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Simple fluids pressure question

  1. Aug 15, 2011 #1
    Hello friends I am again in need of some asistance,

    The Question is as follows:

    The pressure head in a gas main at a point 120 m above sea level is equivalent to 180 mm of water. Assuming that the densities of air and gas remain constant and equal to 1.202 kg m^-3 and 0.561 kg m^-3, respectively, what will be the pressure head in millimeters of water at sea level.

    my attempt was to use the densities given i.e (0.561/1.202) and multiply it by 120m.
    then using the larger number as the height to calculate the pressure in the air, thus using the smaller number as the height used for calculating the pressure from the gas.

    so i end up with the following steps

    0.561*56*981 - pgh (gas)
    1.202*9.81*64 - pgh (air)

    sum those two values up and use the equation for pressure head,

    P/(pg) = (pgh (gas) + pgh (air)) / (1000 *9.81) = 108 mm

    however the answer in the book is 103 mm.

    Any help is welcome,

    Cheers - Iclaudius
  2. jcsd
  3. Aug 15, 2011 #2
    back up you go :)
  4. Aug 16, 2011 #3
    Anyone? :(
  5. Sep 21, 2012 #4
    Let p* be the gas pressure at a height of 120 m above sea level and d* be the density of the gas. Let p be the pressure of air at sea level and d be the density of air. Let z be the elevation ie. z=120 m. Let h= 0.18 m water head at elevation of 120 m and d' be the density of water.
    As the water head is the difference of the gas pressure and the air pressure at z = 120 m,

    For z = 120 m, we can write, p* - (p - dgz) = d'gh............eqn.(1) (p-dgz is the air pressure at height z above the sea level)

    Again for sea level, we can write (p* + d*gz) - p = d'gx.........eqn.(2)...(p* + d*gz is the gas pressure at sea level and let x be the water head at sea level)

    Substracting eqn 2 from eqn 1, we have: dgz - d*gz = d'g(h-x)
    or (d-d*)z =d'(h-x).
    Putting d=1.202, d* = 0.561, d' = 1000 and h = 0.18 we finally get x =0.103m or x = 103mm.

    I hope the steps are clear.
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