
#1
Feb2213, 09:13 PM

P: 179

I need to solve for C. I know it's probably simple but i don't remember how to. This is what I have so far:




#2
Feb2213, 10:33 PM

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PF Gold
P: 4,420

Because the volume flow rate entering is different from the volume flow rate leaving, you need to write down two differential equations, rather than 1:
Volume Input  Volume Output = accumulation for the volume of fluid in the tank Chemical X Input  chemical X Output = accumulation for chemical X in the tank If V(t) is the volume of fluid in the tank at time t, f_{in} is the volumetric flow rate of fluid in, and f_out is the volumetric flow rate of fluid out, what is the differential equation for V? If C(t) is the concentration of chemical X within the tank at time t, and C_in is the concentration of chemical X in the feed to the tank, what is the differential equation for the rate of change of total chemical X in the tank? The next step is to multiply the differential equation for V by C, and subtract the resulting relationship from the mass balance on chemical X. 



#3
Feb2213, 10:46 PM

P: 179

I don't follow.. The way I did it is the way the professor instructed us and the steps match the steps in his example. To solve for C, I now realize from an example in the book that A(0)=35. With that information, I can solve for C.




#4
Feb2313, 07:41 AM

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PF Gold
P: 4,420

Mixture problem. How to solve for C?
OK. I see what you did, and, of course, it is right. But, here's my alternate version to consider:
[tex]\frac{dV}{dt}=f_{in}f_{out}[/tex] [tex]\frac{d(VC)}{dt}=f_{in}C_{in}f_{out}C[/tex] Multiply the first equation by C and subtract it from the second equation: [tex]V\frac{dC}{dt}=f_{in}(C_{in}C)[/tex] where [itex]V=V_0+(f_{in}f_{out})t[/itex] So, [tex]\frac{dC}{(C_{in}C)}=f_{in}\frac{dt}{V_0+(f_{in}f_{out})t}[/tex] 


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