
#1
Feb212, 05:40 AM


#2
Feb212, 06:54 AM

P: 4,570

[EDIT] I made a really stupid mistake by not reading the whole thing properly so I will post reply later. 



#3
Feb212, 07:05 AM

P: 4,570

Ok then.
For this problem I think it may actually be an implicit problem meaning that you can't just represent it in a normal functional manner (like APipe(blah) = blah). This means that you will end up getting an implicit equation instead of an explicit equation. If this is the case you will have to use a root finding algorithm to get answers given that you have values for the other parameters. It would be helpful if you show your work so that we can give you hints and help you understand for yourself what is going on. 



#4
Feb212, 08:23 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,902

Hardcore rearrangement help
Since this question is in basic algebra, not "linear and abstract algebra", I am moving it to "General Mathematics".




#5
Feb212, 09:54 AM

P: 688

I'd try the following (if there is no gross mistake in what follows): multiply up and down the righthand side by [itex]{\sqrt g}[/itex], and pass [itex]g \, C_0 A_{pipe} \sqrt 2[/itex] multiplying to the left; then take [itex]A_{tank} \sqrt{Z_{full}}[/itex] as common factors on the right and pass them dividing to the left:[tex]
\frac {g T C_0 A_{pipe}} {A_{tank}} \sqrt {\frac 2 {Z_{full}}} = 2 \sqrt g  \sqrt{\frac{C_f L P}{A_{pipe}}} [/tex]Now substitute [itex]B^2 = A_{pipe}[/itex], and multiply through by one more [itex]B[/itex]:[tex] \frac {g T C_0} {A_{tank}} \sqrt {\frac 2 {Z_{full}}} B^3  2 \sqrt g B + \sqrt{C_f L P} = 0 [/tex] That's a cubic equation in [itex]B[/itex], which should have at least one real solution (or maybe three): http://en.wikipedia.org/wiki/Cubic_f...cubic_function For each nonnegative solution, take [itex]A_{pipe} = \sqrt B[/itex] as a solution to your original problem. P.S.: Oh well, that's what chiro was saying all along... 


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