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Mechanical Engineering
Thermodynamics

Thermodynamics, hydraulics, machine design

100 Q 2 Topics Take Test
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Difficulty: All Easy Medium Hard 71–80 of 100
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79
Q.71 Medium Thermodynamics
What is the dryness fraction (quality) of steam at a state where internal energy u = 2400 kJ/kg, u_f = 1317.3 kJ/kg, and u_fg = 1753.7 kJ/kg?
A 0.62
B 0.45
C 0.55
D 0.38
Correct Answer:  A. 0.62
EXPLANATION

u = u_f + x × u_fg, so x = (u - u_f)/u_fg = (2400 - 1317.3)/1753.7 = 1082.7/1753.7 ≈ 0.62

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Q.72 Medium Thermodynamics
Which of the following processes is impossible according to the second law of thermodynamics?
A A process where a system loses heat to surroundings at lower temperature
B A process where total entropy of universe remains constant
C A process where a system absorbs heat and converts it completely into work
D A process where entropy of universe decreases
Correct Answer:  D. A process where entropy of universe decreases
EXPLANATION

The second law states that entropy of an isolated system must increase or remain constant (reversible). A decrease in total entropy violates the second law and is impossible.

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Q.73 Hard Thermodynamics
In a gas turbine Brayton cycle, air enters the compressor at 300 K and 100 kPa. The pressure ratio is 8. If γ = 1.4 and R = 287 J/kg·K, find the compressor outlet temperature (assuming isentropic compression).
A 477.5 K
B 520.2 K
C 598.7 K
D 456.3 K
Correct Answer:  B. 520.2 K
EXPLANATION

T_2/T_1 = (P_2/P_1)^((γ-1)/γ) = 8^(0.4/1.4) = 8^0.2857 ≈ 1.734. T_2 = 300 × 1.734 ≈ 520.2 K

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Q.74 Easy Thermodynamics
A reversible adiabatic process for an ideal gas is also known as:
A Isothermal process
B Isobaric process
C Isentropic process
D Isochoric process
Correct Answer:  C. Isentropic process
EXPLANATION

A reversible adiabatic process has constant entropy (dS = 0), making it isentropic. This is a key assumption in many thermodynamic analysis for ideal processes.

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Q.75 Easy Thermodynamics
What is the relationship between specific heats C_p and C_v for an ideal gas?
A C_p = C_v
B C_p - C_v = R
C C_p + C_v = R
D C_p × C_v = R
Correct Answer:  B. C_p - C_v = R
EXPLANATION

The Mayer relation: C_p - C_v = R, where R is the specific gas constant. This relation holds for all ideal gases.

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Q.76 Medium Thermodynamics
In the Rankine cycle, which process involves expansion in a turbine?
A Isobaric process
B Isochoric process
C Isentropic process
D Isothermal process
Correct Answer:  C. Isentropic process
EXPLANATION

In an ideal Rankine cycle, turbine expansion is isentropic (reversible and adiabatic), maximizing work output. Real turbines follow this closely but with some irreversibilities.

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Q.77 Medium Thermodynamics
A piston-cylinder device contains 0.5 kg of steam at 200°C. Heat is removed and the steam condenses to saturated liquid at the same temperature. The latent heat of vaporization at 200°C is 1941 kJ/kg. Calculate heat removed.
A 970.5 kJ
B 1941 kJ
C 3882 kJ
D 485.25 kJ
Correct Answer:  A. 970.5 kJ
EXPLANATION

Q = m × L_fg = 0.5 kg × 1941 kJ/kg = 970.5 kJ heat is removed during condensation

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Q.78 Easy Thermodynamics
What is the first law of thermodynamics in differential form?
A dU = δQ - δW
B dU = δQ + δW
C dU = δQ × δW
D δQ = dU + δW
Correct Answer:  A. dU = δQ - δW
EXPLANATION

The correct form is dU = δQ - δW, where δW = PdV for expansion work. This represents energy conservation in thermodynamic systems.

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Q.79 Hard Thermodynamics
In a diesel engine, if the cutoff ratio is 1.5 and compression ratio is 16, calculate the thermal efficiency. (Assume γ = 1.4)
A 56.2%
B 63.5%
C 52.8%
D 61.2%
Correct Answer:  B. 63.5%
EXPLANATION

Diesel cycle efficiency = 1 - (1/r^(γ-1)) × [(r_c^γ - 1)/(γ(r_c - 1))] where r_c = cutoff ratio. With r=16, r_c=1.5, γ=1.4, efficiency ≈ 63.5%

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Q.80 Easy Thermodynamics
A heat engine operates between temperature reservoirs of 500 K and 300 K. What is the maximum possible efficiency of this engine?
A 40%
B 60%
C 50%
D 70%
Correct Answer:  A. 40%
EXPLANATION

Maximum efficiency is Carnot efficiency = 1 - (T_cold/T_hot) = 1 - (300/500) = 0.4 = 40%

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