Answer to Question #159018 in Physics for besho

Question

Hydraulic engineers in the United States often use, as a units of volume of water, the acre-foot, defined as the volume of water that will cover 1 acre (where 1 acre = 43560 ft²) of land to a depth of 1 ft. A severe thunderstorm dumped 2.1 in. of rain in 30 min on a town of area 43 km². What volume of water, in acre-feet, fell on the town?

Answer 1

By definition, one acre-foot is 43560 ft³. Since one ft is approximately 0.0283 m^3, then one acre-foot is:

$$1\space \text{acre-foot} = 43560\times 0.0283m^3 \approx 1233m^3$$

The volume of water that covered the city is (converted in and km into meters):

$$V = 2.1\space in\times 43km^2 \approx 2.1\cdot 0.0254m\times 43\cdot 10^6m^2 = 2293620m^3$$

In acre-feet it will be:

$$\dfrac{2293620}{1\space \text{acre-foot}} = \dfrac{2293620}{1233}\space \text{acre-feet} \approx 1860\space \text{acre-feet}$$

Answer. $1860\space \text{acre-feet}$.