A hot water heater, similar to those found in residences, has an inside diameter
ID: 1826986 • Letter: A
Question
A hot water heater, similar to those found in residences, has an inside diameter of 30 inches with a vertical height of 5 feet. The average water temperature over time is estimated to be 140?F while the average surrounding air temperature is 70?F. Assume that there is no heat loss through the bottom of the tank and that radiation losses are negligible. Also assume that the convection heat transfer coefficient on the inside of the tank is 4 Btu/h-ft2 ?F while on the outer surface, it is 2 Btu/h-ft2-F. Consider the wall and top of the tank to be mild steel (k=26 Btu/h-ft-?F), 1/16 inch in thickness. What is the total heat loss per hour? If the burner on the heater is 65% efficient in transferring heat to the water in the tank (35% of incoming energy to the burner is lost through the exhaust stack), what is the cost per day of maintaining the water at 140?F when natural gas costs $5.50/MCF? MCF stands for 1000 ft3. Each cubic foot of natural gas contains approximately 1000 Btu.Explanation / Answer
Inner dia = 30 - 2*1/16 in = 29.875 in
Resistance due to convection on inner dia is R_cv1 = 1/(h_i*2*pi*r_i*l) = 1/(4*2*3.14*(30/2/12)*5) = 1/157 = 0.00637
Resistance due to convection on outer dia is R_cv2 = 1/(h_o*2*pi*r_o*l) = 1/(2*2*3.14*(29.875/2/12)*5) = 0.0128
Resistance due to convection from wall = ln(r_o/r_i) / (2*pi*k*l) = ln(30/29.875) / (2*3.14*26*5) = 5.114*10^-6
Total resistance through the walls = 0.00637 + 0.0128 + 5.114*10^-6 = 0.01917
Total heat loss Q = Temp. difference / R = (140 - 70) / 0.01917 = 3651.97 Btu/hr
Energy to burner = 3651.97/0.65 = 5618.4 Btu/hr
Fuel energy = 1000 Btu/ft3
Fuel required = 5618.4 / 1000 = 5.6184 ft3/hr = 5.6184*24 ft3/day = 134.84 ft3/day
Fuel cost = $ 5.5 / 1000 ft3
Cost of fuel per day = 5.5/1000 *134.84 = 0.7416 $
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