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Air-Cooled Heat Exchanger (Fin-Fan)
Rate and size fin-fan air coolers using API 661 guidelines and Briggs & Young correlations.
Fin-Fan Exchanger Parameters
Air-cooled heat exchangers use ambient air to cool process fluids. They consist of:
- Fin Type: Plain, serrated, or studded helical fins.
- Rows & Tubes: Layout arrangement determining the cross-flow velocity.
- Finned Area: Extremely large air-side area to overcome air's low heat transfer coefficient.
- Fan Power: Air flow rate and tube bundle static drop define fan energy.
Sizing Inputs
Calculation Results
Results and visual trends will be displayed here upon completion of the computation.
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 $\eta_o \approx 0.953$. $$\frac{1}{U} = \frac{1}{0.953 \cdot 40} + \frac{25}{25 \cdot 500} = 0.0262 + 0.002 = 0.0282 \implies U \approx 35.4 \text{ W/m²K}$$ $$A_{req} = \frac{500,000 \text{ W}}{35.4 \cdot 0.95 \cdot 43.2} \approx 344 \text{ m² (finned surface area)}$$