Thermal boundary layer and heat transfer coefficient

Velocity and thermal boundary layers, how the wall temperature gradient defines h, the role of Pr, laminar flat-plate results and the heat-momentum analogies.

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Why it matters

The heat transfer coefficient is not a property of the fluid; it is set by a thin layer of fluid next to the wall where temperature changes from the wall value to the free-stream value. Understanding that layer explains why h is high at the leading edge of a plate, why turbulence and fouling-free smooth surfaces matter, why oils give low coefficients, and how heat, momentum and mass transfer are analogous.

Key ideas

Velocity (hydrodynamic) boundary layer. When fluid flows over a surface, the no-slip condition makes the velocity zero at the wall. Viscous effects are confined to a layer of thickness δ (where u reaches 99 % of the free-stream velocity u∞). On a flat plate δ grows with distance x from the leading edge.

Thermal boundary layer. If the wall temperature T_s differs from the free-stream temperature T∞, the temperature changes across a layer of thickness δ_t (where (T_s − T)/(T_s − T∞) = 0.99). It also starts at the leading edge (or where heating starts) and thickens downstream.

How h is defined at the wall. At the wall the fluid is stationary, so heat crosses the first fluid layer by conduction only. Equating Fourier's law in the fluid with Newton's law of cooling: h = −k_f·(∂T/∂y)|_wall / (T_s − T∞). So h is large when the temperature gradient at the wall is steep, which happens when the thermal layer is thin. Roughly h ~ k_f/δ_t.

Consequences.

  • Local h is highest near the leading edge (δ_t → 0) and falls downstream; for a laminar plate h_x ∝ x^(−1/2), and the average over a length L is twice the local value at L.
  • Raising velocity thins both layers (δ ∝ x/√Re_x in laminar flow) and raises h.
  • Turbulence brings fast eddies close to the wall; only a thin viscous sublayer remains, so turbulent h is much higher than laminar h, and grows roughly as u^0.8.
  • Re-starting the boundary layer (interrupted fins, louvres, baffles) keeps h high.

Relative thickness — the Prandtl number. Pr = ν/α compares momentum and heat diffusion. For laminar flow over a flat plate δ/δ_t ≈ Pr^(1/3) (Pr ≥ 0.6):

  • gases (Pr ≈ 0.7): δ_t slightly larger than δ;
  • water (Pr ≈ 2–7) and oils (Pr ≫ 1): δ_t much thinner than δ;
  • liquid metals (Pr ≪ 1): δ_t ≫ δ.

Laminar flat-plate results (Blasius / Pohlhausen). δ = 5x/√Re_x; local Nu_x = 0.332·Re_x^(1/2)·Pr^(1/3); average Nu_L = 0.664·Re_L^(1/2)·Pr^(1/3). Transition near Re_x ≈ 5 × 10⁵. The integral method with cubic profiles gives slightly different constants (δ = 4.64x/√Re_x).

Analogies. Because the momentum and energy equations have the same form, friction and heat transfer are linked: Reynolds analogy St = f/2 (Pr = 1), and the Chilton–Colburn analogy St·Pr^(2/3) = f/2 for 0.6 < Pr < 60. The same applies to mass transfer (concentration boundary layer, Schmidt number).

Pipes. In a tube the boundary layers grow until they meet at the centre; beyond the thermal entry length the flow is thermally fully developed and h becomes constant (Nu = 3.66 or 4.36 in laminar flow). Entry lengths: about 0.05·Re·D (hydrodynamic) and 0.05·Re·Pr·D (thermal) in laminar flow; about 10 D in turbulent flow.

Formulas

h = −k_f·(∂T/∂y)|_(y=0) / (T_s − T∞) Local heat transfer coefficient (W/m²·K); k_f fluid conductivity (W/m·K); y distance normal to the wall (m).

q″ = h·(T_s − T∞) Local heat flux (W/m²).

Re_x = u∞·x/ν Local Reynolds number; ν kinematic viscosity (m²/s), x from leading edge (m).

δ = 5.0·x/√Re_x, δ/δ_t ≈ Pr^(1/3) Laminar flat plate; thicknesses in m.

