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}}