Overall heat transfer coefficient and fouling factors

Overall heat transfer coefficient from series resistances (films, wall, fouling), area basis for tubes, clean versus design U, and the controlling resistance.

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Why it matters

Every heat exchanger is sized from q = U·A·ΔT_m. The overall coefficient U collects the two film coefficients, the tube wall and the fouling deposits into one number. Knowing which resistance controls U tells you where to spend effort (finning the gas side, raising a velocity, choosing a cleaning interval), and the fouling allowance decides how much extra area is bought on day one.

Key ideas

Resistances in series. Heat flowing from the hot fluid to the cold fluid passes through: the hot-side film, any hot-side deposit, the wall, any cold-side deposit and the cold-side film. Each is a thermal resistance; they add. U is the reciprocal of the total resistance per unit of a chosen area: 1/(U·A) = ΣR.

Which area? For a flat wall the area is the same on both sides, so 1/U = 1/h₁ + R_f1 + L/k + R_f2 + 1/h₂. For a tube the inside and outside areas differ, so U must be quoted on a stated area: U_o·A_o = U_i·A_i = 1/ΣR. Process exchangers (TEMA practice) normally quote U on the outside area. For thin-walled tubes the curvature corrections are small and the flat-wall form is often used.

Fouling. Deposits build up on heat transfer surfaces: scale (CaCO₃ from cooling water), corrosion products, biological growth, coke and polymer from hot hydrocarbons, particulate sediment. They add a fouling resistance R_f (m²·K/W, also called fouling factor) that grows with time, often levelling off (asymptotic fouling). Design values (typically 0.0001–0.001 m²·K/W depending on the fluid and temperature) are taken from TEMA tables or company practice — take them from your data book rather than inventing them.

Clean and design (dirty) U. U_c excludes fouling; U_d includes it: 1/U_d = 1/U_c + R_f,total. The ratio U_c/U_d is the extra area needed to cover fouling; over-generous fouling factors can oversize exchangers by 30–50 % and, by lowering velocities, may even encourage fouling. From plant data, R_f = 1/U_measured − 1/U_clean.

Controlling resistance. If one film coefficient is much smaller than the other (gas vs water, viscous oil vs condensing steam), it dominates 1/U, and U is close to that smaller h. Improving the larger h achieves almost nothing; improving the smaller one (higher velocity, fins on that side) pays off. Typical U values: water–water 800–1500, steam condensing–water 1000–3500, oil–water 100–350, gas–gas 10–50 W/m²·K.

Wilson plot. Measuring U at several velocities on one side and plotting 1/U against 1/u^0.8 separates the variable film resistance from the constant ones — a classic way to find h and the fouling resistance experimentally.

Wall temperature. The fraction of the overall ΔT across each resistance equals its share of ΣR. Knowing wall temperatures matters for fouling (some deposits form above a threshold wall temperature), for viscosity corrections and for boiling or condensation regime checks.

Formulas

q = U·A·ΔT q heat rate (W), U overall coefficient (W/m²·K), A area U is based on (m²), ΔT local or mean temperature difference (K).

1/U = 1/h₁ + R_f1 + L/k + R_f2 + 1/h₂ Plane wall; h film coefficients (W/m²·K), R_f fouling resistances (m²·K/W), L wall thickness (m), k wall conductivity (W/m·K).

1/U_o = d_o/(h_i·d_i) + R_fi·d_o/d_i + d_o·ln(d_o/d_i)/(2k) + R_fo + 1/h_o Tube, based on outside area; d_o, d_i outside and inside diameters (m).

U_o·A_o = U_i·A_i Converting between bases.

1/U_d = 1/U_c + R_f,total Design (dirty) and clean coefficients.

R_f = 1/U_dirty − 1/U_clean Fouling resistance from measurements.

Worked examples

Example 1 (standard). Oil flows outside and cooling water inside a steel tube (k = 45 W/m·K) of 25 mm OD and 21 mm ID. h_i = 2000 W/m²·K, h_o = 500 W/m²·K, R_fi = 0.0002 m²·K/W, R_fo = 0.0003 m²·K/W. Find U_o (clean and fouled) and the controlling resistance.

