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Question 1Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

In a reactive material balance, the limiting reactant is the species that

  1. is fed in excess to increase the equilibrium conversion
  2. is consumed completely first as the reaction proceeds Correct answer.
  3. is present in the largest molar amount in the fresh feed
  4. has the highest molecular weight among the reacting species

Explanation

The limiting reactant is identified by comparing the molar feed rate of each reactant with its stoichiometric coefficient. It is the species that would be exhausted first if the reaction went to its maximum extent, and its amount fixes the maximum extent of reaction. Species fed in excess remain after the limiting reactant is consumed.

Question 2Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For a closed system with negligible changes in kinetic and potential energy, the first law of thermodynamics is written as

  1. Q − W = ΔU Correct answer.
  2. Q + W = ΔH
  3. Q = ΔU + Δ(PV)
  4. W = ΔU − Q

Explanation

Using the sign convention that W is work done by the system, the closed-system energy balance reduces to ΔU = Q − W. With negligible kinetic and potential energy changes, only internal energy, heat, and work remain. Rearranging this expression gives Q − W = ΔU.

Question 3Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For an ideal gas, the specific enthalpy depends on

  1. both temperature and pressure
  2. pressure only
  3. temperature only Correct answer.
  4. volume only

Explanation

An ideal gas has no intermolecular forces, so its internal energy is a function of temperature only. Because the product PV equals RT for one mole, enthalpy h = u + RT also depends only on temperature. Pressure and volume changes at constant temperature do not alter ideal-gas enthalpy.

Question 4Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A distillation column is fed 100 kmol/h of a mixture containing 40 mol% A and 60 mol% B. The distillate contains 90 mol% A and the bottoms contains 5 mol% A. What is the distillate flow rate?

  1. 41.2 kmol/h Correct answer.
  2. 60 kmol/h
  3. 58.8 kmol/h
  4. 35 kmol/h

Explanation

An overall mole balance gives D + B = 100, and an A balance gives 0.90D + 0.05B = 40. Substituting B = 100 − D yields 0.90D + 0.05(100 − D) = 40, so 0.85D = 35 and D = 41.2 kmol/h. The remaining 58.8 kmol/h leaves as bottoms.

Question 5Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

The standard heat of reaction for A → B at 298 K is −50 kJ/mol. If 2 mol of A react completely at constant pressure and the reactants and products are both returned to 298 K, how much heat is released?

  1. 25 kJ
  2. 200 kJ
  3. 50 kJ
  4. 100 kJ Correct answer.

Explanation

The reported heat of reaction is −50 kJ per mole of reaction as written. For 2 mol of A reacting completely, the enthalpy change is 2 mol × (−50 kJ/mol) = −100 kJ. The negative sign means 100 kJ is released to the surroundings at constant pressure.

Question 6Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its thermal efficiency?

  1. 0.33
  2. 0.25
  3. 0.50 Correct answer.
  4. 0.67

Explanation

The Carnot efficiency depends only on the absolute temperatures: η = 1 − T_cold/T_hot. Here η = 1 − 300/600 = 0.50, or 50%. No real engine operating between these reservoirs can exceed this value.

Question 7Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

At constant temperature and pressure, a chemical process is spontaneous in the forward direction when

  1. ΔH is negative regardless of the entropy change
  2. ΔG equals zero
  3. ΔG is less than zero Correct answer.
  4. ΔG is greater than zero

Explanation

The Gibbs free energy change at constant T and P measures the maximum non-expansion work available from a process. A negative ΔG means the forward process is spontaneous, while ΔG = 0 indicates equilibrium and ΔG > 0 indicates nonspontaneity. Enthalpy alone is insufficient because entropy also contributes through ΔG = ΔH − TΔS.

Question 8Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

At the azeotropic point of a binary vapor-liquid mixture, which statement is true?

  1. The vapor and liquid phase compositions are identical Correct answer.
  2. The relative volatility is infinite
  3. The two pure components have the same normal boiling point
  4. The excess Gibbs energy is zero

Explanation

At an azeotrope the vapor and liquid phases have the same composition, so x_i = y_i for each component. Because the phases cannot be separated by ordinary distillation at that point, the relative volatility equals one rather than infinity. Azeotropes arise from nonideal solution behavior, so the excess Gibbs energy is generally nonzero.

