Consider a corn yield of 7,200 kg/hectare (equivalent to 120 bushels per acre).

Question

Consider a corn yield of 7,200 kg/hectare (equivalent to 120 bushels per acre). If 25 kg (one bushel) of corn consumed about 20m^3 of water during the growing season, what is the ratio of the weight of corn to the weight of water consumed? Where does most of the water end up? Calculate the minimum quantity of irrigation water required to grow the corn (1 hectare = 10 4 m 2 ) if the corn i s grown in Phoenix, AZ (annual precipitation 7.66 in) or in Lincoln, NE (28.4in). Clearly state any assumptions you make.

Explanation / Answer

Given,

corn yeild=7200 kg/hectare.

25 kg of corn consumed 20 m^3 of water.

20 m^3 = 20000litres =20000kg of water

so the ratio of weight of corn to weight of water is 25kg/20000kg

ratio= 1:800

Most of the water will be absorbed by the soil .

If the corn is grown in Phoenix,minimum quantity of water required is 7200 * 20/25

=5760 m^3 of water.

Annual precipitation is 7.66 in =0.1945 m (Assume for unit area= 1m^2)

then , total precipitation =(0.1945 * 10000)

=1945m^3 of water.

so irrigation water required = Total water required - precipitation

=(5760- 1945)m^3

=3815 m^3 of irrigation water.

If the corn is grown in Lincoln,

the precipitation= 28.4 inches = 0.7213m

then, total precipitation=0.7213 * 10000

=7213 m^3 of water

so water required is only 5760 m^3 so irrigation water is not required.

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