These questions build up on each other 1. Consider the Mississippi River drainag

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These questions build up on each other

1. Consider the Mississippi River drainage basin. The area of the basin is 3x10 km, the average discharge is 17,000 m' s1. Average precipitation in the basin is 800 mm/yr. First, what is the volume of discharge (km') in a year? Answer: 2. Convert the volume of runoff in a year to a "depth of runoff' in a year averaged across the basin. Your answer should be in meters per year Answer: 3. Estimate ET in the Mississippi River basin, using your calculations of P and discharge. Answer in terms of volume of ET (km3) and in termns of depth of water per year (m). Answer volume Answer depth: 4. What percentage of precipitation in the basin returns to the atmosphere via ET? Is this higher or lower than the global average? Answer

Explanation / Answer

Area of basin = 3x 106 km2

Average discharge =17,000 m3s-1

Average precipitation = 800mm/year

1

Volume of discharge = 17,000m3s-1 60x60x24x365 = 5.36112x1011m3/year

1m3=10-9km3, therefore 5.36112x1011m3/year = 536.112km3/year

2

Volume of runoff = (536.112km3/year)/(3x 106 km2)= 1.78707 x 10-4km/year = 0.178707 m/year

3

Average precipitation = 800mm/year = 800 x 10-6km/year = 8 x 10-4km/year

In volume = 8 x 10-4km/year x Area of basin = (8 x 10-4km/year) x (3x 106 km2) = 2400km3/year

Volume of discharge = 536.112km3/year

Average volume of Evapotranspiration per year = (2400- 536.112) = 1863.888km3

Average depth of Evapotranspiration per year = (1863.888km3/3x 106 ) = 6.21296x 10-4km = 6.21296x 10-1m

4

Average depth of Evapotranspiration per year = 6.21296x 10-1m = 621.296mm

Average precipitation = 800mm/year

Percentage of precipitation in the basin returns to atmosphere via ET = (621.296/800)x100 = 77.662%

Global average of pecipitation is 990 millimetres our's is 621.296mm which is less than global average.

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