Question 9Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

The standard Gibbs free energy change for a reaction is −20 kJ/mol at 298 K. Using R = 8.314 J/(mol·K), what is the approximate equilibrium constant?

  1. 3.2 × 10^1
  2. 3.2 × 10^2
  3. 3.2 × 10^3 Correct answer.
  4. 3.2

Explanation

The relation ΔG° = −RT ln K gives ln K = −ΔG°/(RT). Substituting −20,000 J/mol, R = 8.314 J/(mol·K), and T = 298 K yields ln K ≈ 8.07. Therefore K ≈ e^8.07 ≈ 3.2 × 10^3.

Question 10Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

Air at 300 K is compressed reversibly and adiabatically from 1 bar to 10 bar. Treating air as an ideal gas with γ = 1.4, what is the approximate outlet temperature?

  1. 750 K
  2. 450 K
  3. 579 K Correct answer.
  4. 300 K

Explanation

For a reversible adiabatic compression of an ideal gas, T2/T1 = (P2/P1)^((γ−1)/γ). With γ = 1.4, the exponent is 0.2857, so T2 = 300 K × 10^0.2857 ≈ 300 × 1.93 ≈ 579 K. The temperature rises because compression work increases the internal energy.

Question 11Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A mixer receives 200 kg/h of a solution containing 20 wt% NaCl and 100 kg/h of pure water. Assuming complete mixing and no losses, what is the NaCl mass fraction in the mixed product?

  1. 0.067
  2. 0.200
  3. 0.100
  4. 0.133 Correct answer.

Explanation

The NaCl balance gives 0.20 x 200 = 40 kg/h NaCl. The total product mass is 200 + 100 = 300 kg/h, so the mass fraction is 40/300 = 0.133.

Question 12Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

In chemical engineering thermodynamics, which quantity is a state function?

  1. Internal energy Correct answer.
  2. Heat transferred
  3. Entropy generation
  4. Work performed

Explanation

Internal energy depends only on the thermodynamic state of the system, so it is a state function. Heat and work depend on the path taken between states.

Question 13Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

How much sensible heat is required to raise the temperature of 2 kg of liquid water from 20°C to 80°C? Take the specific heat capacity of water as 4.18 kJ/(kg·K).

  1. 125.4 kJ
  2. 1003.2 kJ
  3. 250.8 kJ
  4. 501.6 kJ Correct answer.

Explanation

For a constant specific heat, Q = m Cp ΔT. Here ΔT = 80 - 20 = 60 K, so Q = 2 x 4.18 x 60 = 501.6 kJ.

Question 14Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A fuel gas contains 80 mol% CH4 and 20 mol% N2. It is burned with 20% excess air. Assuming complete combustion, how many moles of air are supplied per mole of fuel gas? Use 21 mol% O2 in air.

  1. 8.00 mol air/mol fuel
  2. 7.62 mol air/mol fuel
  3. 9.14 mol air/mol fuel Correct answer.
  4. 9.60 mol air/mol fuel

Explanation

One mole of fuel gas contains 0.8 mol CH4, which requires 1.6 mol O2. Theoretical air is 1.6/0.21 = 7.62 mol, and 20% excess gives 1.2 x 7.62 = 9.14 mol air.

Question 15Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For an ideal gas undergoing a reversible isothermal expansion from volume V1 to V2, which expression gives the entropy change of the gas?

  1. nCv ln(T2/T1)
  2. nR ln(V2/V1) Correct answer.
  3. nCp ln(T2/T1)
  4. nR ln(P2/P1)

Explanation

For an ideal gas at constant temperature, ΔS = nR ln(V2/V1). The temperature-dependent terms vanish because T2 = T1.

Question 16Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A liquid mixture at 90°C contains 0.30 mole fraction benzene and 0.70 mole fraction toluene. At 90°C, the vapor pressures are 136 kPa for benzene and 54 kPa for toluene. Assuming ideal solution behavior, what is the bubble-point pressure?

