Shell-and-Tube and Cross-Flow Correction Factors
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.
$$\dot{Q} = UAF\,\Delta T_{\text{lm,CF}}$$
$\Delta T_{\text{lm,CF}}$ = 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.
$$P = \frac{T_{c,\text{out}} - T_{c,\text{in}}}{T_{h,\text{in}} - T_{c,\text{in}}} \qquad R = \frac{T_{h,\text{in}} - T_{h,\text{out}}}{T_{c,\text{out}} - T_{c,\text{in}}} = \frac{C_c}{C_h}$$
$P$ = 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.
$$F = \frac{\sqrt{R^2+1}\,\ln\frac{1-P}{1-RP}}{(R-1)\ln\frac{2-P(R+1-\sqrt{R^2+1})}{2-P(R+1+\sqrt{R^2+1})}}$$
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.
$T_h$: 120°C → 80°C. $T_c$: 20°C → 60°C. Find $P$, $R$, and estimate $F$.
$P = (60-20)/(120-20) = 40/100 = 0.4$
$R = (120-80)/(60-20) = 40/40 = 1.0$
From the 1-shell 2-tube-pass chart at $P = 0.4$, $R = 1$: $F \approx 0.93$
Counter-flow $\Delta T_{\text{lm}} = (60-60)/\ln(60/60)$... both $\Delta T$s equal 60°C: $\Delta T_{\text{lm}} = 60$°C.
Effective $\Delta T = F \times 60 = 0.93 \times 60 = 55.8$°C
$\dot{Q} = 300$ kW, $U = 700$ W/(m²·K), counter-flow $\Delta T_{\text{lm}} = 45$°C, $F = 0.85$. Find required area.
$$A = \frac{\dot{Q}}{UF\,\Delta T_{\text{lm}}} = \frac{300{,}000}{700 \times 0.85 \times 45} = \frac{300{,}000}{26{,}775} = 11.2 \text{ m}^2$$
Without the correction ($F = 1$): $A = 300{,}000/(700 \times 45) = 9.5$ m². The multi-pass arrangement needs 18% more area.
$P = 0.9$, $R = 1$. For a 1-shell 2-tube-pass, $F \approx 0.55$. What should you do?
$F < 0.75$ → the exchanger is thermally inefficient. Options:
1. 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.
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
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.