Respuesta :
Explanation:
Expression to calculate gauge pressure is as follows.
[tex]\frac{p_{1}}{\gamma} + \frac{v^{2}_{1}}{2g} + z_{1} = \frac{p_{2}}{\gamma} + \frac{v^{2}_{2}}{2g} + z_{2}[/tex]
where, [tex]\frac{p}{\gamma}[/tex] = pressure head
[tex]\frac{v^{2}}{2g}[/tex] = velocity head z
where, [tex]p_{1}[/tex] = 0, [tex]v_{1}[/tex] = 0, [tex]z_{1}[/tex] = 0, and [tex]z_{2}[/tex] = 6.0 ft
[tex]V_{2} = \frac{Q}{A_{2}} = \frac{4Q}{\pi \times D^{2}_{2}}[/tex]
= [tex]\frac{4 \times 10 ft^{3}/s}{3.14 \times (\frac{9}{12}ft)^{2})}[/tex]
= 22.64 ft/s
Therefore,
[tex]p_{2} = -\gamma z_{2} - \frac{1}{2} \rho V^{2}_{2}[/tex]
= [tex]-62.4 lb/ft^{3} \times 7 ft - \frac{1}{2} \times 1.94 slugs/ft^{3} \times 22.64 ft/s[/tex]
= 414.9 [tex]lb/ft^{2}[/tex]
As [tex]1 lbf/ft^{2} = 0.00694 psi[/tex]
So, 414.9 [tex]lb/ft^{2}[/tex] = [tex]414.9 \times 0.00694[/tex]
= 2.879 psi
Thus, we can conclude that the gauge pressure in the suction pipe is 2.879 psi.
