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ColdFusion85
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[SOLVED] Mixing Problem - Linear First Order ODE
A 500-gallon tank initially contains 50 gallons of brine solution in which 28 pounds of salt have been dissolved. Beginning at time zero, brine containing 2 pounds of salt per gallon is added at the rate of 3 gallons per minute, and the mixture is poured out of the tank at the rate of 2 gallons per minute. How much salt is in the tank when it contains 100 gallons of brine? Hint: The amount of brine in the tank at time t is 50+t.
Let Q(t) be the amount of salt in the mixture at time t. Then dQ/dt = (rate in) - (rate out), or dQ/dt = (2 lbs/gallon)(3 gallons/min) - (Q(t)/50 lbs/gallon)(2 gallons/min). So
dQ/dt = 6-(1/25)*Q(t)
Rearranging, finding the integrating factor, and solving gives
Q(t) = 150 + C*e^(-t/25)
We know at t=0 there is 28 lbs of salt in the tank, so C = -122, and
Q(t) = 150-122e^(-t/25)
The hint tells us that, when 100 gallons of brine are in the tank, 50 seconds have passed. So, the amount of salt in the tank when the tank contains 100 gallons of brine is
Q(50) = 150-122e^(-50/25)
= 133.49 gallons
Is this correct? Does the tank size matter at all? I didn't use it in my calculations, so I was just wondering. Thanks.
Homework Statement
A 500-gallon tank initially contains 50 gallons of brine solution in which 28 pounds of salt have been dissolved. Beginning at time zero, brine containing 2 pounds of salt per gallon is added at the rate of 3 gallons per minute, and the mixture is poured out of the tank at the rate of 2 gallons per minute. How much salt is in the tank when it contains 100 gallons of brine? Hint: The amount of brine in the tank at time t is 50+t.
The Attempt at a Solution
Let Q(t) be the amount of salt in the mixture at time t. Then dQ/dt = (rate in) - (rate out), or dQ/dt = (2 lbs/gallon)(3 gallons/min) - (Q(t)/50 lbs/gallon)(2 gallons/min). So
dQ/dt = 6-(1/25)*Q(t)
Rearranging, finding the integrating factor, and solving gives
Q(t) = 150 + C*e^(-t/25)
We know at t=0 there is 28 lbs of salt in the tank, so C = -122, and
Q(t) = 150-122e^(-t/25)
The hint tells us that, when 100 gallons of brine are in the tank, 50 seconds have passed. So, the amount of salt in the tank when the tank contains 100 gallons of brine is
Q(50) = 150-122e^(-50/25)
= 133.49 gallons
Is this correct? Does the tank size matter at all? I didn't use it in my calculations, so I was just wondering. Thanks.