FE Mechanical · knowledge area 12 · 7–11 questions
FE heat transfer review, three modes, one network
Heat transfer is 7–11 questions on five topics: conduction, convection, radiation, transient processes and heat exchangers. Most problems reduce to a thermal-resistance network or a single energy balance; the work is in choosing the right one.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam7–11
worked problems here15
of 14 knowledge areas12
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What the area covers
The specification names five topics. What each one means on the exam:
Conduction: Fourier's law in plane walls, cylinders and spheres, composite layers, fins and extended surfaces.
Convection: the convection coefficient, dimensionless groups (Reynolds, Prandtl, Nusselt) and the standard correlations for internal and external flow.
Radiation: blackbody and gray-body exchange, emissivity, view factors.
Transient processes: lumped capacitance and the conditions under which it holds.
Heat exchangers: overall coefficient U, log mean temperature difference, effectiveness-NTU.
15 heat transfer problems, every option explained
0 / 15 answered · 0 correct
Question 1 of 15 · Heat Transfer · 7–11 items
An enclosure consists of three surfaces: Surface 1 is a flat circular base of area A1, Surface 2 is a cylindrical side wall of area A2, and Surface 3 is a hemispherical dome of area A3 closing the top. If the view factor from the base to the cylindrical wall is F12, which expression represents the correct view factor from the dome to the base (F31)?
Solution
F11 = 0 → F13 = 1 − F12
A1F13 = A3F31 → F31 = (A1/A3)(1 − F12)
FE Reference Handbook, Heat Transfer: radiation (view factor relations).
F12 describes the wall, not the base-to-dome exchange, and A2 does not appear in the reciprocity for F31.
The area ratio is inverted: A1F13 = A3F31 gives F31 = (A1/A3)F13.
This subtracts a term built from the wrong reciprocity pair; F12 must first be turned into F13 by the summation rule.
This mixes view factors from different surfaces; a summation rule adds factors leaving one surface only.
The flat base sees nothing of itself, so F13 = 1 − F12; reciprocity A1F13 = A3F31 then gives F31.
Question 2 of 15 · Heat Transfer · 7–11 items
A long, straight rectangular aluminum fin (k = 205 W/(m·K)) is 2 mm thick, extends 50 mm from the base, and is 20 mm wide. The convection coefficient is 25 W/(m2·K), and the fin tip is insulated. Using η = tanh(mL)/(mL) with m = √(2h/(kt)), what is the fin efficiency?
Given
Thermal conductivity k = 205 W/(m·K)
Fin thickness t = 2 mm
Fin length L = 50 mm
Fin width w = 20 mm
Convection coefficient h = 25 W/(m2·K)
Find
Fin efficiency η (%)
Solution
m = √(2h/(kt)) = √(2 × 25/(205 × 0.002)) = 11.04 m−1, mL = 11.04 × 0.050 = 0.552
ηf = tanh(mL)/(mL) = 0.502/0.552 = 0.91
FE Reference Handbook, Heat Transfer: fins.
This uses the 20 mm width as the fin length; L is the 50 mm the fin projects from the base.
Dropping the 2 in m = √(2h/(kt)) counts only one face of the fin and overstates its efficiency.
m = √(2h/(kt)) = √(2 × 25/(205 × 0.002)) = 11.0 m−1, mL = 0.552, and η = tanh(0.552)/0.552 = 0.91.
tanh(mL) = 0.50 is only the numerator; the efficiency divides it by mL.
Question 3 of 15 · Heat Transfer · 7–11 items
A composite wall has two layers in series: the first is 0.10 m thick with k = 1.5 W/(m·K), and the second is 0.05 m thick with k = 0.05 W/(m·K). The inner and outer surface convection coefficients are 10 and 15 W/(m2·K), respectively. What is the overall heat transfer coefficient U?
FE Reference Handbook, Heat Transfer: conduction through a composite wall.
Leaving out the 0.05 m insulation layer drops the largest resistance in the wall.
Multiplying thickness by conductivity (L·k) instead of dividing gives the wrong layer resistances.
