Question

A 120-gallon tank initially contains 90 pounds of salt dissolved in 90 gallons of water. Brine containing 2 1b/gal of salt flows into the tank at the rate of 4 cal/min, and the well-stirred mixture flows out of the tank at the rate of 3 gal/min. How much salt does the tank contain when it is full? (At 30 minutes, there is approximately 202 pounds of salt present in the tank.)

Answer 1

Vin = 4 gal/min Cin = 2 lb salt/ gal Tank Initial Mass of salt = 90 lb Initial Volume of water = 90 gal Vout = 3 gal/min Cout-C

Given : Initial Volume of water mixture in tank (V) - Incoming Vol. Flowrate of stream (Vin) - Outgoing Vol. Flowrate of stream (Vout) = 90 gal 4 gal/ min 3 gal / min Concentration of salt in (Cin) = 2 lb salt/gal Here, it is assumed that the tank is perfectly mixed. That means concentration of salt in tank at any time is same as the concentration of salt in the outlet stream. Tank initially contains 90 lb of salt. Here, Amount of water coming in is 4 gal/min and Amount of water leaving is 3 gal/min. Thus, Volume of tank will not remain constant. It will change with time. Lett be any time in minutes. Then Volume at any given time (V(t)) -Initial Volume of tank + +(Vol of water IN - Vol of water OUT) * time 90+ (4-3)*t 90 +t V(t) = Let x be mass of salt in tank at any given time 't' Let C be Concentration of salt in tank at any given time t' Thus, Cout = X(t) Volume at timet General Salt Mass Balance equation is : Amount of salt IN - Amount of salt Out = Rate of Accumulation of salt in tank Here, Amount of salt IN = Amount of salt OUT = Vin * Cin Vour * Cout Rate of accumulation of salt in tank = " Thus, on substituting all the values in general mass balance equation, Vin * Cin - Vout * Cout = But, Cin=2 (Given) Lout 90+t

4*2- 3* 5 = idare dx dt 3x (90+t) = 8 (Differential Equation) This is a linear ODE. For Linear ODE of form x' + p(t) = r(t) Integrating Factor (IF) = Sp(t)dt And Solution is : x(t) = e-F SeFr(t)dt = ce-If Here, p(t) = (90+4) r(t) = 8 Here, Integrating Factor = Srdt = 3*In(90+t) = In (90 +)3 And e-IF = = exp(-In (90 +t)) exp(In(90+ t)3) (90+t) -3 = (90 + t) Thus, Solution is given as : x(t) = c* (90+t)-3 (90+t)- S(90 +t)3 * 8dt + (90+)* *8+ (90+4)* +(004) x(t) = 2(90 + t) + 90++)3 ---> (1) This represents mass of salt in tank at any given time

Now Initial Condition is : When t=0, x=90 lb Thus, On substituting in equation (1), 90 = 2(90 + 0) + (90+0)3 On solving we get, C = -65610000 Thus, On substituting in equation (1), X(t) = 2(90 +t)-65610000 x(t) = 2(907) (90+t) This represents Mass of salt in tank at time t First we need to find time for tank to get full Volume of tank at any time = 90+t (As calculated above) Volume of tank when full = 120 gal (Max volume of tank) Thus, 120 Thus, t = = 90+t 30 minutes Thus, it takes 30 minutes for tank to get full Now, we need to determine mass of salt in tank when it gets full i.e. at t = 30 minutes. At t= 30 minutes, the Mass of salt in tank is : x(t = 30) = 2(90+30) - 90+30)3 = 240 - 37.96875 202.03125. lb salt Thus, tank will contain 202.03125 lb salt when it gets full at 30 minutes