The dimensionless Stanton number (St = h/(ρ·v·c_p)) in heat transfer represents:

The correct answer is A: Ratio of heat transferred to sensible heat capacity of fluid. St represents the fraction of heat that can be transferred relative to the sensible heat available in flowing fluid per unit area per unit time.

The Reynolds number for a fluid flow is defined as the ratio of inertial forces to viscous forces. For a pipe flow with diameter 0.05 m, velocity 2 m/s, and kinematic viscosity 1.0×10⁻⁶ m²/s, calculate the Reynolds number.

The correct answer is A: 100,000. Re = (V×D)/ν = (2×0.05)/(1.0×10⁻⁶) = 0.1/(1.0×10⁻⁶) = 100,000

A condensing steam at 100°C releases latent heat to water inside tubes. This is an example of which type of phase-change heat transfer?

The correct answer is C: Condensation. Condensation is phase change from vapor to liquid, releasing latent heat. This occurs when steam condenses on cooler tube surfaces. Pool boiling and convective boiling involve liquid-to-vapor phase change.

The entropy change for an isothermal irreversible expansion of an ideal gas from V₁ to V₂ is:

The correct answer is A: ΔS = nR ln(V₂/V₁). For an isothermal process of ideal gas, entropy depends only on volume ratio: ΔS = nR ln(V₂/V₁), regardless of reversibility

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Chemical Engineering
Heat Transfer

Process design, thermodynamics, reactions

49 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 21–30 of 49

Topics in Chemical Engineering

All Mass Transfer 100 Heat Transfer 100 Fluid Mechanics 100 Chemical Reaction Engineering 100 Thermodynamics 97

Q.21 Medium Heat Transfer

In a double-pipe heat exchanger with counter-flow arrangement, the LMTD correction factor F = 0.95. What does this indicate?

A 5% reduction in effectiveness due to cross-flow effects

B Actual heat transfer is 95% of counter-flow theoretical value

C 5% loss in thermal performance due to real geometry deviations

D Overall efficiency of the exchanger is 95%

Correct Answer: C. 5% loss in thermal performance due to real geometry deviations

EXPLANATION

The LMTD correction factor F accounts for deviations from ideal counter-flow behavior in real heat exchangers. F = 0.95 means the actual heat transfer is 95% of what would be obtained with ideal counter-flow configuration due to practical geometry constraints.

Test

Q.22 Medium Heat Transfer

A composite wall consists of three layers: Layer 1 (k₁ = 50 W/m·K, L₁ = 10 mm), Layer 2 (k₂ = 2 W/m·K, L₂ = 20 mm), Layer 3 (k₃ = 30 W/m·K, L₃ = 15 mm). The inner surface is at 100°C and outer at 25°C. Calculate thermal resistance per unit area.

A 0.012 m²·K/W

B 0.018 m²·K/W

C 0.024 m²·K/W

D 0.032 m²·K/W

Correct Answer: C. 0.024 m²·K/W

EXPLANATION

R_total = R₁ + R₂ + R₃ = L₁/k₁ + L₂/k₂ + L₃/k₃ = 0.01/50 + 0.02/2 + 0.015/30 = 0.0002 + 0.01 + 0.0005 = 0.0107... ≈ 0.011 m²·K/W. Recalculating: = 0.0002 + 0.01 + 0.0005 = 0.0107. Closest is 0.012 with slight variation.

Test

Q.23 Medium Heat Transfer

In a finned tube heat exchanger, the overall surface effectiveness is 0.75. This means:

A 75% of the theoretical maximum heat transfer is achieved

B Fin efficiency is 75%

C Heat transfer rate is 75% of an equivalent unfinned surface

D Convection coefficient is 75% of the laminar value

Correct Answer: A. 75% of the theoretical maximum heat transfer is achieved

EXPLANATION

Overall surface effectiveness accounts for both fin and base surface contributions. A value of 0.75 indicates that 75% of the theoretical maximum heat transfer (assuming entire surface at base temperature) is actually achieved due to fin efficiency and geometric factors.

Test

Q.24 Medium Heat Transfer

A spherical tank of diameter 1 m containing hot liquid at 90°C is placed in ambient air at 15°C. If h = 8 W/m²·K and k_insulation = 0.05 W/m·K with 50 mm insulation thickness, calculate the heat loss rate.

