Di erential Equations Water Tank Problems

1 Differential Equations Water Tank ProblemsChapter #3 VariationA tank originally contains 100 gal of fresh Water . Then Water containing12lb of salt per 2 gallonis poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the samerate. What is the amount of salt at any instant?dQdt= rate in - rate out= (salt concentration in) x (flow rate in) - (tank salt concentration) x (flow rate out)*NotedQdtshould be in terms oflbminParametersQ(t) : the amount of salt at time t (lbs)Qo: initial amount of salt in the tank Q(0) =Qo= 0salt concentration in =12lbgaltank salt concentration =Qcurrent amount of Water solution=Q100lbgalIn this question flow rate in is the same as flow rate out so we will let the rates be defined as:rate in = rate out = 2galminSubstituting everything into our differential equation, we arrive atdQdt=12lbgal 2galmin Q100lbgal 2galmin= 1lbmin Q50lbminRewriting our equation, we havedQdt+Q50= 11 Multiplying by the integrating factoret/50followed by applying the product rulefor derivatives we havedQdt et/50+Q et/5050= 1 et/50ddt[Q et/50] =et/50 ddt[Q et/50]dt= et/50dtAfter integrating both sides we are left withQ et/50= 50 et/50+CQ= 50 +Ce t/50 From our initial conditionQ(0) = 0we haveC= 50.

2 Our solution to this differentialequation isQ(t) = 50 50e t/50 What is the amount of salt after 10 minutes?Q(10) = 50 50e 10/50 ( )What is the amount of salt after 30 minutes?Q(10) = 50 50e 30/50 ( )2 Chapter #4A tank with a capacity of 500 gal originally contains 200 gal of Water with 100lb of salt insolution. Water containing 1lb of salt per gallon is entering at a rate of 3 gal/min, and themixture is allowed to first thing to note is now the rate out does not match the rate in. The amountof the solution in the tank in increasing(3 gal/min IN - 2 gal/min OUT = INCREASINGby 1 gal/min).This alters our equation for the tank salt concentration. It will nowbetank salt concentration =Qcurrent amount of Water solution=Q200 + 1tdQdt= 1lbgal 3galmin Q200 + 1tlbgal 2galmin= 3lbmin 2200 + 1tQlbminRewriting our equation, we havedQdt+2200 +tQ= 3with the initial condition ofQ(0) = by the integrating factor(200 +t)2followed by applying the productrule for derivatives we havedQdt (200 +t)2+2200 +tQ (200 +t)2= 3 (200 +t)2dQdt (200 +t)2+Q 2(200 +t) = 3(200 +t)2ddt[Q (200 +t)2] = 3(200 +t)2 After integrating both sides we are left withQ (200 +t)2= (200 +t)3+CQ= (200 +t) +C(200 +t) 2 From the initial condition,Q(0) = 100, we getC= 100(200)2.

3 Our solution to thisdifferential equation isQ(t) = 200 +t 100(200)2(200 +t) 2= 200 +t 100(200)2(200 +t)2, t <300We have to consider when the tank will begin to overflow which is after 300 minsince we initially have200gal in our500gal tank which leads to,t < for, Find the concentration (in pounds per gallon) of salt in the tank when it is on thepoint of overflowing. Compare this concentration with the theoretical limiting concentration ifthe tank had infinite capacity. We lett= 300which gets usQ(300) = 484lb of salt, but since it asks for the concen-tration we take our solution and divide it by the amount of Water in the tank andarrive atConcentration =Q(t)200 +t=200 +t 100(200)2(200+t)2200 +t= 1 100(200)2(200 +t)3 Evaluate whent= 300= 484lb/500gal= 121/125lbgalIf the tank had infinite capacity the concentration would then converge to,limt Q(t)200 +t= limt 1 100(200)2(200 +t)3=1lbgal4 Tank with salt coming in from two pipes:A tank with a capacity of 1500 gals originally contains 1000 gals of fresh Water .

4 The first pipecontaining12lb of salt per gallon is entering at a rate of 4 gal/min. The second pipe containing13lb of salt per gallon is entering at a rate of 6 gal/min. The mixture is allowed to flow out ofthe tank at a rate of 5 gal/min. Find the amount of salt in the tank at any time prior to theinstant when the solution begins to * the rate out does not match the rate in. The amount of the solution in thetank in increasing(10 gal/min IN - 5 gal/min OUT = INCREASING by 5 gal/min).Thisalters our equation for the tank salt concentration. It will now betank salt concentration =Qcurrent amount of Water solution=Q1000 + 5tDetermine how much salt is coming inPipe 1 =12lbgal 4galminPipe 2 =13lbgal 6galminTotal Rate In = Pipe 1 + Pipe 2= 2lbmin+ 2lbmin= 4lbminDetermine how much salt is going outRate Out =Q1000 + 5tlbgal 5galmin=5Q1000 + 5tlbminCombining everything our differential equation becomesdQdt= 4lbmin 5Q1000 + 5tlbmindQdt+5Q1000 + 5t= 4 Multiplying by our integrating factor1000 + 5twe arrive todQdt (1000 + 5t) +5Q1000 + 5t (1000 + 5t) = 4 (1000 + 5t)ddt[Q (1000 + 5t)] = 4 (1000 + 5t)Q (1000 + 5t) = 4(1000t+5t22) +CQ=4(1000t+5t22) +C(1000 + 5t)5 BecauseQ(0) = 0, we getC= 0which makes our solutionQ=4(1000t+5t22)(1000 + 5t), t <100 Since our tank overflows after 1500 gallons at timet= 100, to find the amount ofsalt at that instant we evaluateQ(100).

5 Q(100) =5000001500=10003