Question

problem 1) Find the differential equation describing the amount of salt,Qb, in the tank for times t in the interval t>=T. Then solve to obtain Qb(t) for t>= T

A Water Tank Problem with Discontinuous Source A water tank contains V > 0 liters of pure water and Qo grams of salt. At time t = 0 we start pouring water into the tank with a rate r >0 liters per minute with a salt concentration of q> 0 grams per litter, and we let the well-stirred water leave the tank at the same rate. After T > 0 minutes the process is stopped and fresh water is poured into the tank at the same rate as before, r liters per minute, with the water leaving the tank at the same rate. "The problem is to find the amount of salt in the tank, Q, as function of time t. This problem is more difficult than usual because the conditions change abruptly at t = t. At this time the incoming salt concentration changes from q> 0 for t < T, to zero fort > 7. This change will make a discontinuous change in the differential equation describing the amount of salt Q in the tank. We know that Laplace transforms are useful to solve certain differential equations with discontinuous sources. We will use the Laplace transform method in the second part of this project. In the first part of this project we will solve this problem using the Matching Method, which essentially is the following: . We will write a differential equation for the amount of salt Q.() for 0 <t< T and solve it. We fix the free integration constant using the initial condition at t = 0. We will write a differential equation for the amount of salt Qn(t) fort > t and solve it. This solution will contain a free integration constant. We will fix the free constant in Qn by matching both solutions at t = T, that is, using the matching condition QA(T) = Q8(T). Note: Let us review the physical units of the quantities in this problem. Tinits of (0] = grams, Units of (Qo] = grams, Units of (V.) = liters, Units of (r] = minutes, liters grams Units of (n) = Units of (g) minutes liter From the units above we see that grams Units of rol Units of [r/V) Units of (qVo] = grams. minutes minutes Also notice that the units of d, the derivative of the amount of salt Q with respect to time t is grams Units of Q'] minutes 1

Answer 1

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o goflit LED At the time ty eso , & (4) pem be the amount of salt in tank, The differential equation is ho BALC). rem dell - oxP di Slxo Vo 22/2) = 8A1C). doo + at -5/3 Qe =0 d constant E d 11 4. (1) е T Jy Vo q (t) = d de er (c) ㅋ d=8 (2) et le In vocht q (*) e - SA (2) e p It-re) = SA re) e in tank - 8 (t) the amount of salt MO t ostare 819) = 12v. + (0-2v.) é E aléticos {q vo +(8-9vo) é o tyre,