Convective Heat Transfer Coefficient of a Heater Surface: 22-Agric-A3 Heat Engineering, May 2018, Q2

Problem 2: Convective Heat Transfer Coefficient of a Heater Surface (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Heater surface heat flux q₁₂ = 6000 W/m² at heater surface temperature T_s1 = 120°C, cooled by an air stream at T∞ = 70°C; reduced power gives q₂₂ = 2000 W/m² at the same flow conditions.

Find. (a) The average convective heat transfer coefficient ħ; (b) the new heater surface temperature T_s2 after the power is reduced.

Approach. Newton's law of cooling, $q''=\bar{h}(T_s-T_\infty)$, gives ħ directly from part (a)'s data; since only the electrical power (hence q'') changes and the air stream's velocity and the plate geometry do not, ħ itself is unchanged (for forced convection, ħ is set by the flow field and geometry, not by the temperature difference), so the same ħ is reused to solve for the new surface temperature.

Newton's law of cooling: the heat flux leaving the heater surface drives the temperature difference between the surface and the free-stream air.

Average convective heat transfer coefficient. By Newton's law of cooling, $$\bar{h} = \frac{q_1''}{T_{s1}-T_\infty} = \frac{6000\ \text{W/m}^2}{120-70\ \text{K}} = \boxed{120\ \text{W/m}^2\cdot\text{K}}$$
New heater temperature at reduced power. With the flow unchanged, ħ carries over unchanged, so $$T_{s2} = T_\infty + \frac{q_2''}{\bar{h}} = 70 + \frac{2000}{120} = \boxed{86.7^{\circ}\text{C}}$$

Quantity	Result
Average convective heat transfer coefficient, ħ	120 W/m²·K
Heater temperature at q″ = 2000 W/m², T_s2	86.7°C

Concept & Theory

Topic: Newton's law of cooling and the convective heat transfer coefficient.

Key relations: $q'' = \bar{h}(T_s-T_\infty)$, where ħ lumps together the effects of fluid motion, fluid properties and surface geometry into a single coefficient.

Why this works: For a fixed geometry and a fixed flow field (same air velocity across the same plate), the boundary-layer development and hence the local and average Nusselt number depend only on the flow, not on how much the surface happens to be heated; ħ is therefore a property of the flow/geometry pairing, not of q'' or ΔT themselves. Reducing the electrical power changes q'' and, through it, ΔT, but leaves ħ fixed — so a lower q'' must correspond to a proportionally smaller ΔT, and hence a lower (but still above ambient) surface temperature.

Common pitfall: Assuming ħ itself changes with q'' or ΔT for forced convection at constant velocity — that dependence only arises for natural convection, where the buoyancy-driven flow (and hence ħ) is itself generated by ΔT.

Source: Çengel & Ghajar, Heat and Mass Transfer: Fundamentals and Applications, Ch. 1 & 6 (Newton's law of cooling, the convection heat transfer coefficient).

Question 2 of 4: Convective Heat Transfer Coefficient of a Heater Surface

Problem 2: Convective Heat Transfer Coefficient of a Heater Surface (25 points)