A gas mixture contains N₂ (40% mole) and O₂ (60% mole) at 1 atm and 298 K. Calculate the partial pressure of O₂ in the mixture.

The correct answer is A: 0.6 atm. Partial pressure = Total pressure × Mole fraction. P_O₂ = 1 atm × 0.60 = 0.6 atm (Dalton's Law)

In a siphon arrangement, what is the maximum theoretical height from which water can be siphoned up using atmospheric pressure?

The correct answer is C: 10.3 m. The maximum height is approximately 10.3 m (or one atmosphere height), determined by h = P_atm/(ρg) = 101325/(1000 × 9.81) = 10.33 m. Friction losses reduce this in practice.

In a counter-current heat exchanger with equal heat capacities, what is the maximum possible effectiveness?

The correct answer is A: 100% or 1.0. For counter-current arrangement, maximum effectiveness can approach 1.0 (100%) with infinite NTU. For co-current, it's limited to (1-exp(-NTU(1+C_r)))/(1+C_r).

For flow measurement in a Venturi tube with diameter ratio (D₁/D₂) = 2, what is the diameter ratio for minimum vibration and noise?

The correct answer is C: 2.5. Optimal Venturi throat diameter ratio is typically 2.5-3.0 for smooth operation and minimal vibration

A shell-and-tube heat exchanger with one shell pass and two tube passes (1-2 STHE) has a correction factor F. What is the typical range of F values?

The correct answer is A: 0.5 to 0.9. The correction factor F (applied to LMTD) for 1-2 STHE typically ranges from 0.5 to 0.9 depending on temperature ratios and heat capacity ratios. F is always ≤ 1.

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Chemical Engineering

Process design, thermodynamics, reactions

247 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 171–180 of 247

Topics in Chemical Engineering

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

Q.171 Medium Heat Transfer

A condenser operates with steam at 100°C condensing on a tube bank at 20°C. The saturation temperature drop is negligible. Which phase of condensation provides maximum heat transfer rate?

A Initial droplet formation phase

B Dropwise condensation phase

C Filmwise condensation phase

D All phases provide equal heat transfer

Correct Answer: B. Dropwise condensation phase

EXPLANATION

Dropwise condensation provides heat transfer coefficients 5-10 times higher than filmwise condensation because liquid droplets continuously shed, exposing fresh surface to direct contact with steam. However, filmwise condensation is more common industrially due to stability issues with dropwise condensation.

Test

Q.172 Medium Heat Transfer

A heat pipe is used to transfer heat from an electronic component at 80°C to ambient air at 25°C through evaporation and condensation. What is the primary advantage of a heat pipe over conventional conduction cooling?

A Higher thermal conductivity

B Effective thermal conductivity can be 1000 times higher than pure conduction

C Lower cost of fabrication

D Ability to function in vacuum conditions

Correct Answer: B. Effective thermal conductivity can be 1000 times higher than pure conduction

EXPLANATION

Heat pipes achieve very high effective thermal conductivity (often >1000 times that of copper) through the latent heat of evaporation and condensation of working fluid, making them ideal for high-power electronics cooling despite the small cross-sectional area.

Test

Q.173 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.174 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.175 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.176 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.177 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.178 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.179 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.180 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

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