A sequence of three identical barrels each contain 1 m^3 of water. The barrels a
ID: 3142697 • Letter: A
Question
A sequence of three identical barrels each contain 1 m^3 of water. The barrels are connected in series at the base so that the first drains into the second, the second drains into the third, and the third is allowed to drain out. It takes one hour for all three barrels to drain completely. 1. How much water (in m^3) is in each barrel after t hours (0 lessthanorequalto t lessthanorequalto 1)? 2. At what rate (in m^3/hr) is the water flowing out of the third barrel? The second? The first? 3. If there is initially 1 kg of salt mixed into the first barrel, find the amount of salt in the first barrel after t hours. 4. Find the amount of salt in the second barrel after t hours. 5. Find the amount of salt in the third barrel after t hours. 6. Find the concentration of salt (in kg/m^3) in each barrel after t hours. 7. Which barrel has the lowest concentration of salt? What happens to the concentration as t rightarrow 1? Assume there are n identical barrels connected in series in this way, with 1 kg of salt initially in the first barrel. If it takes one hour for all of the barrels to drain, show that the concentration of salt in the k^th barrel is given by C_k(t) = t^k-1.Explanation / Answer
1)
After k hour,
complete water of barrel 3 drains out
complete water of barrel 2 drains into barrel 3
complete water of barrel 1 drains into barrel 2
After 2k hour
complete water of barrel 2 which was in barrel 3 drains out
complete water of barrel 1 which was in barrel 2 drains into barrel 3
After 3k hour
complete water of barrel 1 which has reached to barrel 3 drains out
so draining rate is 3 m3 per hour
water in barrel 1 = 1 - 3t if t <1/3 hours & 0 if t > 1/3 hours
water in barrel 2 = 1 if t < 1/3 hours & 2 - 3t if t > 1/3 hours
water in barrel 3 = 1 if t < 2/3 hours & 3 - 3t if t > 2/3 hours
2)
barrel 1 = 3 if t <1/3 hours & 0 if t > 1/3 hours
barrel 2 = 0 if t < 1/3 hours & 3 if t > 1/3 hours { due to drain in and drain out being same for t < 1/3 hrs}
barrel 3 = 0 if t < 2/3 hours & 3 if t > 2/3 hours
3)
salt in barrel 1 = 1 - 3t if t <1/3 hours & 0 if t > 1/3 hours
4)
salt in barrel 2
when t < 1/3 hours
total influx = 3t
total outflux = 3t * 3t/2
so salt remained = 3t - 4.5t2
when 1/3 < t < 2/3 hours
so salt remained = (3t - 4.5t2) * 3 * (t-1/3)
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