  1. Inside film: d_o/(h_i·d_i) = 0.025/(2000 × 0.021) = 0.000595.
  2. Inside fouling: 0.0002 × 0.025/0.021 = 0.000238.
  3. Wall: 0.025 × ln(25/21)/(2 × 45) = 0.0000484.
  4. Outside fouling: 0.000300. Outside film: 1/500 = 0.002000.
  5. Sum = 0.003182 m²·K/W → U_o = 314 W/m²·K.
  6. Clean: remove fouling terms → 0.002644 → U_o,clean = 378 W/m²·K.
  7. Shares: outside film 62.9 %, inside film 18.7 %, outside fouling 9.4 %, inside fouling 7.5 %, wall 1.5 %.

Answer: U_o ≈ 314 W/m²·K fouled (378 clean); the oil-side film controls.

Example 2 (GATE level). A thin-walled exchanger has h_i = 4000 W/m²·K (water) and h_o = 250 W/m²·K (oil). (a) Find U. (b) Find U if h_i is doubled, and separately if h_o is doubled. (c) A total fouling resistance of 0.0004 m²·K/W is added; by what factor must the area increase to keep the same duty at the same ΔT_m? (d) A clean exchanger had U = 1000 W/m²·K; after a year it measures 800 W/m²·K. Find R_f.

  1. (a) 1/U = 1/h_i + 1/h_o = 0.00025 + 0.004 = 0.00425 → U = 235.3 W/m²·K.
  2. (b) Doubling h_i: 1/U = 0.000125 + 0.004 → U = 242.4 (+3 %). Doubling h_o: 1/U = 0.00025 + 0.002 → U = 444.4 (+89 %).
  3. (c) 1/U_d = 0.00425 + 0.0004 = 0.00465 → U_d = 215.1 W/m²·K; area factor = U_c/U_d = 235.3/215.1 = 1.094.
  4. (d) R_f = 1/U_dirty − 1/U_clean = 1/800 − 1/1000 = 0.00025 m²·K/W.

Answer: (a) 235 W/m²·K; (b) 242 vs 444 W/m²·K — improve the oil side; (c) about 9.4 % more area; (d) R_f = 2.5 × 10⁻⁴ m²·K/W.

Common mistakes

  • Adding coefficients (h₁ + h₂) instead of resistances.
  • Mixing inside and outside areas in one expression without the diameter ratios.
  • Forgetting fouling on one side, or treating the fouling factor as a coefficient (W/m²·K) instead of a resistance (m²·K/W).
  • Trying to raise U by improving the side that already has a high h.
  • Quoting U without saying which area it is based on.

For GATE CH

Expect direct computation of U from film coefficients, wall and fouling; conversion between U_i and U_o; finding R_f from clean and dirty U; "which change increases U most" questions; and wall-temperature calculations from resistance shares. Practise setting out the resistance table with each term's percentage — it answers most conceptual follow-ups immediately.

Quick check

  1. h₁ = 100 and h₂ = 10 000 W/m²·K, thin clean wall. Approximate U?
  2. Units of fouling factor?
  3. U_o = 400 W/m²·K on A_o = 10 m². What is U_i if A_i = 8 m²?
  4. Clean U = 500, dirty U = 400 W/m²·K. R_f?
  5. Which resistance is usually negligible for a thin metal tube?

Answers: 1. 99 W/m²·K (close to the smaller h). 2. m²·K/W. 3. 500 W/m²·K. 4. 0.0005 m²·K/W. 5. The wall.

See it move

All Chemical animations
Explore Heat Transfer Coefficient and Fouling

Adjust the convective heat transfer coefficients and fouling resistance to see how they affect the overall heat transfer coefficient. Notice how fouling can significantly impact efficiency.

Equations used
  • U = 1 / (1/h1 + Rw + 1/h2 + Rf) — U overall heat transfer coefficient, h1 convective heat transfer coefficient on side 1, Rw thermal resistance of the wall, h2 convective heat transfer coefficient on side 2, Rf fouling resistance

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is the overall heat transfer coefficient?Concept

    U is the heat rate per unit area per unit overall temperature difference between the two fluids, q = U·A·ΔT (W/m²·K). It is the reciprocal of the total series resistance per unit area: the two convective films, the wall conduction and any fouling deposits, 1/(UA) = ΣR. For tubes it must be quoted on a stated area (usually the outside), since U_oA_o = U_iA_i.

