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

Suppose a Brine solution containing .2kg of salt per liter runs into a tank initially filled with 500L of water containing 5kg of salt. the brine enters the tank at a rate of 5l/min. and the well mixed solution is flowing out of the tank at a rate of 6L/min. Find the amount of salt A(t) in the tank at time t. My biggest question is since there is already 5kg of salt at t=0 what would my concentration be? Also if you could solve the whole problem to see if i get the same answer, that would be great. Let C(t) be the concentration of salt and V(t) be the volume of brine solution in the tank. Then the amount of salt is given by 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] Note that the brine volume changes with time. Assuming that 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] So the liquid volume in the tank is given by V(t) = V(0) - 1·t = 500 - t [L] with t in min The change of amount of salt equals mass flow rate in minus mass flow rate out. The mass flow rates equal concentration times volumetric flow rate. The brine solution enters at a constant concentration C_in=2kg/L. Because the tank is well mixed the outflowing brine leaves at same concentration as it is 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 => ? 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 (with C in kg/L and t in min) => 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 (with A in kg and t in min)

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