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Pressure after pump

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A pump is on a line. the data are:
    P1=-0.4[bar]
    [itex]\gamma[/itex]=0.8[itex]\cdot[/itex]104[N/m3]
    V1=3[m/s]
    [tex]\frac{D_1}{D_2}=1.4[/tex]
    HP=30[m] (head of pump)
    See attached drawing.

    2. Relevant equations
    Bernoulli equation:
    [tex]H_1+\frac{V_1^2}{2g}+\frac{P_1}{\gamma}+H_P=H_2+ \frac{V_2^2}{2g} +\frac{P_2}{\gamma}[/tex]

    3. The attempt at a solution
    The velocity after the pump:
    [tex]V_2=V_1\frac{D_1^2}{D_2^2}=3\cdot 1.4^2=5.9[/tex]
    Bernoulli equation:
    [tex]\frac{-0.4\times10^5}{8000}+\frac{3^2}{20}+30=0.3+\frac{5.9^2}{20}+\frac{P_2}{\gamma}[/tex]
    [tex]\Rightarrow\frac{P_2}{\gamma}=27.9 \Rightarrow P_2=2.2\times10^5[/tex]
    The answer should be 1.87x105[pa]
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2012 #2

    TSny

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    Homework Helper
    Gold Member

    Hello.
    The negative pressure for P1 is a clue that you're dealing with gauge pressures. [EDIT: However, it's ok to use gauge pressures in the equation. I think your setup is correct. Check the calculation again.]
     
    Last edited: Aug 30, 2012
  4. Aug 30, 2012 #3
    so what? i can put, in bernoulli equation, either absolute or gauge pressures, as i understand, if i use the same type on both sides, which i have done here.
     
  5. Aug 30, 2012 #4

    TSny

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    Homework Helper
    Gold Member

    That's correct. I think you just made a mistake in carrying out the calculation.
     
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