Heat Transfer is 7-11 questions on the FE Mechanical exam. It leans on thermodynamics for energy balances and on fluid mechanics for the flow behaviour that sets convection coefficients, so it is best studied after both rather than in isolation.
The single most useful habit in this area is drawing the thermal resistance network. A composite wall, a pipe with insulation and a heat exchanger all become one-line problems once the resistances are laid out in series and parallel.
Exam weight: NCEES lists Heat Transfer at 7-11 questions (6-10%) of the 110-question FE Mechanical exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Heat Transfer
The specification covers conduction, convection, radiation, and transient processes as well as masses, heat exchangers and thermal resistance. Expect a composite wall or cylinder with a driving temperature difference, a convection coefficient supplied or derived from a Nusselt correlation, a fin efficiency, and a radiation exchange between surfaces.
Heat exchanger items ask for the log mean temperature difference or an effectiveness-NTU result. Transient problems typically use the lumped capacitance method, which is valid only when the Biot number is small enough, and the exam does test whether you check that.
5 Free Heat Transfer Practice Problems
Each problem below comes from the PECivilClick FE Mechanical question bank and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — A. Conduction
Fourier's law of heat conduction states that heat flux is proportional to:
A) The fourth power of temperature
B) The positive temperature gradient
C) The square of temperature difference
D) The negative temperature gradient
Answer: D) The negative temperature gradient
Fourier's law of heat conduction:
\(\boxed{q = -k \frac{dT}{dx}}\)
or
\(\dot{Q} = -kA \frac{dT}{dx}\)
The negative sign indicates heat flows from high to low temperature (opposite to the temperature gradient direction).
Problem 2 — A. Conduction
A wall 0.2 m thick with k = 1.5 W/m·K has surface temperatures of 100°C and 40°C. The heat flux through the wall is:
A) 300 W/m\(^{2}\)
B) 450 W/m\(^{2}\)
C) 600 W/m\(^{2}\)
D) 225 W/m\(^{2}\)
Answer: B) 450 W/m\(^{2}\)
Heat flux through a plane wall:
\(q = \frac{k(T_1 - T_2)}{L}\)
Given: \(k = 1.5\) W/m·K, \(T_1 = 100^{\circ}C\), \(T_2 = 40^{\circ}C\), \(L = 0.2\) m
\(q = \frac{1.5 \times (100 - 40)}{0.2} = \frac{1.5 \times 60}{0.2}\)
\(\boxed{q = 450 \text{ W/m}^2}\)
Problem 3 — A. Conduction
A pipe (k = 50 W/m·K) has inner radius 25 mm and outer radius 30 mm. Steam at 150°C flows inside with \(h_{i}\) = 500 W/m\(^{2}\)·K, and air at 20°C outside with \(h_{o}\) = 20 W/m\(^{2}\)·K. Per meter length, the heat loss is most nearly:
A) 467 W/m
B) 1200 W/m
C) 520 W/m
D) 350 W/m
Answer: A) 467 W/m
Total thermal resistance per unit length (\(L = 1\) m):
\(R_{conv,i} = \frac{1}{h_i \times 2\pi r_i L} = \frac{1}{500 \times 2\pi \times 0.025 \times 1} = 0.01273\) K/W
\(R_{cond} = \frac{\ln(r_2/r_1)}{2\pi k L} = \frac{\ln(30/25)}{2\pi \times 50 \times 1} = 0.000582\) K/W
\(R_{conv,o} = \frac{1}{h_o \times 2\pi r_o L} = \frac{1}{20 \times 2\pi \times 0.030 \times 1} = 0.2653\) K/W
\(R_{total} = 0.01273 + 0.000582 + 0.2653 = 0.2786\) K/W
\(\dot{Q} = \frac{\Delta T}{R_{total}} = \frac{150 - 20}{0.2786}\)
\(\boxed{\dot{Q} \approx 467 \text{ W/m}}\)
Problem 4 — A. Conduction
For a composite wall with two layers in series, the total thermal resistance is:
A) \(R_{total}\) = (R\(_{1}\) + R\(_{2}\))/2
B) 1/\(R_{total}\) = 1/R\(_{1}\) + 1/R\(_{2}\)
C) \(R_{total}\) = R\(_{1}\) + R\(_{2}\)
D) \(R_{total}\) = R\(_{1}\) \(\times\) R\(_{2}\)
Answer: C) \(R_{total}\) = R\(_{1}\) + R\(_{2}\)
For thermal resistances in series (heat flows through each layer sequentially):
\(R_{total} = R_1 + R_2 + R_3 + ...\)
For parallel resistances (heat flows through layers simultaneously):
\(\boxed{\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...}\)
This is analogous to electrical resistances.
