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)?

V = 36.4 m³ = 4.86 gallons

W = 80193.88 lbf = 356720 N = 356.72 KN

Explanation:

We have the following data given in the question:

At = Total area of roof =  1967 ft²

h = Annual Rainfall = 14 inches = 1.17 ft

V = Volume of tank in m³  and gallons = ?

W = Weight of water in N and lbf = ?

So, for volume we know that the area of roof that receives rainfall is 56% of total area and 14 inches of annual rainfall means  that there is a standing height of 14 inches of rain water for a given area, for 1 year.

Area to receive rain = A = 0.56*1967 ft² = 1101.52 ft²

Now,

Volume = V = A * h = 1101.52 ft²)(1.17 ft)

V = 1285.11 ft³

Converting to m³:

V = (1285.11 ft³)(1 m³/35.3147 ft³)

V = 36.4 m³

Converting to gallons:

V = (1285.11 ft³)(1 m³/264.172 gal)

V = 4.86 gal

Now, for the weight of water, we use formula:

W = ρVg

where,

W = weight of water = ?

ρ = Density of water = 1000 kg/m³

V = Volume of tank = 36.4 m³

g = 9.8 m/s²

Therefore,

W = (1000 kg/m³)(36.4 m³)(9.8 m/s²)

W = 356720 N = 356.72 KN

Converting to lbf:

W = (356720 N)(1 lbf/4.44822 N)

W = 80193.88 lbf

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