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Ok, so I am trying to solve this equasion for y
0.33392=(1/0.027)*(y+2*y^2)*((y+2*y^2)/(2.236+4.472*y))^(2/3)
I should probably know how to do this myself (I've done a million and one math courses over the past 5 years :P) but I would rather get a refresher from one of you guys/gals
The problem involves trying to find the height of water in an open channel given the rate of flow and channel cross section. y is the height of water in question.
Thanks!
Hey CivilEngineer1988.
Is a numerical approximation OK for your purposes (i.e. to so many decimal places)?
Hi CivilEngineer1988 !
Don't you see that ((y+2*y^2)/(2.236+4.472*y)) = y/2.236 ?
0.33392=(1/0.027)*(y/2.236)^(2/3)
y = 2.236*(0.33392*0.027)^(3/2) = 0.00191417
Very nice: I'm glad someone is paying attention :)
Sorry, there is something wrong in my precceding answer, because I consider the equation :
0.33392=(1/0.027)*((y+2*y^2)/(2.236+4.472*y))^(2/3)
instead of :
0.33392=(1/0.027)*(y+2*y^2)*((y+2*y^2)/(2.236+4.472*y))^(2/3)
So, after simplification, the equation is not :
0.33392=(1/0.027)*(y/2.236)^(2/3)
but :
0.33392=(1/0.027)*(y+2*y^2)*(y/2.236)^(2/3)
0.33392*0.027*2.236^(2/3) = (1+2y)*y^(5/3)
which can be solved thanks to numerical computation, as chiro suggested first.
y = 0.075210480866976