A 2-m internal diameter spherical tank made of 1.5-cm-thick stainless steel is u
ID: 1718138 • Letter: A
Question
A 2-m internal diameter spherical tank made of 1.5-cm-thick stainless steel is used to store iced water at 0oC. The heat of fusion of water is 333.7 kJ/kg, and the thermal conductivity of stainless steel is 15 W/(m-K). The tank is located in a room the temperature of which is 20oC. The walls of the room are at 20oC. The outer surface of the tank is black, and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients on the inner and outer surfaces of the tank are h1 = 60 W/(m2-K) and h2 = 15 W/(m2-K), respectively. Determine the total thermal resistance of the tank.Determine the amount of ice melt during a 24-hr period.
Explanation / Answer
>> There are three thermal resistances:
1). R1 = Resistance due to convection across inner surface
R1 = (1/hA)
As, A = Area = *d2/4 = *22/4 = 3.142 m2
=> R1 = (1/(60*3.142) = 5.31*10-3 .........(1).........
2). R2 = Resistance due to conduction across stainless stell tank
R2 = (L/kA)
L = Thickness = 1.5 cm = 0.015 m
=> R2 = (0.015/(15*3.142) = 0.32*10-3 .........(2).........
3) R3 = Resistance due to convection across outer surface
R3 = (1/hA)
As, A = Area = *d2/4 = *22/4 = 3.142 m2
=> R3 = (1/(15*3.142) = 21.22*10-3 .........(3).........
As, Net Resistance, R = R1 + R2 + R3
From (1), (2) and (3),
=> R = 5.31*10-3 + 0.32*10-3 + 21.22*10-3 = 26.85*10-3 = 2.685*10-2 .........
=> TOTAL THERMAL RESISTANCE = 2.685*10-2 .....ANSWER.......
>> As, Q = dT/R
=> Q = (20 - 0)/2.685*10-2 = 744.879 W
>> In 24 hr, Q = 744.879*24*3600 = 64.36 MJ
>> Now, Let 'm' amount of ice is melted in 24 hr
=> Q = mh,
where, h = heat of fusion of water is 333.7 kJ/kg
=> 64.36*106 = m*333.7*103
=> m = 192.86 Kg .......ANSWER.........REQUIRED MASS........
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