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## Q. 6.P.3

Water flows at between 3 l and 4 l/s through a 50 mm pipe and is metered by means of an orifice. Suggest a suitable size of orifice if the pressure difference is to be measured with a simple water manometer. What is the approximate pressure difference recorded at the maximum flowrate?

## Verified Solution

Equations 6.19 and 6.21 relate the pressure drop to the mass flowrate. If equation 6.21 is used as a first approximation, $G=C_D A_0 \rho \sqrt{(2 g h)}$.
For the maximum flow of 4 l/s, G = 4 kg/s. The largest practicable height of a water manometer will be taken as 1 m and equation 6.21 is then used to calculate the orifice area $A_0$. If the coefficient of discharge $C_D$ is taken as 0.6, then:

$4.0=0.6 A_0 \times 1000 \sqrt{(2 \times 9.81 \times 1.0)}, A_0=0.0015 m ^2$ and $d_0=0.0438 m$

The diameter, $d_0$, is comparable with the pipe diameter and hence the area correction term must be included and:

$\left[1-\left(A_0 / A_1\right)^2\right]=\left[1-\left(43.8^2 / 50^2\right)^2\right]=0.641$.

Therefore the value of $A_0$ must be recalculated as:

$4.0=0.6 A_0 \times 1000 \sqrt{(2 \times 9.8 \times 1.0) /\left[1-\left(A_0 / A_1\right)^2\right]}$

from which $A_0=0.00195 m ^2$ and d = 0.039 m or $\underline{\underline{39 mm }}$

$\sqrt{\left[1-\left(A_0 / A_1\right)^2\right]}=\sqrt{\left[1-\left(39^2 / 50^2\right)^2\right]}=0.793$

Substituting in equation 6.19:

$4.0=(0.6 \times 0.00195) \times 1000 \sqrt{(2 \times 0.001(-\Delta P) / 0.793)}$

and:              $-\Delta P=12320 N / m ^2$ or $\underline{\underline{12.3 kN / m ^2}}$