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Mechanical Engineering

Thermodynamics, hydraulics, machine design

123 Q 3 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 111–120 of 123
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80
Q.111 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.112 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.113 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.114 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

Test
Q.115 Medium Thermodynamics
In a vapour compression refrigeration cycle, the throttling device (expansion valve) causes:
A Decrease in enthalpy
B Increase in enthalpy
C No change in enthalpy
D Decrease in entropy significantly
Correct Answer:  C. No change in enthalpy
EXPLANATION

The throttling process is isenthalpic (constant enthalpy). The enthalpy remains the same before and after the expansion valve.

Test
Q.116 Medium Thermodynamics
The specific heat capacity at constant volume (Cv) for a monatomic ideal gas is:
A (3/2)R
B (5/2)R
C (7/2)R
D R
Correct Answer:  A. (3/2)R
EXPLANATION

For a monatomic ideal gas with 3 translational degrees of freedom: Cv = (3/2)R. For diatomic: (5/2)R; for polyatomic: (7/2)R.

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Q.117 Medium Thermodynamics
Two bodies at different temperatures are brought into contact in an isolated system. The total entropy of the system will:
A Decrease
B Remain constant
C Increase
D Become zero
Correct Answer:  C. Increase
EXPLANATION

For an irreversible process in an isolated system, total entropy increases (ΔS_total > 0). Heat transfer between bodies at different temperatures is irreversible.

Test
Q.118 Medium Thermodynamics
In a closed system, the total energy balance equation for a time interval is:
A ΔE = Q + W
B ΔE = Q - W
C Q = ΔE + W
D W = Q - ΔE
Correct Answer:  B. ΔE = Q - W
EXPLANATION

First Law: ΔE_system = Q - W, where Q is heat added to system and W is work done by system. This is the IUPAC convention.

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Q.119 Medium Thermodynamics
The polytropic process for an ideal gas follows PV^n = constant. When n = 0, the process is:
A Isobaric
B Isochoric
C Isothermal
D Adiabatic
Correct Answer:  A. Isobaric
EXPLANATION

When n = 0, PV⁰ = P = constant, which is an isobaric process. When n = 1 (isothermal), n = γ (adiabatic), n = ∞ (isochoric).

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Q.120 Medium Thermodynamics
For a real gas, the compressibility factor Z is defined as:
A Z = PV/nRT
B Z = nRT/PV
C Z = P/VT
D Z = V/nRT
Correct Answer:  A. Z = PV/nRT
EXPLANATION

Compressibility factor Z = PV/(nRT). For ideal gas, Z = 1. For real gases, Z ≠ 1 due to molecular interactions.

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