Civil Engineering Hydraulics - re-arranging equation.

**swedishmathsmafia** · November 30th 2011, 12:15 PM

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

**skeeter** · November 30th 2011, 12:38 PM

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 $L \approx 36.7$

**bobbyhayre** · November 30th 2011, 12:42 PM

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))

**David Green** · November 30th 2011, 12:50 PM

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

**swedishmathsmafia** · November 30th 2011, 12:53 PM

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 )

**e^(i*pi)** · November 30th 2011, 01:07 PM

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:

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

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

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

edit: this answer comes out wrong, any corrections gratefully received

**David Green** · November 30th 2011, 01:34 PM

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

**swedishmathsmafia** · November 30th 2011, 02:12 PM

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

**skeeter** · November 30th 2011, 03:16 PM

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.

**David Green** · November 30th 2011, 03:30 PM

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?

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