Heat exchanger calculator: log mean temperature difference, area, ε-NTU method
Enter the fluid temperatures, the heat duty (or the heat capacity rate of a stream) and the overall heat transfer coefficient. The calculator finds the log mean temperature difference for parallel flow and counterflow, the correction for other schemes, the surface area required, the effectiveness ε and the number of transfer units NTU. The rating mode finds the outlet temperatures for a given surface.
The log mean temperature difference
The fluid temperatures change along the surface, so a mean temperature difference is taken for the whole unit. For parallel flow and counterflow it is LMTD = (Δt_max − Δt_min)/ln(Δt_max/Δt_min). In counterflow Δt_max = t₁′ − t₂″ and Δt_min = t₁″ − t₂′, in parallel flow Δt_max = t₁′ − t₂′ and Δt_min = t₁″ − t₂″. If the differences are equal, the LMTD equals them. The surface comes from the heat transfer equation: F = Q/(k·LMTD).
Counterflow is better than parallel flow
At the same inlet and outlet temperatures counterflow gives a larger mean difference and a smaller area. Only in counterflow may the cold stream leave hotter than the hot stream leaves; in parallel flow the outlet temperatures approach each other but never cross. For multipass and crossflow units a correction Δt = ε_Δt·LMTD of counterflow is used, with ε_Δt below 1; it is found through ε-NTU, so the table on the page compares the schemes.
The ε-NTU method
When the outlet temperatures are unknown, the effectiveness method is more convenient. The heat capacity rates are W = G·c, C_min and C_max are the smaller and the larger, C_r = C_min/C_max. The effectiveness ε = Q/(C_min·(t₁′ − t₂′)) is the share of the largest possible heat exchange, and the number of transfer units is NTU = k·F/C_min. For counterflow ε = (1 − exp(−NTU(1 − C_r)))/(1 − C_r·exp(−NTU(1 − C_r))), for parallel flow ε = (1 − exp(−NTU(1 + C_r)))/(1 + C_r). If one stream boils or condenses (C_r = 0), every scheme gives ε = 1 − exp(−NTU).
Design and rating
The design calculation finds the area from given temperatures; the rating calculation finds the outlet temperatures and the duty from a given area. Both give the same result and the page shows their agreement: for parallel flow and counterflow Q = k·F·LMTD. Crossflow with both streams unmixed uses the approximate formula of Kays and London.
FAQ
Why is counterflow better than parallel flow?
At the same inlet and outlet temperatures the mean temperature difference is larger in counterflow, so a smaller area is needed. In addition, in counterflow the cold stream can be heated above the outlet temperature of the hot one, while in parallel flow the outlet temperatures cannot cross.
How do I find the heat exchanger area?
From the heat transfer equation: F = Q/(k·ΔT_m), where ΔT_m is the mean temperature difference (the LMTD with the scheme correction), k the overall coefficient and Q the duty. Enter the four temperatures, one of Q, W₁, W₂ and k: the page gives F for every scheme.
What are NTU and ε?
NTU = k·F/C_min is the dimensionless size of the unit; ε is the share of the largest possible heat exchange that the unit achieves: ε = Q/(C_min·(t₁′ − t₂′)). The ε-NTU method finds the outlet temperatures directly when the surface and the inlet temperatures are known.
What if the scheme is reported as impossible?
Parallel flow and multipass schemes have a limit of effectiveness (1/(1 + C_r) for parallel flow): the hot stream cannot be cooled below where the cold one is heated to. If the temperatures you need lie above the limit, choose counterflow or several units in series.
How do I get the overall coefficient k?
Add the resistances: 1/k = 1/α₁ + Σδ/λ + 1/α₂ + fouling. Find α on the convective heat transfer page and the wall resistance on the multilayer wall page. As a guide: water to water in a shell and tube unit is about 800–1500 W/(m²·°C). For reference.