Heat Transfer: practice problems and exam questions

Conduction through walls, thermal resistance networks, convection and the design of heat exchangers with the log mean temperature difference.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Heat conductionsingle wall · composite wall · wall with convectioncalculations
  2. Heat exchangers (LMTD)LMTD · required area · full designcalculations

Exam tips for Heat Transfer

  • Thermal resistances in series add up, just like electrical resistors.
  • A temperature difference is the same in K and in °C, so differences can stay in °C.
  • For the LMTD, take ΔT at each end of the exchanger; parallel flow and counterflow pair the temperatures differently.

Worked examples

Heat conductionEasy
Wood, k = 0.17150 mm20 °C-9 °CQ̇ = ?

A wall of wood is 150 mm thick and has an area of 6 m². Its faces are held at 20 °C and -9 °C. Find the rate of heat conduction Q̇ through the wall.

k = 0.17 W/(m·K)L = 150 mmA = 6 m²T1 = 20 °CT2 = -9 °C

Find Q̇ (W)

Show hints
  1. Fourier’s law for a plane wall: Q̇ = k·A·ΔT/L.
  2. Convert the thickness to metres so the units of k match.
  3. ΔT = 29 K.
Show the answer and the step-by-step solution

Thermal resistance: steady heat flow through layers in series behaves like current through resistors in series: Q̇ = ΔT/ΣR, with R = L/(kA) for a conducting layer and R = 1/(hA) for a convecting surface. The temperature drop across each layer is proportional to its resistance.

Fourier’s law

ΔT = 20 − (-9) = 29 K
Q̇ = kAΔT/L = 0.17·6·29/(150/1000) = 197.2 W

Answer

Q̇ = 197.2 W
Heat exchangers (LMTD)Medium
lengthT165 °C115 °C10 °C80 °CΔT₁ΔT₂parallel flow

A parallel-flow heat exchanger transfers 60 kW. The hot fluid cools from 165 to 115 °C and the cold fluid heats from 10 to 80 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A needed.

Q̇ = 60 kWU = 500 W/(m²·K)Th: 165 → 115 °CTc: 10 → 80 °C

Find A (m²)

Show hints
  1. Use Q̇ = U·A·ΔTlm and solve for A.
  2. In parallel flow both inlets are at the same end, and both outlets at the other.
  3. ΔTlm = 80.64 K.
Show the answer and the step-by-step solution

LMTD method: the temperature difference between the streams changes along the exchanger, so the rate equation uses its logarithmic mean: Q̇ = U·A·ΔTlm, with ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. Counter-flow gives a larger ΔTlm than parallel flow for the same inlet and outlet temperatures.

End temperature differences

In parallel flow both inlets are at the same end, and both outlets at the other.

ΔT₁ = Th,in − Tc,in = 165 − 10 = 155 K
ΔT₂ = Th,out − Tc,out = 115 − 80 = 35 K

Log-mean temperature difference

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (155 − 35)/ln(155/35) = 80.64 K

Area

A = Q̇/(U·ΔTlm) = 60000/(500·80.64) = 1.488 m²

Answer

A = 1.488 m²

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Heat conduction calculatorLMTD calculator (log mean temperature difference)

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