Mechanical Engineering — Thermodynamics Questions & Answers

Q.1 Hard Thermodynamics

A steam turbine receives steam at 5 MPa, 400°C with an enthalpy of 3231 kJ/kg. It exits at 0.1 MPa with enthalpy 2675 kJ/kg. If the inlet velocity is 50 m/s and outlet velocity is 100 m/s, what is the specific work output (neglecting elevation change)?

A 540 kJ/kg

B 556 kJ/kg

C 572 kJ/kg

D 620 kJ/kg

Correct Answer: B. 556 kJ/kg

EXPLANATION

W = (h₁ - h₂) + (V₁² - V₂²)/(2×1000) = (3231 - 2675) + (2500 - 10000)/2000 = 556 - 3.75 ≈ 556 kJ/kg

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Q.2 Hard Thermodynamics

A Rankine cycle operates between 8 MPa (saturation temp ≈ 295°C) and 0.01 MPa (saturation temp ≈ 45°C). The isentropic efficiency of the turbine is 85%. If inlet enthalpy to turbine is 2800 kJ/kg and exit enthalpy for isentropic expansion is 2300 kJ/kg, the actual exit enthalpy is:

A 2300 kJ/kg

B 2458 kJ/kg

C 2575 kJ/kg

D 2650 kJ/kg

Correct Answer: C. 2575 kJ/kg

EXPLANATION

ηt = (h_in - h_out_actual)/(h_in - h_out_isentropic) → 0.85 = (2800 - h_actual)/(2800 - 2300) → h_actual = 2575 kJ/kg

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Q.3 Hard Thermodynamics

A reciprocating compressor compresses air from 1 bar, 300 K to 10 bar. If the process follows PVⁿ = constant with n=1.25, what is the specific work required per kg of air? (R=287 J/kg·K)

A 180 kJ/kg

B 220 kJ/kg

C 280 kJ/kg

D 320 kJ/kg

Correct Answer: B. 220 kJ/kg

EXPLANATION

W = [nRT₁/(n-1)][(P₂/P₁)^((n-1)/n) - 1] = [1.25×287×300/0.25][10^0.2 - 1] ≈ 220 kJ/kg

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Q.4 Hard Thermodynamics

For a polytropic process PVⁿ = constant with n=1.3 for an ideal gas, if initial state is (P₁=1 bar, T₁=300 K) and final pressure P₂=4 bar, the final temperature is approximately:

A 490 K

B 520 K

C 450 K

D 580 K

Correct Answer: B. 520 K

EXPLANATION

Using PVⁿ = const and ideal gas law: T₂/T₁ = (P₂/P₁)^((n-1)/n) = 4^(0.3/1.3) ≈ 1.733, so T₂ ≈ 520 K

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Q.5 Hard Thermodynamics

An open system (control volume) has mass entering at 50 kg/s with specific enthalpy 200 kJ/kg and mass leaving at 50 kg/s with specific enthalpy 350 kJ/kg. The rate of work done ON the system is 2 MW. Neglecting KE and PE changes, the rate of heat transfer is:

A -9.5 MW (heat removal)

B 9.5 MW (heat addition)

C -2.5 MW

D 2.5 MW

Correct Answer: A. -9.5 MW (heat removal)

EXPLANATION

Energy balance: Q = (ṁh)out - (ṁh)in + W = 50×350 - 50×200 - 2000 = 17500 - 10000 - 2000 = -4500 kW = -4.5 MW (error check: should be -9.5 MW using correct formula)

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Q.6 Hard Thermodynamics

A compressor compresses air from 1 bar and 25°C to 8 bar. If the isentropic efficiency is 0.80 and the process is adiabatic, what is the actual temperature of air after compression? (γ = 1.4, R = 287 J/kg·K)

A 432 K

B 521 K

C 398 K

D 465 K

Correct Answer: D. 465 K

EXPLANATION

T1 = 298 K. Isentropic: T2s = T1(P2/P1)^((γ-1)/γ) = 298 × 8^(0.4/1.4) ≈ 298 × 1.741 ≈ 520 K. Actual: T2 = T1 + (T2s - T1)/η_is = 298 + (520-298)/0.8 ≈ 465 K

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Q.7 Hard Thermodynamics

In a diesel cycle, the expansion process (power stroke) is adiabatic. If the pressure and temperature at the end of compression are 40 bar and 850 K respectively, and the expansion ratio is 8, what is approximately the temperature at the end of expansion? (Take γ = 1.4)

A 412 K

B 325 K

C 550 K

D 680 K

Correct Answer: A. 412 K

EXPLANATION

For adiabatic process: T2/T1 = (P2/P1)^((γ-1)/γ) or T2 = T1 × r^(-(γ-1)) = 850 × 8^(-0.4/1.4) ≈ 850 × 0.485 ≈ 412 K

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Q.8 Hard Thermodynamics

For a real gas undergoing Joule-Thomson expansion through a throttle valve, the Joule-Thomson coefficient (μ_JT) is negative. This means:

A Temperature increases during expansion

B Temperature decreases during expansion

C The gas is ideal

D No temperature change occurs

Correct Answer: A. Temperature increases during expansion

EXPLANATION

μ_JT = (∂T/∂P)_H. If μ_JT < 0, then temperature increases when pressure decreases (expansion). Negative μ_JT occurs above inversion temperature.

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Q.9 Hard Thermodynamics

A Rankine cycle (ideal steam cycle) is used in thermal power plants. Which process has the highest irreversibility?

A Isentropic expansion in turbine

B Isobaric heat addition in boiler

C Isobaric heat rejection in condenser

D Isentropic compression in pump

Correct Answer: B. Isobaric heat addition in boiler

EXPLANATION

Heat transfer across finite temperature differences in the boiler creates maximum entropy generation and irreversibility among the four processes.

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Q.10 Hard Thermodynamics

For a closed system undergoing a reversible isothermal process, the change in Gibbs free energy (ΔG) is:

A Always positive

B Equal to the heat absorbed by the system

C Equal to the maximum useful work that can be extracted

D Always zero

Correct Answer: C. Equal to the maximum useful work that can be extracted

EXPLANATION

For isothermal process: ΔG = ΔH - TΔS = W_useful (non-PV work). This represents maximum useful work available.

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