⚙️ Shell-and-Tube Heat Exchanger Design
Perform preliminary sizing of shell-and-tube heat exchangers. Evaluate required area and LMTD correction factor F for multi-pass shell arrangements.
📝 Configuration
Correction Factor sizing:
Q = U × A × F × LMTD_cf
LMTD_cf = LMTD counterflow baseline
F = correction factor for tube cross-parallel multi-passes
Temperature Cross limit:
If the temperature effectiveness P and capacity ratio R lead to invalid F bounds, it represents a temperature cross too severe for a single shell. An increase in shell passes (N) will be required.
Q = U × A × F × LMTD_cf
LMTD_cf = LMTD counterflow baseline
F = correction factor for tube cross-parallel multi-passes
Temperature Cross limit:
If the temperature effectiveness P and capacity ratio R lead to invalid F bounds, it represents a temperature cross too severe for a single shell. An increase in shell passes (N) will be required.
📊 Results & Visualization
Configure inputs and click Sizing Shell-and-Tube to view results.
📘 Calculation Methodology
Mathematical Model & Theory
Shell-and-tube heat exchangers support high pressures and contain large surface areas in a compact shell. Sizing incorporates multi-pass log-mean temperature correction factors ($F$) and capacity rate considerations:
$$Q = U A F \Delta T_{lm, cf}$$
$$F = f(P, R, N_{shell}) \quad \text{(TEMA correlations for shell passes)}$$
Worked Engineering Example
Problem Statement:
Size a 1-2 shell-and-tube heat exchanger ($U = 400$ W/m²·K) designed to cool water ($C_p = 4180$ J/kg·K, $\dot{m} = 2.5$ kg/s) from 90°C to 45°C using cold water entering at 20°C and exiting at 40°C. Check the correction factor $F$ and required area.
Step-by-step Solution:
1. Calculate heat duty $Q$:
$$Q = 2.5 \times 4180 \times (90 - 45) = 470,250 \text{ W}$$ 2. Calculate LMTD for counter-flow:
$$\Delta T_{lm, cf} = \frac{(90 - 40) - (45 - 20)}{\ln(50 / 25)} = 36.07 \text{ K}$$ 3. Calculate temperature effectiveness $P$ and capacity ratio $R$:
$$P = \frac{40 - 20}{90 - 20} = 0.286, \quad R = \frac{90 - 45}{40 - 20} = 2.25$$ 4. Determine correction factor $F$ (for 1 shell pass, 2 tube passes):
$$F \approx 0.88$$ 5. Calculate required area $A$:
$$A = \frac{Q}{U F \Delta T_{lm, cf}} = \frac{470,250}{400 \times 0.88 \times 36.07} = 37.0 \text{ m}^2$$
Final Result:
Required area is 37.0 m² with correction factor $F = 0.88$.
Size a 1-2 shell-and-tube heat exchanger ($U = 400$ W/m²·K) designed to cool water ($C_p = 4180$ J/kg·K, $\dot{m} = 2.5$ kg/s) from 90°C to 45°C using cold water entering at 20°C and exiting at 40°C. Check the correction factor $F$ and required area.
Step-by-step Solution:
1. Calculate heat duty $Q$:
$$Q = 2.5 \times 4180 \times (90 - 45) = 470,250 \text{ W}$$ 2. Calculate LMTD for counter-flow:
$$\Delta T_{lm, cf} = \frac{(90 - 40) - (45 - 20)}{\ln(50 / 25)} = 36.07 \text{ K}$$ 3. Calculate temperature effectiveness $P$ and capacity ratio $R$:
$$P = \frac{40 - 20}{90 - 20} = 0.286, \quad R = \frac{90 - 45}{40 - 20} = 2.25$$ 4. Determine correction factor $F$ (for 1 shell pass, 2 tube passes):
$$F \approx 0.88$$ 5. Calculate required area $A$:
$$A = \frac{Q}{U F \Delta T_{lm, cf}} = \frac{470,250}{400 \times 0.88 \times 36.07} = 37.0 \text{ m}^2$$
Final Result:
Required area is 37.0 m² with correction factor $F = 0.88$.