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JNTU B.Tech II THERMODYNAMICS Previous Question Paper (October - 2020 - R18)

B.Tech | II | II-I | R18 | October - 2020

Answer PaperPaper Code: 153BZMaximum Marks: 75Time: 2 hours

Answer any five questions

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Main

1

A mass of 8 kg gas expands in a flexible container so that the p-v relationship is in the form of

pv1.2 = constant
. The initial pressure is 1000 kPa and the initial volume is 1 m3. The final pressure is 5 kPa. If the specific internal energy of the gas decreases by 40 kJ/kg, find the heat transfer in magnitude and direction.

15 marks

Answer

Given data:

  • Mass (m) = 8 kg
  • Initial pressure (P1) = 1000 kPa
  • Initial volume (V1) = 1 m3
  • Final pressure (P2) = 5 kPa
  • Polytropic index (n) = 1.2
  • Change in specific internal energy (Δu) = -40 kJ/kg

First, calculate the final volume (V2) using the polytropic relation P1V1n = P2V2n:

V2 = V1 * (P1 / P2)(1/n) = 1 * (1000 / 5)(1/1.2) = 2000.8333 ≈ 89.96 m3

Next, calculate the work done (W) during the polytropic expansion:

W = (P1V1 - P2V2) / (n - 1) = (1000 * 1 - 5 * 89.96) / (1.2 - 1) = (1000 - 449.8) / 0.2 = 2751 kJ

Calculate the total change in internal energy (ΔU):

ΔU = m * Δu = 8 kg * (-40 kJ/kg) = -320 kJ

Apply the First Law of Thermodynamics for a closed system (Q = ΔU + W):

Q = -320 kJ + 2751 kJ = 2431 kJ

The heat transfer is 2431 kJ. Since the value is positive, the heat is transferred to the system (heat addition).

2

A room for four persons has two fans, each consuming 0.18 kW power, and three 100 W lamps. Ventilation air at the rate of 80 kg/h enters with an enthalpy of 84 kJ/kg and leaves with an enthalpy of 59 kJ/kg. If each person puts out heat at the rate of 630 kJ/h, determine the rate at which heat is to be removed by a room cooler, so that a steady state is maintained in the room.

15 marks

Answer

To maintain a steady state in the room, the total rate of heat generation must be balanced by the rate of heat removal by the room cooler and the heat carried away by the ventilation air. The energy balance equation for the room is:

Q̇cooler = Q̇gen + Q̇ventilation

First, calculate the total heat generation rate (Q̇gen) in kW:

  • Fans: 2 × 0.18 kW = 0.36 kW
  • Lamps: 3 × 100 W = 3 × 0.1 kW = 0.3 kW
  • Persons: 4 × 630 kJ/h = 2520 kJ/h. Converting to kW: 2520 / 3600 = 0.7 kW
  • Total Q̇gen = 0.36 + 0.3 + 0.7 = 1.36 kW

Next, calculate the rate of heat removed by the ventilation air (Q̇ventilation). Note that since enthalpy decreases as air leaves, the ventilation air is actually removing heat from the room:

Q̇ventilation = ṁ(hin - hout)

Given ṁ = 80 kg/h = 80/3600 kg/s ≈ 0.02222 kg/s:

Q̇ventilation = (80/3600) kg/s × (84 - 59) kJ/kg = (80/3600) × 25 kW ≈ 0.5556 kW

Finally, calculate the heat to be removed by the room cooler (Q̇cooler):

Q̇cooler = 1.36 kW + 0.5556 kW = 1.9156 kW

The room cooler must remove heat at a rate of approximately 1.916 kW to maintain steady state.

315 marks
a

Prove that energy is a property.

7 marks

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b

Apply steady flow energy equation for nozzle, compressor and condenser.

8 marks

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415 marks
a

Derive the Maxwell’s relations.

7 marks

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b

Sketch the thermodynamic mnemonic diagram and explain its use to obtain Gibbs equations and Maxwell relations.

8 marks

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5

A steam pressure vessel of holding capacity 4 m3 contains a mixture of saturated water and saturated steam at 250°C. The mass of the liquid present is 1 ton. Determine (a) Quality; (b) Specific Volume; (c) Specific Enthalpy; (d) Specific Entropy and (e) Specific Internal Energy of steam.

15 marks

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6

Explain in detail about throttling and free expansion processes with a neat sketch and also mention their applications.

15 marks

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715 marks
a

State Avogadro’s Hypothesis.

7 marks

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b

State and explain Dalton’s law of partial pressure and prove the statement mathematically.

8 marks

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8

How does the vapour compression refrigeration system work? Explain with suitable diagrams.

15 marks

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