Worked problem

Counterflow Heat Exchanger: Log-Mean Temperature Difference

Calculate the LMTD and the heat duty of a counterflow exchanger with hot fluid cooling from 150 °C to 90 °C and cold fluid heating from 30 °C to 70 °C.

Mechanical, HVAC & Fluids · Last reviewed

Given

Counterflow heat exchanger. Hot side: 150 °C in, 90 °C out. Cold side: 30 °C in, 70 °C out. Overall heat-transfer coefficient U = 500 W/m²·K, area A = 10 m².

Find

The log-mean temperature difference and the heat transferred.

Solution

LMTD = (ΔT1 − ΔT2) ÷ ln(ΔT1 ÷ ΔT2)
  1. In counterflow the ends pair hot-in with cold-out and hot-out with cold-in. ΔT1 = 150 − 70 = 80 K. ΔT2 = 90 − 30 = 60 K.
  2. LMTD = (80 − 60) ÷ ln(80 ÷ 60) = 20 ÷ 0.2877 = 69.5 K.
  3. Q = U × A × LMTD = 500 × 10 × 69.5 = 347,600 W, about 348 kW.

Answer

LMTD = 69.5 K | Q ≈ 348 kW

Common mistakes

  • Pairing the wrong ends. For parallel flow, ΔT1 = hot-in − cold-in and ΔT2 = hot-out − cold-out, which gives a smaller LMTD for the same temperatures.
  • Using the arithmetic mean. It overstates the driving force when the two end differences are far apart (here the arithmetic mean is 70 K, close, but the gap grows as the ends diverge).
  • Using the formula when ΔT1 equals ΔT2. The limit is simply ΔT1.

Try a variation: parallel flow

Use the same four temperatures, but with the two fluids flowing in the same direction.

  1. In parallel flow the inlets meet: ΔT1 = 150 − 30 = 120 K and the outlets meet: ΔT2 = 90 − 70 = 20 K.
  2. LMTD = (120 − 20) ÷ ln(120 ÷ 20) = 100 ÷ 1.792 = 55.8 K.
  3. Q = 500 × 10 × 55.8 = 279 kW, compared with 348 kW for counterflow.

With the same area and the same temperatures, counterflow moves about 25% more heat. Counterflow can also bring the cold fluid out hotter than the hot fluid's outlet, which parallel flow cannot.

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Questions

Does this work for shell-and-tube exchangers with several passes?

Multi-pass and cross-flow exchangers apply a correction factor F (less than 1) to the counterflow LMTD.

Do the flow rates matter?

They set the temperature changes through Q = m × cp × ΔT on each side. Here the temperatures are given, so the duty follows directly.

What is the effectiveness-NTU method?

An alternative to the LMTD method that works from the heat-capacity rates and the exchanger area, without needing the outlet temperatures. It suits rating problems where the outlets are unknown.

What if I do not know the outlet temperatures?

Use the effectiveness-NTU method, or iterate with the LMTD method by guessing an outlet temperature until the heat balance matches.

What does fouling do to the calculation?

Deposits add thermal resistance and lower the overall coefficient U. Designs include a fouling allowance, so that the exchanger still meets its duty when it is dirty.

Educational material. Examples use simplified, hypothetical inputs and are not a design, a construction document or professional engineering advice. Confirm requirements with the codes and standards adopted where the work is built and with a licensed engineer.

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