Question

A tank originally holds 500 gal of sugar water with a concentration 1/10 lb/gal. At t=0, there starts a flow of sugar water into the tank with a concentration of 1/2 lb/gal at a rate of 5 gal/min. There is also a pipe at the bottom of the tank removing 10 gal/min from the tank.

Please help me to show the equation. I got different equation than my lecturer

Answer 1

What did you get and what did your lecturer get? Do you know what dimensional analysis is? That will help you immensely with problems of this nature because you'll have a guide for being correct.

Answer 2

But for starters, what's the change in the amount of sugar? It should be \[\LARGE \frac{dQ}{dt}=rate_{ i n }-rate_{out}\] and the units here are lb/min by looking at dQ/dt. So clearly rate in and out must also have the same units since we are adding apples to apples here. If it doesn't match the units then you know you're wrong, which is useful to know to help fix yourself.

Answer 3

My lecturer answer is S(t)= 2.5(100-t) - ((100-t)^2)/50. I have no idea where he got the 100-t

Answer 4

For the rate out, i get s(t)/(50-5t)

Answer 5

So what is the quantity 100-t supposed to represent here?

Answer 6

It looks like at t=0 we have something that resembles our initial volume of water, so do you think that perhaps you can figure it out? But really I suggest instead of trying to deconstruct what your lecturer has done you attempt to construct what you think the answer should be for rate in and rate out and then simply use your lecturer's answer be a check for you.

Show me what you think the rate in should be and be sure to label it with units and I will help you out.

Answer 7

For rate in, i get 2.5 lb/min

Answer 8

But for the rate out, is the answer right?

Answer 9

For the final answer, i got s(t)= [t(t-20)]/(20-2t) + C/(50-5t)