Heat exchanger sizing
Size a counterflow exchanger two ways: from the mean temperature difference given the terminal temperatures, or from the effectiveness-NTU relation given the installed area. The two must agree, and the gap between them is the design margin.
| Quantity | Value | Basis |
|---|---|---|
| Heat duty Q | 627.00 kW | m·cp·ΔT on the hot side |
| Hot-side capacity | 10.450 kW/K | m_h·cp |
| Cold-side capacity | 13.933 kW/K | from the energy balance |
| LMTD | 72.24 K | counterflow terminal differences |
| NTU | 0.976 | UA/C_min |
| Capacity ratio C_r | 0.750 | C_min/C_max |
| Effectiveness ε | 0.4800 | Q/Q_max |
| Area from LMTD | 10.21 m² | Q/(U·LMTD) |
| UA installed | 10.20 kW/K | U × A |
LMTD sizing and effectiveness-NTU answer the same question from opposite ends: one asks for area given temperatures, the other asks for duty given area.
Q = m·cp·ΔT · either stream gives the same answer
(ΔT₁−ΔT₂)/ln(ΔT₁/ΔT₂) · counterflow terminal differences
UA/C_min · dimensionless size
Q/Q_max · Q_max = C_min(T_h,in − T_c,in)
A clear path from inputs to results.
Exchanger sizing is asked two ways: what area do these temperatures require, or what duty does this installed area deliver? LMTD answers the first, effectiveness-NTU the second, and a sound design reconciles them.
How to use
- Enter the four terminal temperatures.
- Set the hot-side mass flow and specific heat.
- Enter the overall heat transfer coefficient and the installed area.
- Read the duty, LMTD, NTU and effectiveness.
- Compare the LMTD-implied area with the installed area to see the margin.
The formula
Reference: Incropera & DeWitt, "Fundamentals of Heat and Mass Transfer", ch. 11.
Worked example
150 → 90 °C hot, 25 → 70 °C cold, 2.5 kg/s water
- Hot 150 → 90 °C; cold 25 → 70 °C.
- 2.5 kg/s at cp 4.18 kJ/kg·K.
- U 850 W/m²K, area 12 m².
- 1
Duty: 2.5 × 4.18 × 60 = 627 kW.
- 2
Counterflow terminal differences: 150 − 70 = 80 K; 90 − 25 = 65 K.
- 3
LMTD: (80 − 65)/ln(80/65) = 72.2 K.
- 4
NTU = 850 × 12 / (2.5 × 4.18 × 1000) = 0.976.
- 5
Effectiveness: 627 / (10.45 × 125) = 0.48.
627 kW · LMTD 72.2 K · NTU 0.98 · effectiveness 0.48.
Interpreting the result
An effectiveness of 0.48 means the exchanger captures less than half of the thermodynamic maximum. Doubling the area would push NTU to about 2 and effectiveness to roughly 0.7 — real but diminishing. Beyond NTU 3 the curve flattens, which is why very large exchangers are rarely economic.
Assumptions & limitations
- Steady state with constant fluid properties and no phase change.
- No fouling resistance; in service, fouling can cut U by 30–50%.
- No heat loss to ambient and no axial conduction.
- Counterflow arrangement; parallel flow gives lower effectiveness for the same NTU.
Frequently asked questions
Why counterflow rather than parallel flow?
Because counterflow keeps the terminal temperature differences larger, which raises the LMTD and the effectiveness for the same area. Parallel flow also cannot cool the hot stream below the cold outlet.
How much fouling margin should I add?
Typically 10–25% extra area, depending on the fluid. Fouling factors are tabulated by TEMA, and the margin should be explicit in the design rather than hidden in a fudged U.
What is a temperature cross?
When the cold outlet would exceed the hot inlet, which is impossible in a single-pass exchanger. It signals that the specified duty cannot be achieved with the chosen arrangement.
What if the fluids have very different capacities?
Then C_r is small and the effectiveness approaches 1 − exp(−NTU), the same as a condenser or evaporator where one stream is nearly isothermal.
Methodology last reviewed 2026-10-18. Results are informational and not financial or engineering advice — see our disclaimer.