Question

Problem One (33%) – water usage calculation in a confined aquifer

A confined aquifer with a storativity of 7.8 x 10^-5 extends over an area of 50 km². During the wet season the aquifer has been recharged and the storage in that area has been increased by 8 ML.

1.      How much has the potentiometric surface raised?

2.      If the aquifer was unconfined with a specific yield of 18%, what would be the volume recharged (in ML) for the same amount of raise in the water table if we consider that the specific retention is effectively 0%?

Answer 1

(a) For confined Aquifer

Storativity is the volume of wate released from storage per unit surface area per unit decline in the hydraulic head.

volume of water increased in the storage = 8 Million litte = 8000 m^3

Surface area, A = 50 KM^2 = 50*10^6 m^2

Storativity = volume of water added/(surface area*increased in hydraulic head)

7.8*10^-5 = 8000/(50*10^6 * dh)

Increased in hydraulic gradient, dh = 8000/3900 = 80/39 m = 2.051 m

So, Potentiometric head raised = 2.051m (Ans)

(b) For unconfined aquifer

Specific yeild is the amount of water released from unconfined aquifer per unit surface area per unit decline in hydraulic head.

Specific retention=0% ie there is no retentive water holded by the soil and all the water is yeilding.

Specific yeild =18%

Surface area, A = 50 KM^2=50*16^6 m^2

Potentiometric head raise= 80/39 m

Specific yealid, 0.18= Volume recharged/(Surface Area* Potentiometric head raised)

0.18 = Volume reachrged/(50*10^6*80/39)

Volume recharged = 18.4615*10^6 m^3 (Ans)