💨 Air-Cooled Heat Exchanger
Rate and size air-cooled heat exchangers (fin-fan coolers) with API 661 guidelines and Briggs-Young correlations.
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Heat Exchangers
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Sizing Inputs
Calculation Results
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Calculation Methodology
Mathematical Model & Theory
Air-cooled heat exchangers (fin-fan coolers) force air across a bank of externally finned tubes carrying process fluid. For sizing, the air flow rate is calculated from the heat load and temperature rise:
$$\dot{m}_{air} = \frac{Q_d}{c_{p,air} (T_{a,out} - T_{a,in})}$$
Fin efficiency ($\eta_f$) accounts for the temperature drop along the fin height:
$$\eta_f = \frac{\tanh(m \cdot h_f)}{m \cdot h_f}, \quad m = \sqrt{\frac{2 h_{air}}{k_{fin} t_{fin}}}$$
Overall surface efficiency ($\eta_o$) is computed from bare area ($A_b$) and fin area ($A_f$):
$$\eta_o = 1 - \frac{A_f}{A_{tot}} (1 - \eta_f)$$
The overall heat transfer coefficient ($U$) based on outer tube surface is:
$$\frac{1}{U} = \frac{1}{\eta_o h_{air}} + R_{wall} + \frac{D_{tube,out}}{D_{tube,in} \cdot h_{process}}$$
Academic References:
- API Standard 661, Air-Cooled Heat Exchangers for General Refinery Service.
- Briggs, D. E., and Young, E. H. (1963). Convective Heat Transfer and Pressure Drop of Muffled Finned Tube Bundles.
- Shah, R. K., & Sekulić, D. P. (2003). Fundamentals of Heat Exchanger Design.
Worked Engineering Example
Problem Statement:
Size an air-cooled lube-oil cooler transferring $500$ kW of thermal duty. Lube oil ($h_{proc} = 500$ W/m²K) enters the tubes at $120$°C and exits at $60$°C. Ambient air enters at $35$°C. The exchanger has 4 rows of finned tubes, 20 tubes per row, with $25$ mm tube outer diameter, $6$ m tube length, fin pitch of $394$ fins/m, and fin height of $12.7$ mm.
Step-by-Step Solution:
1. Calculate process Log-Mean Temp Difference (LMTD):
The air outlet temperature rise is determined based on air flow design to yield $T_{a,out} \approx 51.5$°C. $$dT_1 = 120 - 51.5 = 68.5^\circ\text{C}, \quad dT_2 = 60 - 35 = 25^\circ\text{C}$$ $$\text{LMTD} = \frac{68.5 - 25}{\ln(68.5/25)} \approx 43.2^\circ\text{C}$$ Using crossflow correction factor $F_c = 0.95$, corrected LMTD $= 41.0$°C.
2. Find Fin Efficiency ($\eta_f$):
For plain fins ($\text{pitch} = 394/\text{m}$, height $= 12.7$ mm, $k_{fin} = 200$ W/mK), $h_{air}$ is rated at $40$ W/m²K. $$m = \sqrt{\frac{2 \cdot 40}{200 \cdot 0.0004}} = 31.62 \text{ m}^{-1}$$ $$m \cdot h_f = 31.62 \cdot 0.0127 = 0.401$$ $$\eta_f = \frac{\tanh(0.401)}{0.401} \approx 0.950$$
3. Compute Overall U and Required Surface:
Overall surface efficiency $\eta_o \approx 0.953$. $$\frac{1}{U} = \frac{1}{0.953 \cdot 40} + \frac{25}{25 \cdot 500} = 0.0262 + 0.002 = 0.0282 \implies U \approx 35.4 \text{ W/m²K}$$ $$A_{req} = \frac{500,000 \text{ W}}{35.4 \cdot 0.95 \cdot 43.2} \approx 344 \text{ m² (finned surface area)}$$
Size an air-cooled lube-oil cooler transferring $500$ kW of thermal duty. Lube oil ($h_{proc} = 500$ W/m²K) enters the tubes at $120$°C and exits at $60$°C. Ambient air enters at $35$°C. The exchanger has 4 rows of finned tubes, 20 tubes per row, with $25$ mm tube outer diameter, $6$ m tube length, fin pitch of $394$ fins/m, and fin height of $12.7$ mm.
Step-by-Step Solution:
1. Calculate process Log-Mean Temp Difference (LMTD):
The air outlet temperature rise is determined based on air flow design to yield $T_{a,out} \approx 51.5$°C. $$dT_1 = 120 - 51.5 = 68.5^\circ\text{C}, \quad dT_2 = 60 - 35 = 25^\circ\text{C}$$ $$\text{LMTD} = \frac{68.5 - 25}{\ln(68.5/25)} \approx 43.2^\circ\text{C}$$ Using crossflow correction factor $F_c = 0.95$, corrected LMTD $= 41.0$°C.
2. Find Fin Efficiency ($\eta_f$):
For plain fins ($\text{pitch} = 394/\text{m}$, height $= 12.7$ mm, $k_{fin} = 200$ W/mK), $h_{air}$ is rated at $40$ W/m²K. $$m = \sqrt{\frac{2 \cdot 40}{200 \cdot 0.0004}} = 31.62 \text{ m}^{-1}$$ $$m \cdot h_f = 31.62 \cdot 0.0127 = 0.401$$ $$\eta_f = \frac{\tanh(0.401)}{0.401} \approx 0.950$$
3. Compute Overall U and Required Surface:
Overall surface efficiency $\eta_o \approx 0.953$. $$\frac{1}{U} = \frac{1}{0.953 \cdot 40} + \frac{25}{25 \cdot 500} = 0.0262 + 0.002 = 0.0282 \implies U \approx 35.4 \text{ W/m²K}$$ $$A_{req} = \frac{500,000 \text{ W}}{35.4 \cdot 0.95 \cdot 43.2} \approx 344 \text{ m² (finned surface area)}$$