  1. 118.6 kPa
  2. 190.0 kPa
  3. 100.0 kPa
  4. 78.6 kPa Correct answer.

Explanation

Bubble-point pressure for an ideal mixture is P = x1 P1sat + x2 P2sat. Substituting gives 0.30 x 136 + 0.70 x 54 = 78.6 kPa.

Question 17Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For an exothermic gas-phase reaction at equilibrium, what happens when the temperature is increased at constant pressure?

  1. The equilibrium constant increases and the reaction shifts toward products.
  2. The equilibrium constant remains unchanged because pressure is constant.
  3. The reaction shifts toward products because the forward rate increases.
  4. The equilibrium constant decreases and the reaction shifts toward reactants. Correct answer.

Explanation

For an exothermic reaction, increasing temperature favors the endothermic reverse reaction. By Le Chatelier's principle and the van't Hoff relation, K decreases and the equilibrium shifts toward reactants.

Question 18Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

Steam enters an adiabatic turbine at 3 MPa and 400°C with h = 3231 kJ/kg and s = 6.923 kJ/(kg·K). It expands isentropically to 10 kPa, where saturated liquid has hf = 191.8 kJ/kg, hfg = 2392.8 kJ/kg, sf = 0.6493 kJ/(kg·K), and sfg = 7.5009 kJ/(kg·K). What is the turbine work per kilogram of steam?

  1. 837 kJ/kg
  2. 2,000 kJ/kg
  3. 1,038 kJ/kg Correct answer.
  4. 1,500 kJ/kg

Explanation

First find quality: x = (s2 - sf)/sfg = (6.923 - 0.6493)/7.5009 = 0.836. Then h2 = hf + x hfg = 191.8 + 0.836 x 2392.8 = 2193 kJ/kg, so w = h1 - h2 = 3231 - 2193 = 1038 kJ/kg.

Question 19Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

Which of the following is a valid Maxwell relation derived from the Helmholtz free energy A(T, V)?

  1. (∂S/∂P)_T = (∂V/∂T)_P
  2. (∂H/∂P)_T = (∂V/∂T)_S
  3. (∂T/∂V)_S = (∂P/∂S)_V
  4. (∂S/∂V)_T = (∂P/∂T)_V Correct answer.

Explanation

From dA = -S dT - P dV, the mixed partial derivatives give (∂S/∂V)_T = (∂P/∂T)_V. The other options either omit a required minus sign or are not Maxwell relations.

Question 20Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For a real gas undergoing an isenthalpic throttling process, what is the sign of the Joule-Thomson coefficient at the inversion temperature?

  1. Negative
  2. Positive
  3. Zero Correct answer.
  4. Undefined

Explanation

The Joule-Thomson coefficient μ = (∂T/∂P)_H is zero at the inversion temperature. Below it μ is positive (cooling), and above it μ is negative (heating).

Question 21Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A continuous evaporator receives 1000 kg/h of an aqueous solution and removes 300 kg/h of water as vapor. No other streams leave the evaporator except the concentrated liquid product. At steady state, what is the mass flow rate of the liquid product?

  1. 1,300 kg/h
  2. 1,000 kg/h
  3. 300 kg/h
  4. 700 kg/h Correct answer.

Explanation

At steady state there is no accumulation, so total mass entering equals total mass leaving. The outlets are 300 kg/h of water vapor and the liquid product, so the product flow is 1000 − 300 = 700 kg/h.

Question 22Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

In a closed system at equilibrium with both a liquid phase and a vapor phase present, which condition must hold for every component i?

  1. T^L > T^V while P^L = P^V
  2. x_i^L = y_i^V for each component i
  3. μ_i^L = μ_i^V only for the more volatile component
  4. μ_i^L = μ_i^V for each component i Correct answer.

Explanation

At phase equilibrium, the chemical potential of every component must be equal in all phases present. Thus μ_i^L = μ_i^V for each component i, and temperature and pressure are also uniform at equilibrium.

Question 23Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For an ideal gas mixture at total pressure P, the partial pressure of component i is expressed in terms of its vapor-phase mole fraction y_i as which of the following?

