Pressure after Pump with -0.4[bar], 0.8x104[N/m3], 3[m/s], D1/D2=1.4, HP=30[m]

[Karol]

Homework Statement

A pump is on a line. the data are:
P₁=-0.4[bar]
$\gamma$=0.8$\cdot$10⁴[N/m³]
V₁=3[m/s]
$$\frac{D_1}{D_2}=1.4$$
H_P=30[m] (head of pump)
See attached drawing.

Homework Equations

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

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]

 

[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.]

 

[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.

 

[TSny]

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