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ε-NTU Method Heat Exchanger Check Calculation Example

Aug 13, 2026 123 views ~6 min read Technical Knowledge
When outlet temperatures are unknown and only equipment capacity is known, the LMTD method requires repeated trial calculation. This article demonstrates the ε-NTU method on the same operating condition: from K·A and the two-side heat capacity flow rates, the effectiveness and actual heat transfer are calculated in one step, suitable for checking existing equipment.

I. Known Conditions

  • Overall heat transfer capacity K·A = 1500 W/K;
  • Hot side C_hot = 3000 W/K (i.e., m·cp);
  • Cold side C_cold = 5000 W/K;
  • Hot inlet 180°C, cold inlet 25°C.

II. Calculate NTU and Heat Capacity Ratio

Take the smaller one Cmin = 3000 W/K:

NTU = K·A / Cmin = 1500 / 3000 = 0.5
Cr = Cmin / Cmax = 3000 / 5000 = 0.6

III. Substitute into the Counter-Flow Effectiveness Formula

Counter flow: ε = (1 − exp(−NTU(1−Cr))) / (1 − Cr·exp(−NTU(1−Cr))).

First compute the exponential term: NTU·(1−Cr) = 0.5 × 0.4 = 0.2, exp(−0.2) ≈ 0.8187.

ε = (1 − 0.8187) / (1 − 0.6 × 0.8187) = 0.1813 / 0.5088 ≈ 0.356

IV. Calculate the Actual Heat Transfer

Maximum possible heat transfer Qmax = Cmin·(T_hot_in − T_cold_in) = 3000 × (180 − 25) = 465000 W.

Q = ε · Qmax = 0.356 × 465000 ≈ 165.5 kW

That is, this equipment actually transfers about 165 kW under the current operating condition. To improve efficiency, increasing K·A (raising NTU) or adjusting Cr (e.g., increasing cold-side flow to better match the two sides) will both raise the effectiveness.

The benefit of ε-NTU: no need to guess outlet temperatures; given the equipment capacity, you can measure what it can deliver.

Related Reading

Keywords: ε-NTU method heat exchanger check effectiveness calculation calculation example number of transfer units
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