  2. 2.Explain the concept of fouling factor in heat exchangers.Concept

    The fouling factor is a measure of the resistance to heat transfer caused by the accumulation of unwanted materials on the heat transfer surfaces. It is expressed in m²·K/W and is used to account for the decrease in heat transfer efficiency over time due to fouling. The fouling factor is added to the overall heat transfer resistance to ensure the heat exchanger is designed to handle the expected fouling.

  3. 3.How is the overall heat transfer coefficient calculated for a heat exchanger?Concept

    Add the resistances in series and invert. For a plane or thin wall: 1/U = 1/h₁ + R_f1 + L/k + R_f2 + 1/h₂, with h the film coefficients, R_f the fouling resistances and L/k the wall resistance. For a tube based on the outside area: 1/U_o = d_o/(h_i·d_i) + R_fi·d_o/d_i + d_o·ln(d_o/d_i)/(2k) + R_fo + 1/h_o. The film coefficients come from correlations (Dittus–Boelter, Nusselt condensation, etc.) and the fouling factors from TEMA or plant data.

  4. 4.Why is it important to consider fouling factors in the design of heat exchangers?Application

    Considering fouling factors in the design of heat exchangers is important because fouling can significantly reduce the efficiency of heat transfer over time. By accounting for fouling, engineers can design heat exchangers that maintain their performance even as fouling occurs, ensuring reliability and reducing maintenance costs.

  5. 5.What happens if the fouling factor is underestimated in a heat exchanger design?Application

    If the fouling factor is underestimated, the heat exchanger may not perform as expected over time. The heat transfer rate will decrease more than anticipated, leading to insufficient heating or cooling. This can result in increased operational costs, reduced process efficiency, and potentially the need for more frequent maintenance or replacement.

  6. 6.In what situations might the overall heat transfer coefficient be particularly low?Application

    The overall heat transfer coefficient might be particularly low in situations where there is significant fouling, low convective heat transfer coefficients, or thick walls with low thermal conductivity. These factors increase the total resistance to heat transfer, reducing the overall heat transfer coefficient.

  7. 7.How does increasing the flow rate of a fluid affect the overall heat transfer coefficient in a heat exchanger?Application

    Increasing the flow rate of a fluid generally increases the convective heat transfer coefficient, which can lead to an increase in the overall heat transfer coefficient. This is because higher flow rates enhance turbulence, improving the heat transfer efficiency on the fluid side.

  8. 8.Calculate the overall heat transfer coefficient for a heat exchanger (thin plane wall) with h₁ = 500 W/(m²·K), h₂ = 300 W/(m²·K), wall resistance R_w = 0.002 m²·K/W and fouling resistance R_f = 0.001 m²·K/W.Numerical

    1/U = 1/h₁ + R_w + 1/h₂ + R_f = 0.002 + 0.002 + 0.003333 + 0.001 = 0.008333 m²·K/W, so U = 1/0.008333 = 120 W/(m²·K). The 300 W/m²·K film is the largest single resistance (40 %), so that side is where improvement would pay off most.

  9. 9.A heat exchanger has an overall heat transfer coefficient of 150 W/(m²·K). If the fouling factor increases by 0.001 m²·K/W, what is the new overall heat transfer coefficient?Numerical

    The new overall heat transfer coefficient (U_new) can be calculated using the formula: 1/U_new = 1/U_old + ΔR_f, where ΔR_f is the increase in fouling factor. Given U_old = 150 W/(m²·K) and ΔR_f = 0.001 m²·K/W, 1/U_new = 1/150 + 0.001 = 0.00667 + 0.001 = 0.00767. Therefore, U_new = 1/0.00767 = 130.39 W/(m²·K).

  10. 10.Explain how the overall heat transfer coefficient is used in the design of heat exchangers.Application

    The overall heat transfer coefficient is used in the design of heat exchangers to determine the required surface area for a given heat transfer rate. By knowing the desired heat transfer rate and the temperature difference between the fluids, engineers can use the formula Q = U·A·ΔT to calculate the necessary surface area (A) to achieve the desired performance. This ensures the heat exchanger is appropriately sized for its intended application.

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