Question

The dark surface of a ceramic stove top may be approximated as a blackbody. The "burners", which are integral with the stove top, are heated from below by electric resistance heaters. Consider a burner of diameter D = 200 mm operating at a uniform surface temperature of Ts = 250°C in ambient (environment) air at T? = 20°C. Without a pot or pan on the burner, what are the rates of heat loss by radiation and convection from the burner? If the efficiency associated with the energy transfer from the heaters to the burners is 90%, what is the electric power requirement?

Answer 1

To calculate the rates of heat loss by radiation and convection from the burner, use the formulas Q = εσA(Ts^4 - T?^4) and Q = hA(Ts - T?) respectively. The electric power requirement can be calculated using P = Q/η.

Step-by-step explanation:

To calculate the rates of heat loss by radiation and convection from the burner, we need to use the Stefan-Boltzmann law for radiation and Newton's law of cooling for convection.

For radiation, we can calculate the rate of heat loss using the formula:

Q = εσA(Ts^4 - T?^4)

where Q is the rate of heat loss, ε is the emissivity of the burner (assumed to be 1 for a blackbody), σ is the Stefan-Boltzmann constant (5.67x10^-8 W/m^2 K^4), A is the surface area of the burner, Ts is the surface temperature of the burner, and T? is the ambient temperature.

For convection, we can calculate the rate of heat loss using the formula:

Q = hA(Ts - T?)

where Q is the rate of heat loss, h is the heat transfer coefficient for convection, A is the surface area of the burner, Ts is the surface temperature of the burner, and T? is the ambient temperature.

To calculate the electric power requirement, we can use the formula:

P = Q/η

where P is the power requirement, Q is the rate of heat loss, and η is the efficiency of the energy transfer from the heaters to the burners.