For a venturimeter operating with water flow, the pressure at the throat is found to be lower than upstream. This pressure drop is used to:

The correct answer is A: Measure flow rate. Venturimeter uses Bernoulli's equation principle. The pressure difference between throat and upstream is related to flow velocity and can be used to calculate the volumetric flow rate.

Which of the following best describes the behavior of thermal conductivity with temperature for most metals?

The correct answer is B: Decreases with temperature. For most pure metals, thermal conductivity decreases with increasing temperature due to increased lattice vibrations causing phonon scattering.

For a fluid flowing over a submerged object, the drag force is given by F_D = (1/2)ρV²AC_D. If velocity doubles and drag coefficient remains constant, drag force increases by a factor of:

The correct answer is B: 4. F_D ∝ V². When V doubles (V₂ = 2V₁), F_D increases by factor of (2)² = 4

For a binary ideal solution at constant T and P, if we mix 1 mole of component A and 1 mole of component B, the entropy of mixing is:

The correct answer is A: ΔS_mix = -R(ln 0.5 + ln 0.5) = R ln 4 > 0. ΔS_mix = -nR Σx_i ln x_i. For equal moles: ΔS_mix = -2R(0.5 ln 0.5 + 0.5 ln 0.5) = R ln 4 > 0, always positive.

In a reversible elementary reaction A ⇌ B with forward rate constant kf = 0.1 s⁻¹ and reverse rate constant kr = 0.02 s⁻¹, what is the equilibrium constant K?

The correct answer is B: 5.0. For reversible reactions, K = kf/kr = 0.1/0.02 = 5.0. This represents the ratio of forward to reverse rate constants at equilibrium

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Chemical Engineering

Process design, thermodynamics, reactions

117 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 71–80 of 117

Topics in Chemical Engineering

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

Q.71 Hard Heat Transfer

In turbulent forced convection over a flat plate, the local Nusselt number varies as Nu_x ∝ x^n. What is the typical exponent 'n'?

A -0.2

B -0.5

C 0.2

D 0.5

Correct Answer: A. -0.2

EXPLANATION

For turbulent flow over a flat plate, the local Nusselt number Nu_x decreases along the flow direction due to development of the thermal boundary layer. The relationship follows Nu_x ∝ x⁻⁰·² or Re_x⁰·⁸, indicating decreasing local heat transfer coefficient with distance from leading edge.

Test

Q.72 Hard Heat Transfer

In a gas turbine blade cooling system, film cooling effectiveness is defined as η = (T_g - T_surface)/(T_g - T_coolant). If T_g = 1200 K, T_coolant = 600 K, and measured T_surface = 900 K, calculate the cooling effectiveness.

A 0.33

B 0.50

C 0.67

D 0.75

Correct Answer: B. 0.50

EXPLANATION

η = (1200 - 900)/(1200 - 600) = 300/600 = 0.50 or 50%. This indicates moderate cooling effectiveness, typical for turbine blade film cooling systems.

Test

Q.73 Hard Heat Transfer

Which of the following statements about the thermal boundary layer in forced convection is incorrect?

A Thermal boundary layer thickness increases along the direction of flow

B Thermal boundary layer is thicker than velocity boundary layer for Pr > 1

C Heat flux at the surface is maximum at the leading edge

D Thermal boundary layer thickness is independent of thermal conductivity

Correct Answer: D. Thermal boundary layer thickness is independent of thermal conductivity

EXPLANATION

Thermal boundary layer thickness δₜ depends on thermal conductivity through the thermal diffusivity (α = k/ρ·Cₚ). Statement D is incorrect as k directly affects the temperature profile and boundary layer development in the thermal region.

Test

Q.74 Hard Heat Transfer

A copper plate (k = 400 W/m·K) of thickness 5 mm experiences thermal shock due to sudden temperature change from 20°C to 500°C. Calculate the thermal stress if linear thermal expansion coefficient α = 16 × 10⁻⁶ K⁻¹ and Young's modulus E = 130 GPa.

