Suppose you have an electric hot water heater for your house which is an aluminu

Question

Suppose you have an electric hot water heater for your house which is an aluminum cylinder which has a 0.58 m radius and is 1.2 m high. The walls are 1.0 cm thick. The thermal conductivity of Aluminum is 217 W/(m K). Assume that the temperature of the hot water inside the hot water heater is kept at a constant 90 C, and the external temperature is 27 C. Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the heater for one week? Suppose that you wrap the hot water heater on all sides with a 10 cm thick blanket of fiberglass insulation which has a thermal conductivity of 0.04 W/(m K). Assume the inner surface of the fiberglass insulation is at 90 C and the outer surface is at 27 C, and the total surface area is still what you calculated in part A.

Explanation / Answer

Surface area including top and bottom = 2 R H + 2 R^2
= (2 x x 0.58 x 1.2) + (2 x x 0.58 x 0.58)
= 6.487 m^2 = surface area

Q = k A /w
where Q is power loss (Watts)
k = conductivity; A = surface area; = temperature difference; w = wall thickness

Q = 217 x 6.487 x (90-27) / 0.01 = 8868377 Watts
Energy = power x time
= 8868377x 3600 x 24 x 7 (seconds in 1 week)
= 5.364 x 10^12 Joules per week

Conductivities are in the ratio of 217 / 0.04 i.e 5425:1
Also, thickness of fiberglass is 10 times that of aluminium.
Heat loss will be in the ratio 54250 : 1 since 'A' and ' ' have remained constant

Heat loss will be 5.364 x 10^12 / 54251
= 98.88 MJ per week

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