Calculating Salt Solution Concentration: Calculus & Non-Calculus Solutions

[eldrick]

There is a tank that contains 10,000 gallons of a weak salt solution. The salt concentration is 1300 parts per million. Now we start adding 130 gallons per minute of water and also remove 130 gallons per minute of salt solution. Thererore the salt solution concentration gradually decreases to zero parts per million. Assume the tank is agitated and well mixed.

How long does it take the salt solution concentration to decrease to 100 parts per million?

How long ... to 10 parts per million?

How long ... to 1 part per million?

I tried a decay solution as a logarithmic function ( which I believe is not quite correct ) but couldn't come up with correct calculus method.

Can someone offer a calculus & non-calculus solution ?

 

[tiny-tim]

Hi eldrick!

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[eldrick]

I tried this initially as a non-calculus solution ( but told not quite correct ) :

initial salt content is

10,000 * 1300/1,000,000 = 13 units

we want time salt concentration is

1) 10 per million or total salt content is

10,000 * 10/1,000,000 = 0.1 units

2) 1 per million or total salt content is

10,000 * 1/1,000,000 = 0.01 units

how do we find the answers ?

- 0'00 : total salt content is
10,000 * 1300/1,000,000 = 13 units

- 1'00 : total salt content is
(10,000 * 1300/1,000,000) - (130 * 1300/1,000,000) = 1300/1,000,000 * (10,000 - 130) = 12.831 units

without wasting time on endless factorising :

at any given time t, total salt is

13 * ( (10,000 - 130)/10,000 )^t

or simplify

13 * 0.987^t

so, for answers

1 ) 0.1 = 13 * 0.987^t

-> t = 371.987 minutes

1 ) 0.01 = 13 * 0.987^t

-> t = 547.954 minutes

The problem appears to be initial formula which is based on salt content at 1 minute & not instantaneous

 

[tiny-tim]

Hi eldrick!

(try using the X² tag just above the Reply box )

I haven't completely checked your other figures, but yes your 0.987^t is correct.

(the calculus method, for the proportion p, would have been dp/dt = -0.013p)

 

[eldrick]

Very kind of you Tim

I'd appreciate it if you would show me how you obtained the differential

 

[tiny-tim]

dp/dt = -0.013p ?

A fixed proportion of the water leaves per second.

So a fixed proportion of the salt leaves per second.

That's English for "dp/dt is a constant times p".