Suppose a brine containing 0.8 kg of salt per liter runs into a tank initially f

Question

Suppose a brine containing 0.8 kg of salt per liter runs into a tank initially filled with 500 L of water containing 4 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (a) Find the concentration, in kilograms per liter, of salt in the tank after 5 min. Round to 3 decimal places. [Hint: Let A denote the number of kilograms of salt in the tank at t minutes after the process begins and use the following fact. rate of increase in A = rate of input - rate of exit. answer should be in Kg/L

Explanation / Answer

Let C(t) be the concentration of salt and V(t) be the volume of brine at any instant t. The amount of salt A(t) is given as A(t) = C(t)·V(t) Hence the initial concentration of salt is: C(0) = A(0)/V(0) = 5/500 = 0.01 [kg/L] Volume of brine changes with time. Assuming the density of brine solution does not vary with salt concentration, the change of brine volume equals volumetric flow rate into the tank minus flow rate out of the tank: dV/dt = Q_in - Q_out = 5 - 6 = -1 [kg/L] Thus the liquid volume in the tank is given by V(t) = V(0) - 1·t = 500 - t [L] with t in min C_in=2kg/L (:.Because the tank is well mixed the outflowing brine leaves same concentration in the tank) dA/dt = C_in·Q_in - C·Q_out in terms of C and V d(C·V)/dt = C_in·Q_in - C·Q_out => C·(dV/dt) + V·(dC/dt) = C_in·Q_in - C·Q_out => C·(-1) + (500 - t)·(dC/dt) = 2·5 - C·6 =>(500 - t)·(dC/dt) = 10 - 5·C = 5·(2 - C) =>1/(2 - C) dC = 5/(500 - t) dt => - ln(2 - C) = -5·ln(500 - t) + k (:.k is the constant of integration) C = 2 - e^(5·ln(500 - t) - k) = 2 - e^(-k)·e^(ln( (500 - t)5 ) = 2 - K·(500 - t)5 (with K = e^(-k)) Apply initial condition C(0) = 0.01 0.01 = 2 - K·(500 - 0)5 K = (2 - 0.01)/5005 = 6.368×10?¹4 C(t) = 2 - 6.368×10?¹4·(500 - t)5 putting values in first eq A(t) = C(t)·V(t) = (2 - 6.368×10?¹4·(500 - t)5) · (500 - t) = 1000 - 2·t - 6.368×10?¹4·(500 - t)6

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