Heat transfer carries 8–12 questions. Conduction, convection and radiation are on the outline, but its equipment subtopic (double pipe, shell and tube, LMTD, NTU) is the chemical part, and this page stays with it.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam8–12
problems on this page16
of 17 knowledge areas09
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What the specification covers
The outline lists five subtopics. In our words:
Conduction through walls, tubes and insulation.
Convection, natural and forced, through the usual dimensionless correlations.
Radiation between surfaces and from flames or furnace walls.
Heat-transfer coefficients: local, overall and the fouling allowance.
Exchanger equipment and design: double pipe, shell and tube, flow arrangement, the log-mean ΔT and the number of transfer units.
16 heat transfer problems
0 / 16 answered · 0 correct
Question 1 of 16 · Heat Transfer · 8–12 items
A single-effect evaporator is used to concentrate an aqueous solution. The required evaporation rate of water is 0.5 kg/s. The solution boils constantly at 105°C, and its latent heat of vaporization is 2200 kJ/kg. The evaporator is heated by a thermal oil stream flowing countercurrently through a heat exchanger jacket. The oil enters the jacket at 180°C and leaves at 130°C. If the overall heat transfer coefficient is 2000 W/(m2 · K) and sensible heat demands of the feed are negligible, what is the required heat transfer area?
Given
Evaporation rate = 0.5 kg/s
Boiling temperature = 105 °C
Latent heat λ = 2200 kJ/kg
Oil inlet temperature = 180 °C
Oil outlet temperature = 130 °C
Overall coefficient U = 2000 W/(m2·K)
Find
Heat transfer area A (m2)
Solution
q = ṁλ = 0.5 kg/s × 2,200 kJ/kg = 1,100 kW
ΔT1 = 180 − 105 = 75 °C, ΔT2 = 130 − 105 = 25 °C → LMTD = (75 − 25)/ln(75/25) = 45.5 °C
FE Reference Handbook, Heat Transfer: heat exchangers (log mean temperature difference).
q = 0.5 kg/s × 2,200 kJ/kg = 1,100 kW; with the boiling side at 105 °C the LMTD is (75 − 25)/ln 3 = 45.5 °C, so A = 1.1 × 106/(2,000 × 45.5) = 12.1 m2.
10.8 m2 assumes the solution boils at 100 °C (pure water at 1 atm) instead of the stated 105 °C, so ΔT1 = 80 and ΔT2 = 30 °C.
7.33 m2 uses only the hot-end difference (180 − 105 = 75 °C) as the driving force, which overstates it.
5.25 m2 takes a base-10 logarithm in the LMTD ((75 − 25)/log 3 = 104.8 °C); the log mean uses ln.
0.0121 m2 keeps the duty in kW (1,100) while U is in W/(m2·K), so the area is a thousand times too small.
Question 2 of 16 · Heat Transfer · 8–12 items
A counterflow shell-and-tube heat exchanger is designed to transfer 100 kW of heat duty. The operating conditions and fluid parameters are provided in the table below. The separating tube wall has a thickness of 2 mm and a thermal conductivity of 50 W m−1 K−1. Assuming planar wall resistance for simplicity and treating the area basis as identical, calculate the required heat transfer area.
FE Reference Handbook, Heat Transfer: heat exchangers (overall coefficient with fouling, LMTD).
1.06 m2 adds the two film coefficients (800 + 1,200) as if they were conductances in parallel; resistances in series add, not coefficients.
3.46 m2 uses only the hot-end difference (120 − 50 = 70 °C) as the driving force instead of the log mean (47.2 °C).
4.50 m2 leaves out both fouling resistances (0.0003 m2·K/W in total): U = 471 W/(m2·K) instead of 413.
1/U = 1/800 + 1/1,200 + 0.0002 + 0.0001 + 0.002/50 = 0.00242, so U = 413 W/(m2·K); counterflow LMTD = (70 − 30)/ln(70/30) = 47.2 °C, and A = 100,000/(413 × 47.2) = 5.13 m2.
6.66 m2 uses the parallel-flow end differences (90 and 10 °C); the exchanger is counterflow.
