15.21 An island has an area of 9.5 square miles. the annual net precipitation is

Question

15.21 An island has an area of 9.5 square miles. the annual net precipitation is 30 inches. The infiltration rate is 8%, and the porosity of the island formation is 10%. (a) What is the safe maximum withdrawal of groundwater in ft3, acre-ft, and gallon (without groundwater depletion)? (b) if one person one day consume 150 gallon water, how many people can this amount of water support, in the unit of person-days. (c) in one day, what is the maximum number of people per sq. mi? per acre? (d)if the water table is lowered 1 in by over-pumping, how much water does this represent? in gallon, and in percentage of the "safe" maximum allowable consumption? and number of people? (e) One inch draw down of freshwater induce how much raising of the saltwater in the ground? 1 mile=5280 ft 1 sq. Mi.=5280x5280 sq. ft. (ft2) = 640 acre 1 cubic foot=7.48 gal 1 ft3 = 7.48 gal

Explanation / Answer

Area of an island = 9.5 sq mile = 9.5*63360^2 => 3.8* 10^10 sq inches

Let assume that layer (upto the infiltration level) be homogeneous and isotropic then,

Volume of water coulumn having inflitration = (3.8*10^10)* 30 * 8/10 * 10/100 => 9.1 * 10^11 cubic inch = 14999080565.14 ltr => 3962337670.28 =>3.9 * 10^9 gallons of water

A) estimated yield/recharge percentages ratio, when baseflow is reduced to 0, which is clearly the least conservative assumption, varied in the range 20-38% (by todd), thus, safe yielding or withdrawl of water can be

=9.6*10^8 gallon for annual basis

Per day=9.6*10^8/365 = 2.6 * 10^6 gallon

B) per person one day consumption is 150 gallon per day

then no. of person can use this water = 18,093 approx

C) per day,the maximum number of people per sq. mi= 18,093 / 9.5 =>1905 approx

D) This question is subjective and most of its sense has been covered in B option. if the value of lowering of level or volume of overdraft was given, it might be possible to solve it.

E) If ground water level already had attained an equilibrium with hydrostatic pressure, the reduce in level will cause to intrude saline water fron sea having exactly same volume of water that has been overdrafted.

thus, volume of 1 inch width is= 3.8*10^10)* 8/10 * 10/100= 1.9 *10^9 cubic inches => 132077922.3=> 1.3*10^8 gallon of water (if we assume same density/ volume of both the water).

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