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## Q. 6.P.10

What size of orifice would give a pressure difference of 0.3 m water gauge for the flow of a petroleum product of density 900 kg/m³ at 0.05 m³/s in a 150 mm diameter pipe?

## Verified Solution

As in previous problems, equations 6.19 and 6.21 may be used to calculate the flow through an orifice. In this problem the size of the orifice is to be found so that the simpler equation will be used in the first instance.

$G=C_D A_0 \rho \sqrt{(2 g h)}$            (equation 6.21)

$G=(0.05 \times 900)=45.0 kg / s$

$\rho=900 kg / m ^3$

h = 0.3 m of water or $(0.3 / 0.9)=0.333$ m of petroleum product

$C_D=0.62$ (assumed)

∴              $45.0=\left(0.62 \times A_0 \times 900\right) \sqrt{(2 \times 9.81 \times 0.333)}$

Thus:            $A_0=0.3155 m ^2$ and $d_0=0.2 m$

This orifice diameter is larger than the pipe size so that it was clearly wrong to use the simpler equation.
Thus:           $G=C_D A_0 \rho \sqrt{\left[2 g h /\left(1-\left(A_0 / A_1\right)^2\right)\right]}$            (equation 6.19)

$A_1=(\pi / 4)(0.15)^2=0.0177 m ^2$

∴                  $45.0=\left(0.62 \times A_0 \times 900\right) \sqrt{\left[2 \times 9.81 \times 0.33 /\left(1-\left(A_0 / 0.0177\right)^2\right)\right]}$

Thus:                      $A_0=0.154 m ^2$ and $d_0=\underline{\underline{0.14 m }}$