Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolve

Question

Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentration of 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallons per minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons per minute. The correct differential equations with initial conditions for the amounts, x(t) and y(t), of salt in tanks A and B respectively, at time t are:
A) dx/dt=2-x/40+y/5,dy/dt=x/40-y/3,x(0)=20,y(0)=5
B) dx/dt=2-3x/40+y/15,dy/dt=3x/40-y/5,x(0)=20,y(0)=5
C) dx/dt=4-3x/40+y/15,dy/dt=3x/40-y/5,x(0)=20,y(0)=5
D) dx/dt=4-x/40+y/5,dy/dt=x/40-y/3,x(0)=20,y(0)=5

Explanation / Answer

correct answer is option (b) solution: x(0)=20, y(0)=5 for x: rate in=(0.5 salt/gallon)*(4 gallon/ min)+(y(t) pounds/30 gallons)*((2 gallon/min) =2 + y/15 rate out=(x(t) pounds/80 gallons)*((6 gallon/min) =3*x(t)/40 dQ dt = (rate in) - (rate out) dx dt=(2+ y(t)/15)-(3*x(t)/40) =2-3x/40+y/15 for y: rate in=(x(t) pounds/80 gallons)*((6 gallon/min) =3x/40 rate out=(y(t) pounds/30 gallons)*((4 gallon/min)+(2 gallon/min into tank A)) =y/5 dQ dt = (rate in) - (rate out) dy dt=(3*x(t)/40)-y(t)/5 =3x/40-y/5

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