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 }}