Heat Exchanger Effectiveness–NTU Calculator
The effectiveness–NTU (ε-NTU) method computes heat exchanger performance from three dimensionless numbers: effectiveness ε = q/qmax, NTU = U·A/Cmin, and capacity ratio Cr = Cmin/Cmax. Ideal for rating (outlet temperatures unknown) and sizing (finding NTU/area from target ε) — no temperature iteration as in the LMTD method.
Scope limitsThis tool focuses fully on the ε-NTU method: 5 flow arrangements, ε from NTU, inverse NTU from target ε, outlet temperatures of both fluids, and area sizing. For log mean temperature difference, the Bowman–Mueller–Nagle correction factor F, and heat duty Q = U·A·F·LMTD, use our LMTD calculator.
Effectiveness ε from NTU & Cr
Dimensionless number, ≥ 0.
Between 0 and 1. Enter 0 for a condenser/evaporator.
Effectiveness ε = q / qmax
–
| As a percentage | – |
| Maximum ε of this arrangement (NTU → ∞) | – |
NTU from target effectiveness (design/sizing)
0 < ε < 1, and below the arrangement’s maximum ε.
Required NTU
–
Rating: heat rate & outlet temperatures
Enter the capacity rates of both fluids (C = ṁ·cp, e.g. W/K) and their inlet temperatures. The calculator determines Cmin/Cmax automatically.
| Cmin / Cmax | – / – W/K |
| Cr | – |
| qmax = Cmin·(Th,in−Tc,in) | – W |
| Effectiveness ε | – |
| Actual q = ε·qmax | – W |
| Th,out — hot outlet temperature | – °C |
| Tc,out — cold outlet temperature | – °C |
Sizing: exchanger area from target ε
| Required NTU | – |
| Area A = NTU·Cmin / U | – m² |
ε–NTU correlations per flow arrangement
All correlations below are the canonical forms from Fundamentals of Heat and Mass Transfer (Incropera):
| Flow arrangement | Effectiveness correlation |
|---|---|
| Counter-flow | ε = [1 − e−NTU·(1−Cr)] / [1 − Cr·e−NTU·(1−Cr)]; for Cr = 1: ε = NTU/(1+NTU) |
| Parallel-flow | ε = [1 − e−NTU·(1+Cr)] / (1 + Cr) |
| Phase change (Cr = 0) | ε = 1 − e−NTU — same for all arrangements |
| Cross-flow, Cmin unmixed | ε = 1 − e−(1−e−Cr·NTU)/Cr |
| Cross-flow, Cmin mixed | ε = (1/Cr)·[1 − e−Cr·(1−e−NTU)] |
Verified examples
Example 1 — ε from NTU. Counter-flow exchanger with NTU = 2 and Cr = 0.5: ε = [1 − e−1] / [1 − 0.5·e−1] ≈ 0.7746003 (with e−1 ≈ 0.3678794412). Meaning the exchanger transfers 77.46% of the thermodynamically possible maximum heat rate.
Example 2 — rating. Ch = 1000 W/K, Cc = 2000 W/K, Th,in = 120 °C, Tc,in = 20 °C, NTU = 2 (counter-flow): Cmin = 1000 W/K, Cr = 0.5, qmax = 100,000 W, ε ≈ 0.7746003, q ≈ 77,460 W, Th,out ≈ 42.54 °C, Tc,out ≈ 58.73 °C.
Example 3 — sizing. Target ε = 0.8, Cr = 0.5, counter-flow → NTU ≈ 2.1972. With Cmin = 1000 W/K and U = 500 W/m²K, the required area A = NTU·Cmin/U ≈ 4.3944 m².
ε-NTU vs LMTD — when to use which?
| Situation | Right method |
|---|---|
| All four terminal temperatures known, want to compute duty or area | LMTD — q = U·A·F·LMTD (LMTD calculator) |
| Outlet temperatures not yet known (rating an existing exchanger) | ε-NTU — compute NTU, find ε, then q = ε·qmax |
| Designing: target ε known, find size (NTU/area) | Inverse ε-NTU — the "NTU from target ε" / "Sizing" tabs |
| One fluid changes phase (condenser/evaporator) | ε-NTU with Cr = 0 (or LMTD, both are easy) |
Frequently asked questions
What is the effectiveness-NTU method?
