Problem 4: Chilled water is being kept in a cylindrical tank with radius 1.5m an
ID: 2073623 • Letter: P
Question
Problem 4: Chilled water is being kept in a cylindrical tank with radius 1.5m and height 5m. The tank is well-insulated at the bottom (it is oriented such that one of its flat sides is laying on the ground). You measure the surface temperature of the tank to be 5C. If the convective heat transfer coefficient is 20W/(m2K), and the average ambient temperature is 18C, find: (a) The heat leakage, q, into the tank, in Watts. (b) The amount of energy leaked into the tank, Q, in a one-year period (assume 365 days per year), in Joules. (c) To operate the chilled water system you must power a water chiller with an estimated coefficient of performance () of 3.0. The electrical energy required to do this is therefore Q/. If the cost of electricity is $0.095/kWh, how much does this un-insulated tank cost to operate, per year? (d) You decide to put an insulating blanket around the tank, reducing heat leakage by 95%. The blanket costs $1500. Is this a wise investment? Explain your answer.
Explanation / Answer
Cylinder radius = 1.5m = diameter = D=3 m.
Height =L= 5 m.
Its bottom face is well insulated, so no heat leak from the bottom surface. Heat leak is only from top portion and circumference of the cylinder.
Effective heat transfer area =A=0.785*D2 + 3.14*D*L = 0.785*3*3+3.14*3*5 = 54.165 m2.
Convective heat transfer coefficient = h = 20 W/m2K.
Cylinder surface temperature = Ts = 5 C.
Ambient temperature = T0 = 18 C.
a)
Het leakage into the tank = Q = hA*(T0 - Ts) = 20*54.165*(18-5) = 14083 W. =
b)
Amount of heat leak into the tank in one year period = 14083*365*24*60*60=4.44*1011 J.
c)
Chiller coefficient of performance = C.O.P = Desired effect/work input = 3.
Electrical energy required to do this operation = W = Desired effect/3 = 14083/3 = 4694.3 W = 4.6943 KW.
The cost of electricity = $ 0.095 / KWh.
Heat to be removed from the tank in one year in KW-h = 14083*365*24 = 123367080 W-hr = 123367.08 kW-hr
The total cost in one year = 123367.08*0.095 = $ 11719.8 .
d)
Due to the blanket insulation, the heat loss is reduced by 95%.
Heat loss into cylinder = 0.05*14083=704.15 W.
The blanket cost = $ 1500.
Work required to remove heat from cylinder = 704.15/3 = 234.7 W = 0.2347 KW.
Amount of heat to be removed from cylinder =0.2347*365*24 = 2056.118 KW-hr.
Annual cost = 2056.118*0.095 = $ 195.33.
Annual cost = $ 195.33.
When we compare the annual electrical cost for without insulation and with insulation, it is concluded that with insulation, there is a drastic reduction in electrical charges, thereby significant financial savings are possible. Hence I strongly recommend insulation.