Question

Integrating Factors (equation soon to come)...
Constants: alpha and k
Initial Conditions: t=0, Cr=Cr*, Ca=Ca(0) 'naught'
Final Conditions: t, Cr(t), Ca*

Answer 1

\[\frac{ dC _{r} }{ dt }+\frac{ 1 }{ \alpha }C_r=0.5kC_a\]

Answer 2

I think I am doing the integrating factor wrong

Answer 3

\[e^{\int\limits_{?}^{?}\frac{ 1 }{ \alpha }dt}\]what are the limits? 0 to t ??

Answer 4

$$\frac{dC_r}{dt}+\frac1\alpha C_r=\frac12kC_\alpha$$What exactly is \(C_a\)? What is \(r\)?

Answer 5

two different unknowns. I need to solve for Cr(t) in my final answer. If you must know, Cr is the concentration of the reaction product and Ca is the concentration of the reactants

Answer 6

So they are both constants here? Not functions of \(t\)?

Answer 7

They are dependent on t

Answer 8

if \[\frac{ dY(t) }{ dt }+P(t)Y=Q(t)\]Then\[Y(t)=Cr\]\[{1 \over \alpha}=P(t)\]and you get the pattern for Q(t)

Answer 9

So is this a partial differentiation equation? If it's not,$$\frac{dC_r}{dt}+\frac1\alpha C_r=\frac12kC_a$$We want to multiply by some integration factor \(\mu\) so that our left-hand side is just the product rule, i.e.$$\mu\frac{dC_r}{dt}+\frac1\alpha\mu C_r=\frac{d}{dt}[\mu C_r]\\\mu\frac{dC_r}{dt}+\frac1\alpha\mu C_r=\mu\frac{dC_r}{dt}+\frac{d\mu}{dt}C_r$$Subtracting \(\mu\frac{dC_r}{dt}\) from both sides we find:$$\frac1\alpha\mu C_r=\frac{d\mu}{dt}C_r\\\frac1\alpha\mu=\frac{d\mu}{dt}$$... which is a separable differentiation equation in \(\mu\). Let's solve it by separating our variables:$$\frac1\alpha dt=\frac1\mu d\mu\\\int\frac1\alpha dt=\int\frac1\mu du\\\frac1\alpha t=\log\mu\\\mu=e^{\frac1\alpha t}$$Now we have our integration factor. Multiply throughout:$$e^{\frac1\alpha t}\frac{dC_r}{dt}+\frac1\alpha e^{\frac1\alpha t}C_r=\frac12k C_ae^{\frac1\alpha t}\\\frac{d}{dt}\left[e^{\frac1\alpha t}C_r\right]=\frac12kC_ae^{\frac1\alpha t}\\e^{\frac1\alpha t}C_r=\int\frac12kC_ae^{\frac1\alpha t} dt=\frac12kC_a\int e^{\frac1\alpha t} dt$$

Answer 10

Evaluating that integral, we get $$e^{\frac1\alpha t}C_r=\frac12\alpha kC_ae^{\frac1\alpha t}+C\\C_r=\frac12\alpha kC_a+Ce^{-\frac1\alpha t}$$

Answer 11

The solution is supposed to be.... which I am supposed to derive.

Answer 12

The solution is what I found there. I just didn't use shortcut formulae for solving first-order linear equations like you wanted to. If that's what you want, consider that the solution for \(\dfrac{dy}{dt}+P(t)y=Q(t)\) can be written:$$y=\frac1\mu\int \mu Q(t)\,dt$$where$$\mu=e^{\int P(t)\,dt}$$
In our case, we have \(P(t)=\frac1\alpha\) and \(Q(t)=\frac12kC_a\), yielding $$\mu=e^{\int\frac1\alpha dt}=e^{\frac1\alpha t}$$and thus $$y=\frac1{e^{\frac1\alpha t}}\int\frac12kC_ae^{\frac1\alpha t}dt=e^{-\frac1\alpha t}\left(\frac12\alpha kC_ae^{\frac1\alpha t}+C\right)=\frac12\alpha kC_a+Ce^{-\frac1\alpha t}$$where \(C\) is our constant of integration.

Answer 13

\[ C_r(t)=\frac{ 0.5k{\alpha} }{ 1+0.5k{\alpha} }C_{a_0}+(\frac{ C_{a_0} }{ 1+0.5k{\alpha} }-C_a^*)e^{-\frac{ (1+0.5k{\alpha})t }{ \alpha }}+(C_r^*+Ca^*-C_{a_0})e^{\frac{ -t }{ \alpha }}\]

Answer 14

It's all there this time

Answer 15

Thanks for helping, I have learned how to solve for the integrating factor. I'm going to use that and continue on. I think my chemical engineering teacher is breaking some rules but his notes are so hard to interpret. Thanks for your help

Answer 16

No problem! I'm not sure where the other terms came from -- perhaps from solving for the initial conditions? Are there are other equations to solve?

Answer 17

Ya he uses the initial and final conditions of Cr and Ca I posted in the header. These come into play when you integrate both sides, but I think this is assuming the integrating factor is treated as a constant.

Answer 18

he shows this in his notes...

Answer 19

for the left side

Answer 20

\[\int\limits_{Y(t_0)}^{Y(t)} d[Y*\mu]=Y(t)\mu(t)-Y(t_0)\mu(t_0)\]