Pressure after pump

1. Aug 30, 2012

Karol

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

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

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

Attached Files:

• Pump.png
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2. Aug 30, 2012

TSny

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
3. Aug 30, 2012

Karol

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. Aug 30, 2012

TSny

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

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