Library / Heat Transfer / HTR-08
Heat Transfer
Heat exchanger duty (LMTD method)
Heat transferred in a heat exchanger from its overall coefficient, area and end temperature differences.
- Ref
- HTR-08
- Rev
- v1
- Field
- Heat Transfer
- Result
- Q̇ [kW]
Here ΔTlm = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂). Steady parallel or counterflow exchanger, constant overall U and negligible external losses. Pair local hot and cold temperatures at each end. Equal terminal differences give that common difference.
The average temperature difference that drives heat flow in an exchanger is a logarithmic mean of the differences at the two ends.
Calculator
- Log-mean ΔT43.28 K
- Arithmetic mean ΔT45 K
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The log mean is always at most the arithmetic mean, so using the arithmetic mean overestimates the duty.
- If the two ends have the same difference the log mean equals that difference.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kW) | Local step | Change in result |
|---|---|---|---|---|---|---|
| A Area | 0.1 to 1,000 m² | 2.164 | 21,640 | 21,640 | +10% | +10 % |
| U Overall coefficient | 20 to 5,000 W/(m²·K) | 10.39 | 2,597 | 2,586 | +10% | +10 % |
| ΔT₂ Temperature difference at end 2 | 2 to 150 K | 102.3 | 589.3 | 487 | +10% | +4.35 % |
| ΔT₁ Temperature difference at end 1 | 5 to 200 K | 83.72 | 537.7 | 453.9 | +10% | +5.49 % |
Sweep of Overall coefficient
| Overall coefficient (W/(m²·K)) | Heat duty (kW) |
|---|---|
| 20 | 10.39 |
| 39.88 | 20.71 |
| 79.53 | 41.3 |
| 158.6 | 82.36 |
| 316.2 | 164.2 |
| 630.6 | 327.5 |
| 1,257 | 653.1 |
| 2,507 | 1,302 |
| 5,000 | 2,597 |
Sweep one variable
Variables and how each one contributes
Heat duty
What the equation gives you. Everything else feeds it.
Overall coefficient
Overall heat transfer coefficient. Directly proportional.
Area
Heat transfer area. Directly proportional.
Temperature difference at end 1
Hot minus cold temperature at one end. Raises ΔTlm.
Temperature difference at end 2
Hot minus cold temperature at the other end. Raises ΔTlm.
Worked example with the values above
- ΔTlm = (60 − 30) / ln(60/30) = 43.28 K.
- U A = 500 × 12 = 6,000 W/K.
- Q = U A ΔTlm = 259.7 kW.
Assumptions and limits
- Steady parallel or counterflow exchanger.
- Constant U and no external heat loss.
Common mistakes
- Pairing the wrong end temperatures.
- Using the LMTD for cross-flow without a correction factor.
Statistics and uncertainty
U is uncertain by ±15–25% because of fouling, which dominates the uncertainty in Q.
Where it is used
- Radiator, condenser and intercooler sizing.
- Fuel and oil cooler design.
Maths you need
- Overall coefficient: U
- Logarithms: ln