Question 760319: A circular swimming pool is 18.6 ft in diameter and 4.2 ft deep.
a. Determine the volume, v, of the pool in gallons if one gallon is 231 cubic inches.
b. The pool's filter pump can circulate 2500 gal per hour. how many hours, t, do I need to run the filter in order to to filter the number of gallons contained in the pool?
c. one pound of chlorine shock treatment can treat 10000 gal. how much of the shock treatment, n, should I use?
I would really appreciate if you can help me with this question, i'm stuck with it for two days now
thank you
Found 2 solutions by stanbon, rothauserc:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A circular swimming pool is 18.6 ft in diameter and 4.2 ft deep.
a. Determine the volume, v, of the pool in gallons if one gallon is 231 cubic inches.
Volume of pool = area of base * depth = (pi*9.3^2)*4.2 = 1141.21 cubic ft
Note: 1 cubic ft = 12^3 = 1728 cubic inches
231 in^3/1728 cu in^3 = 0.1337 ft^3
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# of gallons = 1141.21 ft^3/0.1337 ft^3 = 8536.84
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b. The pool's filter pump can circulate 2500 gal per hour. how many hours, t, do I need to run the filter in order to to filter the number of gallons contained in the pool?
Ans: 8536.84/2500 = 3.4147 hours
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c. one pound of chlorine shock treatment can treat 10000 gal. how much of the shock treatment, n, should I use?
x/8536.84 = 1 lb/10,000
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x = 8536.84/10,000 lb = 0.8537 lbs
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Cheers,
Stan H.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! Part a)
convert the diameter and depth of the pool to inches
diameter = 18.6 * 12 = 223.2 inches and radius is 1/2 of this = 111.6 inches
depth = 4.2 * 12 = 50.4 inches
volume (v) = pi * 111.6^2 * 50.4 = 1.972802 * 10^6 cubic inches
v (in gallons) = 1972802 / 231 = 8540 gallons
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Part b)
t = 8540 / 2500 = 3.4 hours necessary to run filter
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Part c)
n = 8540 / 10000 = .85 lbs of chlorine is necessary
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