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🔀 Heat Exchanger Fouling Factor

Calculate clean vs. fouled overall heat transfer coefficients, required surface area oversizing percentage, and evaluate fouling resistance effects.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated
Heat Exchanger Fouling Factor Tools
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📅 Released Jun 2026
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📝 Configuration

🔥 Fluid Selection
Equations:
U_design = 1/(1/U_clean + Rf_total)
CF = U_design / U_clean
Overdesign = (U_clean/U_design − 1) × 100%

📊 Results

Select fluids and click Compute.

📘 Methodology

TEMA Fouling Factors

The Tubular Exchanger Manufacturers Association (TEMA) publishes recommended fouling resistances Rf for common process fluids. These values represent the thermal resistance of the fouling deposit expected to accumulate during normal service between cleaning cycles.

Design Coefficient

The design (fouled) overall heat transfer coefficient is U_design = 1/(1/U_clean + Rf_total). The cleanliness factor CF = U_design/U_clean indicates how much the exchanger performance degrades. Typical CF values range from 0.80 to 0.95.

Overdesign

The required overdesign percentage = (U_clean/U_design − 1) × 100% represents the extra surface area needed to compensate for fouling. Engineers should balance overdesign cost against cleaning frequency to optimize total lifecycle cost.

📘 Calculation Methodology: TEMA Heat Exchanger Fouling Resistances & Cleanliness

Mathematical Model & Theory

Fouling deposits (scaling, corrosion, biological growth) add thermal resistance $R_f$, reducing overall heat transfer coefficient from clean $U_{clean}$ to service $U_{fouled}$:

$$\frac{1}{U_{fouled}} = \frac{1}{U_{clean}} + R_{f,inside} + R_{f,outside}, \quad CF = \frac{U_{fouled}}{U_{clean}} \times 100\%$$
$$\text{Over-surface Safety Margin: } \frac{A_{fouled} - A_{clean}}{A_{clean}} = \frac{U_{clean}}{U_{fouled}} - 1$$

Assumptions

  • Uniform deposit layer thickness across tube length.
  • TEMA standard recommended fouling factors for industrial service fluids.

Academic References

  1. TEMA Standards: Tubular Exchanger Manufacturers Association, 10th Edition.
  2. Hewitt, G. F.: Heat Exchanger Design Handbook, Begell House.

Worked Engineering Example

Problem Statement:
A condenser with $U_{clean} = 2200\text{ W/m}^2\cdot\text{K}$ accumulates cooling water scaling ($R_{f,i} = 0.0002\text{ m}^2\cdot\text{K/W}$) and steam condensate fouling ($R_{f,o} = 0.00005\text{ m}^2\cdot\text{K/W}$). Calculate $U_{fouled}$ and cleanliness factor.

Step-by-step Solution:
1. $1/U_{clean} = 1/2200 = 0.0004545\text{ m}^2\cdot\text{K/W}$.
2. $1/U_{fouled} = 0.0004545 + 0.0002 + 0.00005 = 0.0007045\text{ m}^2\cdot\text{K/W}$.
3. $U_{fouled} = 1 / 0.0007045 \approx 1419.4\text{ W/m}^2\cdot\text{K}$.
4. Cleanliness factor: $CF = (1419.4 / 2200) \times 100\% = 64.5\%$.
Final Result:
Fouled overall coefficient is $\mathbf{1419.4\text{ W/m}^2\cdot\text{K}}$ ($CF = \mathbf{64.5\%}$, requiring +55% surface area margin).