Heat exchangers (LMTD): practice problems with step-by-step solutions

Free Heat exchangers (LMTD) practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: LMTD, required area, full design.

Free practice problemEasy
lengthT115 °C100 °C100 °C15 °CΔT₁ΔT₂counter-flow

In a counter-flow heat exchanger the hot fluid cools from 115 to 100 °C while the cold fluid heats from 15 to 100 °C. Find the log-mean temperature difference ΔTlm.

Th: 115 → 100 °CTc: 15 → 100 °C
K
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What you practise

  1. EasyLMTD
  2. Mediumrequired area
  3. Hardfull design

Key formulas

Heat duty
Q = U·A·ΔTlm
Log mean temperature difference
ΔTlm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2)
Stream balance
Q = ṁcp(Tin − Tout)

Learn Heat exchangers (LMTD) from scratch

Before this: Energy balance

1What a heat exchanger does

A heat exchanger moves heat from a hot fluid to a cold fluid through a metal wall, for example a tube with hot oil inside and cooling water outside.

How much heat it can move depends on three things: how well the wall passes heat, how much wall there is, and how big the temperature difference between the fluids is.

Q = U · A · ΔTlm
  • Q: the heat duty in W.
  • U: the overall heat transfer coefficient in W/(m²·K). A larger U means a better exchanger.
  • A: the heat transfer area in m².
  • ΔTlm: the log-mean temperature difference, the right “average” temperature difference.

The streams can flow in the same direction (parallel flow) or in opposite directions (counter-flow).

hc⇉hc⇄

In the figure, the left exchanger is parallel flow and the right one is counter-flow. Counter-flow keeps a larger temperature gap along its length, so it transfers more heat for the same area.

2Why a log mean

The temperature difference between the fluids is not the same everywhere. Along the exchanger the hot fluid cools and the cold fluid warms, so the gap changes from one end to the other.

Tx120 °C50 °C60 °C20 °CΔT₁ΔT₂

Call the differences at the two ends ΔT1 and ΔT2. The gap does not change in a straight line, so the plain average is not right. The correct average is the log mean:

ΔTlm = (ΔT1 − ΔT2) / ln(ΔT1 / ΔT2)

Which temperatures face each other at each end depends on the arrangement:

  • Counter-flow: ΔT1 = Th,in − Tc,out and ΔT2 = Th,out − Tc,in.
  • Parallel flow: ΔT1 = Th,in − Tc,in and ΔT2 = Th,out − Tc,out.

The log mean always lies a little below the arithmetic mean. If ΔT1 = ΔT2, ΔTlm is simply that value.

Self-check Counter-flow: the hot fluid goes from 100 to 60 °C, the cold fluid from 20 to 50 °C. What are the end differences?
  1. 40 K and 30 K
  2. 50 K and 10 K
  3. 50 K and 40 K
  4. 80 K and 10 K

Answer: 50 K and 40 K. In counter-flow the hot inlet faces the cold outlet (100 − 50 = 50 K) and the hot outlet faces the cold inlet (60 − 20 = 40 K). 80 K and 10 K is the parallel-flow pairing.

3The method step by step

  1. Write down the four temperatures and check whether the flow is counter or parallel.
  2. If an outlet temperature is missing, find it first with an energy balance: Q = ṁhcp,h(Th,in − Th,out) = ṁccp,c(Tc,out − Tc,in).
  3. Find the end differences ΔT1 and ΔT2 with the correct pairing.
  4. Calculate ΔTlm with the natural logarithm.
  5. For the area: A = Q/(U·ΔTlm). Put Q in W (kW × 1000) when U is in W/(m²·K).
Self-check The end differences are ΔT₁ = 40 K and ΔT₂ = 20 K. What is ΔTlm?
  1. 13.86 K
  2. 20 K
  3. 28.85 K
  4. 30 K

Answer: 28.85 K. ΔTlm = (40 − 20)/ln(40/20) = 20/0.693 = 28.85 K. The arithmetic mean, 30 K, is close but too high.

4A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

A counter-flow heat exchanger transfers 100 kW. The hot fluid cools from 120 to 60 °C and the cold fluid heats from 20 to 50 °C. The overall heat transfer coefficient is 500 W/(m²·K). Find the heat transfer area A.

1

Sketch the temperatures

Tx120 °C50 °C60 °C20 °CΔT₁ΔT₂

Counter-flow, so the hot inlet (120 °C) faces the cold outlet (50 °C).

2

End differences

ΔT1 = 120 − 50 = 70 K
ΔT2 = 60 − 20 = 40 K
3

Log-mean temperature difference

ΔTlm = (70 − 40) / ln(70/40) = 30 / 0.5596 = 53.61 K
4

Area

Q = 100 kW = 100 000 W.

A = Q / (U · ΔTlm) = 100 000 / (500 · 53.61) = 3.73 m²
5

Check the answer

ΔTlm = 53.61 K lies between 40 and 70 K and just below their average of 55 K, as it should.

In parallel flow the ends would be 100 K and 10 K, giving ΔTlm = 39.09 K and A = 5.12 m². Counter-flow needs less area.

Answer

A ≈ 3.73 m² (ΔTlm = 53.61 K).

