Training Heat Exchangers — LMTD Method LMTD for Parallel-Flow and Counter-Flow
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LMTD for Parallel-Flow and Counter-Flow

24 min Heat Exchangers — LMTD Method

The log-mean temperature difference (LMTD) is the effective driving force for heat transfer in a heat exchanger: ΔTlm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). Combined with the overall heat-transfer coefficient U and the surface area A, the LMTD method gives q = UAΔTlm. This approach is most convenient when all four terminal temperatures are known or can be found from energy balances, making it the standard method for sizing new heat exchangers.

LMTD for Parallel-Flow and Counter-Flow

The Log Mean Temperature Difference (LMTD) is the proper average driving force for heat transfer across a heat exchanger. It accounts for the fact that the temperature difference varies along the length.

Heat Exchanger Design Equation

$$\dot{Q} = UA\,\Delta T_{\text{lm}}$$

$U$ = overall heat transfer coefficient, $A$ = heat transfer area, $\Delta T_{\text{lm}}$ = log mean temperature difference.

LMTD Formula

$$\Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$

Parallel-flow: Both fluids enter the same end.
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{in}}$   $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{out}}$

Counter-flow: Fluids enter from opposite ends.
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}}$   $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}}$

If $\Delta T_1 = \Delta T_2$, then $\Delta T_{\text{lm}} = \Delta T_1$ (avoid 0/0).

Example 1 — Counter-Flow LMTD

Hot fluid: 150°C → 90°C. Cold fluid: 30°C → 70°C. Counter-flow. Find $\Delta T_{\text{lm}}$.

  1. $\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 150 - 70 = 80$°C
  2. $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 90 - 30 = 60$°C
  3. $$\Delta T_{\text{lm}} = \frac{80 - 60}{\ln(80/60)} = \frac{20}{\ln 1.333} = \frac{20}{0.2877} = 69.5\text{°C}$$
Example 2 — Parallel-Flow LMTD

Same temperatures as Example 1 but parallel-flow. Find $\Delta T_{\text{lm}}$.

  1. $\Delta T_1 = T_{h,\text{in}} - T_{c,\text{in}} = 150 - 30 = 120$°C
  2. $\Delta T_2 = T_{h,\text{out}} - T_{c,\text{out}} = 90 - 70 = 20$°C
  3. $$\Delta T_{\text{lm}} = \frac{120 - 20}{\ln(120/20)} = \frac{100}{\ln 6} = \frac{100}{1.7918} = 55.8\text{°C}$$
  4. Counter-flow LMTD (69.5) > parallel-flow (55.8)
  5. so counter-flow needs less area for the same duty — this is why counter-flow is generally preferred.
Example 3 — Sizing a Heat Exchanger

$\dot{Q} = 200$ kW, $U = 800$ W/(m²·K), counter-flow $\Delta T_{\text{lm}} = 50$°C. Find the required area.

  1. Calculate the log mean temperature difference (LMTD).
  2. $$A = \frac{\dot{Q}}{U \cdot \Delta T_{\text{lm}}} = \frac{200{,}000}{800 \times 50} = 5.0 \text{ m}^2$$
Example 4 — Finding the Outlet Temperature

Counter-flow HX: $U = 500$ W/(m²·K), $A = 10$ m². Hot water: $\dot{m} = 1$ kg/s, $c_p = 4180$, $T_{h,\text{in}} = 80$°C. Cold water: $\dot{m} = 1.5$ kg/s, $c_p = 4180$, $T_{c,\text{in}} = 15$°C. Find outlet temperatures.

  1. $C_h = 4180$ W/K, $C_c = 6270$ W/K.
  2. Energy balance: $4180(80 - T_{h,o}) = 6270(T_{c,o} - 15)$
  3. $\dot{Q} = UA \Delta T_{\text{lm}} = 5000 \cdot \Delta T_{\text{lm}}$
  4. This requires iteration.
  5. Guess $T_{h,o} = 50$°C:
  6. $\dot{Q} = 4180 \times 30 = 125{,}400$ W
  7. $T_{c,o} = 15 + 125{,}400/6270 = 35$°C
  8. $\Delta T_1 = 80 - 35 = 45$, $\Delta T_2 = 50 - 15 = 35$
  9. $\Delta T_{\text{lm}} = (45-35)/\ln(45/35) = 10/0.2513 = 39.8$
  10. $\dot{Q}_{\text{calc}} = 5000 \times 39.8 = 199{,}000$ W
  11. Too high vs. 125,400.
  12. Adjust: try $T_{h,o} = 56$°C:
  13. $\dot{Q} = 4180 \times 24 = 100{,}320$ W; $T_{c,o} = 31$°C
  14. $\Delta T_1 = 49$, $\Delta T_2 = 41$, $\Delta T_{\text{lm}} = 44.9$, $\dot{Q}_{\text{calc}} = 224{,}400$ W. Still too high.
  15. The iteration converges around $T_{h,o} \approx 44$°C giving $\dot{Q} \approx 150$ kW. In practice, the $\varepsilon$-NTU method (next module) avoids this iteration entirely.

Practice Problems

1. Counter-flow: $T_h$ 100→60, $T_c$ 20→50. Find $\Delta T_{\text{lm}}$.
2. Parallel-flow: same temps as #1. $\Delta T_{\text{lm}}$?
3. $\dot{Q} = 500$ kW, $U = 1200$, $\Delta T_{\text{lm}} = 40$. Required $A$?
4. $U = 600$, $A = 8$ m², $\Delta T_{\text{lm}} = 25$°C. Find $\dot{Q}$.
5. Why is $\Delta T_{\text{lm}}$ always between $\Delta T_1$ and $\Delta T_2$?
6. Counter-flow with $\Delta T_1 = \Delta T_2 = 30$°C. $\Delta T_{\text{lm}}$?
Show Answer Key

1. $\Delta T_1 = 100-50 = 50$, $\Delta T_2 = 60-20 = 40$. $\Delta T_{\text{lm}} = 10/\ln(50/40) = 10/0.2231 = 44.8$°C

2. $\Delta T_1 = 100-20 = 80$, $\Delta T_2 = 60-50 = 10$. $\Delta T_{\text{lm}} = 70/\ln 8 = 70/2.079 = 33.7$°C

3. $A = 500{,}000/(1200 \times 40) = 10.4$ m²

4. $\dot{Q} = 600 \times 8 \times 25 = 120{,}000$ W = 120 kW

5. The log mean is always between the two values (it’s a weighted average).

6. $\Delta T_{\text{lm}} = 30$°C (when equal, LMTD = the common value).

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