A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?

The area of the roof = 1967 ft^2

Area of water to be possibly captured = 56% of 1967 ft^2 = 0.56 x 1967 = 1101.52 ft^2

Rainfall depth = 14 inches = 14/12 ft = 1.167 ft

Volume of water = (area of water) x (depth of water) = 1101.52 ft^2 x 1.167 ft = 1285.47 ft^3

35.315 ft^3 = 1 m^3

1285.47 ft^3 = $x$ m^3

==> 1285.47/35.315 = 36.4 m^3

1 ft^3 = 7.481 gallon

1285.47 ft^3 = $x$ gallon

==> 1285.47 x 7.481 = 9616.6 gallon

Weight of water W = ρv

where ρ is the density of water = 62.4 lbs/ft^3

v is the volume of water = 1285.47 ft^3

substituting values,

W = 62.4 x 1285.47 = 80213.33 lbs

but,

1 lb = 4.45 N

therefore,

80213.33 lbs = 80213.33/4.45 = 18025.47 N

7 0

2 years ago