Differential Equations: Linear Models

[Arshad_Physic]

Homework Statement

A 500-gallon tank originally contains 350 gallons of pure water. A 2-pound per gallon brine solution is pumped into the tank at 10 gallons per minute and the well-stirred mixture is pumped out at one-half that rate. How much salt is in the tank at th emoment of overflow?

Answer: 510 lb

Homework Equations

dA/dt = [(rate in)(Concentration in)] - [(rate out)(concentration out)]

The Attempt at a Solution

Here is what I did.

dA/dt = (2lb/gal * 10 gal/min) - [A/(350+5t) * 5]

dA/dt = 20 - A/(70+t)

Rearrange:

dA/dt + A/(70+t) = 20

Initial condition: A (0) = 0

e^{intgral[1/(70+t)]} = 70 + t

A(t) = [1/(70+t)] * [10* integral(t+70)dt + c

A(t) = 5(t+70) + c/(t+70)

A(0) = 0;

c = -24,500

Thus; A(t) = 5(t+70) - 24500/(t+70)

A (10)/500 = 400 - 24500/80

= 93.75 lb/gal

This answer is incorrect, as the answer should be 510 lb.

PLEASE HELP!

Thanks!

 

[HallsofIvy]

Why did you evaluate at t= 10? If there were initially 350 gallons in the 500 gallon tank, we need to add 150 gallons. With water coming in at a net 5 gallons per minute, that will take 150/5= 30 minutes, not 10. Also, of course, you give A in pounds per gallon. To find the number of pounds in the tank when it overflows, multiply that by 500 gallons.

 

[Arshad_Physic]

hi - thanks for help! :)

But I am still confused;

I now understand why t = 30 minutes. So, if I take t = 30, then I get:

Thus; A(t) = 5(t+70) - 24500/(t+70)

A (30)/500 = 500 - 24500/100

= 500 - 245 = 255 lb/gal.

But, if I multiply this 255 lb/gal by 500, I get ridiculously large number! :) I get 130050 lb, which is incorrect..

*I just observed that if I multiply 255 by 2, then I get the CORRECT answer: 510. But I get the wrong units. Also, it doesn't make sense to multiply 255 by 2 :)

So, its the LAST part that's creating problem for me now!