Question

A well 3 meters in diameter has its normal water level 3 meters below the ground level. By pumping water level in the well is depressed to 10 meters below the ground level. In 4 hours the water rises by 5 meters. Calculate the specific yield of the well.

(a) 2.213 m^3/hr

(b) 5 m^3/hr

(c) 1.242 m^3/hr

(d) 3.224 m^3/hr

The question was asked during an online exam.

My question comes from Open Well and Dug Well in division Ground Water for Irrigation through Wells and Tubewells of Irrigation Engineering

Answer 1

Correct answer is (a) 2.213 m^3/hr

The explanation is: Specific yield (K) is calculated from the formula, K = 2.303 (A/T) log (H1/H2) where A is the area of the well, T is the total time of recuperation to bring the water level from depth H1 to H2.

Given values, A = 3.14/4 x 9 = 7.065 sq. m, T = 4 hours, H1 = 7m, H2 = 2m

K = 2.303 x 7.065 x 1/4 x log (7/2) = 2.213 m^3/hr under a head of one meter.