Question 3 of 16 · Heat Transfer · 8–12 items
When sizing the surface of a shell-and-tube film condenser, an engineer mistakenly ignores the condensate-film resistance while calculating the overall heat transfer coefficient. What is the effect on the calculated required area?
Solution
1/U = 1/hfilm + Rwall + Rf + 1/hcw
Dropping 1/hfilm raises U; A = Q/(U·ΔTlm) falls, so the condenser is undersized.
Dropping a series resistance always changes U; the condensate film is often the largest resistance in a condenser.
The direction is wrong: a missing resistance inflates U and shrinks the computed area.
Leaving out the film resistance overstates U, and A = Q/(U·ΔTlm) then comes out too small, so the condenser underperforms.
An infinite area would need zero U or zero driving force; omitting a resistance makes U larger, not zero.
An oversized result would come from adding a resistance, not from leaving one out.
Question 4 of 16 · Heat Transfer · 8–12 items
A process engineer evaluates a heat exchanger heating liquid water (cp = 4.18 kJ/(kg · K)) from 20°C to 80°C, requiring a heat duty of 1254 kW. Saturated steam enters the hot side at 100°C and exits as subcooled liquid at 90°C. The engineer models the hot side duty using the sensible heat equation, Q = ṁhot · cp,vapor · (100 - 90), arriving at a calculated duty of only 15 kW. This severely underpredicts the actual heat transferred to the cold stream. Assuming zero environmental heat loss, what is the primary cause of this severe calculation discrepancy?
Solution
Qhot = ṁ[hfg + cp,liq(100 − 90 °C)]
For steam condensing the latent term dominates; a sensible-only balance (even with cp,vapor) misses almost all of the duty.
LMTD belongs to the rate equation Q = UAΔTlm; the duty from an energy balance needs no temperature-difference method.
The steam enters saturated, so there is no superheat; the missing energy is the phase change.
U affects the area, not the duty; the energy balance does not involve U.
Saturated steam condensing releases hfg (about 2,257 kJ/kg at 100 °C), far more than the sensible heat of 10 °C of subcooling, so ṁhfg must be added.
Using liquid cp for the subcooling step is a small correction; it cannot close a gap of more than 1,200 kW.
Question 5 of 16 · Heat Transfer · 8–12 items
A thick-walled process tube has an inner radius of 0.02 m and an outer radius of 0.03 m. The thermal conductivity of the tube material is 50 W/(m · K). The inside convective heat transfer coefficient is 500 W/m2 · K, and the outside convective coefficient is 1000 W/m2 · K. Determine the overall heat transfer coefficient based strictly on the outside surface area (Uo).
An exothermic reaction takes place in a continuous stirred-tank reactor (CSTR) operating at a steady temperature of 350 K. The feed enters at 350 K, meaning no sensible heat change is required for the reacting fluid. A cooling jacket with a constant water temperature of 290 K removes the generated heat. Using the process data provided below, determine the total required heat transfer area for the cooling jacket.
Parameter
Value
Reactor volume
4.8 m3
Reaction rate at 350 K
1.5 kmol/(m3·h)
Heat of reaction
−90,000 kJ/kmol
Clean overall heat transfer coefficient
500 W/(m2·K)
Area safety margin for fouling
25 % increase over clean area
Given
Reactor temperature T = 350 K
Feed temperature Tfeed = 350 K
Cooling water temperature Tw = 290 K
Reactor volume V = 4.8 m3
Reaction rate at 350 K −rA = 1.5 kmol/(m3·h)
Heat of reaction ΔHR = −90,000 kJ/kmol
Clean overall heat transfer coefficient U = 500 W/(m2·K)
286 kW divides qmax by ε; the actual duty can never exceed the maximum possible.
200 kW is the thermodynamic maximum, reached only by an infinitely large counterflow exchanger (ε = 1).
ε = q/qmax, so q = 0.7 × 200 kW = 140 kW.
60 kW is (1 − ε)·qmax, the share of the maximum that the exchanger fails to transfer.
Question 9 of 16 · Heat Transfer · 8–12 items
In a double pipe heat exchanger, the hot fluid enters at 90°C and exits at 60°C. The cold fluid enters at 20°C and exits at 40°C. What is the log-mean temperature difference (LMTD) for counter-current flow?
