💨 Air Cooler / Finned Tube

Design and rate finned-tube air coolers. Model plate or spiral fins with Briggs & Young correlation and cross-flow effectiveness.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated

Heat Transfer

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📅 Released Jun 2026

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Finned Tube geometries

Finned surfaces significantly enhance air-side thermal conductance. Briggs and Young Nusselt correlations are applied along with:

Fins spacing ($s = p_f - t_f$): High fin densities enhance heat transfer but block cross-flow areas.
Air pressure drop ($\Delta P_{air}$): Calculated via Robinson & Briggs. Crucial for fan sizing.
Schmidt Rayon equivalent: Determines the temperature profile along the aluminum fin height.

📝 Configuration

🚀 Process Air Stream (Crossflow)

Air Mass Flow ($m_a$) [kg/s]:

Air Inlet Temp ($T_{in,a}$) [°C]:

Density ($\rho_a$) [kg/m³]:

Specific Heat ($C_{p,a}$) [J/kg·K]:

Viscosity ($\mu_a$) [Pa·s]:

Conductivity ($k_a$) [W/m·K]:

🚀 Tube-side Process Fluid

Fluid Mass Flow ($m_t$) [kg/s]:

Fluid Inlet Temp ($T_{in,t}$) [°C]:

Density ($\rho_t$) [kg/m³]:

Specific Heat ($C_{p,t}$) [J/kg·K]:

Viscosity ($\mu_t$) [Pa·s]:

Conductivity ($k_t$) [W/m·K]:

🏗️ Tubes Bundle Geometry

Tube OD ($d_o$) [mm]:

Tube ID ($d_i$) [mm]:

Tubes Length ($L$) [m]:

Tubes Wall Cond. [W/m·K]:

Number of Rows ($N_{rows}$):

Tubes per Row ($N_{t,row}$):

Transverse Pitch ($P_T$) [mm]:

Longitudinal Pitch ($P_L$) [mm]:

Number of Tube Passes:

🥞 Fins Specifications

Fins Layout type:

Fin Material Cond. ($k_f$) [W/m·K]:

Fin Thickness ($t_f$) [mm]:

Fin Pitch ($p_f$) [mm]: Note: 2.54 mm = 10 FPI

Fin Height ($h_f$) [mm]:

📊 Results & Verification

Configure process flows and geometry and click Sizing Air Cooler HX to view results.

📘 Calculation Methodology: Finned Tube Heat Exchanger Rating & Overall UA

Mathematical Model & Theory

Finned tubes enhance gas-side thermal conductance by increasing external surface area $A_o$, accounting for fin efficiency $\eta_f$ and overall surface efficiency $\eta_o$:

$$\eta_o = 1 - \frac{A_f}{A_o}(1 - \eta_f), \quad \frac{1}{UA} = \frac{1}{\eta_o h_o A_o} + \frac{\ln(d_o/d_i)}{2\pi k_{pipe} L} + \frac{1}{h_i A_i}$$

$$q = UA \cdot LMTD \cdot F$$

Assumptions

Uniform fin thickness and base contact.
Constant convective heat transfer coefficients.

Academic References

Kern, D. Q., & Kraus, A. D.: Extended Surface Heat Transfer, McGraw-Hill.
Incropera, F. P. et al.: Heat Transfer, Ch. 11.

Worked Engineering Example

Problem Statement:
A finned bank has $A_o = 25\text{ m}^2$, $\eta_o = 0.85$, $h_o = 60\text{ W/m}^2\cdot\text{K}$, $A_i = 3.0\text{ m}^2$, $h_i = 1200\text{ W/m}^2\cdot\text{K}$ (negligible wall resistance). Find overall $UA$.

Step-by-step Solution:
1. $R_o = 1 / (0.85 \times 60 \times 25) = 1 / 1275 = 0.000784\text{ K/W}$.
2. $R_i = 1 / (1200 \times 3.0) = 1 / 3600 = 0.000278\text{ K/W}$.
3. $R_{total} = 0.000784 + 0.000278 = 0.001062\text{ K/W} \implies UA = 1 / 0.001062 = 941.6\text{ W/K}$.
Final Result:
Overall conductance is $UA = \mathbf{941.6\text{ W/K}}$.