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

Explanation: Energy Eq.: m(u2 -u1) = 1Q2 – 1W2 = -500 – 1W2

State 1: T1,P1, v1 = 0.04226 m

^{3}/kg ; u1 = 390.52 kJ/kg

=> V1 = mv1 = 0.211 m

^{3}

State 2: T2 , x2 ⇒ u2 = 227.03 + 0.5 × 162.16 = 308.11 kJ/kg,

v2 = 0.000817 + 0.5 × 0.03524 = 0.018437 m

^{3}/kg

=> V2 = m(v2) = 0.0922 m

^{3}

work = -500 – m(u2 – u1) = -500 – 5 × (308.11 – 390.52) = -87.9 kJ.

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

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

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.

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

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

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.

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

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 m

^{3}/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 m

^{3}

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.

a) 115 L

b) 125 L

c) 135 L

d) 145 L

View Answer

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 m

^{3}/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 m

^{3}/kg

V2 = m(v2) = 0.916 × 0.1255 = 0.115 m

^{3}= 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

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 m

^{3}/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 m

^{3}/kg, P2 = P1 = 681 kPa this is still 2-phase

Work = P1(V2 – V1) = 681 × 5 (0.02 – 0.01) = 34.1 kJ.

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

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

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.

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

Explanation: Ideal gas PV = mRT

P2V2 = mRT2; V2 = mR T2 / P2 = 3×0.287×600 / 300 = 1.722 m

^{3}

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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