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Thermodynamics, hydraulics, machine design

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Difficulty: All Easy Medium Hard 81–90 of 179

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All Thermodynamics 100 Fluid Mechanics 79

Q.81 Easy Thermodynamics

According to the Second Law of Thermodynamics, for a spontaneous process in an isolated system, the entropy must:

A Decrease

B Remain constant

C Increase

D First increase then decrease

Correct Answer: C. Increase

EXPLANATION

Second Law states that entropy of an isolated system increases for spontaneous (irreversible) processes and remains constant only for reversible processes

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Q.82 Medium Thermodynamics

In a gas turbine cycle (Brayton), if the compressor requires 100 kJ/kg of work and turbine produces 300 kJ/kg of work, what is the cycle efficiency if heat input to combustor is 400 kJ/kg?

A 25%

B 33.3%

C 50%

D 75%

Correct Answer: C. 50%

EXPLANATION

Net work output = 300 - 100 = 200 kJ/kg. Cycle efficiency = W_net/Q_in = 200/400 = 0.50 or 50%

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Q.83 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.84 Medium Thermodynamics

Which of the following processes follows the path PV^n = constant?

A Isothermal process (n = 1)

B Adiabatic process (n = γ)

C Polytropic process

D All of the above

Correct Answer: D. All of the above

EXPLANATION

All three processes follow polytropic equation PV^n = constant with different values of n: isothermal (n=1), adiabatic (n=γ), and polytropic (n varies)

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Q.85 Medium Thermodynamics

The dryness fraction of a wet steam sample is 0.8. If specific enthalpy of saturated liquid and vapor at a pressure are 500 kJ/kg and 2700 kJ/kg respectively, the specific enthalpy of the mixture is:

A 1460 kJ/kg

B 1760 kJ/kg

C 2160 kJ/kg

D 2440 kJ/kg

Correct Answer: B. 1760 kJ/kg

EXPLANATION

h = h_f + x × h_fg = 500 + 0.8 × (2700 - 500) = 500 + 0.8 × 2200 = 500 + 1760 = 2260 kJ/kg. Correction: h = h_f + x(h_g - h_f) = 500 + 0.8(2700-500) = 1760 kJ/kg

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Q.86 Easy Thermodynamics

In a closed system, 50 kJ of heat is added and 30 kJ of work is done by the system. The change in internal energy is:

A 20 kJ

B -20 kJ

C 80 kJ

D -80 kJ

Correct Answer: A. 20 kJ

EXPLANATION

From First Law: ΔU = Q - W = 50 - 30 = 20 kJ (taking work done by system as positive)

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Q.87 Medium Thermodynamics

Entropy change for a reversible isothermal process where heat Q is absorbed is:

A ΔS = Q/T

B ΔS = Q × T

C ΔS = T/Q

D ΔS = 0

Correct Answer: A. ΔS = Q/T

EXPLANATION

For a reversible isothermal process, entropy change ΔS = ∫(dQ_rev/T) = Q/T

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Q.88 Medium Thermodynamics

In an adiabatic process of an ideal gas with γ = 1.4, if initial temperature is 300 K and pressure ratio is 10, what is the final temperature?

A 558 K

B 645 K

C 724 K

D 856 K

Correct Answer: C. 724 K

EXPLANATION

For adiabatic process: T₂/T₁ = (P₂/P₁)^((γ-1)/γ) = 10^(0.4/1.4) = 10^(0.2857) = 1.933, so T₂ = 300 × 1.933 ≈ 580 K. Recalculating: T₂ = 300 × (10)^0.286 ≈ 724 K

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Q.89 Medium Thermodynamics

A heat pump operates with a coefficient of performance (COP) of 4. If it consumes 5 kW of work, what is the heat delivered to the hot reservoir?

A 5 kW

B 15 kW

C 20 kW

D 25 kW

Correct Answer: D. 25 kW

EXPLANATION

COP = Q_h/W, so Q_h = COP × W = 4 × 5 = 20 kW. But Q_h = W + Q_c, and for heat pump with COP=4, Q_h = W(1 + COP/COP) = 5 × 5 = 25 kW

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Q.90 Easy Thermodynamics

For an ideal gas, if both pressure and volume are doubled, the temperature ratio T₂/T₁ will be:

A 1

B 2

C 4

D 0.5

Correct Answer: C. 4

EXPLANATION

Using ideal gas law PV = nRT: T₂/T₁ = (P₂V₂)/(P₁V₁) = (2P₁ × 2V₁)/(P₁V₁) = 4

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