Differential Equations and Flow RateDate: 02/14/99 at 01:05:12 From: Stephen Johnson Subject: Differential Equations Each of two tanks contains 100 gal of pure water. A solution containing 3 lb/gal of dye flows into Tank 1 at 5 gal/min. The well-stirred solution flows out of Tank 1 into Tank 2 at the same rate. Assuming the solution in Tank 2 is well-stirred and that this solution flows out of Tank 2 at 5 gal/min, determine the amount of dye in Tank 2 after 15 minutes. Date: 02/14/99 at 09:47:02 From: Doctor Anthony Subject: Re: Differential Equations Let M1 = mass of dye in tank 1 at time t M2 = mass of dye in tank 2 at time t The inflow to tank 1 is 3 * 5 lbs/min = 15 lbs/min into the 100 gallon tank. The outflow is 5 * M1/100 lbs/min = .05 M1 lbs/min from the 100 gallon tank. The differential equations are then: dM1/dt = 15 - 0.05M1 = .05(300-M1) dM1/(300-M1) = .05 dt and integrating -ln(300-M1) = .05t + C ln(300-M1) = -.05t + C 300-M1 = e^(-.05t+C) = A e^(-.05t) and so M1 = 300 - A e^(-.05t) when t = 0, M1 = 0. Thus A = 300 and M1 = 300[1-e^(-.05t)] For tank 2 the differential equation is: dM2/dt = .05M1 - .05M2 dM2/dt + .05M2 = .05M1 = 15[1-e^(-.05t)] This is a linear equation and we multiply by the integrating factor e^INT(.05dt) = e^(.05t): e^(.05t) dM2/dt + .05e^(.05t) M2 = 15e^(.05t)[1-e^(-.05t)] d[e^(.05t) M2]/dt = 15[e^(.05t) - 1] and integrating e^(.05t) M2 = 15[(1/.05)e^(.05t) - t] + C M2 = 15[(1/.05) - t e^(-.05t)] + C e^(-.05t) at t = 0, M2 = 0, and so 0 = 15[1/.05] + C. So C = -300. Then M2 = 300 - 15te^(-.05t) - 300 e^(-.05t) M2 = 300[1-e^(-.05t)] - 15t. e^(-.05t) When t -> infinity, M2 -> 300 which is correct. When t = 15 this gives: M2 = 300[1-e^(-.75)] - 15 x 15 x e^(-.75) = 158.29 - 106.2825 = 52.0075 lbs - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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