Question

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.

Short answer

Answer: The heat flux between the surface of the pond and the air is \(400\mathrm{\ W/m^2}\).

Step-by-step solution

1: Understand given parameters

We have the following information: - Air temperature: \(T_\text{air} = 20^{\circ}\mathrm{C}\) - Convection heat transfer coefficient: \(h = 20\mathrm{\ W/m^2K}\) - Surface temperature of the pond: \(T_\text{surface} = 40^{\circ}\mathrm{C}\)

2: Determine the temperature difference

We need to calculate the temperature difference between the surface of the pond and the air. We can find it as follows: \(\Delta T = T_\text{surface} - T_\text{air}\)

3: Perform the calculation of the temperature difference

Use the given temperature values to find the temperature difference: \(\Delta T = 40^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C} = 20^{\circ}\mathrm{C}\)

4: Apply the convection heat transfer equation

Now we will apply the convection heat transfer equation, which is given by: \(q = h \times A \times \Delta T\) However, since we are interested in the heat flux, we need the heat transfer rate per unit area. Thus, we will use the following form of the equation: \(q'' = h \times \Delta T\)

5: Calculate the heat flux

We can now calculate the heat flux using the given convection heat transfer coefficient and the calculated temperature difference: \(q'' = 20\mathrm{\ W/m^2 K} \times 20^{\circ}\mathrm{C} = 400\mathrm{\ W/m^2}\) The heat flux between the surface of the pond and the air is equal to \(400\mathrm{\ W/m^2}\).