Thread: Civil Engineering Hydraulics - re-arranging equation.

1. Civil Engineering Hydraulics - re-arranging equation.

Hi,
I am having difficulty isolating the 'L' in the following equation. It is an equation for Total Head between two reservoirs.

10 = [(1/12.1)*(0.032*1000-L)/(0.2^5)*(0.045^2)] + [(1/12.1)*(0.032*L)/(0.2^5)*(0.045/2)^2]

Any help would be appreciated

Thanks

2. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

Originally Posted by swedishmathsmafia
Hi,
I am having difficulty isolating the 'L' in the following equation. It is an equation for Total Head between two reservoirs.

10 = [(1/12.1)*(0.032*1000-L)/(0.2^5)*(0.045^2)] + [(1/12.1)*(0.032*L)/(0.2^5)*(0.045/2)^2]

Any help would be appreciated

Thanks
$a = 1/12.1$

$b = (0.2^5) \cdot (0.045^2)$

$c = (0.2^5) \cdot \left(\frac{0.045}{2} \right)^2$

$10 = \frac{a(32-L)}{b} + \frac{a(.032L)}{c}$

$10bc = ac(32-L) + ab(.032L)$

$10bc = 32ac - acL + .032abL$

$10bc - 32ac = .032abL - acL$

$10bc - 32ac = L(.032ab - ac)$

$L = \frac{c(10b - 32a)}{a(.032b-c)}$

Assuming I interpreted you syntax correctly, I get $L \approx 36.7$

3. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

Sorry i have the answer for this question which is L= 542m if that helps, but i also require the process.

I think a clearer away of looking at this equation might be:

10 = 1/12.1 x 0.032/0.2^5 x (L x 0.045^2 + (0.045/2)^2 x (1000-L))

4. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

I got L = 32, but I could be wrong?

5. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

Annoyingly no the answer given is 543.2m.

Let me rewrite the equation

10=(1/12.1*(0.032*(1000-L))/〖0.2〗^5 *〖0.045〗^2 )+(1/12.1*(0.032*L)/〖0.2〗^5 *(0.045/2)^2 )

6. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

Originally Posted by swedishmathsmafia
Hi,
I am having difficulty isolating the 'L' in the following equation. It is an equation for Total Head between two reservoirs.

10=(1/12.1*(0.032*(1000-L))/〖0.2〗^5 *〖0.045〗^2 )+(1/12.1*(0.032*L)/〖0.2〗^5 *(0.045/2)^2 )
Thanks
$10 = \dfrac{\frac{0.032}{12.1}(1000-L)}{0.2^5 \cdot 0.045^2} + \dfrac{\frac{0.0032L}{12.1}} {0.2^5 \cdot \frac{0.045^2}{4}}$

$10 = \dfrac{32- 0.032L}{12.1 \cdot 0.2^5 \cdot 0.045^2}+ \dfrac{0.032L}{12.1 \cdot 0.2^5 \cdot \frac{0.045^2}{4}}$

Let's define some constants:

$a = 12.1 \cdot 0.2^5 \cdot 0.045^2$

$b = \dfrac{12.1 \cdot 0.2^5 \cdot 0.045^2}{4}$

$10 = \dfrac{32-0.032L}{a} +\dfrac{0.032L}{b}$

Multiply through by the LCD which is ab:

$10ab = b(32-0.032L) + 0.032aL = 32b - 0.032bL + 0.032aL$

$10ab - 32b = 0.032aL - 0.032bL = 0.032L(a-b)$

$L = \dfrac{10ab}{0.032(a-b)}$

7. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

I might want to double check the source of your answer if I was you?

I am not saying definately that the answer given is incorrect, but mistakes can and do get made when typing answers to questions.

So far I get an answer of 617.7m

8. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

the Source answer is from my lecturer, and if you substitute it in for L, you get the correct Head loss of 10.

9. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

Originally Posted by swedishmathsmafia
the Source answer is from my lecturer, and if you substitute it in for L, you get the correct Head loss of 10.
you will not get a value of 10 with the original expression you posted ...

10 = [(1/12.1)*(0.032*1000-L)/(0.2^5)*(0.045^2)] + [(1/12.1)*(0.032*L)/(0.2^5)*(0.045/2)^2]
... my advice to you is to learn Latex so your equation can be posted correctly.

10. Re: Civil Engineering Hydraulics Equation, Rearranging help needed!!

I agree with Skeeter, the smallest answer I could calculate with any reasonableness to the problem was 25.

There must be an error or something?