Question

The original well in a confined aquifer is pumped at a rate of 220 gal/min. Measurement of drawdown in two observation wells shows that after 1270 min of pumping, no further drawdown is occurring. Well 1 is 26 ft from the pumping well and has a head of 29.34 ft above the top of the aquifer. Well 2 is 73ft from the pumping well and has ahead of 32.56 ft above the top of the aquifer. Use the Thiem equation to find the aquifer transmissivity. We must first convert the pumping rate of 220 gal/min to an equivalent rate in cubic feet per day. We make this conversion, even though steady-state conditions were reached, before a full day (1440 min) of pumping occurred

Answer 1

Transmissivity is
`186.96* 10^6 ft^2/sec`

Step-by-step explanation:

So the well 1 has

r1=26ft

h1=29.34ft

The well 2 has

r2=73 ft

h2=32.56 ft

Converting the rate of flow from gal/min to ft3/day

`Q=220(gal)/(min)*(1 ft^3)/(7.48 gal)*(1440 min)/(1 day)\\Q=42400(ft^3)/(day)`

Now the transmissivity is is given as

`T=\frac{\dot{Q}}{2\pi(h_2-h_1)}ln((r_2)/(r_1))`

Now by substituting values

`T=\frac{\dot{Q}}{2\pi(h_2-h_1)}ln((r_2)/(r_1))\\T=(42400)/(2\pi(32.56-29.34))ln((73)/(26))\\T=(42400)/(6.44\pi )\left(\ln \left(73\right)-\ln \left(26\right)\right)\\T=2164 ft^2/day`

Now converting it to ft^2/sec

`T=2164 ft^2/day\\T=186.96* 10^6 ft^2/sec`

Transmissivity is
`186.96* 10^6 ft^2/sec`