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Consider a pan placed on top of electric range to cook spaghetti. The bottom section of the pan is 0.4 cm thick and has a diameter of 18 cm. The electric heating unit on the range top consumes 800 W of power during cooking, and 80 percent of the heat generated in the heating element is transferred uniformly to the pan. Assuming constant thermal conductivity and steady state operation. The heat transfer can be approximated as being one- dimensional through the bottom surface to the top. Take the direction normal to the bottom surface to be the x-axis.

Required:

a. Perform an energy balance on a thin element of the pan during a small-time interval △t , and indicate which terms may be neglected or ignored and why?

b. Assuming constant thermal conductivity, obtain the differential equation that describes the variation of the temperature in the bottom section of the pan during steady state operation.

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a. To perform an energy balance on a thin element of the pan during a small-time interval Δt, we consider the heat transfer and energy generation terms. Neglecting any potential heat loss to the surroundings, the energy balance equation for the thin element can be written as:

Energy in - Energy out = Energy generated,

where the terms can be expressed as:

Heat transferred into the element: Qin = kAΔT/Δx,

where k is the thermal conductivity, A is the cross-sectional area of the thin element, ΔT is the temperature difference across the element, and Δx is the thickness of the thin element.

Heat transferred out of the element: Qout = kAΔT/Δx,

where the same terms apply as in Qin.

Energy generated: Qgen = PΔt,

where P is the power consumed by the electric heating unit, and Δt is the time interval.

In this case, neglecting any potential heat loss to the surroundings is reasonable, assuming that the pan is well-insulated.

b. Assuming constant thermal conductivity and steady-state operation, the differential equation that describes the variation of the temperature in the bottom section of the pan can be obtained by equating the heat transferred into the element to the energy generated.

Since the heat transfer is one-dimensional through the bottom surface, we can express the thermal conductivity as k = (Qin - Qout) / (AΔT/Δx), where ΔT/Δx is the temperature gradient across the thin element.

Substituting the values for Qin, Qout, and Qgen, we have:

(kAΔT/Δx - kAΔT/Δx) = PΔt.

Simplifying the equation, we get:

0 = PΔt,

which indicates that the energy generated by the heating unit is balanced by the heat transfer within the pan.
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