Answer:
[tex]P = 2.246\times 10^{5}\,Pa[/tex]
Explanation:
The horizontal pipe is modelled after the Bernoulli's Principle. The fluid speeds of at each stage of the horizontal pipe are, respectively:
[tex]v_{1} = \frac{7.2\times 10^{-3}\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.08\,m)^{2} }[/tex]
[tex]v_{1} \approx 1.433\,\frac{m}{s}[/tex]
[tex]v_{2} = \frac{7.2\times 10^{-3}\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.04\,m)^{2} }[/tex]
[tex]v_{2} \approx 5.730\,\frac{m}{s}[/tex]
The absolute pressure of water is:
[tex]\frac{2.40\times 10^{5}\,Pa}{\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +\frac{\left(1.433\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} = \frac{P}{\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}}} \right)} + \frac{\left(5.730\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]P = 2.246\times 10^{5}\,Pa[/tex]
