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

Q = 0.118
`m^(3)`/s

Step-by-step 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*\left [ (S_(m))/(S_(w))-1 \right ]`

`h =0.167*\left [ (13.6)/(1)-1 \right ]`

h = 2.1042 m

Now, average velocity is v =
`C_(v)`
`√(2.g.h)`

=
`1.05* √(2* 9.81* 2.1042)`

= 6.74 m/s

Area of the pipe, A =
`(\pi )/(4)* d^(2)`

=
`(\pi )/(4)* 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