Calculate logarithmic mean temperature difference (LMTD) or solve for ΔT1 and ΔT2 from the other two temperature differences in °C, °F, or K.
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Logarithmic Mean Temperature Difference (LMTD) Formula
The logarithmic mean temperature difference is calculated from the two end-point temperature differences in a heat exchanger.
LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}If the two end temperature differences are equal, the formula approaches this limit:
LMTD = \Delta T_1 = \Delta T_2
When solving for a missing end temperature difference, the calculator solves the same LMTD equation numerically:
LMTD = \frac{x - \Delta T}{\ln(x / \Delta T)}- LMTD = logarithmic mean temperature difference
- ΔT1 = temperature difference at one end of the heat exchanger
- ΔT2 = temperature difference at the other end of the heat exchanger
- ln = natural logarithm
- x = unknown end temperature difference being solved
The calculator uses positive temperature differences. If you enter a negative value, the magnitude is used in the LMTD calculation.
- Calculate LMTD: enter ΔT1 and ΔT2. The calculator applies the main LMTD formula.
- Calculate ΔT1: enter ΔT2 and LMTD. The calculator solves the LMTD equation for the missing ΔT1.
- Calculate ΔT2: enter ΔT1 and LMTD. The calculator solves the LMTD equation for the missing ΔT2.
- Unit conversion: temperature differences are converted by scale only. A 1 K difference equals a 1 °C difference. A 1 °F difference equals 5/9 °C.
End Temperature Difference Setup
Use the table below to identify ΔT1 and ΔT2 before entering values.
| Heat exchanger flow type | ΔT1 | ΔT2 |
|---|---|---|
| Counterflow | Hot fluid inlet temperature minus cold fluid outlet temperature | Hot fluid outlet temperature minus cold fluid inlet temperature |
| Parallel flow | Hot fluid inlet temperature minus cold fluid inlet temperature | Hot fluid outlet temperature minus cold fluid outlet temperature |
For temperature-difference units, use scale conversion only.
| Temperature difference unit | Conversion to °C difference | Example |
|---|---|---|
| °C | No conversion | 20 °C = 20 °C |
| K | Same size as °C difference | 20 K = 20 °C difference |
| °F | Multiply by 5/9 | 36 °F = 20 °C difference |
Example Problems
Example 1: Calculate LMTD
You have ΔT1 = 60 °C and ΔT2 = 30 °C.
LMTD = \frac{60 - 30}{\ln(60 / 30)}LMTD = \frac{30}{\ln(2)} = 43.280852\ ^\circ CThe LMTD is about 43.28 °C.
Example 2: Calculate a Missing End Difference
You have ΔT2 = 30 °C and LMTD = 43.280852 °C. The missing value is ΔT1.
43.280852 = \frac{\Delta T_1 - 30}{\ln(\Delta T_1 / 30)}Solving this equation gives:
\Delta T_1 = 60\ ^\circ C
FAQ
Why does the calculator not add 32 when converting °F?
The inputs are temperature differences, not actual temperatures. A temperature difference in °F converts by multiplying by 5/9. The +32 offset is only used when converting an absolute temperature reading from °F to °C.
Can LMTD be larger than both ΔT1 and ΔT2?
No, not for valid positive end temperature differences. The LMTD falls between ΔT1 and ΔT2. For example, if ΔT1 is 60 °C and ΔT2 is 30 °C, the LMTD is about 43.28 °C.
What happens if ΔT1 and ΔT2 are the same?
If ΔT1 and ΔT2 are equal, the standard formula creates a 0 divided by 0 form. The limit is the common temperature difference. So if both values are 25 °C, the LMTD is 25 °C.
