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2 problems of fluids help

  1. Dec 3, 2004 #1
    16. The normal atmosphercic pressure is 1.013 x 10^5. A storm causes that the height of the barometer of mercury reduce 20.0 mm form the normal height. What is the atmosferic preasurre ? ( the density of mercury is 13.59 g/cm3


    50. Out of a fire extingisher comes out water under air presurre. What manometric pressure is required in the tank for the water burst to have a speed of 30 m/s when the water level is 0.5 m below the water pump.


    as hard as i try i really cant solve is problems can anyone help me???
     
  2. jcsd
  3. Dec 3, 2004 #2
    i got the first one , can anyone help me with the other one?
     
  4. Dec 4, 2004 #3

    Andrew Mason

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    It is not entirely clear on the question but it seems that the air pressure has to lift the water .5 m and send it out at 30 m/sec.

    Think of pressure as an energy density (energy/unit volume):
    [tex]P = \frac{\delta E}{\delta V}[/tex]

    In order to lift an element [itex]\delta m = \rho \delta v[/itex] of water h = .5 m. and accelerate it to v = 30 m/sec energy of:

    [tex]\delta E = \delta mgh + \frac{1}{2}\delta mv^2[/tex]

    is required. So an energy density of:

    [tex]\frac{\delta E}{\delta V} = \rho gh + \frac{1}{2}\rho v^2 = P[/tex]

    [tex]P = \rho (gh + \frac{1}{2}v^2)[/tex]

    [tex]P = 10^3 (9.8 \times .5 + .5 \times 900)[/tex]

    [tex]P = 4.55 \times 10^2 KPa[/tex] or about 4.55 atm. (4.55 bars)

    AM
     
    Last edited: Dec 4, 2004
  5. Dec 4, 2004 #4
    thanks andrew but i have another question

    so i did P = 10^3(9.8*0.5 +900*0.5) = and it gives me another answer diffrent that yours it gives me 4.55 x 10^5 half of what you got , why did u multiply it by two ??
     
  6. Dec 4, 2004 #5

    Andrew Mason

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    Just forgot to multiply by the .5. See edited reply above

    AM
     
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