Training Heat Exchangers — ε-NTU Method LMTD vs. ε-NTU — When to Use Which
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LMTD vs. ε-NTU — When to Use Which

24 min Heat Exchangers — ε-NTU Method

LMTD vs. ε-NTU — When to Use Which

Both methods solve the same physics. The choice between them depends on what’s known and what’s unknown.

Method Selection Guide

Use LMTD when: All four temperatures (inlet and outlet of both streams) are known or can be found from the energy balance without iteration. This is a sizing problem: find the required area $A$.

Use ε-NTU when: Only the inlet temperatures are known and $A$ is given (or to be found for a target outlet). This is a rating problem: find what a given HX can do. It avoids the iteration needed by LMTD.

Summary of Key Formulas

LMTD method: $\dot{Q} = UAF\,\Delta T_{\text{lm,CF}}$

ε-NTU method: $\dot{Q} = \varepsilon C_{\min}(T_{h,\text{in}} - T_{c,\text{in}})$, where $\varepsilon = f(NTU, C_r)$

Connection: Both give the same answer. $NTU = UA/C_{\min}$ links area to effectiveness.

Example 1 — Sizing Problem (LMTD Natural)

Counter-flow HX. $T_h$: 150→80°C, $T_c$: 20→70°C. $U = 600$. Find area.

All 4 temps known → LMTD is direct.

$\dot{Q} = C_h(70) = C_c(50)$ (from energy balance, determines flow rates)

$\Delta T_1 = 150-70 = 80$, $\Delta T_2 = 80-20 = 60$

$\Delta T_{\text{lm}} = (80-60)/\ln(80/60) = 20/0.2877 = 69.5$°C

$\dot{Q} = C_h \times 70$ (need a flow rate to get $\dot{Q}$). Let’s say $\dot{Q} = 210$ kW:

$A = 210{,}000/(600 \times 69.5) = 5.04$ m²

Example 2 — Rating Problem (ε-NTU Natural)

An existing counter-flow HX has $U = 500$, $A = 10$ m². Oil ($C = 3000$, $T_{h,\text{in}} = 100$°C) and water ($C = 8000$, $T_{c,\text{in}} = 20$°C). Find outlet temps.

$C_{\min} = 3000$, $C_r = 3000/8000 = 0.375$

$NTU = 500 \times 10/3000 = 1.667$

Counter-flow: $\varepsilon = \frac{1 - e^{-1.667(0.625)}}{1 - 0.375\,e^{-1.667(0.625)}} = \frac{1-e^{-1.042}}{1-0.375e^{-1.042}}$

$e^{-1.042} = 0.353$

$\varepsilon = (1-0.353)/(1-0.375 \times 0.353) = 0.647/0.868 = 0.746$

$\dot{Q} = 0.746 \times 3000 \times 80 = 178{,}900$ W

$T_{h,\text{out}} = 100 - 178{,}900/3000 = 40.4$°C

$T_{c,\text{out}} = 20 + 178{,}900/8000 = 42.4$°C

Try this with LMTD — you’d need to guess $T_{h,\text{out}}$, compute LMTD, check $\dot{Q}$, and iterate!

Example 3 — Sizing with ε-NTU

Design a counter-flow HX: $C_h = 4000$, $C_c = 6000$. Oil enters at 120°C, must exit at 60°C. Water enters at 15°C. $U = 700$. Find $A$.

$\dot{Q} = 4000 \times (120-60) = 240{,}000$ W

$C_{\min} = 4000$, $\dot{Q}_{\max} = 4000 \times (120-15) = 420{,}000$ W

$\varepsilon = 240{,}000/420{,}000 = 0.571$

$C_r = 4000/6000 = 0.667$

$NTU = \frac{1}{0.667-1}\ln\frac{0.571-1}{0.571 \times 0.667-1} = \frac{1}{-0.333}\ln\frac{-0.429}{-0.619}$

$= -3.0 \times \ln(0.693) = -3.0 \times (-0.366) = 1.098$

$A = NTU \times C_{\min}/U = 1.098 \times 4000/700 = 6.27$ m²

Practice Problems

1. All 4 temperatures known. Which method is more direct?
2. Only inlet temps and area known. Which method?
3. Counter-flow: $\varepsilon = 0.60$, $C_r = 0.4$. Find NTU.
4. $NTU = 1.0$, $C_r = 0$. Find $\varepsilon$.
5. $U = 1000$, $A = 4$ m², $C_{\min} = 2000$. NTU? If counter-flow with $C_r = 0.3$, $\varepsilon$?
6. Can ε-NTU be used for sizing? Explain.
Show Answer Key

1. LMTD (no iteration needed)

2. ε-NTU (avoids iteration)

3. $NTU = (1/(0.4-1))\ln((0.6-1)/(0.6 \times 0.4-1)) = (-1/0.6)\ln(0.4/0.76) = -1.667 \times (-0.642) = 1.07$

4. $\varepsilon = 1 - e^{-1} = 0.632$

5. $NTU = 2.0$. $\varepsilon = (1-e^{-2 \times 0.7})/(1-0.3e^{-1.4}) = (1-0.247)/(1-0.0740) = 0.753/0.926 = 0.813$

6. Yes — find $\varepsilon$ from required temperatures, then invert the ε-NTU formula to get NTU, then $A = NTU \cdot C_{\min}/U$.

ε-NTU for Shell-and-Tube and Cross-Flow Practice Test — Heat Exchangers (ε-NTU)