5Common mistakes

  • Pairing the temperatures for parallel flow in a counter-flow exchanger, or the other way round.
  • Using the arithmetic mean instead of the log mean.
  • Using the temperature change of one stream (for example 120 − 60) as ΔT1. ΔT is always hot minus cold at the same end.
  • Leaving Q in kW while U is in W/(m²·K). The area comes out 1000 times too small.
  • Using log₁₀ instead of ln.
Self-check Q = 50 kW, U = 500 W/(m²·K) and ΔTlm = 25 K. What area is needed?
  1. 0.004 m²
  2. 0.25 m²
  3. 4 m²
  4. 4000 m²

Answer: 4 m². A = 50 000/(500 · 25) = 4 m². Leaving Q in kW gives 0.004 m², 1000 times too small. 0.25 m² comes from turning the fraction upside down.

6How exam questions look

  • Easy: find ΔTlm for a counter-flow exchanger from the four temperatures.
  • Medium: find the area A for a given duty and U, in counter-flow or parallel flow, as in the example.
  • Hard: a full design. The water outlet temperature is not given: find Q from the hot stream, the water outlet from the energy balance, then ΔTlm and A.

Some questions use a condenser: the hot side condenses at one constant temperature. Then both end differences use that temperature, and counter or parallel flow gives the same ΔTlm.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
lengthT100 °C80 °C85 °C30 °CΔT₁ΔT₂counter-flow

In a counter-flow heat exchanger the hot fluid cools from 100 to 80 °C while the cold fluid heats from 30 to 85 °C. Find the log-mean temperature difference ΔTlm.

Th: 100 → 80 °CTc: 30 → 85 °C

Find ΔTlm (K)

Show hints
  1. Find the temperature difference between the two streams at each end of the exchanger.
  2. In counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.
  3. ΔT₁ = 15 K, ΔT₂ = 50 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 counter-flow the hot inlet faces the cold outlet, and the hot outlet faces the cold inlet.

ΔT₁ = Th,in − Tc,out = 100 − 85 = 15 K
ΔT₂ = Th,out − Tc,in = 80 − 30 = 50 K

Log-mean temperature difference

ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂) = (15 − 50)/ln(15/50) = 29.07 K

Answer

ΔTlm = 29.07 K
Example 2Medium
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
Example 3Hard
lengthT105 °C10 °C?ΔT₁ΔT₂condensing steam

Steam condenses at a constant 105 °C on the outside of the tubes of a condenser with a heat transfer area of 3 m² and U = 1500 W/(m²·K). Cooling water (1 kg/s, cp = 4.18 kJ/(kg·K)) enters the tubes at 10 °C. Find the water outlet temperature Tc,out.

Ts = 105 °CA = 3 m²U = 1500 W/(m²·K)ṁ = 1 kg/scp = 4.18 kJ/(kg·K)Tc,in = 10 °C

Find Tc,out (°C)

Show hints
  1. The outlet temperature is unknown, so the LMTD cannot be computed directly. Combine Q̇ = ṁcp(Tc,out − Tc,in) with Q̇ = U·A·ΔTlm.
  2. This gives ln((Ts − Tc,in)/(Ts − Tc,out)) = U·A/(ṁ·cp).
  3. U·A/(ṁ·cp) = 1.077.
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.

Combine the energy balance and the rate equation

With ΔT₁ = Ts − Tc,in and ΔT₂ = Ts − Tc,out, the difference ΔT₁ − ΔT₂ equals the water temperature rise, so the rise cancels.

ṁcp(ΔT₁ − ΔT₂) = U·A·(ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂)
ln(ΔT₁/ΔT₂) = U·A/(ṁcp) = 1500·3/(1·4180) = 1.077

Outlet temperature

ΔT₂ = ΔT₁·e−1.077 = 95·0.3408 = 32.37 K
Tc,out = 105 − 32.37 = 72.63 °C

Answer

Tc,out = 72.63 °C

Heat transferred: Q̇ = 261.8 kW.

Questions and answers

What is Heat exchangers (LMTD)?

A heat exchanger moves heat from a hot fluid to a cold fluid through a metal wall, for example a tube with hot oil inside and cooling water outside.

How much heat it can move depends on three things: how well the wall passes heat, how much wall there is, and how big the temperature difference between the fluids is.

How do I solve Heat exchangers (LMTD) problems step by step?

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.

What are common mistakes in Heat exchangers (LMTD)?
  • The temperature difference decays exponentially: T<sub>s</sub> − T<sub>c,out</sub> = (T<sub>s</sub> − T<sub>c,in</sub>)·e−UA/(ṁc<sub>p</sub>).
What do the Heat exchangers (LMTD) exercises cover?
  • Easy: LMTD
  • Medium: required area
  • Hard: full design

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.

How practising works

  • 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.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

Topics before and after this one

Learn first

Energy balance

Calculators

LMTD calculator (log mean temperature difference)

Courses with this topic

Heat TransferFE Chemical exam practice

More topics in Chemical

Mass balanceEnergy balanceIdeal gas lawReaction kineticsReactor sizing (CSTR & PFR)

Printable worksheet

Heat exchangers (LMTD) worksheet with answers
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