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Math Help - Transform this formula

  1. #1
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    Transform this formula

    I'd be grateful with some help transforming formula in the attached pic for H

    I am not sure from the way it is written whether the root sign covers both D and H

    My solution has it as H = Q(squared) x S x L / 0.0071 x (D5)squared

    Am I wrong if so where and why?

    TIA


    PS the formula are for calculating flow rates of gas and the drop in pressure over a length of pipe
    Attached Thumbnails Attached Thumbnails Transform this formula-pole-formula.jpeg  
    Last edited by Cardigan; February 3rd 2013 at 10:18 AM.
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  2. #2
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    Re: Transform this formula

    The actual formula is Q=0.0071(hd^5/sl)^(1/2)

    (hd^5/sl)^(1/2)= Q/0.0071

    hd^5/sl= (Q/0.0071)^2

    hd^5= sl(Q/0.0071)^2

    h=(sl(Q/0.0071)^2)/d^5

    h=(slq^2/(0.0071^2 * d^5)

    I hope it's clear, I don't know how to use LaTex, hehe.

    Source:http://www.mech.hku.hk/bse/MEBS6000/...1_04_steam.pdf (slide 10)
    Thanks from Cardigan
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  3. #3
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    Re: Transform this formula

    Hi

    Thanks for that.

    I have attached here a pic of the formula shown in your attachment and can see now how it differs from the formula in my original attachment (it's root everything and not just the top line).

    I have attached here a pic of the correct formula I want to transform for H.

    Your symbol ^ does that always mean to the power of? After d for example in the forumla I published it looks like d x 5?
    Attached Thumbnails Attached Thumbnails Transform this formula-pole_1.jpeg  
    Last edited by Cardigan; February 3rd 2013 at 11:40 PM.
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  4. #4
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    Re: Transform this formula

    Whereas I am sure that the transformed formula given by Cesc1 above here I would be most grateful if someone could tell me where I went wrong in my transformation of the formula shown on the pic attached here showing my results and workings

    Thanks
    Attached Thumbnails Attached Thumbnails Transform this formula-poles_h.jpg  
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  5. #5
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    Re: Transform this formula

    Quote Originally Posted by Cardigan View Post
    Whereas I am sure that the transformed formula given by Cesc1 above here I would be most grateful if someone could tell me where I went wrong in my transformation of the formula shown on the pic attached here showing my results and workings

    Thanks
    Cesc1's result and your result are nearly(!) indentical.

    Cesc1 wrote: h=(slq^2)/(0.0071^2 * d^5)

    I've added the missing bracket.

    You wrote: h=(q^2 * s * l)/(0.0071 * d^5)

    You forgot to square the denominator in the 2nd step.
    Thanks from Cardigan
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  6. #6
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    Re: Transform this formula

    Thanks for the help

    I'd be grateful, for my education and better understanding, if someone could explain to me why the factor 0.0071 which is outside the root sign gets squared when it moves across (see my workings on my pic in post 4 here)?
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  7. #7
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    Re: Transform this formula

    Where have I gone wrong?

    q = flow (m3/h)
    d = diameter of pipe (mm)
    h = pressure drop (mbar)
    l = length of pipe (m)
    s = specific gravity of gas (density of gas / density of air)

    h = ( q^2 * s * l ) / ( 0.0071^2 * d^5 )

    q = 6 m3/h
    d = 20mm
    l = 19M
    s = 0.58

    The result is 2.459 i.e. the pressure drops by 2.2459 mb, which on the face of it looks fine, the problem is that if I reduce the flow rate the loss of pressure over the length of the pipe drops. Which in theory means that if I start with 21mb gas pressure and have a lower flow rate I end up with a higher pressure at the end of the pipe; that can't be right can it?
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