LMTD Calculator | Log Mean Temperature Difference & Heat Exchanger Design
Calculate the log mean temperature difference (LMTD) for counter-flow, parallel-flow, and cross-flow heat exchangers. Computes the LMTD correction factor F, overall heat transfer coefficient U, heat transfer area A, NTU-effectiveness, and heat duty Q = U·A·LMTD.
What Is the LMTD Calculator | Log Mean Temperature Difference & Heat Exchanger Design?
The Log Mean Temperature Difference (LMTD) is the effective average driving temperature difference in a heat exchanger where the temperature difference varies along the length. For counter-flow, ΔT₁ = T_h,in − T_c,out and ΔT₂ = T_h,out − T_c,in. If the terminal differences are equal, LMTD equals either difference. The correction factor F accounts for cross-flow and multi-pass configurations. The NTU-effectiveness method solves problems where outlet temperatures are unknown.
Formula
LMTD = (ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) · Q = U·A·F·LMTD · ε = (1−e^(−NTU(1−Cr)))/(1−Cr·e^(−NTU(1−Cr)))
How to Use
- 1
Select the appropriate tab: LMTD Method, NTU-Effectiveness, or Heat Balance.
- 2
In LMTD Method: enter all four temperatures T_h,in, T_h,out, T_c,in, T_c,out.
- 3
Select flow configuration (counter-flow or parallel-flow) and enter U and area A.
- 4
Click Calculate to get LMTD, correction factor F, and heat duty Q.
- 5
In NTU-Effectiveness: enter hot and cold inlet temperatures, heat capacity rates ṁ·cp, and NTU.
- 6
Results include effectiveness ε, total heat transfer Q, and both outlet temperatures.
- 7
In Heat Balance tab: enter Q, U, LMTD, and F to find the required exchanger area A.
Select the LMTD Method tab to find LMTD and heat duty from four temperatures. Use NTU-Effectiveness for design problems with known flow rates and inlet temperatures. Use Heat Balance to find required area from Q, U, and LMTD.
Example Calculation
Example 1 — Counter-flow water-to-water: T_h,in = 120°C, T_h,out = 80°C, T_c,in = 20°C, T_c,out = 60°C. ΔT₁ = 120−60 = 60°C, ΔT₂ = 80−20 = 60°C → LMTD = 60°C. Q = 500×10×1×60 = 300 000 W. Example 2 — NTU method: T_h,in = 120°C, T_c,in = 20°C, C_h = 2000 W/K, C_c = 3000 W/K, NTU = 2. C_min = 2000, Cr = 2000/3000 = 0.667. ε = (1−e^(−2×0.333))/(1−0.667·e^(−2×0.333)) ≈ 0.714. Q = 0.714×2000×100 = 142 800 W.
Understanding LMTD | Log Mean Temperature Difference & Heat Exchanger Design
LMTD vs flow configuration
| Parameter | Counter-flow | Parallel-flow | Cross-flow |
|---|---|---|---|
| ΔT₁ definition | T_h,in − T_c,out | T_h,in − T_c,in | Depends on passes |
| ΔT₂ definition | T_h,out − T_c,in | T_h,out − T_c,out | Depends on passes |
| Max cold outlet T | T_h,in (theoretical) | T_h,out | Between limits |
| Effectiveness | Highest | Lower | Intermediate |
| Correction factor F | 1.0 (by definition) | 1.0 (single pass) | < 1 (use F chart) |
| Typical ε (NTU = 2) | 0.83 (Cr = 0.5) | 0.64 (Cr = 0.5) | 0.72 (Cr = 0.5) |
Typical overall heat transfer coefficients U
| Heat exchanger type | Hot side | Cold side | Typical U (W/m²·K) |
|---|---|---|---|
| Shell-and-tube | Water | Water | 1 000–2 000 |
| Shell-and-tube | Steam condensing | Water | 2 000–5 000 |
| Shell-and-tube | Organic solvent | Water | 300–700 |
| Plate heat exchanger | Water | Water | 3 000–7 000 |
| Air cooler (finned tube) | Process gas | Air | 25–60 |
| Double-pipe | Water | Water | 800–1 500 |
Heat exchanger design concepts
- ›LMTD method: Best for rating existing exchangers where all four temperatures are known. Q = U·A·F·LMTD gives heat duty directly if U and A are known.
- ›NTU-ε method: Best for design when inlet temperatures and flow rates are known but outlet temperatures are unknown. NTU = UA/C_min; effectiveness ε = Q/Q_max.
- ›Correction factor F: For multi-pass or cross-flow arrangements, LMTD must be multiplied by F < 1 to account for departure from true counter-flow. F = 1 only for pure counter-flow or pure parallel-flow.
- ›Fouling resistance: Real heat exchangers accumulate scale and deposits. Design U is reduced from clean U: 1/U_design = 1/U_clean + R_f,hot + R_f,cold. Typical R_f ≈ 0.0001 m²·K/W for water.
Frequently Asked Questions
When should I use LMTD vs NTU-effectiveness?
Use the LMTD method when all four terminal temperatures are known (rating problem). Use the NTU-ε method when inlet temperatures and flow rates are known but outlets are not (design problem). LMTD requires iteration if outlet temperatures are unknown; NTU-ε solves directly.
Why is counter-flow more effective than parallel-flow?
In counter-flow, the hot stream meets the coldest cold stream at one end, maintaining a larger temperature difference throughout. This allows the cold outlet to exceed the hot outlet temperature. In parallel-flow, both fluids enter at the same end, and the temperature difference collapses quickly — limiting heat transfer.
What is the NTU (number of transfer units)?
NTU = UA/C_min is a dimensionless measure of heat exchanger size. A larger NTU means the exchanger can transfer more heat for the same inlet conditions. NTU = 1 means moderate performance; NTU > 3 gives diminishing returns as ε approaches its maximum asymptote.
What does the LMTD correction factor F represent?
For a pure counter-flow or parallel-flow exchanger, F = 1. For cross-flow and multi-pass shell-and-tube designs, F < 1 accounts for the fact that not all fluid pathways experience the optimum temperature difference. Values of F < 0.75 are generally uneconomical and suggest adding more passes.
How do I find the heat transfer area if I know Q and U?
Use the Heat Balance tab: A = Q / (U·F·LMTD). You need Q (heat duty in W), U (overall heat transfer coefficient in W/m²·K), the LMTD (from four temperatures), and F (correction factor). This is the fundamental sizing equation for heat exchangers.
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