Question

Water flovs in a pipe of diameter 150 mm. The velocity of the water is measured at a certain spot which reflects the average flow velocity. A pitot static tube has a meter coefficient of C = 1,05 and is joined to a mercury manometer indicating a reading of 167 mm. Determine the flow rate of the water.

Answer 1

Answer:

Q = 0.118 $m^{3}$/s

Explanation:

Given :

diameter of the pipe, d = 150 mm

                                       = 0.15 m

Pitot tube co efficient, $C_{v}$ = 1.05

manometer reading is given, x = 167 mm

                                                   = 0.167 m

From manometer reading,we can find the difference between the manometer height, h

 $h =x\times\left [ \frac{S_{m}}{S_{w}}-1 \right ]$

$h =0.167\times\left [ \frac{13.6}{1}-1 \right ]$

h = 2.1042 m

Now, average velocity is v = $C_{v}$$\sqrt{2.g.h}$

                                            = $1.05\times \sqrt{2\times 9.81\times 2.1042}$

                                            = 6.74 m/s

Area of the pipe, A = $\frac{\pi }{4}\times d^{2}$

                                = $\frac{\pi }{4}\times 0.15^{2}$

                                = 0.0176 $m^{2}$

Therefore, flow rate is given by, Q = A.v

                                                          = 0.0176 X 6.74

                                                          = 0.118$m^{3}$/s