Nu_x = h_x·x/k_f = 0.332·Re_x^(1/2)·Pr^(1/3), h̄_L = 2·h_x(at x = L) Laminar, Re_x < 5 × 10⁵, properties at the film temperature.

St·Pr^(2/3) = f/2 (Fanning f) Chilton–Colburn analogy; St = h/(ρ·c_p·u∞).

L_h ≈ 0.05·Re·D, L_t ≈ 0.05·Re·Pr·D Laminar entry lengths in a tube (m).

Worked examples

Example 1 (standard). Air flows at 4 m/s over a flat plate (ν = 1.6 × 10⁻⁵ m²/s, k = 0.0263 W/m·K, Pr = 0.70). At x = 0.3 m find δ, δ_t, the local h, and the average h over the first 0.3 m. Find the heat loss from a 0.3 m × 1 m plate held 50 K above the air.

  1. Re_x = u∞·x/ν = 4 × 0.3/1.6 × 10⁻⁵ = 75 000 → laminar.
  2. δ = 5.0·x/√Re_x = 5 × 0.3/273.9 = 5.48 mm.
  3. δ_t = δ/Pr^(1/3) = 5.48/0.888 = 6.17 mm (thicker than δ since Pr < 1).
  4. h_x = 0.332·Re_x^(1/2)·Pr^(1/3)·k/x = 0.332 × 273.9 × 0.888 × 0.0263/0.3 = 7.08 W/m²·K.
  5. Average: h̄ = 2 × 7.08 = 14.15 W/m²·K.
  6. q = h̄·A·ΔT = 14.15 × 0.3 × 50 = 212 W.

Answer: δ ≈ 5.5 mm, δ_t ≈ 6.2 mm, h_x ≈ 7.1 W/m²·K, h̄ ≈ 14.2 W/m²·K, q ≈ 212 W. (The crude estimate k/δ_t = 4.3 W/m²·K has the right order but is not a design value.)

Example 2 (GATE level). In a laminar boundary layer the temperature profile is approximated by (T − T_s)/(T∞ − T_s) = 1.5·(y/δ_t) − 0.5·(y/δ_t)³ for y ≤ δ_t. At a station where δ_t = 4 mm, k_f = 0.03 W/m·K, T_s = 90 °C and T∞ = 30 °C, find h and the wall heat flux.

  1. Write θ = (T − T_s)/(T∞ − T_s); dθ/dy at y = 0 is 1.5/δ_t.
  2. (∂T/∂y)|₀ = (T∞ − T_s)·1.5/δ_t = (−60) × 1.5/0.004 = −22 500 K/m.
  3. h = −k_f·(∂T/∂y)|₀/(T_s − T∞) = −0.03 × (−22 500)/60 = 11.25 W/m²·K. (In general h = 1.5·k_f/δ_t for this profile.)
  4. q″ = h·(T_s − T∞) = 11.25 × 60 = 675 W/m².

Answer: h = 11.25 W/m²·K; q″ = 675 W/m² from the wall into the fluid.

Common mistakes

  • Saying heat "convects" across the wall layer: at y = 0 the fluid is still, so the wall flux is pure conduction through the fluid; convection sets the gradient.
  • Using the solid's conductivity in h = −k·(∂T/∂y)/ΔT; it is the fluid's.
  • Assuming δ_t = δ for every fluid; it is only so when Pr ≈ 1.
  • Confusing local and average h on a laminar plate (factor 2).
  • Thinking h is constant along a surface; it varies, and correlations give local or averaged values.

For GATE CH

Expect questions that give a temperature or velocity profile and ask for h or wall shear, ratio questions on δ/δ_t from Pr, how local h varies with x, local versus average h, and analogy-based questions (find h from a friction factor or vice versa). Practise differentiating a profile at the wall carefully with signs, and recall the x^(−1/2) laminar dependence.

Quick check

  1. For laminar flow on a plate, how does local h change when x is quadrupled?
  2. Oil with Pr = 1000: is δ_t larger or smaller than δ?
  3. What is the heat transfer mechanism in the fluid right at the wall?
  4. Local h at x = L is 10 W/m²·K (laminar). Average h over 0 to L?
  5. Write the Chilton–Colburn analogy.

Answers: 1. It halves. 2. Smaller (δ/δ_t ≈ 10). 3. Conduction. 4. 20 W/m²·K. 5. St·Pr^(2/3) = f/2.

