Question

The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum main pressure (85 psig), the water discharge rate is 1620 gpm through a 2.5-inch open fire hydrant. The water main, in which the velocity is very small, is situated 8 ft below the hydrant discharge point. Determine the friction loss from the main to the discharge point. Assume atmospheric pressure is 15 psia.

Answer 1

Explanation:

The given data is as follows.

      $P_{1}$ = 85 psig,   $P_{2}$ = $P_{atm}$ = 15 psia

        Q = 1620 gpm,  d = 2.5 inch,     l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

                        $h_{l} = \frac{4fl \nu^{2}}{2gD}$  ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that $R_{l} = \frac{\rho \nu d}{\mu}$

where,  $\mu_{water}$ = $10^{-3}$ kg/ms

As it is known that 1 gpm = $\frac{1}{3.67} m^{3}/hr$

So,  $1 m^{3}/hr$ = 3.67 gpm

Therefore,   Q = $1620 \times \frac{1}{3.67}$

                        = $441.4168 m^{3}/hr$

                         = 0.1226 $m^{3}/s$

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

                V = $\frac{Q}{\frac{\pi}{4}d^{2}}$

                    = $\frac{0.1226}{0.785 \times (0.0635)^{2}}$

                    = 38.73 m/s

Hence, we will calculate Reynold number as follows.

             $R_{l}$ = $\frac{1000 \times 38.73 \times 0.0635}{10^{-3}}$

                             = 2459355

As $R_{l}$ > 2000 then, it means that flow is turbulent.

As, f = 0.079 $R^{-0.25}_{l}$

        = 0.001994

Putting all the values into equation (1) formula as follows.    

                          $h_{l} = \frac{4fl \nu^{2}}{2gD}$

                                     = $\frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}$

                                      = $1.04069 \times 10^{5} m$

Thus, we can conclude that friction loss from the main to the discharge point is $1.04069 \times 10^{5} m$.