Thread: chemical solution

a tank initially contains salt in the pores of inert materials and 10 gal of fresh water. The salt dissolved at a rate per min of 2 times the difference between 3lb/gal and the concentration of the brine. The two gal of fresh water enters the tank per min. How much salt will be dissolved in the 1st 10 min?

Can I get the initial value of salt from the equation?

what equation should i use?
dE/dt = 2(3-C)
where C = E/(100+2t) or C = (X-E)/(100+2t)
x=initial value of salt
E=dissolve salt

Originally Posted by sean123

a tank initially contains salt in the pores of inert materials and 10 gal of fresh water. The salt dissolved at a rate per min of 2 times the difference between 3lb/gal and the concentration of the brine. The two gal of fresh water enters the tank per min. How much salt will be dissolved in the 1st 10 min?

Can I get the initial value of salt from the equation?

what equation should i use?
dE/dt = 2(3-C)
where C = E/(100+2t) or C = (X-E)/(100+2t)
x=initial value of salt
E=dissolve salt

The red one. Only the dissolved salt contributes to the concentration of the brine, not the salt still inside the material.

ok. thank u! how would i know the initial value of salt?

Originally Posted by sean123

ok. thank u! how would i know the initial value of salt?

At t = 0, all the salt is inside the inert material (and you're told that initially the water is fresh) .....

uhmm sir...
I have the same problem as well... using the D.E. dE/dt = 2(3-C) and C = E/100+2t, i got

dE/dt=2(3-E/100+2t)
dE = 6dt - E/50+t dt so...
E(e^ln(50+t) = ∫6e^ln(50+t) dt
E(50+t) = ∫6(50+t) dt
E(50+t) = ∫(300 + 6t)dt
E(50+t) = 300t + 3t^2 +c
E = (300t + 3t^2 +c)/50+t... that's how i got my equation... then from there, when t = 0, E = 0, i got C = 0.
any mistakes on my solution??? please tell me so i can fix it.. thank you...

Originally Posted by Mrnerd

uhmm sir...
I have the same problem as well... using the D.E. dE/dt = 2(3-C) and C = E/100+2t, i got

dE/dt=2(3-E/100+2t)
dE = 6dt - E/50+t dt so...
E(e^ln(50+t) = ∫6e^ln(50+t) dt
E(50+t) = ∫6(50+t) dt
E(50+t) = ∫(300 + 6t)dt
E(50+t) = 300t + 3t^2 +c
E = (300t + 3t^2 +c)/50+t... that's how i got my equation... then from there, when t = 0, E = 0, i got C = 0.
any mistakes on my solution??? please tell me so i can fix it.. thank you...

That looks OK. Confirmed here: solve dy/dx + y/(50 + x) = 6 and y(0) = 0 - Wolfram|Alpha