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Pressure after pump

  • Thread starter Karol
  • Start date
  • #1
1,380
22

Homework Statement


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.

Homework 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]

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]
 

Attachments

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,233
2,751
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:
  • #3
1,380
22
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.
 
  • #4
TSny
Homework Helper
Gold Member
12,233
2,751
That's correct. I think you just made a mistake in carrying out the calculation.
 

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