Consider the two tank apparatus shown in the figure. Each tank has capacity 700
ID: 900150 • Letter: C
Question
Consider the two tank apparatus shown in the figure. Each tank has capacity 700 liters and initially contains 50 liters of fresh water. At time t = 0. the well-stirred mixing process begins. Suppose that the concentration of brine flowing into Tank 1 via the top tube is 0.8 kilograms per liter, and that the flow rates are r_1 = r_3 = 5 liters per minute, and r_2 -=10 liters per minute. Determine the volume of solution in each tank as a function of time, t, in minutes. Determine the time interval of interest. (The process stops when a tank is full or empty.) Stopping time is minutes. Let Q_1 (t) and Q_2(t) denote the amounts of salt (in kilograms) in the tanks at time t (in minutes). Derive the initial value problem with Q_1(t) and Q_2(t) as dependent variables describing the mixing process. Enter Q_1(t) as Q1 (t) and Q_2(t) as Q_2(t).Explanation / Answer
In tank 1 :
volume to water entering the tank in t mins = 5t L + 10t
volume of water leaving the tank in t mins = 10t L
volume of water already present in the tank = 50 L
Net volume of the water in the tank = 50L + {(5t + 10t) - 10t} = 55t
In tank 2:
volume to water entering the tank in t mins = 10t L
volume of water leaving the tank in t mins = 5t + 10t L
volume of water already present in the tank = 50 L
Net volume of the water in the tank = 50 L -{10t - (5t + 10t)} = 45t L
(a) V1(t) = 55t L /min
V2 (t) = 45t L/ min
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The process will stop when a tank is full
55t = 700 L
t = 12.73 min
stopping time is 12.73 min
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Volume of solution in tank 1 at time t = 55t
concentration of the salt = 55 *0.8 = 44t kg
concentration per minute[Q1(t)] = 44t/t = 44 kg/min
In tank 2:
concentration of salt [Q2(t)] = 45 * 0.8 = 36 kg/min
Q1(0) = 0 kg Q2(g) = 0 kg
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