Problem #2: You have a hot tub that will filled with 550 gallons water at a temp
ID: 702896 • Letter: P
Question
Problem #2: You have a hot tub that will filled with 550 gallons water at a temperature of 50 F. The hot tub is 4 ft in depth. You want to increase the temperature of the bathtub by heating the hot tub with a heater that produces 5.5 kW of power. Write an energy balance on the hot tub and calculate how long it will take to increase the temperature from 50 F to 103 F. (use Cpatek 4184 J/kg K) Now imagine that the hot tub is at 104 F and is losing heat to cold air above the water. The air above is 30 F and the loss of heat that can be expressed as Q-hA (T Ts) where h20 W/m2-K. Write out a new energy balance and solve for Tlt). How long will it take for the hot tub to drop to 80 F? Now include both the heater and the loss of heat. Write out an energy balance on the system and solve for the steady-state temperature (hint -there is no flow in or out so you can consider this a batch reactor but it is not at steady-state.) a. b. c.Explanation / Answer
550 gallons =2.08198 m3
depth =4 ft =1.2192m
density of water =1000kg/m3
so water in hot tub = 2.08198*1000=2081.98 kg
50oF=283.15 K
103oF=312.594 K
a)
heater produces 5.5kW of power
power supply * time =energy supply
Q*t=mCpdT
5.5*1000*t =2081.98*4184*(312.594-283.15)
t=46633.96 sec = 12.95 hr
considering no heat loss
b)
hot tub is at 104 F
the air is at 30 F
Q=hA(T-Ts) where h = 20 W/m2-K
amount of heat need to remove from water for temp dropping from 104 F (313.15K) to 80 F(299.817K)
Q=2081.98*4184*(313.15-299.817)
Q = 116143820.6J
heat transfer between water & air
HT area =volume/depth =2.08198 /1.2192=1.7076m2
30 F=272.039K
Q=hA(T-Ts)
Q=20*1.7076(313.15-272.039)
Q=1404.022W
time required to remove 116143820.6 by rate of 1404.022W
t=82722.22sec=22.97 hr
c) to answer this one you nned to consider ammount of heat supply by heater + rate heat loss by convection with air
Q net = 5.5*1000*12.95*3600-1404.022=256408596J
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