2 – U1 + 1W2 = 3(435.097 – 209.45) + 264.2 = 941 kJ.

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