  1. P_i = y_i P Correct answer.
  2. P_i = P/y_i
  3. P_i = y_i/P
  4. P_i = x_i P

Explanation

Dalton's law for an ideal gas mixture states that each component's partial pressure equals its mole fraction in the gas phase multiplied by the total pressure. Therefore P_i = y_i P.

Question 24Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A gas stream flowing at 100 kmol/h contains 5.0 mol% CO2. It enters an absorber where 90% of the CO2 is removed. How many kmol/h of CO2 leave in the treated gas?

  1. 9.5 kmol/h
  2. 5.0 kmol/h
  3. 0.5 kmol/h Correct answer.
  4. 4.5 kmol/h

Explanation

The feed CO2 rate is 0.050 × 100 = 5.0 kmol/h. Removing 90% leaves 10% of that CO2, or 0.10 × 5.0 = 0.5 kmol/h in the treated gas.

Question 25Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

In an adiabatic mixing process, 50 kg/h of liquid water at 80°C is mixed with 150 kg/h of liquid water at 20°C. Assuming constant specific heat and no heat losses, what is the outlet temperature?

  1. 40°C
  2. 35°C Correct answer.
  3. 30°C
  4. 50°C

Explanation

An adiabatic energy balance gives 50(80) + 150(20) = (50 + 150)T_out. Solving gives T_out = (4000 + 3000)/200 = 35°C.

Question 26Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A Carnot refrigerator absorbs heat from a cold region at 270 K and rejects heat to a warm region at 300 K. What is its coefficient of performance as a refrigerator?

  1. 0.90
  2. 9.0 Correct answer.
  3. 10.0
  4. 1.11

Explanation

For a Carnot refrigerator, COP = T_c/(T_h − T_c). Substituting gives 270/(300 − 270) = 270/30 = 9.0.

Question 27Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

One mole of ideal gas A and one mole of ideal gas B, initially at the same temperature and pressure, are mixed isothermally and isobarically. What is the total entropy change of mixing?

  1. −2R ln 2
  2. R ln 2
  3. 0
  4. 2R ln 2 Correct answer.

Explanation

For ideal gases, ΔS_mix = −R Σ n_i ln y_i. With one mole of each gas, y_A = y_B = 0.5, so ΔS_mix = −R(ln 0.5 + ln 0.5) = 2R ln 2.

Question 28Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A fresh feed of 100 kmol/h contains 90 mol% A and 10 mol% inert I. The reactor feed is maintained at 20 mol% I. Unreacted A is recycled, and a purge taken from the recycle loop has the same composition as the reactor feed. At steady state, what is the total purge rate?

  1. 10 kmol/h
  2. 40 kmol/h
  3. 50 kmol/h Correct answer.
  4. 20 kmol/h

Explanation

The fresh inert rate is 0.10 × 100 = 10 kmol/h, and inert leaves only in the purge. Since the purge contains 20 mol% inert, the total purge rate is 10/0.20 = 50 kmol/h.

Question 29Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For a pure real gas, which expression correctly relates the fugacity coefficient φ to the chemical potential μ and the ideal-gas chemical potential μ^ig at the same temperature and pressure?

  1. φ = P/P^sat
  2. φ = RT/P
  3. ln φ = (μ^ig − μ)/(RT)
  4. ln φ = (μ − μ^ig)/(RT) Correct answer.

Explanation

The fugacity coefficient measures departure from ideal-gas behavior through the residual chemical potential. For a pure gas, ln φ = (μ − μ^ig)/(RT), which is also the residual molar Gibbs energy divided by RT.

Question 30Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A reaction has standard enthalpy change ΔH° = +50 kJ/mol and equilibrium constant K = 0.10 at 298 K. Using the van't Hoff equation and R = 8.314 J/(mol·K), estimate K at 308 K, assuming ΔH° is constant.

  1. 0.19 Correct answer.
  2. 0.52
  3. 0.10
  4. 0.052

Explanation

The van't Hoff equation gives ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1). Substitution gives ln(K2/K1) ≈ 0.655, so K2 ≈ 0.10 × 1.93 ≈ 0.19.