Interview questions

All Heat Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is a thermal boundary layer in the context of heat transfer?Concept

    A thermal boundary layer is a region adjacent to a heated or cooled surface where the temperature gradient exists due to heat transfer between the surface and the fluid. It is characterized by a gradual change in temperature from the surface temperature to the bulk fluid temperature. The thickness of this layer depends on the fluid properties, flow velocity, and surface conditions.

  2. 2.Explain the significance of the heat transfer coefficient in thermal systems.Concept

    The heat transfer coefficient is a measure of the heat transfer rate per unit area per unit temperature difference between a surface and a fluid. It is crucial in designing and analyzing thermal systems as it helps determine the efficiency of heat exchangers, radiators, and other heat transfer equipment. A higher heat transfer coefficient indicates more efficient heat transfer.

  3. 3.How does the thermal boundary layer affect the heat transfer coefficient?Concept

    The thermal boundary layer affects the heat transfer coefficient by influencing the temperature gradient at the surface. A thinner boundary layer typically results in a higher temperature gradient, leading to a higher heat transfer coefficient. Conversely, a thicker boundary layer reduces the temperature gradient and thus the heat transfer coefficient.

  4. 4.Why is the concept of the thermal boundary layer important in the design of heat exchangers?Application

    The thermal boundary layer is important in heat exchanger design because it directly impacts the heat transfer rate. Understanding the boundary layer helps engineers optimize the surface area and flow conditions to enhance heat transfer efficiency. Proper management of the boundary layer can lead to more compact and cost-effective heat exchanger designs.

  5. 5.What happens to the thermal boundary layer when the flow velocity of a fluid increases?Application

    Higher velocity (higher Re) thins both the velocity and thermal boundary layers; in laminar flow on a plate δ ∝ x/√Re_x. A thinner thermal layer means a steeper temperature gradient at the wall, and since h = −k_f(∂T/∂y)_wall/(T_s − T∞), h increases — roughly as u^0.5 in laminar flow and u^0.8 in turbulent flow. Above a critical Re the layer becomes turbulent, which raises h much further at the cost of more friction.

  6. 6.In what scenarios would you prefer a high heat transfer coefficient, and why?Application

    A high heat transfer coefficient is preferred in scenarios where rapid heat exchange is required, such as in cooling systems for electronic devices, automotive radiators, and industrial heat exchangers. A high coefficient ensures efficient heat removal or addition, leading to better performance and energy efficiency of the system.

  7. 7.How does surface roughness affect the thermal boundary layer and heat transfer coefficient?Application

    Surface roughness can disrupt the thermal boundary layer, leading to increased turbulence and mixing of fluid layers. This disruption can enhance the heat transfer coefficient by increasing the temperature gradient at the surface. However, excessive roughness may also lead to increased pressure drop and energy consumption.

  8. 8.Calculate the heat transfer rate if the heat transfer coefficient is 500 W/m²·K, the surface area is 2 m², and the temperature difference is 30 K.Numerical

    The heat transfer rate (Q) can be calculated using the formula: Q = h·A·ΔT, where h is the heat transfer coefficient, A is the surface area, and ΔT is the temperature difference. Substituting the given values: Q = 500 W/m²·K × 2 m² × 30 K = 30,000 W or 30 kW.

  9. 9.If the thermal boundary layer thickness is reduced by half, how does it affect the heat transfer coefficient?Application

    Since h is set by the wall temperature gradient and h ~ k_f/δ_t for a given profile shape (for example h = 1.5k_f/δ_t for the cubic profile), halving δ_t roughly doubles h. This is consistent with the laminar plate result h_x ∝ x^(−1/2) while δ_t ∝ x^(1/2). The exact factor depends on whether the profile shape stays similar.

  10. 10.A flat plate is exposed to air at 25°C. If the surface temperature is 75°C and the heat transfer coefficient is 25 W/m²·K, what is the heat flux?Numerical

    The heat flux (q) can be calculated using the formula: q = h·ΔT, where h is the heat transfer coefficient and ΔT is the temperature difference. Here, ΔT = 75°C - 25°C = 50°C. Substituting the given values: q = 25 W/m²·K × 50 K = 1250 W/m².

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