Problem 5 — A. Conduction
The thermal resistance for conduction through a plane wall is:
A) R = 1/(hA)
B) R = L/(kA)
C) R = kA/L
D) R = hA
Answer: B) R = L/(kA)
Thermal resistance for a plane wall:
\(\boxed{R = \frac{L}{kA}}\)
where:
- \(L\) = wall thickness (m)
- \(k\) = thermal conductivity (W/m·K)
- \(A\) = cross-sectional area (m\(^{2}\))
Note: Convection resistance is \(R = \frac{1}{hA}\)
Using the FE Reference Handbook for Heat Transfer
The Heat Transfer section holds the conduction shape factors, the thermal resistance expressions for plane and cylindrical geometries, fin efficiency charts and the Nusselt correlations. The cylindrical resistance expression with its natural logarithm of the radius ratio is the one worth locating before exam day, because writing it from memory is where sign and ratio errors creep in.
Four Mistakes That Cost Points
- Adding parallel thermal resistances in series. Layers stacked through the heat flow path add directly; parallel paths combine as reciprocals. Mis-wiring the network is the dominant error in composite-wall problems, and both totals are on the answer list.
- Using Celsius in a radiation calculation. The Stefan-Boltzmann law requires absolute temperature raised to the fourth power. A Celsius substitution does not give a slightly wrong answer, it gives a wildly wrong one, which at least makes it easy to catch if you sanity-check magnitude.
- Applying lumped capacitance without checking the Biot number. Lumped capacitance assumes negligible internal temperature gradient, valid for a small Biot number. Applying it to a thick, low-conductivity body is precisely the trap a transient question is built around.
- Using the wrong area for a cylinder. Conduction through a cylindrical wall uses the log mean area, not the inner or outer surface area. Convection at the surface uses that surface's actual area. Using one where the other belongs is a designed distractor.
Frequently Asked Questions
How many heat transfer questions are on the FE Mechanical exam?
NCEES specifies 7-11 questions out of 110, about 6-10 percent. That is the same weight as Material Properties and Processing, and slightly less than Thermodynamics at 10-15.
Do I need to memorise Nusselt number correlations?
No. The handbook provides correlations for the standard geometries and flow regimes. What you must do is identify the regime first, since a correlation for laminar internal flow gives a wrong answer when applied to turbulent flow, and the exam supplies the numbers needed to tell them apart.
How are heat exchangers tested?
Usually through the log mean temperature difference method for a straightforward duty calculation, or through effectiveness-NTU when outlet temperatures are unknown. Know which method each situation calls for; that decision is often the whole question.
Should I study heat transfer before or after thermodynamics?
After. Heat transfer questions frequently open with an energy balance that assumes thermodynamic fluency, and convection depends on flow behaviour from fluid mechanics. Studying it first means learning three subjects at once.
Keep Going
These topics feed into each other on the exam:
- FE Mechanical Thermodynamics practice problems — 10-15 questions on the exam
- FE Mechanical Fluid Mechanics practice problems — 10-15 questions on the exam
- FE Mechanical Material Properties and Processing practice problems — 7-11 questions on the exam
Browse every knowledge area from the free FE Mechanical practice problem hub, see what the full bank covers on the FE Mechanical exam prep page, or plan your schedule with the FE study timeline.