Question 31Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A pure liquid in equilibrium with its vapor in a closed vessel has two phases present. According to the Gibbs phase rule, how many intensive degrees of freedom remain?

  1. One; specifying temperature fixes the saturation pressure. Correct answer.
  2. Zero; both temperature and pressure are fixed by the phase equilibrium.
  3. Three; composition adds an additional independent variable.
  4. Two; temperature and pressure can be varied independently.

Explanation

For a pure component, C = 1, and with two phases present P = 2, so the Gibbs phase rule gives F = C - P + 2 = 1. Thus only one intensive property can be chosen independently; for example, setting temperature fixes the saturation pressure.

Question 32Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

What is the standard enthalpy of formation of O2(g) at 298.15 K and 1 bar?

  1. -393.5 kJ/mol
  2. -285.8 kJ/mol
  3. +141.0 kJ/mol
  4. 0 kJ/mol Correct answer.

Explanation

The standard enthalpy of formation is defined for formation of one mole of a substance from its elements in their standard states. Oxygen gas is itself an element in its standard state, so its standard enthalpy of formation is zero by definition.

Question 33Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A tank initially contains 500 kg of brine. Feed enters at 20 kg/min and discharge leaves at 15 kg/min; there is no reaction. What are the accumulation rate and the tank mass after 10 min?

  1. +5 kg/min; 550 kg Correct answer.
  2. +35 kg/min; 850 kg
  3. +5 kg/min; 500 kg
  4. -5 kg/min; 450 kg

Explanation

The accumulation rate is feed minus discharge, so 20 - 15 = +5 kg/min. Over 10 min the tank gains 50 kg, giving 500 + 50 = 550 kg.

Question 34Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A liquid oil stream of 2.5 kg/s is cooled from 150°C to 90°C in a heat exchanger. If the oil has a constant specific heat of 2.0 kJ/(kg·K) and no phase change occurs, what is the heat duty?

  1. 900 kW
  2. 300 kW Correct answer.
  3. 600 kW
  4. 150 kW

Explanation

For a single liquid stream with constant specific heat, Q = m_dot cp ΔT. Here ΔT = 60 K, so the duty magnitude is 2.5 × 2.0 × 60 = 300 kW.

Question 35Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

Two moles of an ideal gas with Cv = 12.5 J/(mol·K) are heated at constant volume from 300 K to 600 K. What is the entropy change of the gas?

  1. 25.0 J/K
  2. 34.7 J/K
  3. 8.7 J/K
  4. 17.3 J/K Correct answer.

Explanation

At constant volume, the entropy change of an ideal gas is ΔS = n Cv ln(T2/T1). Substitution gives ΔS = 2 × 12.5 × ln(600/300) = 25 × 0.693 = 17.3 J/K.

Question 36Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A crystallizer receives 1000 kg/h of a solution containing 30 wt% dissolved solute. It produces pure solid crystals and a mother liquor containing 20 wt% solute. What mass flow of crystals is produced?

  1. 300 kg/h
  2. 125 kg/h Correct answer.
  3. 175 kg/h
  4. 250 kg/h

Explanation

Let C be crystals and L be mother liquor. Total balance gives C + L = 1000, and solute balance gives C + 0.20L = 300. Solving gives C = 125 kg/h.

Question 37Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A steady reactor converts A and B to C by A + B → C. The standard heat of reaction is -50 kJ per mole of A reacted. If 8 mol/s of A react and the reactor is kept isothermal, what is the required heat-transfer rate?

  1. 400 kW supplied
  2. 400 kW removed Correct answer.
  3. 16 kW removed
  4. 50 kW removed

Explanation

The reaction releases 50 kJ for each mole of A reacted. With 8 mol/s reacting, the heat release is 8 × 50 = 400 kW, so an isothermal reactor must remove 400 kW.

Question 38Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

For the gas-phase equilibrium A(g) ⇌ 2B(g), Kp = 1.0 at the reaction temperature and total pressure is 1 bar. Pure A is fed and equilibrium is reached isothermally. What is the equilibrium conversion of A?

