Two tanks A and B, are connected by a tube which runs from the bottom of A to th
ID: 2962138 • Letter: T
Question
Two tanks A and B, are connected by a tube which runs from the bottom of A to the top of B (the top of tank B is below the bottom of tank A). Initially each tank contains 100L of pure water. If a brine solution containing 1 gm/L of salt enters the top of tank A at the rate of 5 L/min, and leaves at the same rate, whereas the flow rate out of the bottom of tank B is 4 L/min, write down equations for the amount of salt in tank A and tank B after t min (Hint: you will need variables x(t) and y(t) to specify the amount of salt in each tank at time t min). Do not attempt to solve the equations!
Explanation / Answer
Let x(t) grams and y(t) grams to be the amount of salt in each tank at time t min.
Balancing mass flow of brine around tank A.
rate of inflow of brine = rate of change in amount of salt in tank A + rate of outflow of brine
rate of inflow of brine = (1 gm/L )* (5 L/min)
rate of change in amount of salt in tank A =dx(t)/dt
rate of outflow of brine = concentration of brine in A at time t * rate of at which solution leaves A = (x(t)/100 L )* (5 L/min) = x(t) / 20 (g/min)
(1 gm/L )* (5 L/min) = dx(t)/dt + x(t) / 20
5 = dx(t)/dt + x(t) / 20
dx(t)/dt + 0.05* x(t) -5 = 0
For tank B,
rate of inflow of brine in B = rate of change in amount of salt in tank B + rate of outflow of brine
rate of inflow of brine in tank B = rate of outflow of brine from tank A = x(t) / 20 (g/min)
rate of change in amount of salt in tank B =dy(t)/dt
rate of outflow of brine = concentration of brine in B at time t * rate of at which solution leaves B = (y(t)/100 L )* (4 L/min) = y(t) / 25 (g/min)
x(t) / 20 (g/min) = dy(t)/dt + y(t) / 25
0.05*x(t) = dy(t)/dt + 0.04*y(t)
dy(t)/dt + 0.04* y(t) - 0.05*x(t) = 0
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