LMTD for Parallel-Flow and Counter-Flow
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.
$$\dot{Q} = UA\,\Delta T_{\text{lm}}$$
$U$ = overall heat transfer coefficient, $A$ = heat transfer area, $\Delta T_{\text{lm}}$ = log mean temperature difference.
$$\Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)}$$
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).
Hot fluid: 150°C → 90°C. Cold fluid: 30°C → 70°C. Counter-flow. Find $\Delta T_{\text{lm}}$.
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{out}} = 150 - 70 = 80$°C
$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{in}} = 90 - 30 = 60$°C
$$\Delta T_{\text{lm}} = \frac{80 - 60}{\ln(80/60)} = \frac{20}{\ln 1.333} = \frac{20}{0.2877} = 69.5\text{°C}$$
Same temperatures as Example 1 but parallel-flow. Find $\Delta T_{\text{lm}}$.
$\Delta T_1 = T_{h,\text{in}} - T_{c,\text{in}} = 150 - 30 = 120$°C
$\Delta T_2 = T_{h,\text{out}} - T_{c,\text{out}} = 90 - 70 = 20$°C
$$\Delta T_{\text{lm}} = \frac{120 - 20}{\ln(120/20)} = \frac{100}{\ln 6} = \frac{100}{1.7918} = 55.8\text{°C}$$
Counter-flow LMTD (69.5) > parallel-flow (55.8), so counter-flow needs less area for the same duty — this is why counter-flow is generally preferred.
$\dot{Q} = 200$ kW, $U = 800$ W/(m²·K), counter-flow $\Delta T_{\text{lm}} = 50$°C. Find the required area.
$$A = \frac{\dot{Q}}{U \cdot \Delta T_{\text{lm}}} = \frac{200{,}000}{800 \times 50} = 5.0 \text{ m}^2$$
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.
$C_h = 4180$ W/K, $C_c = 6270$ W/K.
Energy balance: $4180(80 - T_{h,o}) = 6270(T_{c,o} - 15)$
$\dot{Q} = UA \Delta T_{\text{lm}} = 5000 \cdot \Delta T_{\text{lm}}$
This requires iteration. Guess $T_{h,o} = 50$°C:
$\dot{Q} = 4180 \times 30 = 125{,}400$ W
$T_{c,o} = 15 + 125{,}400/6270 = 35$°C
$\Delta T_1 = 80 - 35 = 45$, $\Delta T_2 = 50 - 15 = 35$
$\Delta T_{\text{lm}} = (45-35)/\ln(45/35) = 10/0.2513 = 39.8$
$\dot{Q}_{\text{calc}} = 5000 \times 39.8 = 199{,}000$ W
Too high vs. 125,400. Adjust: try $T_{h,o} = 56$°C:
$\dot{Q} = 4180 \times 24 = 100{,}320$ W; $T_{c,o} = 31$°C
$\Delta T_1 = 49$, $\Delta T_2 = 41$, $\Delta T_{\text{lm}} = 44.9$, $\dot{Q}_{\text{calc}} = 224{,}400$ W. Still too high.
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
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).