A swimming pool is heated by a gas-fired heater that supplies 200,000 Btu/hr to

Question

A swimming pool is heated by a gas-fired heater that supplies 200,000 Btu/hr to the water. The pool is rectangular, 15 ft by 30 ft, with an average depth of 5 ft. The water circulating system keeps the water temperature quite uniform, even while the heater is operating. The heat loss from the pool has been found to be approximately equal to 5000 Btu/hr for each degree Fahrenheit difference between the temperature of the water and that of the ambient air. Write the differential equation that describes how the water temperature varies with time when the heater is turned on and the ambient air temperature is 50 degrees Fahrenheit.

Explanation / Answer

Keep in mind that Btu's are units of energy, E.

dEin/dt = 200,000

Vpool = 15*30*5 = 2250 ft3

dEout = 5000*(Tpool - Tambient)

Tambient = 50 °F

Note that we are not given the temperature of the pool explicitly.  We must derive this from the energy in the pool.  However, we can find this by the relation,

Q = mcT

where Q is the energy in the water, m is the mass of water, c is the specific heat (4.166 J/(g*K) * 1055 Btu/J * 1 K/1.8 °F = 2442 Btu/(g*°F)), and T is the change in temperature.  We can obtain the mass of the pool by assuming the water has a density of 1 g/cm3.  

m = V = (1 g/cm3)(28317 cm3/1 ft3)(2250 ft3) = 6.37e7 g

When we multiply this by c and plug in, we obtain,

Q = (2442*6.37e7)T = (1.56e11)T

Let's define T to be Tpool - Tambient = Tpool - 50.  And we can say that Q = E since they are both measuring the same energy.

E = (1.56e11)(Tpool - Tambient)

Or,

Tpool - Tambient = E/1.56e11

If we say 

dE/dt = dEin/dt - dEout/dt,

Then,

dE/dt = 200000 - 5000*(Tpool - Tambient)

dE/dt = 200000 - 5000*E/1.56e11

dE(t)/dt = 200000 - 3.21e-8*E(t), E(0) = 0

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