A water treatment plant is built with three cylindrical tanks to contain the water for a town. Each tank has a radius of 15 feet and a depth of 25 feet and there are about 7.5 gallons in a cubic foot of water.
Engineers know that the amount of water lost due to evaporation is directly proportional to the surface area of the holding tanks and that the local climate causes water to evaporate at the rate of about 1/10 gallons of water per hour for each square foot of water exposed to the air. About how many total gallons of waste water may enter the three tanks each day to maintain full capacity?
A. 2.1 x 10^2
B. 1.7 x 10^3
C. 5 x 10^3
D. 5 x 10^4
2006-08-12 18:09:23 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Ok,
The area of each tank is pi*15^2 (square feet)
Water loss rate is (1/10 gallon/(hr x sq ft)) x 24 hr/day or 2.4 gal/sq ft/day
Water loss from each tank is 2.4 * pi * 15^2 = 1.7 x 10^3
Since there are 3 tank 3 x 1.7 x 10^3 = 5 x 10^3 gallons/day
2006-08-12 18:34:02 · answer #1 · answered by Roadkill 6 · 0⤊ 0⤋
The total area exposed is 3 x 15^2 x pi = 2.1*10^3 sq ft.
The evaporation rate is .1 gal per sq ft per hr, or 2.1*10^2 gal per hour for the 3 tanks. One day has 24 hours, so the replacement rate for the three tanks is 2.1 x 24 x 10^2 gal/day or 5 x 10^3 gal per day,
Answer C
2006-08-12 18:26:52 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
This ain't Yahoo! Tests
2006-08-12 18:15:36 · answer #3 · answered by somedude 1 · 0⤊ 0⤋
c
2006-08-12 18:26:19 · answer #4 · answered by Anonymous · 0⤊ 0⤋
why r u so smart, and I'm not?
2006-08-12 18:41:16 · answer #5 · answered by The "Spence" 2 · 0⤊ 0⤋