<Text-field layout="Heading 1" style="Heading 1">Outline of Lesson 5</Text-field>5.A One-Tank Mixing Problem5.B Variable Volume Mixing Problem5.C Two-Tank Mixing Problem
<Text-field layout="Heading 2" style="Heading 2">Problem Statement</Text-field>A tank initially contains 40 gal of sugar water having a concentration of 3 lb of sugar for each gallon of water. At time zero, sugar water with a concentration of 4 lb of sugar per gal begins pouring into the tank at a rate of 2 gal per minute. Simultaneously, a drain is opened at the bottom of the tank so that the volume of the sugar-water solution in the tank remains constant.(a) How much sugar is in the tank after 15 minutes?(b) How long will it take the sugar content in the tank to reach 150 lb? 170 lb?(c) What will be the eventual sugar content in the tank?
<Text-field layout="Heading 2" style="Heading 2">Solution</Text-field>The mathematical formulation of this problem must express the physical requirement thatNiMqJiUiZEciIiIlI2R0RyEiIg== ( amount of sugar in tank) = ( rate sugar is added to tank ) - ( rate sugar is removed from tank ) Let NiMtJSJ4RzYjJSJ0Rw== denote the amount of sugar (pounds) in the tank at time NiMlInRH (minutes). Then, the rates in and out are respectivelyrate_in := 4 * 2;rate_out := (x(t)/40) * 2;so that the governing ODE isode := diff( x(t), t ) = rate_in - rate_out;The amount of sugar in the tank initially, that is, when NiMvJSJ0RyIiIQ==, gives the initial conditionic := x(0)=40 * 3;The ODE in this IVP is first-order and linear. The integrating factor ismu(t) = intfactor( ode );The solution to the IVP issol := dsolve( { ode, ic }, x(t), [linear] );
<Text-field layout="Heading 3" style="Heading 3">(a)</Text-field>The amount of sugar (in pounds) in the tank after 15 minutes iseval( sol, t=15. );
<Text-field layout="Heading 3" style="Heading 3">(b)</Text-field>The tank will contain 150 pounds of sugar at a time NiMlInRH (in minutes) satisfyingeq150 := eval( sol, x(t)=150 );Thus, the desired time is found by the calculationst150 := solve( eq150, t ):t[`150 lbs`] = t150;`` = evalf(t150);Repeating the same steps for the time when 170 pounds of sugar are in the tank leads to the equationeq170 := eval( sol, x(t)=170 );whose solution ist170 := solve( eq170, t ):t[`170 lbs`] = t170; `` = evalf(t170); This complex-valued solution is clearly not physically realistic. A quick inspection of the solution, graphed in Figure 5.1,plot( rhs(sol), t=0..120, title="Figure 5.1" );shows that the amount of sugar in the tank reaches a steady-state limit that is well below 170 pounds. Therefore, at no time is there ever 170 pounds of sugar in the tank.
<Text-field layout="Heading 3" style="Heading 3">(c)</Text-field>In (b) it was noted that the amount of sugar in the tank levels off below 170 pounds. The exact limit can be determined from the solution by looking at the limit as NiMlInRH NiMsJCUiPkchIiI= NiMlKWluZmluaXR5Rw==, that is, atsteady_state := map( Limit, sol, t=infinity );whose value isvalue( steady_state );Note that NiMvJSJ4RyIkZyI= is an equilibrium solution for this ODE. However, be careful to avoid the common error of concluding that the limit is 160 pounds because NiMvJSJ4RyIkZyI= is an equilibrium solution. (Recall the logistic growth model, Lesson 3, Section B , which has two equilibria.)