# Application of Differential Equations

• Jul 12th 2010, 03:08 AM
confusedgirl
Application of Differential Equations
The rate at which a solid dissolves varies directly as the amount of undissolved solid present in the solvent and as the difference between the saturation concentration and the instantaneous concentration of the substance.
a saturated solution of salt in H2O will hold approximately 3 lb salt/gal. A block of salt weighing 60 lb is placed into a vessel containing 100 gal of H20. In 5 mins., 20 lb of salt were dissolved. (a) How much salt will be dissolved in one hour? (b) When will 45 lb of salt be dissolved?
• Jul 12th 2010, 05:19 AM
Ackbeet
Your first sentence allows you to set up the DE. First thing to do is assign variable names to relevant quantities. Then you want to write out your DE. What do you get when you do that?
• Jul 12th 2010, 05:22 AM
confusedgirl
DE = k (a-x)(Cs-Ci)dt is that correct?
• Jul 12th 2010, 05:25 AM
Ackbeet
Hmm. That might be right, but it's a bit confusing. Set up your variables first. Try constructing a list like this:

$t$ = time
$C_{s}$ = saturation concentration = const
.
.
.

What do you get?
• Jul 12th 2010, 05:32 AM
confusedgirl
k = constant of proportionality
a = total solid
x = dissolved solid
Cs = saturation concentration
Ci = instantaneous concentration

may i ask what is instantaneous concentration? is constant?
• Jul 12th 2010, 06:29 AM
Ackbeet
Ok, nice list. Some thoughts, then:

1. The instantaneous concentration is going to be related to the amount of dissolved solid somehow, in, I think, a linear fashion. That is, I think you're going to have $x = m C_{i} + b$, where $m$ and $b$ are constants.
2. The instantaneous concentration at time $t$ is defined as the amount of dissolved solid divided by the total volume of the brine, in your case. The units would be lb./gal. It definitely changes with time. While you're going to try to find $x(t)$ as the solution of your differential equation, I think $C_{i}$ is going to play an important role.
3. The DE you posted in post #3 is almost correct. Hint: look at the LHS.
• Jul 16th 2010, 05:19 AM
confusedgirl
im still having trouble solving this problem, but i already have a solution but still can't figure out what i'm doing wrong

dQ/dt = kQ(Cs -Ci)
where dQ/dt = the rate of which the solid dissolves
k = constant of proportionality
Q= number of salt at any given time
Cs = saturation concentration = 3 lb/gal
Ci = Q/V= Q/100

dQ/dt = kQ(3-Q/100)
through bernoulli's equation i'll be able to get
1/Q = 1/300 + ce^(-3kt)
when t=0 Q=60
therefore c=1/75
so, the equation would be 1/Q =1/300 + (1/75)e^(-3kt)
when t=5 Q=60-20=40
k= -0.0323718772
so, when t=60 Q=?
1/Q = 1/300 +(1/75)e^[(-3)(-0.0323718772)(60)]
1/Q = 4.469
Q=.224

b. 1/15 = 1/300 + (1/75)e^[(-3)(-0.0323718771)t]
t = 16.07 mins

i keep on rechecking my solution but still have the wrong answer
a. Q= .885
b. t = 18.21
• Jul 16th 2010, 05:33 AM
Ackbeet
The problem is your DE. You're confusing dissolved salt with undissolved salt. Let Q be the quantity of dissolved salt. Then you're going to have

dQ/dt = k(60-Q)(Cs-Q/100), I think.

Now, before you plunge ahead and solve this DE (which might be wrong!), I think you should justify this DE to me. Explain why it's correct or incorrect.
• Jul 16th 2010, 05:36 AM
confusedgirl
well, i thought of that also, but im not sure which one is correct but i don't know how to solve that differential equation.
• Jul 16th 2010, 07:09 AM
Ackbeet
Don't worry about solving the DE: we can deal with that one later. Right now, the important thing is to make sure we're solving the right problem. Take this sentence: "The rate at which a solid dissolves varies directly as the amount of undissolved solid present in the solvent and as the difference between the saturation concentration and the instantaneous concentration of the substance." and translate that directly into a differential equation, or compare to the one I gave you. Does it make sense from that perspective?
• Jul 16th 2010, 08:28 AM
confusedgirl
The equation u gave me was the right one. I already solved it. thank u!
• Jul 16th 2010, 08:30 AM
Ackbeet
You're very welcome. Have a good one!
• Jul 18th 2010, 02:33 AM
Mrnerd
excuse me... i have the same problem as well... can you show me the solution on the D.E. dQ/dt = k(60-Q)(Cs-Q/100). thank you...
• Jul 18th 2010, 03:16 AM
Prove It
$\frac{dQ}{dt} = k(60 - Q)\left(C_S - \frac{Q}{100}\right)$

$\frac{dt}{dQ} = \frac{1}{k(60 - Q)\left(C_S - \frac{Q}{100}\right)}$

$t = \frac{1}{k}\int{\frac{1}{(60 - Q)\left(C_S -\frac{Q}{100}\right)}$.

Now solve the integral using Partial Fractions.