Question

Suppose you have an electric hot water heater for your house which is an aluminum cylinder which has a 0.56 m radius and is 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.

Part A:

What is the surface area of the cylinder?

Part B:

How much energy is lost through the walls of the hot water heater in one week? (Assume thinner surface of the heater is 90 C and the outer surface is 27 C.)

Part 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?

Part D:

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.

Part E:

Assume you pay $0.10 per kW-hour for electricity. How much would it cost just to keep the hot water inside the fiberglass-wrapped heater for one week?

Answer 1

A) Surface area including top and bottom = 2 ? R H + 2 ? R^2
= (2 x ? x 0.56 x 1.4) + (2 x ? x 0.56 x 0.56)
= 4.926 + 1.970
= 6.896 m^2 = surface area

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

Q = 217 x 6.896 x (90-27) / 0.01 = 9427522 Watts
Energy = power x time
= 9427522 x 3600 x 24 x 7 (seconds in 1 week)
= 5.702 x 10^12 Joules per week

C) 1KWh = 1000watts x 3600 sec = 3.6 x 10^6 Joules
Energy loss in KWh:
= 5.702 x 10^12 / 3.6 x 10^6
= 1.584 x 10^6 KWh

Cost = $0.10 x 1.584 x 10^6
= $158389

D) 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 5.702 x 10^12 / 54251
= 105.1 MJ per week

E) Cost will be $158389 / 54251
= $2.92