This includes only the two conduction layers; the inner and outer film resistances also belong in the series sum.
Rtotal = 1/10 + 0.10/1.5 + 0.05/0.05 + 1/15 = 1.233 m2·K/W, so U = 1/1.233 = 0.81 W/(m2·K).
Question 4 of 15 · Heat Transfer · 7–11 items
Air flows parallel to a flat plate of length L = 2.0 m and width W = 1.0 m with a free-stream velocity of u∞ = 5.0 m/s. The free-stream temperature is T∞ = 300 K, and the plate surface is held at Ts = 400 K. Take the critical Reynolds number for transition as Rec = 5 × 105. Using the appropriate correlation below (laminar: NuL = 0.664 ReL0.5 Pr1/3; mixed: NuL = (0.037 ReL0.8 − 871) Pr1/3) and air properties from the table, what is the total convective heat loss from one side of the plate?
FE Reference Handbook, Heat Transfer: forced convection (flat plate).
This uses the mixed correlation with properties at the surface temperature; properties belong at the film temperature, where the flow is laminar.
The mixed correlation applies only when ReL exceeds 5 × 105; at the film temperature the whole plate is laminar.
At the 350 K film temperature ReL = 5.0 × 2.0/20.92 × 10−6 = 4.78 × 105 < 5 × 105, so the laminar correlation applies: Nu = 407, h = 6.1 W/(m2·K), q = 1.22 kW.
Evaluating properties at the free-stream temperature makes ReL exceed 5 × 105 and wrongly switches to the mixed correlation.
This counts both faces of the plate (2 × 1.223 kW); the question asks for one side.
Question 5 of 15 · Heat Transfer · 7–11 items
What is the critical radius of insulation for a cylindrical pipe with insulation having a thermal conductivity of 0.05 W/(m·K) and exposed to air with a convection heat transfer coefficient of 10 W/(m²·K)?
Given
Insulation conductivity k = 0.05 W/(m·K)
Convection coefficient h = 10 W/(m2·K)
Find
Critical radius rcr (mm)
Solution
rcr = k/h = 0.05 W/(m·K)/10 W/(m2·K) = 0.005 m = 5.0 mm
FE Reference Handbook, Heat Transfer: critical insulation radius.
For a cylinder rcr = k/h = 0.05/10 = 0.005 m = 5.0 mm.
2k/h is the critical radius of a sphere, not a cylinder.
4k/h = 20.0 mm is the critical diameter of a sphere; the question asks for the critical radius of a cylinder, k/h.
h/k = 200 is the inverse ratio and has units of 1/m, not length.
Question 6 of 15 · Heat Transfer · 7–11 items
Two parallel plates with emissivities ε1 = 0.8 and ε2 = 0.4 face each other. If the view factor is 1, what is the effective emissivity for radiation heat transfer between these surfaces?
An engineer is evaluating a 1-shell-pass, 2-tube-pass heat exchanger. The shell-side (hot) fluid cools from 200 °F to 140 °F, while the tube-side (cold) fluid heats from 80 °F to 135 °F. Based on common design practice and the correction factor (F) values below for this configuration, what is the main conclusion about this design?
F(P = 0.46, R ≈ 1.1) ≈ 0.65 < 0.75 → add a shell pass
FE Reference Handbook, Heat Transfer: heat exchangers (LMTD correction factor).
A cold exit below the hot exit avoids a temperature cross at the outlets, but F still falls steeply here.
The method can be applied, but the low F shows the exchanger sits on the steep part of the curve and needs more shells.
P = 0.46 is below the asymptote near 0.50, so the LMTD and F are both defined.
R above 1 does not make a design acceptable; the F-factor at the operating P is what decides it.
R = 60/55 = 1.09 and P = 55/120 = 0.46 give F ≈ 0.65–0.73, below the usual 0.75–0.80 floor where a 1-2 exchanger becomes unreliable.