  1. 0.33
  2. 0.20
  3. 0.45 Correct answer.
  4. 0.50

Explanation

Let conversion be X for one mole of A initially. At equilibrium, A = 1 - X and B = 2X with total 1 + X, so Kp = 4X²/[(1 - X)(1 + X)] = 1 at 1 bar. Solving gives X = 1/√5 ≈ 0.45.

Question 39Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

A single-effect evaporator concentrates 10,000 kg/h of an aqueous solution from 10 wt% solids to 50 wt% solids. Feed enters at 60°C and the contents boil at 100°C. Neglecting heat losses and using cp = 4.18 kJ/(kg·K) for feed and ΔHvap = 2257 kJ/kg at 100°C, what heat duty is required?

  1. 5,016 kW
  2. 9,640 kW
  3. 5,480 kW Correct answer.
  4. 4,640 kW

Explanation

The product is 1000/0.50 = 2000 kg/h, so vapor is 8000 kg/h. Sensible heating is 10,000 × 4.18 × 40 = 1.672×10^6 kJ/h, and vaporization is 8000 × 2257 = 1.806×10^7 kJ/h; the total is about 5,480 kW.

Question 40Chemical Engineering Fundamentals (material and energy balances, thermodynamics)

Using standard enthalpies of formation ΔHf°(CO,g) = -110.5 kJ/mol, ΔHf°(CO2,g) = -393.5 kJ/mol, and ΔHf°(O2,g) = 0, what is ΔH° for CO(g) + ½O2(g) → CO2(g)?

  1. -110.5 kJ/mol
  2. -504.0 kJ/mol
  3. -283.0 kJ/mol Correct answer.
  4. +283.0 kJ/mol

Explanation

For a reaction, ΔH° equals the sum of formation enthalpies of products minus reactants. Thus ΔH° = -393.5 - (-110.5 + 0) = -283.0 kJ/mol.

Question 41Fluid Mechanics and Heat Transfer

A vessel contains a gas at a pressure above atmospheric. A U-tube manometer filled with mercury (relative density 13.6) shows a difference of 200 mm between the two mercury levels. Taking g = 9.81 m/s² and the density of water as 1000 kg/m³, what is the gauge pressure of the gas in kPa?

  1. 267 kPa
  2. 2.67 kPa
  3. 13.3 kPa
  4. 26.7 kPa Correct answer.

Explanation

The manometer measures gauge pressure through the hydrostatic relation Δp = ρgh. For mercury, ρ = 13.6 × 1000 kg/m³, so Δp = 13 600 × 9.81 × 0.200 ≈ 26 700 Pa, or 26.7 kPa.

Question 42Fluid Mechanics and Heat Transfer

Water (density 1000 kg/m³, viscosity 1.0 mPa·s) flows through a 50 mm internal diameter pipe at a volumetric flow rate of 2.0 L/s. What is the Reynolds number?

  1. 5.1×10⁴ Correct answer.
  2. 2.0×10³
  3. 2.5×10⁴
  4. 1.0×10⁵

Explanation

The mean velocity is Q/A = 0.002/(π×0.025²) ≈ 1.02 m/s. Then Re = ρVD/μ = 1000×1.02×0.050/0.001 ≈ 5.1×10⁴, which is turbulent.

Question 43Fluid Mechanics and Heat Transfer

In convective heat transfer, the Prandtl number represents the ratio of which two quantities?

  1. Inertial force to viscous force
  2. Buoyancy force to viscous force
  3. Momentum diffusivity to thermal diffusivity Correct answer.
  4. Thermal diffusivity to momentum diffusivity

Explanation

The Prandtl number is Pr = ν/α = μc_p/k, the ratio of momentum diffusivity to thermal diffusivity. It indicates how quickly momentum diffuses relative to heat in a fluid.

Question 44Fluid Mechanics and Heat Transfer

Water flows through a 100 m long, 100 mm diameter commercial steel pipe at a mean velocity of 2.0 m/s. The Darcy friction factor is 0.020. What is the head loss due to friction? (g = 9.81 m/s²)

  1. 2.04 m
  2. 8.16 m
  3. 0.41 m
  4. 4.08 m Correct answer.

