Question

Water is pumped from a lake through an 9-in-diameter pipe at a rate of 10 ft3/s. If viscous effects are negligible, what is the gauge pressure in the suction pipe (the pipe between the lake and the pump) at an elevation of 7 ft above the lake?

Answer 1

Explanation:

Expression to calculate gauge pressure is as follows.

    $\frac{p_{1}}{\gamma} + \frac{v^{2}_{1}}{2g} + z_{1} = \frac{p_{2}}{\gamma} + \frac{v^{2}_{2}}{2g} + z_{2}$

where,   $\frac{p}{\gamma}$ = pressure head

             $\frac{v^{2}}{2g}$ = velocity head z

where,   $p_{1}$ = 0, $v_{1}$ = 0, $z_{1}$ = 0, and $z_{2}$ = 6.0 ft

          $V_{2} = \frac{Q}{A_{2}} = \frac{4Q}{\pi \times D^{2}_{2}}$

                      = $\frac{4 \times 10 ft^{3}/s}{3.14 \times (\frac{9}{12}ft)^{2})}$

                      = 22.64 ft/s

Therefore,

             $p_{2} = -\gamma z_{2} - \frac{1}{2} \rho V^{2}_{2}$

                      = $-62.4 lb/ft^{3} \times 7 ft - \frac{1}{2} \times 1.94 slugs/ft^{3} \times 22.64 ft/s$

                    = 414.9 $lb/ft^{2}$

As $1 lbf/ft^{2} = 0.00694 psi$

So,                414.9 $lb/ft^{2}$ = $414.9 \times 0.00694$

                                    = 2.879 psi

Thus, we can conclude that the gauge pressure in the suction pipe is 2.879 psi.