Question 8 of 15 · Heat Transfer · 7–11 items
A counterflow heat exchanger cools 10 kg/s of oil (cp = 2.0 kJ/(kg·K)) from 90 °C to 50 °C using water (cp = 4.0 kJ/(kg·K)) that enters at 20 °C and leaves at 30 °C. If the overall heat transfer coefficient is U = 0.5 kW/(m2·K), what is the required surface area?
This uses the parallel-flow end differences (70 K and 20 K); the exchanger is counterflow.
Using the water's cp of 4.0 for the oil doubles the duty; the oil has cp = 2.0 kJ/(kg·K).
Question 9 of 15 · Heat Transfer · 7–11 items
A composite wall consists of an outer brick layer, a middle section containing insulation and parallel wooden studs, and an inner drywall layer. Heat flows 1-dimensionally from the inside to the outside environment. Represent series heat-transfer paths with thermal-resistance networks: for conduction Rcond = L/(k · A) and for convection Rconv = 1/(h · A); combine resistances (series and parallel). Based on the variables defined in the table below, which of the following expressions correctly represents the equivalent conductive thermal resistance of the solid composite wall (excluding the air convective boundary layers)?
Layer
Description
Thermal Resistance Variable
Outer
Brick layer
R1
Middle
Insulation
R2
Middle
Wooden studs
R3
Inner
Drywall
R4
Solution
Req = R1 + (R2R3)/(R2 + R3) + R4
Brick and drywall are in series with the middle layer, where insulation and studs are parallel paths: Rtotal = R1 + (R2 ∥ R3) + R4.
The parallel pair is combined correctly, but series resistances add directly, not as reciprocals.
This pairs the layers backwards: insulation and studs are side by side, not one after the other.
This puts all four layers in parallel; brick and drywall carry the full heat flow in series.
Adding reciprocals gives a conductance-like sum and treats every layer as parallel.
Question 10 of 15 · Heat Transfer · 7–11 items
A 5 mm diameter steel sphere (k = 50 W/(m·K), ρ = 7800 kg/m3, cp = 450 J/(kg·K)) initially at 200 °C is suddenly immersed in a fluid at 20 °C with h = 250 W/(m2·K). What is the Biot number for this transient process?
Given
Sphere diameter D = 5 mm
Conductivity k = 50 W/(m·K)
Density ρ = 7800 kg/m3
Specific heat cp = 450 J/(kg·K)
Convection coefficient h = 250 W/(m2·K)
Find
Biot number Bi
Solution
Lc = V/As = D/6 = 0.005/6 = 8.33 × 10−4 m
Bi = hLc/k = 250 × 8.33 × 10−4/50 = 0.0042 (< 0.1, lumped model valid)
FE Reference Handbook, Heat Transfer: transient conduction (lumped capacitance).
For a sphere Lc = V/As = D/6 = 0.833 mm, so Bi = hLc/k = 250 × 0.000833/50 = 0.0042.
This uses Lc = D/3; V/As for a sphere is r/3 = D/6.
This uses the radius as Lc; the lumped-capacitance length is V/As = r/3.
This uses the full diameter as the characteristic length.
Question 11 of 15 · Heat Transfer · 7–11 items
For a lumped-capacitance thermal system, how is the time constant τ defined?
Solution
(T − T∞)/(Ti − T∞) = e−t/τ, τ = ρVcp/(hA)
FE Reference Handbook, Heat Transfer: transient conduction (lumped capacitance).
This is the reciprocal of τ, the decay rate in 1/s.
Multiplying capacitance by conductance does not give a time; the units are not seconds.
k/(ρcp) is the thermal diffusivity α, in m2/s, not a time constant.
The time constant is thermal capacitance ρVcp divided by convective conductance hA; the temperature falls to 37% of the initial difference after one τ.
Question 12 of 15 · Heat Transfer · 7–11 items
What is the Nusselt number for fully developed laminar flow in a circular tube with constant heat flux at the wall?
Solution
Fully developed laminar flow in a circular tube: NuD = 4.36 (uniform q''), 3.66 (uniform Ts). FE Reference Handbook, Heat Transfer: forced convection (internal flow).
