Question

Consider a steel pan used to boil water on top of an electric range. The bottom section of the pan is \(L=0.5 \mathrm{~cm}\) thick and has a diameter of \(D=20 \mathrm{~cm}\). The electric heating unit on the range top consumes $1250 \mathrm{~W}$ of power during cooking, and 85 percent of the heat generated in the heating element is transferred uniformly to the pan. Heat transfer from the top surface of the bottom section to the water is by convection with a heat transfer coefficient of \(h\). Assuming constant thermal conductivity and onedimensional heat transfer, express the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation. Do not solve.

Short answer

The mathematical formulation of the heat conduction problem during steady operation in a steel pan on top of an electric range is given by the differential equation: $$ \frac{dT}{dx} = -\frac{h}{k}(T_{s} - T_{w}) $$ with the following boundary conditions: 1. At the bottom surface of the pan, \(x = 0\): $$ T(0) = T_{bottom} $$ 2. At the top surface of the pan, \(x = L\): $$ T(L) = T_{s} $$

Step-by-step solution

Heat transfer rate calculation

First, let's calculate the heat transfer rate (energy per unit time) from the electric range to the pan. The heating element transfers \(85\%\) of the power consumed. Therefore, the heat transfer rate, \(q_{in}\) is given by: $$ q_{in} = 0.85 \times 1250 = 1062.5\,\text{W} $$

Energy balance at the bottom section of the pan

During steady operation, the amount of energy entering the bottom section of the pan is equal to the amount of energy leaving it. So, the heat conduction through the thickness of the pan (`L`) equals the convection heat transfer at the top surface from the pan to the water. We will use Fourier's law, which states that the rate of heat conduction through a solid material depends on the temperature gradient and the thermal conductivity of the material: $$ q_{cond} = -kA\frac{dT}{dx} $$ Here, \(q_{cond}\) is the heat conduction rate, \(k\) is the thermal conductivity, \(A\) is the cross-sectional area (\(\pi\frac{D^2}{4}\)), and \(\frac{dT}{dx}\) is the temperature gradient. Next, we will use Newton's law of cooling, which describes the rate of heat transfer due to convection: $$ q_{conv} = hA(T_{s} - T_{w}) $$ Here, \(q_{conv}\) is the convection heat transfer rate, \(h\) is the heat transfer coefficient, \(T_s\) is the temperature at the top surface of the pan, and \(T_w\) is the temperature of the water. Since the heat conduction through the thickness of the pan equals the convection heat transfer at the top surface, we have: $$ q_{cond} = q_{conv} $$ Combining Fourier's law and Newton's law of cooling: $$ -kA\frac{dT}{dx} = hA(T_{s} - T_{w}) $$ This is the equation that we are interested in.

Differential Equation and Boundary Conditions

Now let's express the mathematical formulation of this heat conduction problem during steady operation. We will write down the differential equation and the boundary conditions separately. Differential Equation: The differental equation we derived above is: $$ -kA\frac{dT}{dx} = hA(T_{s} - T_{w}) $$ This equation describes the heat transfer through the thickness of the pan and can be simplified for steady-state: $$ \frac{dT}{dx} = -\frac{h}{k}(T_{s} - T_{w}) $$ Boundary Conditions: We need two boundary conditions for the problem, which can be expressed as follows: 1. At the bottom surface of the pan, \(x = 0\): $$ T(0) = T_{bottom} $$ 2. At the top surface of the pan, \(x = L\): $$ T(L) = T_{s} $$ We have successfully provided the mathematical formulation (the differential equation and the boundary conditions) of this heat conduction problem during steady operation, as required by the exercise.