Understanding the Convective Heat Transfer Problem

The problem asks us to find the local convective heat transfer coefficient, denoted by h, for a fluid flowing over a heated flat plate. We are given the fluid temperature (T_∞), the plate surface temperature (T_s), the temperature distribution within the fluid's boundary layer near the surface, and the thermal conductivity (k) of the fluid.

• Fluid Temperature, T_∞ = 30°C
• Plate Surface Temperature, T_s = 100°C
• Temperature Distribution, T(y) = 30 + 70exp(-y) °C
• Thermal Conductivity, k = 1.0 W/mK
• We need to find h in W/m²K.

Calculating Local Heat Flux using Fourier's Law

The convective heat transfer coefficient relates the surface heat flux (q”) to the temperature difference between the surface and the free stream fluid using Newton's Law of Cooling: q” = h(T_s - T_∞).

At the surface (y=0), the heat transfer from the plate to the fluid is essentially conduction across the fluid boundary layer. Therefore, we can use Fourier's Law of Conduction to find the local heat flux:

q” = -k $\frac{\partial T}{\partial y}\bigg|_{y=0}$

First, let's find the temperature gradient $\frac{\partial T}{\partial y}$ from the given temperature distribution:

T(y) = 30 + 70exp(-y)

Taking the derivative with respect to y:

$\frac{\partial T}{\partial y} = \frac{d}{dy}(30 + 70\text{exp}(-y))$

$\frac{\partial T}{\partial y} = 0 + 70 \times (-\text{exp}(-y))$

$\frac{\partial T}{\partial y} = -70\text{exp}(-y)$

Now, evaluate this gradient at the surface, where y = 0:

$\frac{\partial T}{\partial y}\bigg|_{y=0} = -70\text{exp}(-0)$

Since exp(0) = 1:

$\frac{\partial T}{\partial y}\bigg|_{y=0} = -70 \times 1 = -70$ °C/m

Now, substitute the value of k and the gradient into Fourier's Law:

q” = -(1.0 W/mK) $\times$ (-70 °C/m)

q” = 70 W/m²

Determining the Convective Heat Transfer Coefficient

We have calculated the local heat flux q” as 70 W/m². Now we can use Newton's Law of Cooling to find the local convective heat transfer coefficient, h.

q” = h(T_s - T_∞)

Substitute the known values:

70 W/m² = h (100°C - 30°C)

70 W/m² = h (70°C)

Solving for h:

h = $\frac{70 \text{ W/m}^2}{70 \text{ °C}}$

h = 1 W/m²K

The calculated value for the local convective heat transfer coefficient is 1 W/m²K.