Bonus: A system of two 2000L holding tanks is set up with a pipe bringing water
ID: 3146537 • Letter: B
Question
Bonus: A system of two 2000L holding tanks is set up with a pipe bringing water to the first tank, another pipe taking water from the first tank to the second, and a final pipe draining away from the second tank, with each pipe moving 100L/hr. Usually, the system holds nothing but pure water, with the tanks kept half-full. One day (or so the story is told), an apprentice spilled a 10kg bag of salt at the mouth of the pipe feeding into the first tank. To add further injury to the original injury, the bag itself flew from his grasp and partially clogged the pipe draining from the second tank (don't ask how - it just did). If the pile of salt is being dissolved by the water at the rate of half of the remaining pile every three hours and if the drainage pipe now only carries 90L/hr, what is the amount of salt in the second tank after t hours? As usual, assume everything mixes instantlyExplanation / Answer
Here given that each pipe moving is equal to 100 Lhr
and also time t seconds
salt =10 grams
For solution
Consider 2 pipes for 3 hrs
Let A = salt in 1st tank
B = salt in 2nd tank
Now we have to find the amount of salt in the second tank after t seconds
For this first we have to consider
dA / dt = 100 [ 10 /2 (t/3)] - 90 A / [5000 + 10 t].........................................(1)
and
dB / dt = 90 A / [5000 + 10 t] - 100 B / [5000 - 10 t ].........................................(2)
By solving (1) we get A .
By solving (2) we get B
dA / dt = 100 [ 10 /2 (t/3)] - 90 A / [5000 + 10 t]
=1000 / 2t/3 - 90 A / [5000+10t]
= 20
dB / dt = 90 A / [5000 + 10 t] - 100 B / [5000 - 10 t ]
= 1000 t/3 - 100 B / [5000 -10t]
= 1000 gm
1000gm of salt solution in second tank after t hours