$5\mathrm{.}7,$ p$.173$

The "ground water reservoir" in Fig. $5\mathrm{.}31$ is the same thing as the "underground tank." The

water surface is open to atmospheric pressure.

"against a total dynamic head of $50$ m $"$ means that $h_p=50m.$

Draw the EGL and HGL along the pipeline, showing minor losses as well as frictional head loss. Q$=125$ L$/$s$.$Figure $5\mathrm{.}18$ Water System for

Problem $5\mathrm{.}7.$

Use $Q=AV$ to find the velocity in each pipe.

Given $C=100,$ use the Hazen$-$Williams equation to calculate frictional head losses.

Neglecting minor losses, use the energy equation to calculate the water surface elevation in the

elevated reservoir; use this to calculate the depth of water in the reservoir.