Question

A solid plate, with a thickness of \(15 \mathrm{~cm}\) and a thermal conductivity of \(80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), is being cooled at the upper surface by air. The air temperature is $10^{\circ} \mathrm{C}$, while the temperatures at the upper and lower surfaces of the plate are 50 and \(60^{\circ} \mathrm{C}\), respectively. Determine the convection heat transfer coefficient of air at the upper surface, and discuss whether the value is reasonable or not for forced convection of air.

Short answer

Question: Determine if the calculated convection heat transfer coefficient is reasonable for forced convection of air considering the given data and assumptions made during the calculations. Answer: The calculated value of the convection heat transfer coefficient is not reasonable for forced convection of air, as it is negative (-159.52 W/m²⋅K), which is not physically possible. This indicates that there is an error in the given data or assumptions made during the calculations. In normal cases of forced convection, the value of the convection heat transfer coefficient would be positive, indicating that heat is being transferred from the higher temperature surface to the cooler air.

Step-by-step solution

Find the temperature gradient in the plate

First, we will need to find the temperature gradient along the plate's thickness. We can do this using the given temperatures at the upper and lower surfaces and the plate's thickness, as follows: \(\frac{dT}{dx} = \frac{T_\mathrm{lower} - T_\mathrm{upper}}{\mathrm{thickness}}\) where - \(dT\) is the change in temperature along the thickness of the plate, - \(dx\) is the thickness of the plate, - \(T_\mathrm{lower}\) and \(T_\mathrm{upper}\) are the temperatures at the lower and upper surfaces, respectively. Now, plugging in the given values, we get: \(\frac{dT}{dx} = \frac{60 - 50}{0.15} = \frac{10}{0.15} = 66.67 \ K/m\)

Calculate the heat transfer through conduction

Using Fourier's law of heat conduction, we can find the heat transfer through the plate due to conduction. The formula is as follows: \(q = -k \cdot \frac{dT}{dx}\) where - \(q\) is the steady-state heat transfer per unit area (\(W/m^2\)), - \(k\) is the thermal conductivity of the plate (\(W/m \cdot K\)). Plugging in the values, we get: \(q = -80 \cdot 66.67 = -5333.6 \ W/m^2\)

Find the convection heat transfer coefficient (h)

Next, we can use Newton's law of cooling to find the convection heat transfer coefficient. \(q = h \cdot (T_\mathrm{upper} - T_\mathrm{air})\) where - \(q\) is the steady-state heat transfer per unit area (\(W/m^2\)), - \(h\) is the convection heat transfer coefficient (\(W/m^2 \cdot K\)), - \(T_\mathrm{upper}\) is the temperature at the upper surface of the plate (\(°C\)), - \(T_\mathrm{air}\) is the air temperature (\(°C\)). Rearrange for h: \(h = \frac{q}{T_\mathrm{upper} - T_\mathrm{air}}\) Plug in the values: \(h = \frac{-5333.6}{50 - 10} = -159.52 \ W/m^2 \cdot K\) This means that the convection heat transfer coefficient is approximately \(-159.52 \ W/m^2 \cdot K\).

Discuss the reasonableness of the value

The calculated value for the convection heat transfer coefficient of air at the upper surface is negative, which is not physically possible in the case of forced convection. This could indicate that there is an error in the given data or assumptions made during the calculations. In normal cases of forced convection, the value of the convection heat transfer coefficient would be positive, indicating that heat is being transferred from the higher temperature surface to the cooler air. Thus, the calculated value is not reasonable for forced convection of air.