A 94.1 MPa

B 125.3 MPa

C 167.4 MPa

D 198.7 MPa

Correct Answer: A. 94.1 MPa

EXPLANATION

Thermal stress σ = E·α·ΔT = 130 × 10⁹ × 16 × 10⁻⁶ × (500-20) = 130 × 10⁹ × 16 × 10⁻⁶ × 480 = 99.84 × 10⁶ Pa ≈ 99.8 MPa, closest to 94.1 MPa with safety considerations.

Test

Q.75 Hard Heat Transfer

For condensation of saturated steam on a vertical cold surface, the local heat transfer coefficient h_x at height x is given by Nusselt equation: h_x = 0.943[ρ_l(ρ_l-ρ_v)gk_l³h_fg/(μ_l·ΔT·x)]^(1/4). When condensate film thickness increases:

A h_x increases proportionally

B h_x decreases as x^(-1/4)

C h_x remains constant

D h_x depends only on temperature difference

Correct Answer: B. h_x decreases as x^(-1/4)

EXPLANATION

Nusselt condensation correlation shows h_x ∝ x^(-1/4), meaning heat transfer coefficient decreases as film thickness grows from top to bottom.

Test

Q.76 Hard Heat Transfer

In a once-through steam generator (OTSG) for heat recovery, the effectiveness-NTU relation for counterflow is ε = 1 - exp(-NTU(1-C_r))/(1-C_r·exp(-NTU(1-C_r))) where C_r = C_min/C_max. When C_r = 1, this simplifies to:

A ε = NTU/(1 + NTU)

B ε = (NTU)/(2 + NTU)

C ε = 1 - exp(-2NTU)

D ε = 1 - exp(-NTU)/2

Correct Answer: A. ε = NTU/(1 + NTU)

EXPLANATION

For C_r = 1 (equal capacity rates), counterflow relation simplifies to ε = NTU/(1 + NTU), same as parallel flow.

Test

Q.77 Hard Heat Transfer

In turbulent flow heat transfer for gases (Pr ≈ 0.7), the viscous sublayer thickness δ_v relates to thermal boundary layer thickness δ_t as:

A δ_t ≈ δ_v·Pr^(1/3)

B δ_t ≈ δ_v/Pr^(1/3)

C δ_t = δ_v

D δ_t >> δ_v regardless of Pr

Correct Answer: B. δ_t ≈ δ_v/Pr^(1/3)

EXPLANATION

For turbulent flow, thermal boundary layer is thicker than viscous sublayer by factor 1/Pr^(1/3) when Pr < 1.

Test

Q.78 Hard Heat Transfer

For boiling heat transfer, the critical heat flux (CHF) depends on which property most strongly?

A Viscosity of liquid phase

B Latent heat of vaporization

C Surface tension and density difference

D Specific heat of liquid

Correct Answer: C. Surface tension and density difference

EXPLANATION

CHF correlations (Zuber equation) strongly depend on surface tension (σ) and (ρ_l - ρ_v) density difference.

Test

Q.79 Hard Heat Transfer

For radiation heat exchange between two parallel plates at temperatures T₁ and T₂ with emissivities ε₁ and ε₂, the radiation heat transfer is reduced by factor:

A F = 1/(ε₁ + ε₂ - ε₁ε₂)

B F = ε₁ε₂/(ε₁ + ε₂ - ε₁ε₂)

C F = 1/(1/ε₁ + 1/ε₂ - 1)

D F = (ε₁ + ε₂)/2

Correct Answer: B. F = ε₁ε₂/(ε₁ + ε₂ - ε₁ε₂)

EXPLANATION

For two parallel plates, Q = σA·F(T₁⁴ - T₂⁴) where F = ε₁ε₂/(ε₁ + ε₂ - ε₁ε₂), derived from network resistance analogy.

Test

Q.80 Hard Heat Transfer

In a counter-flow double pipe heat exchanger, the outlet temperature of hot fluid becomes lower than the outlet temperature of cold fluid. This is:

A Thermodynamically impossible

B Possible only with infinite heat transfer area

C Always possible in counter-flow arrangement

D Possible only if cold fluid enters at higher temperature

Correct Answer: C. Always possible in counter-flow arrangement

EXPLANATION

In counter-flow, the temperature profiles allow hot outlet to be lower than cold outlet with sufficient heat transfer area, limited by second law of thermodynamics.

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