Explanation

Darcy–Weisbach gives h_f = f(L/D)(V²/2g). Substituting 0.020 × (100/0.100) × (2.0²/(2×9.81)) gives about 4.08 m.

Question 45Fluid Mechanics and Heat Transfer

Compared with a Venturi meter of the same throat-to-pipe area ratio, a sharp-edged orifice meter has:

  1. A higher permanent pressure loss because of the vena contracta and eddy formation downstream Correct answer.
  2. A higher discharge coefficient because the edge is sharp
  3. The same permanent pressure loss because both obey Bernoulli's equation
  4. A lower permanent pressure loss because the vena contracta reduces turbulence

Explanation

A sharp-edged orifice creates a vena contracta and a downstream eddy zone, so much of the kinetic energy is dissipated. The Venturi meter has a gradual contraction and expansion, giving a higher discharge coefficient and a smaller permanent pressure loss.

Question 46Fluid Mechanics and Heat Transfer

For a centrifugal pump, net positive suction head available (NPSH_A) is defined as:

  1. The total discharge head minus the vapor pressure head of the liquid
  2. The absolute suction pressure head at the pump inlet minus the vapor pressure head of the liquid Correct answer.
  3. The gauge suction pressure head plus the pump head
  4. The difference between the pump head and the friction head in the suction line

Explanation

NPSH_A is the absolute head at the pump suction flange, including velocity head, minus the liquid vapor pressure head. It is compared with NPSH_R, and NPSH_A must be larger to avoid cavitation.

Question 47Fluid Mechanics and Heat Transfer

In a counterflow heat exchanger, hot oil enters at 150°C and leaves at 90°C, while cold water enters at 30°C and leaves at 70°C. What is the log-mean temperature difference?

  1. 75.0°C
  2. 80.0°C
  3. 69.5°C Correct answer.
  4. 60.0°C

Explanation

For counterflow, the terminal differences are 150−70 = 80°C and 90−30 = 60°C. The LMTD is (80−60)/ln(80/60) = 69.5°C, not the arithmetic average.

Question 48Fluid Mechanics and Heat Transfer

A thin-walled pipe carrying a hot fluid is to be insulated. The insulation has thermal conductivity 0.15 W/(m·K) and the outside convective heat transfer coefficient is 10 W/(m²·K). What is the critical radius of insulation, and what happens if the insulation radius is below this value?

  1. 15 mm; adding insulation decreases heat loss
  2. 1.5 mm; adding insulation increases heat loss
  3. 15 mm; adding insulation increases heat loss Correct answer.
  4. 67 mm; adding insulation decreases heat loss

Explanation

The critical radius is r_c = k/h = 0.15/10 = 0.015 m, or 15 mm. Below this radius, adding insulation increases the outer area faster than it adds conduction resistance, so heat loss rises.

Question 49Fluid Mechanics and Heat Transfer

A small gray sphere is placed at the center of a large evacuated enclosure whose walls are black. What is the view factor from the sphere to the enclosure walls?

  1. 0.5
  2. 0
  3. It depends on the ratio of the sphere diameter to the enclosure diameter
  4. 1 Correct answer.

Explanation

For any convex body completely enclosed by another surface, all radiation leaving the body must strike the enclosure. Therefore the view factor from the small sphere to the enclosure is exactly 1.

Question 50Fluid Mechanics and Heat Transfer

A furnace wall is 200 mm thick and has thermal conductivity 1.5 W/(m·K). The inner surface is at 1000°C and the outer surface is at 200°C. What is the steady-state heat flux through the wall?

  1. 6 kW/m² Correct answer.
  2. 600 kW/m²
  3. 0.6 kW/m²
  4. 60 kW/m²

Explanation

Fourier's law for a plane wall gives q″ = kΔT/L. With k = 1.5 W/(m·K), ΔT = 800 K, and L = 0.200 m, q″ = 1.5×800/0.200 = 6000 W/m², or 6 kW/m².