- #1
jordan123
- 16
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Differential Equation mixing problem!
A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?
This is what I've been doing.
dA/dt = inflow - out flow
= (.06)(5) - (A/500)(5)
= .3 - (A/100)
= (30 - A)/100
etc but I feel I must be setting it up wrong or something ?? But it never gets the correct answer which is.
4.9%
If anyone can point out what's up that would be much appreciated!
Homework Statement
A vat with 500 gallons of beer contains 4% alcohol (by volume). Beer with 6% alcohol is pumped into the vat at a rate of 5 gal/min and the mixture is pumped out at the same rate. What is the percentage of alcohol after an hour?
The Attempt at a Solution
This is what I've been doing.
dA/dt = inflow - out flow
= (.06)(5) - (A/500)(5)
= .3 - (A/100)
= (30 - A)/100
etc but I feel I must be setting it up wrong or something ?? But it never gets the correct answer which is.
4.9%
If anyone can point out what's up that would be much appreciated!