Question 5.13: For the configuration shown, (Fig. 5.35) calculate the minim...
For the configuration shown, (Fig. 5.35) calculate the minimum or just sufficient head H in the vessel and the corresponding discharge which can pass over the plate. (Take C_{v}=1, C_{d} = 0.8)
For a point \text { ‘O' } on the trajectory,
x=u_{1} t
and z=\frac{1}{2} g t^{2}
where t is the time taken for any liquid particle to reach the point \text { ‘O' } after being ejected from the orifice with a velocity u_{1}. Eliminating t from the two equations, we get,
z=\frac{g}{2} \frac{x^{2}}{u_{1}^{2}}
which shows that the trajectory must be parabolic.
Again, for C_{v}=1,
u_{1}=\sqrt{2 g H}
Therefore, z=\frac{x^{2}}{4 H}
or H=\frac{x^{2}}{4 z}
Hence, the minimum head H to pass over c
=(1)^{2} / 4 \times(0.5)=0.5 m
The corresponding minimum discharge
Q=\sqrt{2 \times 9.81 \times 0.5} \times\left(2 \times 10^{-4}\right) \times 0.8 m ^{3} / s
=0.0005 m ^{3} / s