FE Reference Handbook, Heat Transfer: log mean temperature difference.
70.0 °C is the largest possible difference (hot inlet − cold inlet, 90 − 20), not a mean over the exchanger.
50.0 °C is just the larger terminal difference, not a mean over the exchanger.
45.0 °C is the arithmetic mean of 50 and 40 °C; the log mean is always slightly smaller.
Counter-current ends: 90 − 40 = 50 °C and 60 − 20 = 40 °C; LMTD = (50 − 40)/ln(50/40) = 44.8 °C.
Question 10 of 16 · Heat Transfer · 8–12 items
What is the heat transfer rate through a plane wall with thermal conductivity of 0.5 W/(m·K), area of 2 m², thickness of 0.1 m, and temperature difference of 40°C?
Given
Thermal conductivity k = 0.5 W/(m·K)
Area A = 2 m2
Thickness L = 0.1 m
Temperature difference ΔT = 40°C
Find
Heat transfer rate q (W)
Solution
q = kAΔT/L = 0.5 W/(m·K) × 2 m2 × 40 K/0.1 m = 400 W
FE Reference Handbook, Heat Transfer: conduction (Fourier's law) through a plane wall.
4,000 W takes the wall as 0.01 m thick, a factor-of-ten slip when converting the thickness.
q = kAΔT/L = 0.5 W/(m·K) × 2 m2 × 40 K/0.1 m = 400 W.
200 W is the heat flux in W/m2; it still has to be multiplied by the 2 m2 area.
4 W multiplies by the thickness instead of dividing by it; conduction falls as the wall gets thicker.
Question 11 of 16 · Heat Transfer · 8–12 items
Which dimensionless number represents the ratio of convective to conductive heat transfer in a fluid?
Solution
Nu = hL/k = convective heat transfer/conductive heat transfer across the fluid layer
FE Reference Handbook, Heat Transfer: dimensionless numbers.
Re = ρVL/μ compares inertial with viscous forces; it describes flow, not the heat transfer ratio.
Nu = hL/k compares convection at a surface with pure conduction through the fluid layer of the same length.
Pr = cpμ/k compares momentum diffusivity with thermal diffusivity, a fluid property.
Gr compares buoyancy with viscous forces and drives natural convection.
Question 12 of 16 · Heat Transfer · 8–12 items
In natural convection, which dimensionless parameter represents the ratio of buoyancy forces to viscous forces?
Solution
Gr = gβ(Ts − T∞)L3/ν2 (buoyancy/viscous); Ra = Gr·Pr
FE Reference Handbook, Heat Transfer: natural (free) convection.
Gr = gβ(Ts − T∞)L3/ν2 is the ratio of buoyancy to viscous forces that governs natural convection.
Pr = ν/α relates momentum and thermal diffusivity; it is a fluid property with no buoyancy in it.
Nu = hL/k is the dimensionless heat transfer coefficient, the result of a natural-convection correlation rather than the buoyancy ratio.
Re compares inertial with viscous forces in forced flow; in natural convection there is no imposed velocity.
Question 13 of 16 · Heat Transfer · 8–12 items
A process engineer sized a shell-and-tube heat exchanger using the arithmetic mean temperature difference (AMTD) instead of the log-mean temperature difference (LMTD). With U and fluid properties constant, how does this choice affect the calculated heat transfer area?
A = Q/(U·ΔTm): a larger ΔTm gives a smaller area, so the exchanger falls short of its duty (equal only when ΔT1 = ΔT2).
F is a separate multipass correction applied to either mean; leaving it out is a different error from choosing AMTD.
The two means coincide only when the two end differences are equal; otherwise AMTD exceeds LMTD in any flow arrangement.
The inequality runs the other way: the arithmetic mean is at least the log mean, so the area is undersized, not oversized.
AMTD = (ΔT1 + ΔT2)/2 is never smaller than LMTD; a larger ΔT in A = Q/(U·ΔT) gives an area too small for the duty.
