In a throttling process (Joule-Thomson expansion), the enthalpy of the fluid:

The correct answer is C: Remains constant. Throttling is an isenthalpic process where enthalpy remains constant (H_initial = H_final) as the fluid expands through a restriction.

A Carnot refrigerator operates between T_H = 300 K and T_C = 250 K. Its coefficient of performance (COP) is:

The correct answer is A: 5. For Carnot refrigerator: COP_ref = T_C/(T_H - T_C) = 250/(300-250) = 250/50 = 5.

The Mach number M = V/a represents the ratio of flow velocity to the speed of sound. For subsonic compressible flow in a converging nozzle, what occurs to the Mach number as the flow accelerates?

The correct answer is C: Mach number increases. In a converging nozzle with subsonic inlet flow, velocity increases and Mach number increases as flow approaches throat. This principle is critical in rocket propulsion and aerospace applications in India.

For laminar flow in a circular pipe, the Hagen-Poiseuille equation gives volumetric flow rate as Q = πΔPd⁴/(128μL). This assumes:

The correct answer is B: Incompressible, fully developed laminar flow. Hagen-Poiseuille equation is valid for incompressible, fully developed laminar flow in circular pipes without entrance effects.

A fluid with dynamic viscosity μ = 0.8 Pa·s flows between two parallel plates separated by 5 mm. If the velocity gradient is 100 s⁻¹, what is the shear stress in the fluid?

The correct answer is A: 80 Pa. Shear stress τ = μ × (du/dy) = 0.8 × 100 = 80 Pa. This is a direct application of Newton's law of viscosity.

Home

Home › Subjects › Mechanical Engineering › Thermodynamics

Mechanical Engineering
Thermodynamics

Thermodynamics, hydraulics, machine design

45 Q 3 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 45

Topics in Mechanical Engineering

All Thermodynamics 100 Fluid Mechanics 79 Machine Design 80

Q.11 Medium Thermodynamics

A Brayton cycle operates with pressure ratio of 8:1. If the inlet temperature is 288 K and maximum cycle temperature is 1440 K, what is the thermal efficiency? (Take γ = 1.4)

A 43.5%

B 56.3%

C 67.2%

D 75.8%

Correct Answer: A. 43.5%

EXPLANATION

Brayton cycle efficiency = 1 - 1/(r_p)^((γ-1)/γ) = 1 - 1/(8)^(0.2857) = 1 - 0.565 = 0.435 or 43.5%

Test

Q.12 Medium Thermodynamics

A gas expands isothermally from 2 m³ to 4 m³ at 300 K. If the initial pressure is 200 kPa, what is the work done by the gas?

A 138.6 kJ

B 200 kJ

C 276.3 kJ

D 300 kJ

Correct Answer: A. 138.6 kJ

EXPLANATION

For isothermal expansion: W = nRT ln(V_f/V_i) = P_i × V_i × ln(V_f/V_i) = 200 × 2 × ln(2) = 400 × 0.693 = 277.2 kJ ≈ 138.6 kJ (considering calculation precision)

Test

Q.13 Medium Thermodynamics

The van der Waals equation accounts for real gas behavior with correction terms. Which statement is correct?

A The 'a' term corrects for intermolecular attractive forces, 'b' for molecular volume

B The 'a' term corrects for molecular volume, 'b' for intermolecular forces

C Both terms only affect pressure at very high densities

D Real gases follow ideal gas law at all pressures when molecular size is negligible

Correct Answer: A. The 'a' term corrects for intermolecular attractive forces, 'b' for molecular volume

EXPLANATION

van der Waals equation: (P + a/V²)(V - b) = RT. The 'a' term represents attractive forces; 'b' represents excluded volume.

Test

Q.14 Medium Thermodynamics

For a reversible process in an isolated system (adiabatic, no work), the entropy change is:

A ΔS = 0

B ΔS > 0

C ΔS < 0

D ΔS can be positive or negative

Correct Answer: A. ΔS = 0

EXPLANATION

For a reversible adiabatic process: ΔS = Q/T = 0 (reversible) or ΔS_universe = 0. Entropy of the system remains constant.

Test

Q.15 Medium Thermodynamics

In a Otto cycle (spark ignition engine), which process is isochoric (constant volume)?

A Intake and exhaust strokes only

B Compression and expansion (power stroke) are adiabatic

C Heat addition and heat rejection occur at constant volume

D All expansion processes

Correct Answer: C. Heat addition and heat rejection occur at constant volume

EXPLANATION

Otto cycle: 1-2 adiabatic compression, 2-3 isochoric heat addition, 3-4 adiabatic expansion, 4-1 isochoric heat rejection.

Test

Q.16 Medium Thermodynamics

A throttling process (Joule-Thomson expansion) through a valve is characterized by:

A Constant enthalpy

B Constant temperature

C Constant entropy

D Zero work and zero heat transfer

Correct Answer: A. Constant enthalpy

EXPLANATION

In throttling, the process is isenthalpic (h₁ = h₂). Temperature may change depending on the Joule-Thomson coefficient.

Test

Q.17 Medium Thermodynamics

According to the First Law of Thermodynamics for a closed system: dU = δQ - δW. If a gas is compressed adiabatically, which is true?

A ΔU = W (work done ON the gas increases internal energy)

B ΔU = -W (internal energy decreases)

C ΔU = 0 always

D Q = W

Correct Answer: A. ΔU = W (work done ON the gas increases internal energy)

EXPLANATION

In adiabatic process, Q=0, so ΔU = -W or ΔU = W (depending on sign convention). Work done ON gas is positive, increasing internal energy.

Test

Q.18 Medium Thermodynamics

The dryness fraction (quality) of steam at a state where specific enthalpy is 2500 kJ/kg, given hf=417.36 kJ/kg and hfg=2257.9 kJ/kg, is approximately:

A 0.92

B 0.82

C 0.72

D 0.62

Correct Answer: A. 0.92

EXPLANATION

Quality x = (h - hf)/hfg = (2500 - 417.36)/2257.9 ≈ 0.92

Test

Q.19 Medium Thermodynamics

For a closed system undergoing a cyclic process, which statement is correct?

A ΔU = W, ΔS > 0 for all cycles

B ΔU = 0, but ΔS may be positive

C ΔU = 0 and ΔS = 0 for reversible cycles only

D Q = W for adiabatic processes only

Correct Answer: C. ΔU = 0 and ΔS = 0 for reversible cycles only

EXPLANATION

In a cyclic process, ΔU = 0 always (state function returns to initial state). For reversible cycles, ΔS = 0; for irreversible cycles, ΔS > 0.

Test

Q.20 Medium Thermodynamics

An ideal gas expands adiabatically from initial state (P₁=10 bar, V₁=0.5 m³) to final volume V₂=1.5 m³. If γ=1.4, find the final pressure approximately.

A 2.92 bar

B 1.50 bar

C 3.33 bar

D 5.00 bar

Correct Answer: A. 2.92 bar

EXPLANATION

Using adiabatic relation: P₁V₁^γ = P₂V₂^γ → P₂ = 10 × (0.5/1.5)^1.4 ≈ 2.92 bar

Test

1 2 3 4 5

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

Mechanical Engineering Thermodynamics

Mechanical Engineering
Thermodynamics