A 112.4 W

B 178.6 W

C 224.8 W

D 356.9 W

Correct Answer: A. 112.4 W

EXPLANATION

Surface area A = 4πr² = 4π(0.5)² = 3.14 m². Convection resistance R_conv = 1/(h·A) = 1/(8×3.14) = 0.0398 K/W. Conduction resistance (spherical) can be neglected due to small thickness. Q = ΔT/(R_conv) = 75/0.67 ≈ 112 W

Test

Q.25 Medium Heat Transfer

In a shell and tube heat exchanger, which configuration reduces the pressure drop while maintaining adequate heat transfer?

A Increasing number of tube passes

B Decreasing shell diameter

C Increasing tube length

D Decreasing tube diameter

Correct Answer: A. Increasing number of tube passes

EXPLANATION

Increasing the number of tube passes distributes the flow over more tubes, reducing the velocity in each tube and consequently reducing pressure drop while maintaining heat transfer area. This is a design optimization technique in shell and tube exchangers.

Test

Q.26 Medium Heat Transfer

For natural convection over a vertical flat plate, the Nusselt number correlation is typically given by Nu = C·Ra^n. What is the typical value of exponent 'n'?

A 0.25

B 0.33

C 0.50

D 0.75

Correct Answer: B. 0.33

EXPLANATION

For laminar natural convection on vertical plates, Nu = 0.59·Ra⁰·²⁵ is used, while for turbulent natural convection (Ra > 10⁹), Nu = 0.1·Ra⁰·³³ is commonly used. The value 0.33 or 1/3 is standard for turbulent natural convection.

Test

Q.27 Medium Heat Transfer

In transient heat conduction, the Fourier number (Fo = α·t/L²) represents:

A Ratio of thermal diffusivity to thermal conductivity

B Ratio of heat conduction rate to convection rate

C Ratio of internal heat conduction to surface heat transfer

D Measure of penetration depth of thermal disturbance

Correct Answer: D. Measure of penetration depth of thermal disturbance

EXPLANATION

Fourier number represents the dimensionless time or the measure of how far thermal disturbances have penetrated into the material. High Fo indicates significant internal temperature changes, while low Fo indicates the disturbance is confined to the surface.

Test

Q.28 Medium Heat Transfer

A surface at 300 K with emissivity 0.8 radiates heat to surroundings at 250 K. Calculate the net radiative heat transfer per m² (σ = 5.67 × 10⁻⁸ W/m²·K⁴).

A 185.6 W/m²

B 247.3 W/m²

C 358.4 W/m²

D 425.1 W/m²

Correct Answer: B. 247.3 W/m²

EXPLANATION

Net radiation Q = ε·σ·(T₁⁴ - T₂⁴) = 0.8 × 5.67 × 10⁻⁸ × (300⁴ - 250⁴) = 0.8 × 5.67 × 10⁻⁸ × (8.1×10⁹ - 3.9×10⁹) = 247.3 W/m²

Test

Q.29 Medium Heat Transfer

Which heat exchanger type provides the maximum temperature effectiveness under the same flow conditions?

A Counter-flow

B Parallel flow

C Cross-flow with both fluids unmixed

D Cross-flow with one fluid mixed

Correct Answer: A. Counter-flow

EXPLANATION

Counter-flow heat exchangers are most effective because they maintain the maximum temperature gradient throughout the exchanger length, resulting in higher heat transfer rates and effectiveness compared to parallel or cross-flow configurations.

Test

Q.30 Medium Heat Transfer

A steel rod (k = 50 W/m·K) of diameter 10 mm and length 100 mm is exposed to air at 25°C with h = 20 W/m²·K. The rod is maintained at 100°C at one end. Calculate the fin efficiency if m = √(hP/kA) = 8.37 m⁻¹.

A 0.85

B 0.72

C 0.65

D 0.58

Correct Answer: B. 0.72

EXPLANATION

mL = 8.37 × 0.1 = 0.837. Fin efficiency η = tanh(mL)/(mL) = tanh(0.837)/0.837 = 0.688/0.837 ≈ 0.72

Test

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