Question 14 of 16 · Heat Transfer · 8–12 items
Surface superheat (ΔTexcess)
Observed boiling regime
5 °C to 30 °C
Nucleate boiling
30 °C to 120 °C
Transition boiling
120 °C to 400 °C
Film boiling
An engineer applies the Rohsenow correlation to estimate the heat flux for water boiling on a metal surface with a surface superheat (ΔTexcess) of 250 °C. Based on the boiling-curve regime limits in the table, what fundamental error has the engineer made?
Solution
Boiling curve: natural convection (ΔTexcess < 5 °C) → nucleate (up to the critical heat flux) → transition → film boiling. Rohsenow's correlation, q″ ∝ ΔTexcess3, holds only in the nucleate regime; at 250 °C the surface is in film boiling.
Rohsenow is for saturated pool boiling, and no subcooling is described; the regime is the problem.
Rohsenow is a nucleate-boiling correlation; at 250 °C of superheat the surface is blanketed by vapor (film boiling), which needs a film-boiling correlation.
Natural convection lies below about 5 °C of superheat, where no bubbles form; 250 °C is far beyond it.
The transition regime spans 30–120 °C in the table; 250 °C sits in film boiling.
Rohsenow is written in terms of liquid properties (μl, cp,l, Prl) and hfg, so using them is correct, not an error.
Question 15 of 16 · Heat Transfer · 8–12 items
A shell-and-tube exchanger cools a process gas with cooling water. The gas-side film coefficient is 50 W/(m2·K) and the water-side coefficient is 5,000 W/(m2·K); wall and fouling resistances are negligible. Which single change raises the overall coefficient U the most?
2hgas: 1/100 + 1/5,000 = 0.0102 → U = 98.0 · 2hwater: U = 49.75
The smaller coefficient controls U; that is why finned tubes go on the gas side.
More water flow raises hwater by about 20.8 = 1.74, but the water film is not the controlling resistance: U gains under 0.5 %.
The wall resistance is already negligible, so a thinner wall changes U by practically nothing.
The water film is only 1 % of the resistance: U rises from 49.5 to 49.75 W/(m2·K), about 0.5 %.
1/U = 1/50 + 1/5,000 gives U = 49.5 W/(m2·K); with hgas = 100 it becomes 98.0, almost double, because the gas film holds 99 % of the resistance.
Question 16 of 16 · Heat Transfer · 8–12 items
In a countercurrent exchanger, hot oil (ṁ = 2.0 kg/s, cp = 2.2 kJ/(kg·K)) entering at 150 °C heats water (ṁ = 1.5 kg/s, cp = 4.18 kJ/(kg·K)) from 25 °C to 55 °C. Heat losses are negligible. What is the oil outlet temperature?
Given
Oil: ṁ = 2.0 kg/s, cp = 2.2 kJ/(kg·K), Tin = 150 °C
Water: ṁ = 1.5 kg/s, cp = 4.18 kJ/(kg·K), 25 °C → 55 °C
q = 1.5 × 4.18 × (55 − 25) = 188.1 kW; the oil cools by 188.1/(2.0 × 2.2) = 42.75 K, so it leaves at 107.3 °C.
120.0 °C assumes the oil cools by the same 30 K that the water warms; with a different ṁcp the temperature changes differ.
127.5 °C uses the water's heat capacity (4.18) for the oil, so the oil's temperature drop comes out about half the true value.
192.8 °C adds the oil's 42.75 K change instead of subtracting it; the hot stream cools.
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Common traps
The multipass correction. The log mean is exact only for pure counter- or co-current flow (end pairing is on FE Mechanical heat transfer); multipass and crossflow exchangers need the factor F:
Q = U·A·F·ΔTlm
Area basis of U. An overall coefficient is quoted on the inside or the outside tube area; use the area it was based on.
Treating fouling as a coefficient. A fouling factor is a resistance; it adds to the other resistances in series, it does not multiply U.
Effectiveness without Cmin. The ceiling on duty is set by the stream with the smaller heat-capacity rate; take the larger and the effectiveness comes out too low.
Sources: the Heat Transfer chapter of the FE Reference Handbook; Incropera et al., Fundamentals of Heat and Mass Transfer.
How it connects
Heat duties come from the energy balance, and evaporators and condensers tie into mass transfer. Conduction, fins and radiation fundamentals are covered once for the site on FE Mechanical heat transfer; this page stays with process equipment.