3.66 is the value for a uniform wall temperature, not a uniform heat flux.
For fully developed laminar flow in a circular tube with uniform wall heat flux, NuD = 48/11 = 4.36.
7.54 is for laminar flow between infinite parallel plates at uniform temperature.
8.23 is for parallel plates with uniform heat flux, not a circular tube.
Question 13 of 15 · Heat Transfer · 7–11 items
What is the shape of the steady temperature distribution in a plane wall with uniform heat generation?
Solution
d2T/dx2 + q̇/k = 0 → T(x) = −q̇x2/(2k) + C1x + C2
A logarithmic profile belongs to radial conduction in a cylinder without generation.
With uniform generation, d2T/dx2 = −q̇/k is constant, so T(x) is a second-order polynomial with its peak at the center.
A linear profile belongs to a plane wall without heat generation.
Exponential profiles appear in fins and transient decay, not in steady conduction with uniform generation.
Question 14 of 15 · Heat Transfer · 7–11 items
A counterflow heat exchanger has an overall heat transfer coefficient of 500 W/(m2·K) and an area of 10 m2. Hot fluid enters at 80 °C at 2 kg/s (cp = 4.2 kJ/(kg·K)), and cold fluid enters at 30 °C at 1 kg/s (cp = 4.2 kJ/(kg·K)). What are the NTU and the effectiveness of the exchanger?
The thermal resistance of a cylindrical pipe with inner radius r1, outer radius r2, length L, and thermal conductivity k is given by:
Solution
R = ln(r2/r1)/(2πkL)
FE Reference Handbook, Heat Transfer: conduction through a cylindrical wall.
Radial conduction through a cylindrical shell gives a logarithmic temperature profile and this resistance.
This is the resistance to axial conduction along the pipe wall, not radial conduction through it.
This has no conductivity in it; it looks like a convection resistance 1/(hA) with h missing.
The plane-wall formula ignores that the area grows with radius; it is a fair approximation only for thin walls.
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Common traps
Checked against the Heat Transfer chapter of the FE Reference Handbook and Incropera's Fundamentals of Heat and Mass Transfer:
Radiation needs absolute temperature, to the fourth power. Subtracting temperatures first and then raising to the fourth power is a different, wrong number. Convert to K or °R, raise each to the fourth power, then subtract.
h is not U. The convection coefficient h describes one surface. The overall coefficient U combines both films and the wall, and it must be referenced to an area: inner and outer U of a thick tube differ.
Draw the network before adding numbers. Layers in the heat path are in series; side-by-side materials across the same path are in parallel. Writing the network first prevents adding conductances where resistances belong.
Check the Biot number before using lumped capacitance. The lumped model assumes a uniform temperature inside the body, which holds only when internal conduction resistance is small next to surface convection.
LMTD end differences. In counterflow, pair the hot inlet with the cold outlet; in parallel flow, inlet with inlet. Shell-and-tube and crossflow units need the correction factor on top.
Mass-based specific heat. The Handbook notes that its thermal diffusivity expression holds only when the specific heat is per unit mass.
Emissivity is a surface property, the view factor is geometry
Two surfaces with perfect emissivity still exchange little radiation if they barely see each other. Keep the two effects separate: emissivities go into the surface resistances, the view factor into the space resistance.
Heat transfer and its neighbors
The heat-exchanger energy balance is the same first-law balance used in thermodynamics; convection correlations take their Reynolds numbers from fluid mechanics; conductivities and specific heats come from material properties. A sensor's time constant in measurements and controls is a lumped-capacitance problem in disguise.
This page owns heat transfer fundamentals for the whole site. Process exchangers with fouling, condensers, reboilers and evaporators are on FE Chemical heat transfer; HVAC coils, chillers and cooling towers on PE HVAC&R equipment and components.
Use LMTD when all four end temperatures are known or easy to find. Use effectiveness-NTU when the outlet temperatures are the unknowns, because it avoids iterating.
How many heat transfer questions are on the FE Mechanical exam?
7–11, the same range as material properties and processing.
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