## Elevated Storage Tank

Consider a pump delivers water from a lake up to an elevated storage tank through a 10 in pipe diameter that is 2,000 ft long. If the Hazen-Williams roughness coefficient is 100 and the pump’s discharge rate is 500 gpm, what is the friction loss? Ignore minor losses and assume turbulent flow.

Hint
Hazen-Williams Equation:
$$Q=k_1CAR_{H}^{0.63}S^{0.54}$$$where $$Q$$ is the discharge rate, $$k_1$$ is 0.849 for SI units or 1.318 for USCS, $$C$$ is the Hazen-Williams roughness coefficient, $$A$$ is the cross sectional flow area, $$R_H$$ is the hydraulic radius, and $$S$$ is the energy grade line slope. Hint 2 $$S=\frac{h_f}{L}$$$
where $$h_f$$ is the friction loss, and $$L$$ is the total length.
Hazen-Williams Equation:
$$Q=k_1CAR_{H}^{0.63}S^{0.54}$$$where $$Q$$ is the discharge rate, $$k_1$$ is 0.849 for SI units or 1.318 for USCS, $$C$$ is the Hazen-Williams roughness coefficient, $$A$$ is the cross sectional flow area, $$R_H$$ is the hydraulic radius, and $$S$$ is the slope of energy grade line. Remember, the hydraulic radius is: $$R_H=\frac{cross \: sectional\:area}{wetted\:perimeter}=\frac{D_H}{4}$$$
where $$D_H$$ is the hydraulic diameter.
Since the energy grade line slope is:
$$S=\frac{h_f}{L}$$$where $$h_f$$ is the friction loss, and $$L$$ is the total length. The friction loss is: $$\frac{Q}{k_1CA(D_{H}/4)^{0.63}}=(\frac{h_f}{L})^{0.54}$$$
$$h_f=L[\frac{Q}{k_1CA(D_{H}/4)^{0.63}}]^{\frac{1}{0.54}}$$$$$=(2,000ft)[\frac{\frac{500gal}{min}\cdot \frac{0.134ft^3}{1gal}\cdot \frac{1min}{60sec}}{(1.318)(100)(\frac{\pi (10/12ft)^2}{4})(\frac{(10/12ft)}{4})^{0.63}}]^{1.85}$$$
$$=(2,000ft)[\frac{1.117ft^3/s}{(131.8)(0.5451ft^2)(0.372ft)}]^{1.85}=(2,000ft)[\frac{1.117ft^3/s}{26.73ft^3}]^{1.85}$$$$$=(2,000ft)(0.04179)^{1.85}=(2,000ft)(0.00281)=5.6\:ft$$$
Keep in mind that the non-applicable units don’t cancel out completely when using the Hazen-Williams equation. It’s important to intentionally alter the units on the final answer. You can always verify the answer via an online calculator.
5.6 ft