Training Heat Exchangers — LMTD Method Shell-and-Tube and Cross-Flow Correction Factors
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Shell-and-Tube and Cross-Flow Correction Factors

24 min Heat Exchangers — LMTD Method

Shell-and-tube and cross-flow heat exchangers have temperature profiles that differ from pure counter-flow, requiring a correction factor F to account for the reduced effectiveness: q = UAFΔTlm,CF. The correction factor depends on two dimensionless parameters (P and R) and is read from charts or computed from closed-form expressions. Understanding when F drops too low (below about 0.75) helps engineers avoid impractical designs and select the right exchanger configuration.

Shell-and-Tube and Cross-Flow Correction Factors

Real heat exchangers are rarely pure counter-flow. Shell-and-tube exchangers have multiple passes, and cross-flow exchangers have perpendicular streams. We handle these by applying a correction factor $F$ to the counter-flow LMTD.

Corrected LMTD

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

$\Delta T_{\text{lm,CF}}$ = LMTD calculated as if pure counter-flow. $F$ = correction factor ($0 < F \le 1$). $F = 1$ for true counter-flow; $F < 1$ for multi-pass or cross-flow.

Correction Factor Parameters

$$P = \frac{T_{c,\text{out}} - T_{c,\text{in}}}{T_{h,\text{in}} - T_{c,\text{in}}} \qquad R = \frac{T_{h,\text{in}} - T_{h,\text{out}}}{T_{c,\text{out}} - T_{c,\text{in}}} = \frac{C_c}{C_h}$$

$P$ = temperature effectiveness of the cold side (0 to 1). $R$ = capacity rate ratio. $F$ is read from charts or computed from formulas for each configuration.

$F$ for 1-Shell, 2-Tube-Pass

$$F = \frac{\sqrt{R^2+1}\,\ln\frac{1-P}{1-RP}}{(R-1)\ln\frac{2-P(R+1-\sqrt{R^2+1})}{2-P(R+1+\sqrt{R^2+1})}}$$

If $R = 1$: $F = \frac{P\sqrt{2}}{(1-P)\ln\frac{2-P(2-\sqrt{2})}{2-P(2+\sqrt{2})}}$

Design rule: if $F < 0.75$, consider adding shells or changing configuration.

Example 1 — 1-Shell, 2-Tube-Pass

$T_h$: 120°C → 80°C. $T_c$: 20°C → 60°C. Find $P$, $R$, and estimate $F$.

  1. $P = (60-20)/(120-20) = 40/100 = 0.4$
  2. $R = (120-80)/(60-20) = 40/40 = 1.0$
  3. From the 1-shell 2-tube-pass chart at $P = 0.4$, $R = 1$: $F \approx 0.93$
  4. Counter-flow $\Delta T_{\text{lm}} = (60-60)/\ln(60/60)$... both $\Delta T$s equal 60°C: $\Delta T_{\text{lm}} = 60$°C.
  5. Effective $\Delta T = F \times 60 = 0.93 \times 60 = 55.8$°C
Example 2 — Sizing with $F$

$\dot{Q} = 300$ kW, $U = 700$ W/(m²·K), counter-flow $\Delta T_{\text{lm}} = 45$°C, $F = 0.85$. Find required area.

  1. $$A = \frac{\dot{Q}}{UF\,\Delta T_{\text{lm}}} = \frac{300{,}000}{700 \times 0.85 \times 45} = \frac{300{,}000}{26{,}775} = 11.2 \text{ m}^2$$
  2. Without the correction ($F = 1$): $A = 300{,}000/(700 \times 45) = 9.5$ m².
  3. The multi-pass arrangement needs 18% more area.
Example 3 — When $F$ Is Too Low

$P = 0.9$, $R = 1$. For a 1-shell 2-tube-pass, $F \approx 0.55$. What should you do?

  1. $F < 0.75$ → the exchanger is thermally inefficient.
  2. Options:
  3. 1.
  4. Use 2 shells in series (each with lower $P$) → higher overall $F$.
    2. Switch to pure counter-flow if possible.
    3. Accept larger area and cost penalty.
  5. With 2 shells: each shell has $P_1 = 0.684$ (from the 2-shell formula), which gives $F \approx 0.91$ — much better.

Practice Problems

1. $T_h$: 100→70, $T_c$: 25→55. Find $P$ and $R$.
2. If $F = 0.90$ and counter-flow $\Delta T_{\text{lm}} = 40$, what is the effective $\Delta T$?
3. $\dot{Q} = 150$ kW, $U = 500$, $F = 0.88$, $\Delta T_{\text{lm,CF}} = 35$. Find $A$.
4. For pure counter-flow, what is $F$?
5. $P = 0.3$, $R = 2$, 1-shell 2-tube. Estimate $F$ from the fact that low $P$ gives $F$ close to 1.
6. Why is $F$ always $\le 1$?
Show Answer Key

1. $P = 30/75 = 0.4$, $R = 30/30 = 1.0$

2. $0.90 \times 40 = 36$°C

3. $A = 150{,}000/(500 \times 0.88 \times 35) = 9.74$ m²

4. $F = 1.0$

5. $F \approx 0.95$ (low $P$ means moderate duty, so multi-pass penalty is small).

6. Counter-flow is thermodynamically the most effective arrangement. Any departure (multi-pass, cross-flow) reduces the effective temperature difference.

📐 LMTD Correction Factor (F)
P parameter
R parameter
F (1-shell, 2-tube passes)
Effective ΔT = F×LMTDᴄᶠ
LMTD for Parallel-Flow and Counter-Flow Fouling and Design Considerations