Question #278245

The flat base of the bottom of an electric water heater is made of a 2 mm thick steel plate. thermal conductivity of 40 W/(m. oC), and an insulating plate with a thickness of 4 mm and thermal conductivity of 0.06 W/(m.oC). Between the steel and insulating plates there is an electrical resistance (RE) that dissipates a power of 800 W.

Consider the steady state situation with boiling water at a temperature of 100 oC and with a coefficient of heat transfer by convection of 3000 W/(m2. oC), while the ambient air, which is in contact with the insulator, it remains at a temperature of 25oC and a convection heat transfer coefficient of 10 W/(m2.oC).

Considering a perfect thermal contact between the electrical resistance and the steel and insulating sheets and that the base of the heater has an area of 0.018 m2, determine:

a) The temperature (TR) of the electrical resistance

b) The rate of heat lost to the air through of the insulating plate

ç). The rate of heat lost to water through the steel plate

Expert's answer

Heat transfer occurs mainly by conduction, convection, and radiation. In conduction, the energy from one end to another is transferred without the actual transfer of matter.

Conduction mostly takes place in solids. Metals have a large number of free electrons. These free electrons are responsible for carrying energy from one end to another. Heat energy is transferred from the hotter end to the colder end.

The amount of heat transferred between a material with its two ends at two different temperatures T_{1} and T_{2} is given by,

"dQdt=kA(T1\u2212T2)l\nd\nQ\nd\nt\n=\nk\nA\n(\nT\n1\n-\nT\n2\n)"

k is the thermal conductivity of the material.

A is the area of the cross-section of the material.

l is the length of the material.

T_{1}=60°*C*

T_{2}=35°C

l= 220 mm=0.22m

Given k=0.51 w/m°*C*

Hence, the rate of heat transfer per m^{2} of the surface area of the wall is 57.954 W/m^{2}.

Learn more about our help with Assignments: Chemical Engineering

## Comments

## Leave a comment