The effectiveness-NTU method computes heat exchanger performance from three dimensionless numbers: effectiveness ε = q/qmax (actual heat rate divided by the thermodynamically possible maximum), NTU = U·A/Cmin (Number of Transfer Units), and capacity ratio Cr = Cmin/Cmax. This method is used when the fluid outlet temperatures are not yet known, so no temperature iteration is needed as in the LMTD method.
When should I use ε-NTU vs LMTD?
Use LMTD when all four terminal temperatures (inlet and outlet of both fluids) are known and you want to compute the duty or the required area. Use the ε-NTU method when the outlet temperatures are not yet known — for example rating an existing exchanger (whose U·A is known) or designing from a target effectiveness. For LMTD calculations, use our LMTD calculator.
What is NTU?
NTU (Number of Transfer Units) = U·A/Cmin, dimensionless. NTU measures how large the exchanger is relative to the flow’s capacity rate: small NTU means a small exchanger compared to its flow (low effectiveness), large NTU means a large exchanger (effectiveness approaching the arrangement’s maximum).
What is the capacity ratio Cr?
Cr = Cmin/Cmax, ranging from 0 to 1, with C = mass flow rate × specific heat (e.g. W/K) for each fluid. Cr = 0 means one fluid changes phase (condenser or evaporator) so its temperature is constant; Cr = 1 means both fluids have equal capacity (balanced flow).
Which flow arrangement is most effective?
For the same NTU and Cr, counter-flow always gives the highest effectiveness, so it is the most area-efficient for the same duty. Parallel-flow is always lower and is limited to εmax = 1/(1+Cr).
What happens when Cr = 0?
Cr = 0 occurs when one fluid changes phase (condensation or evaporation) or its temperature is held constant, so Cmax = infinity. Under this condition effectiveness does not depend on flow arrangement: ε = 1 − e^(−NTU) for all geometries.
How about the special case Cr = 1 in counter-flow?
The general counter-flow formula has a zero denominator at Cr = 1, but the analytical limit exists: ε = NTU/(1+NTU). Example: NTU = 3 and Cr = 1 give ε = 0.75. This calculator uses that analytical limit, not division by zero.
What is the maximum parallel-flow effectiveness?
Parallel-flow can never reach ε = 1; its limit is εmax = 1/(1+Cr). Example: Cr = 0.5 gives εmax = 2/3 ≈ 0.6667 — targets above this value are impossible and the calculator will reject them.
Worked example of effectiveness?
A counter-flow exchanger with NTU = 2 and Cr = 0.5: ε = [1 − e^(−1)] / [1 − 0.5·e^(−1)] ≈ 0.7746003 (with e^(−1) ≈ 0.3678794412). Meaning the exchanger transfers 77.46% of the thermodynamically possible maximum heat rate.
How is NTU computed from a target effectiveness?
For counter-flow, parallel-flow, and phase change, exact closed-form analytical inverses are available — for example, target ε = 0.8 with Cr = 0.5 in counter-flow needs NTU ≈ 2.1972. For cross-flow, ε increases monotonically with NTU, so its inverse is computed with honest bisection (200 iterations).
How are the outlet temperatures of both fluids computed?
From ε we get q = ε·Cmin·(Th,in − Tc,in), then the energy balance of each fluid: Th,out = Th,in − q/Ch and Tc,out = Tc,in + q/Cc. Example: Ch = 1000 W/K, Cc = 2000 W/K, 120/20 °C, NTU = 2 counter-flow → q ≈ 77,460 W, Th,out ≈ 42.54 °C, Tc,out ≈ 58.73 °C.
Are this calculator’s results certified engineering data?
No. This calculator uses canonical correlations from the heat transfer literature (Incropera) for estimation and study. For designs that will be built or certified, results must be validated by a competent engineer.
ε–NTU correlations per the canonical Incropera forms; single-pass cross-flow uses the exact analytical solution; cross-flow inverse uses bisection (200 iterations). For estimation and study — not certified engineering data. Zero external requests: all computation runs in your browser.
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