An LMTD calculator for heat exchanger sizing
This LMTD calculator works out the log mean temperature difference from the four terminal temperatures of a heat exchanger, applies the correction factor F for shell-and-tube units, and then solves Q = U·A·F·LMTD for the area, the heat duty or the overall heat transfer coefficient. It handles counterflow, parallel flow and shell-and-tube exchangers with one to four shell passes.
Use it for quick sizing, to check a vendor’s datasheet or to study for a heat-transfer exam. Enter temperatures in °C or °F and duty in kW or Btu/h. The results also show the capacity rates, effectiveness and NTU, so you can cross-check with the ε-NTU method. Everything runs in your browser.
How to use the heat exchanger calculator
- Choose SI or US units and the flow arrangement.
- Enter the hot inlet and outlet and the cold inlet and outlet temperatures.
- Pick what to solve for: area A, heat duty Q or the overall coefficient U.
- Give the heat duty directly, or let the calculator find it from one stream’s mass flow and specific heat (Q = ṁ·cp·ΔT).
- Read the answer, the LMTD, F and the temperature-profile plot, and press Copy results for your calculation sheet.
LMTD / Heat Exchanger Calculator features
Exact LMTD
Log mean temperature difference for counter and parallel flow, with the equal-ΔT limit handled.
Shell-and-tube F-factor
The Bowman–Mueller–Nagle equation behind the TEMA charts, for 1 to 4 shell passes.
Three ways to solve
Area, duty or U, with duty entered or computed from a stream’s ṁ·cp·ΔT.
ε-NTU cross-check
Capacity rates, effectiveness and NTU from the same numbers.
SI and US units
Switching units converts every input, so you never mix °C with Btu.
LMTD formula and a worked example
For counterflow, ΔT₁ = Thot,in − Tcold,out and ΔT₂ = Thot,out − Tcold,in; for parallel flow, ΔT₁ = Thot,in − Tcold,in and ΔT₂ = Thot,out − Tcold,out. Then LMTD = (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂).
Example: oil cools from 150 to 90 °C while water warms from 30 to 70 °C in counterflow. ΔT₁ = 80 K, ΔT₂ = 60 K, so LMTD = 20 / ln(80/60) = 69.5 K. To transfer 500 kW with U = 300 W/(m²·K), the area is 500,000 / (300 × 69.5) = 24.0 m². The same duty in parallel flow has an LMTD of only 55.8 K and needs 29.9 m², which is why counterflow is preferred.
The F correction factor for shell-and-tube exchangers
In a shell-and-tube exchanger the flow is partly counter and partly parallel, so the mean temperature difference is F × the counterflow LMTD. F depends on P = (t₂ − t₁)/(T₁ − t₁), the cold-side effectiveness, and R = (T₁ − T₂)/(t₂ − t₁), the ratio of capacity rates. For the example in a 1-2 exchanger P = 0.333 and R = 1.5, giving F ≈ 0.91 and an area of 26.3 m² instead of 24.0 m².
F falls steeply where the outlet temperatures approach or cross. Below about 0.75 a design is sensitive to small errors; adding shell passes (two shells in series) recovers F. When the equation has no solution the duty is impossible with that number of shells.
| Service | Typical U, W/(m²·K) | Btu/(h·ft²·°F) |
|---|---|---|
| Water to water | 800–1,500 | 140–260 |
| Water to oil | 100–350 | 20–60 |
| Steam condensing to water | 1,000–3,500 | 180–600 |
| Gas to gas | 10–40 | 2–7 |
Assumptions and limits
- Steady flow, constant U and constant specific heats; heat loss to the surroundings ignored.
- If one stream boils or condenses at constant temperature, every arrangement gives F = 1.
- U values in the table are rough guides; use measured or calculated film coefficients and fouling factors for design.
LMTD / Heat Exchanger Calculator FAQ
How do I use an LMTD calculator for a heat exchanger?
Enter the four inlet and outlet temperatures, choose the flow arrangement and what to solve for, then give either the duty and U (to get area) or U and area (to get duty). The calculator applies Q = U·A·F·LMTD.
What is the log mean temperature difference?
It is the effective average temperature difference driving heat transfer along an exchanger: (ΔT₁ − ΔT₂) / ln(ΔT₁/ΔT₂). It is always less than the arithmetic mean of the two end differences.
When do I need the LMTD correction factor F?
For shell-and-tube and cross-flow exchangers. Pure counterflow and parallel flow have F = 1.
Why is counterflow better than parallel flow?
Counterflow keeps a more even temperature difference along the length, giving a higher LMTD and allowing the cold outlet to exceed the hot outlet, so it needs less area for the same duty.
What if ΔT₁ equals ΔT₂?
The formula becomes 0/0, and the LMTD is simply that common temperature difference. The calculator handles this case.
Can I use °F and Btu/h?
Yes. Choose US units: temperatures in °F, duty in Btu/h, U in Btu/(h·ft²·°F) and area in ft². Switching converts the values you have typed.