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Math Help - Thermodynamic formula - transposition help please

  1. #1
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    Thermodynamic formula - transposition help please

    Hi all,

    I've recently decided to get back into education having been out for years....I was ok at Math's back in the day!!

    I'm studying thermodynamics at HND level and have come across a formula I just can't seem to get my head around.

    I have the answer from the worked example solution but can't get how he got there. any help much appreciated

    Formula for calculating amount of flash steam generated from hot condensate

    (mg + mf ) hf1 = mg x hg2 + mf x hf2 .....we're looking for mg - the mass of gas/steam released

    the figures are as follows

    (mg+mf) = 550 -- the amount of condensate supplied (mf is water content, mg is vapour content)

    hf1 = 670 -- enpalthy value of condensate going in

    hg2 = 2697 -- enpalthy value of steam in vessel at a different pressure

    hf2 = 467 -- enpalthy of water contenet of steam in vessel

    In order to remove one of the two unknown's mf which is not given = (550 - mg)

    Putting into formula

    550 x 670 = mg x 2697 + (550 - mg)467

    it's at this point that the formula needs to be transposed for mg and I have spent ages with no success.

    Thanks in advance guys.
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  2. #2
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    Re: Thermodynamic formula - transposition help please

    (mg + mf ) hf1 = mg x hg2 + mf x hf2 .....we're looking for mg - the mass of gas/steam released
    I'm just guessing with the variables you provided w/regard to subscripts and such ...

    mg \cdot hf_1 + mf \cdot hf_1 = mg \cdot hg_2 + mf \cdot hf_2

    mg \cdot hf_1 - mg \cdot hg_2 = mf \cdot hf_2 - mf \cdot hf_1

    mg(hf_1 - hg_2) = mf(hf_2 - hf_1)

    mg = \frac{mf(hf_2 - hf_1)}{hf_1 - hg_2}
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