Question

Ambient air flows over a heated slab having flat, top surface at \( y = 0 \). The local temperature (in Kelvin) profile within the thermal boundary layer is given by \( T(y) = 300 + 200 \exp(-5y) \), where \( y \) is the distance measured from the slab surface in meter. If the thermal conductivity of air is 1.0 W/m.K and that of the slab is 100 W/m.K, then the magnitude of temperature gradient \( \left| \frac{dT}{dy} \right| \) within the slab at \( y = 0 \) is \(\underline{\hspace{1cm}}\) K/m (round off to the nearest integer).

Hint:

The temperature gradient in a boundary layer is obtained by differentiating the temperature profile with respect to the distance.

Answer 1

Correct Answer: 10

Step 1: Find the Temperature Gradient in Air

Taking the derivative of the temperature profile:

$$\frac{dT}{dy} = \frac{d}{dy}(300 + 200e^{-5y})$$

$$\frac{dT}{dy} = 200 \times (-5) \times e^{-5y}$$

$$\frac{dT}{dy} = -1000e^{-5y}$$

Step 2: Evaluate Temperature Gradient in Air at y = 0

At the surface ($y = 0$):

$$\left(\frac{dT}{dy}\right)_{air, y=0} = -1000e^{0} = -1000 \text{ K/m}$$

Step 3: Apply Heat Flux Continuity at the Interface

At the interface between the slab and air, the heat flux must be continuous (no heat generation or accumulation at the interface):

$$q_{slab} = q_{air}$$

$$-k_{slab}\left(\frac{dT}{dy}\right){slab} = -k{air}\left(\frac{dT}{dy}\right)_{air}$$

Step 4: Calculate Temperature Gradient in Slab at y = 0

$$k_{slab}\left(\frac{dT}{dy}\right){slab, y=0} = k{air}\left(\frac{dT}{dy}\right)_{air, y=0}$$

$$100 \times \left(\frac{dT}{dy}\right)_{slab, y=0} = 1.0 \times (-1000)$$

$$\left(\frac{dT}{dy}\right)_{slab, y=0} = \frac{-1000}{100} = -10 \text{ K/m}$$

Step 5: Find the Magnitude

$$\left|\frac{dT}{dy}\right|_{slab, y=0} = |-10| = 10 \text{ K/m}$$

Answer: The magnitude of temperature gradient within the slab at y = 0 is 10 K/m.