Engineering
Heat Exchanger LMTD (Log-Mean Temperature Difference) & Area sizing formula
Calculate counter-current heat exchanger Log-Mean Temperature Difference LMTD = (dT1 - dT2) / LN(dT1 / dT2) with L'Hopital equal-pinch guard and required surface area.
Logarithmic mean for an ideal counterflow model
Use terminal differences hot inlet minus cold outlet, and hot outlet minus cold inlet. This example assumes steady counterflow, a constant overall heat-transfer coefficient and negligible heat loss to the surroundings. Other flow arrangements may require a correction factor; the area calculation here uses a factor of one.
Both terminal differences must be positive. Equal differences give that common value as the limiting logarithmic mean. The implementation uses the arithmetic mean only when relative separation is below 1e-8 to avoid cancellation, and rounds the displayed result only after this calculation.
Temperature differences in kelvins and Celsius degrees have the same numerical size. Duty is in watts, U in W/(m²·K), and area in m². Input temperatures and duty must also be consistent with an independent energy balance; this exercise does not establish a complete exchanger design.
| Stream / Metric | Cells & Inputs | Counter-Current Values | Design Output |
|---|---|---|---|
| Hot Stream (In -> Out) | B2 = 140 C, B3 = 80 C | Cools by 60 K | Hot terminal ends |
| Cold Stream (In -> Out) | B4 = 60 C, B5 = 80 C | dT1 = 140-80 = 60 K | dT2 = 80-60 = 20 K | Arithmetic mean = 40.00 K (Too high!) |
| True LMTD Driving Force | (60 - 20) / LN(60 / 20) | 40 / LN(3) = 36.41 K | 9% lower than arithmetic mean |
| Required Area A = Q / (U*LMTD) | Duty B6=182048 W, U B7=500 W/(m²·K) | 182048 / (500 × 36.4096) | 10.00 m^2 heat-transfer area |
L'Hopital-safe Counter-Current LMTD formula (K or C)
Handles both unequal terminal approaches (60 K & 20 K -> 36.41 K) and balanced pinch (40 K & 40 K -> 40 K) without error.
=IF(OR(COUNT(B2:B5)<>4,B2<=B5,B3<=B4,B2<B3,B5<B4),"Invalid inputs",LET(deltaHot,B2-B5,deltaCold,B3-B4,meanDelta,IF(ABS(deltaHot-deltaCold)<=0.00000001*MAX(deltaHot,deltaCold),(deltaHot+deltaCold)/2,(deltaHot-deltaCold)/LN(deltaHot/deltaCold)),ROUND(meanDelta,2)))Required heat-exchanger surface area A = Q / (U * LMTD) (m^2)
Divides thermal duty Q (Watts) by overall heat-transfer coefficient U (W/(m^2*K)) times exact LMTD.
=IF(OR(COUNT(B2:B7)<>6,B2<=B5,B3<=B4,B2<B3,B5<B4,B6<0,B7<=0),"Invalid inputs",LET(deltaHot,B2-B5,deltaCold,B3-B4,meanDelta,IF(ABS(deltaHot-deltaCold)<=0.00000001*MAX(deltaHot,deltaCold),(deltaHot+deltaCold)/2,(deltaHot-deltaCold)/LN(deltaHot/deltaCold)),ROUND(B6/(B7*meanDelta),2)))Try the example
Use the original values or change them to test your understanding.