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 0.56 m radius and is 1.8 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. A)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? The Joules are 6.87 *10^12. B)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. C)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?

Explanation / Answer

given radius =0.54m...so surface area of cylinder=p1*0.54^2=0.916 m2......... b)we have Q=KA(T2-T1)/L=217*0.916*(90-27)/10^-2=1.2522 MW ....energy lost in one week=1.2522*24*7=2.1 *10^5 KWH............ c)cost for req electricity=0.1*2.1*10^5=21000$........ d)the total surface is still equal to that calculated in par t A which is 0.916 m2....... here Q along insulation=KA(T2-T1)/L=0.04*0.916*(90-27)/10^-1=23 W......... for 1 week energy=23 *24*7=3.878 KWH........... e) if o.1 $ per KWH is paid total cost =0.1*3.878=0.3878 $

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.