Question 6.P.25: Water is flowing through a 150 mm diameter pipe and its flow...

For the pitot tube:

u=\sqrt{2 g h}              (equation 6.10)

where h is the manometer reading in m of the same fluid which flows in the pipe.

∴         h=(100 / 1000) \times((13.6-1.0) / 1.0)=1.26 m of water

The velocity at the pipe axis, u=\sqrt{(2 \times 9.81 \times 1.26)}=4.97 m/s
For turbulent flow, the Prandtl one-seventh power law can be used to give:

u_{ av }=0.82 \times u_{ axis }             (equation 3.60)

∴                             u_{ av }=0.82 \times 4.97=4.08 m/s

For the orifice meter, the average velocity is:

u=\sqrt{\frac{2\left(P_1-P_2\right) v}{1-\left(A_2 / A_1\right)^2}}

A_2=(\pi / 4) \times 0.05^2=0.00196  m ^2,  A_1=(\pi / 4) \times 0.15^2=0.0177  m ^2,

v=0.001  m ^3 / kg

∴                            u_{ av }=6.78 m / s

The coefficient of discharge = (u_{ av } rom pitot)/(u_{ av } from orifice meter).

=(4.08 / 6.78)=\underline{\underline{0.60}}