Introduction & Context

The overall heat-transfer coefficient U quantifies how easily heat flows across a multi-layer thermal circuit consisting of inside convection, inside fouling, wall conduction, outside fouling, and outside convection. A single value of U (referenced to a chosen heat-transfer area) is required for sizing heat exchangers, boilers, condensers, and any process equipment where energy is exchanged between two fluids. Accurate prediction of U directly governs the required heat-transfer area and therefore the capital cost and operability of the unit.

Methodology & Formulas

Geometry conversion
All input dimensions are converted to consistent SI units: \[ d_o = \frac{d_{o,\text{mm}}}{1000} \quad ; \quad d_i = \frac{d_{i,\text{mm}}}{1000} \quad ; \quad x = \frac{x_{\text{mm}}}{1000} \]
Resistance network (per unit of outer area)
The total thermal resistance is the sum of five series resistances:
- Inside convection resistance (referred to A_o): \[ R_{\text{conv},i} = \frac{1}{h_i}\frac{d_o}{d_i} \]
- Inside fouling resistance (referred to A_o): \[ R_{\text{foul},i} = R_{f,i}\frac{d_o}{d_i} \]
- Conduction through the cylindrical wall (referred to A_o): \[ R_{\text{cond}} = \frac{d_o\ln(d_o/d_i)}{2k_{\text{wall}}} \]
- Outside fouling resistance: \[ R_{\text{foul},o} = R_{f,o} \]
- Outside convection resistance: \[ R_{\text{conv},o} = \frac{1}{h_o} \]
Total resistance: \[ R_{\text{total}} = R_{\text{conv},i} + R_{\text{foul},i} + R_{\text{cond}} + R_{\text{foul},o} + R_{\text{conv},o} \]
Overall heat-transfer coefficient
The coefficient based on the outer surface area is: \[ U = \frac{1}{R_{\text{total}}} \]

Parameter	Recommended / Limiting Value	Remarks
Fouling factor (per side)	≤ 0.00035 m² K W⁻¹	TEMA maximum for process design
Flow regime	Re ≥ 10,000	Assumed turbulent flow for both streams

The overall heat-transfer coefficient U links the total heat duty Q to the heat-exchanger area A and the effective temperature difference ΔT_LM through Q = U A ΔT_LM. It is calculated from the sum of all thermal resistances:

Inside film resistance: 1/h_i
Outside film resistance: 1/h_o
Tube-wall conduction resistance: (D_o ln(D_o/D_i))/(2 k)
Fouling resistances: R_fi and R_fo

Combine them into: 1/U_o = 1/h_o + R_fo + (D_o ln(D_o/D_i))/(2 k) + (D_o/D_i)(1/h_i + R_fi).

Typical ranges for clean services are:

Water-water: 800–1500 W m^-2 K^-1
Steam condensers: 1500–5000 W m^-2 K^-1
Light organics: 300–700 W m^-2 K^-1
Heavy hydrocarbons: 50–200 W m^-2 K^-1

Always verify against plant data; fouling can cut these values in half.

Fouling deposits add extra thermal resistance, so U drops and the exchanger may no longer meet its duty. When you include R_f in the resistance equation, the calculated U becomes smaller, requiring either:

A larger area at design stage, or
More frequent cleaning to restore performance.

Use the outside diameter as the reference area; all resistances are converted to an outside basis. The tube-wall term already accounts for the diameter ratio, so simply compute 1/U_o as shown and then multiply U_o by A_o to get the duty.

Measure inlet/outlet temperatures and flow rates under steady conditions.
Calculate actual Q = m C_p ΔT for the limiting fluid.
Compute ΔT_LM with the measured temperatures.
Solve U = Q/(A ΔT_LM).
Compare to design U; a drop >20% usually signals fouling or maldistribution.

Worked Example – Overall Heat Transfer Coefficient for a Cooling-Water Heat Exchanger Tube

A process engineer is sizing a single-pass shell-and-tube heat exchanger that cools a hydrocarbon stream with cooling water. To predict the required tube length, the overall heat-transfer coefficient U must be calculated for a plain carbon-steel tube under clean service conditions.

Knowns

Outside diameter, d_o = 25 mm
Inside diameter, d_i = 21 mm
Wall thickness, x = 2 mm
Thermal conductivity of steel, k = 16 W m^-1 K^-1
Inside (water-side) convection coefficient, h_i = 5000 W m^-2 K^-1
Outside (process-side) convection coefficient, h_o = 4000 W m^-2 K^-1
Inside fouling resistance, R_f,i = 0.00018 m² K W^-1
Outside fouling resistance, R_f,o = 0.00009 m² K W^-1

Step-by-Step Calculation

Convert diameters to meters: d_o = 0.025 m, d_i = 0.021 m, x = 0.002 m.
Compute the inside convective resistance per unit outside area: \[ R_{\text{conv,i}} = \frac{1}{h_i} \cdot \frac{d_o}{d_i} = \frac{1}{5000} \cdot \frac{0.025}{0.021} = 0.000238 \; \text{m}^2 \text{K W}^{-1} \]
Compute the inside fouling resistance referred to the outside area: \[ R_{\text{foul,i}} = R_{f,i} \cdot \frac{d_o}{d_i} = 0.00018 \cdot \frac{0.025}{0.021} = 0.000214 \; \text{m}^2 \text{K W}^{-1} \]
Compute the conductive resistance of the tube wall (log-mean area ≈ arithmetic mean for thin tubes): \[ R_{\text{cond}} = \frac{x}{k} \cdot \frac{d_o}{d_{\text{lm}}} \approx \frac{0.002}{16} \cdot \frac{0.025}{0.023} = 0.000136 \; \text{m}^2 \text{K W}^{-1} \]
Take the outside fouling resistance directly: \[ R_{\text{foul,o}} = 0.00009 \; \text{m}^2 \text{K W}^{-1} \]
Compute the outside convective resistance: \[ R_{\text{conv,o}} = \frac{1}{h_o} = \frac{1}{4000} = 0.00025 \; \text{m}^2 \text{K W}^{-1} \]
Sum all resistances to obtain the total thermal resistance: \[ R_{\text{total}} = 0.000238 + 0.000214 + 0.000136 + 0.00009 + 0.00025 = 0.000929 \; \text{m}^2 \text{K W}^{-1} \]
Calculate the overall heat-transfer coefficient based on the outside surface: \[ U_o = \frac{1}{R_{\text{total}}} = \frac{1}{0.000929} = 1077 \; \text{W m}^{-2} \text{K}^{-1} \]

Final Answer

U = 1077 W m^-2 K^-1 (based on outside area)

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle