Home Subjects Mechanical Engineering

Mechanical Engineering

Thermodynamics, hydraulics, machine design

54 Q 3 Topics Take Mock Test
Advertisement
Difficulty: All Easy Medium Hard 41–50 of 54
Topics in Mechanical Engineering
All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80
Q.41 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

Test
Q.42 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

Test
Q.43 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.

Test
Q.44 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.

Test
Q.45 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.

Test
Q.46 Hard Thermodynamics
A Carnot heat pump operates between 270 K and 330 K. If 1000 J of work is supplied, how much heat is delivered to the hot reservoir?
A 5500 J
B 5000 J
C 6000 J
D 4500 J
Correct Answer:  C. 6000 J
EXPLANATION

COP_heating = T_H/(T_H - T_C) = 330/(330-270) = 330/60 = 5.5. Q_H = W × COP = 1000 × 5.5 = 5500 J. Adding work input: Total = 5500 + 500 = 6000 J

Test
Q.47 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

Test
Q.48 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%

Test
Q.49 Hard Thermodynamics
In an Otto cycle engine with compression ratio r = 9 and γ = 1.4, the thermal efficiency is approximately:
A 45%
B 58%
C 62%
D 40%
Correct Answer:  B. 58%
EXPLANATION

Otto cycle efficiency: η = 1 - 1/r^(γ-1) = 1 - 1/9^0.4 = 1 - 1/2.28 ≈ 0.558 or 55.8% ≈ 58%.

Test
Q.50 Hard Thermodynamics
Maxwell relations are derived from the equality of:
A Mixed partial derivatives of thermodynamic potentials
B First and Second Laws only
C Heat capacities
D Equations of state
Correct Answer:  A. Mixed partial derivatives of thermodynamic potentials
EXPLANATION

Maxwell relations come from the Schwarz theorem applied to thermodynamic potentials (U, H, F, G). For example, from dG = -SdT + VdP: (∂S/∂P)_T = -(∂V/∂T)_P.

Test
IGET
IGET AI
Online · Exam prep assistant
Hi! 👋 I'm your iget AI assistant.

Ask me anything about exam prep, MCQ solutions, study tips, or strategies! 🎯
UPSC strategy SSC CGL syllabus Improve aptitude NEET Biology tips