Please solve the following correctly. The following equations and conversion fac
ID: 287821 • Letter: P
Question
Please solve the following correctly.
The following equations and conversion factors may be useful: Conversions: 3.785 liters/gallon7·48 gallons/cubic foot; 3.28 feet/meter 2.30 225T 4TAS Cooper Jacob equations Transmissivity-K*m. Darcy velocity q-Ki: Well efficiency (%)-theoretical s measured s)*100 T-transmissivity, S-storativity, Q pumping rate: s-drawdown (delta s is drawdown per log cycle)r-distance from pumping well; to-time intercept where s-0; h hydraulic head; m-saturated aquifer thickness Problem 1 (10 pts). A public water supply well is pumped into a reservoir at a steady and continuous rate of 200 liters per minute for a total of 24 hours. The reservoir can store a maximum of 500,000 gallons and had 150,000 gallons in it before the well started pumping. The town withdrew its daily average of 99,000 gallons per day during the time the well was pumping. The average daily evaporation rate is 5,000 gallons from the reservoir, which occurred fully during the pumping period. How many more gallons of water could be pumped to fill the reservoir when the well stopped pumping?Explanation / Answer
The public water well supplies a total of 200*60*24 litres of water per day, which is equal to 76089.82 gallons.
So the total water in the resevoir should be 150000+76089.82 gallons. But in the same time there is a outflux of water which equals to 99000+5000 gallons of water per day.
Therefore at the end of the day the water level in the reservoir is 226089.82 - 104000. Which is equal to 122089.82 gallons of water.
Therefore the number of gallons that can be pumped into the reservoir when the well stopped pumping is 500000 - 122089.82 gallons. which is equal to 377910.18 gallons of water.