Heat Exchanger Calculator (LMTD / NTU)
Size a recuperator without phase change: choose the arrangement (counterflow or parallel flow) and the method. The LMTD method gives the log-mean temperature difference and the heat duty from the four temperatures; the ε-NTU method gives the outlet temperatures and effectiveness from inlet temperatures, capacity rates and k·A - live with every input.
Calculation
Model: recuperator without phase change, constant fluid properties, area-averaged overall heat transfer coefficient k. No crossflow, no multi-pass units, no evaporation or condensation. Sizing tool, not a substitute for a detailed design with a locally varying k.
Results
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Formulas and fundamentals
LMTD method
The LMTD method states the transferred heat duty as:
The log-mean temperature difference is:
With the terminal temperature differences ΔT₁ and ΔT₂; for equal terminal differences the limit ΔT_m = ΔT₁ applies. In counterflow the terminal differences are hot-in against cold-out and hot-out against cold-in; in parallel flow hot-in against cold-in and hot-out against cold-out.
ε-NTU method
The ε-NTU method uses the capacity rates of both streams:
From C_min and C_max follow the capacity ratio and the number of transfer units:
The effectiveness ε depends on NTU, Cr and the arrangement:
The heat duty is:
The outlet temperatures follow from the energy balance.
Consistency and the counterflow advantage
Both methods are consistent: the heat duty from ε-NTU also satisfies Q = k·A·ΔT_m with the LMTD formed from the resulting temperatures. For a given k·A, counterflow always reaches a higher effectiveness than parallel flow and even allows the cold outlet temperature to exceed the hot outlet temperature.
Worked example
A counterflow recuperator cools a hot stream from 100 to 60 °C and heats a cold stream from 20 to 40 °C. The terminal differences are ΔT₁ = 100 − 40 = 60 K and ΔT₂ = 60 − 20 = 40 K.
This gives ΔT_m = (60 − 40)/ln(60/40) = 49.33 K. With k·A = 1000 W/K the heat duty is Q = 1000 · 49.33 = 49.3 kW.
In the ε-NTU example with equal capacity rates C = 1000 W/K (Cr = 1) and k·A = 1000 W/K, NTU = 1 and in counterflow ε = NTU/(1+NTU) = 0.50. The heat duty Q = 0.50 · 1000 · (100 − 20) = 40 kW leads to outlet temperatures of 60 °C hot and 60 °C cold.
Frequently asked questions
When to use LMTD and when ε-NTU?
If all four temperatures are known, the LMTD method is direct: it gives the heat duty from k·A. If only the inlet temperatures and the capacity rates are known (the typical sizing case), the ε-NTU method is preferable because it gives the outlet temperatures without iteration.
Why is counterflow better than parallel flow?
In counterflow the driving temperature difference stays more uniform along the length, so the log-mean temperature difference is larger. For a given k·A, counterflow therefore transfers more heat and reaches a higher effectiveness. Only in counterflow can the cold outlet temperature exceed the hot outlet temperature.
What is the effectiveness ε?
It is the ratio of the actual heat duty to the thermodynamically maximum possible duty Q_max = C_min·(T_hot,in − T_cold,in). ε lies between 0 and 1 and depends only on NTU, the capacity ratio Cr and the arrangement.
What does NTU mean?
NTU (number of transfer units) is the dimensionless quantity NTU = k·A/C_min. It measures the transfer capability of the unit relative to the smaller capacity rate. Large NTU means a large or highly effective unit that approaches the maximum effectiveness.
What is the calculation valid for?
For recuperators without phase change, with constant fluid properties and a constant, area-averaged overall heat transfer coefficient k. Evaporators, condensers, crossflow and multi-pass units are not covered. The schematic temperature profile connects the exact endpoints linearly.
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