Question 5.14: Two identical orifices are mounted on one side of a vertical...
Two identical orifices are mounted on one side of a vertical tank (Fig. 5.36). The height of water above the upper orifice is 3 m. If the jets of water from the two orifices intersect at a horizontal distance of 8 m from the tank, estimate the vertical distance between the two orifices. Calculate the vertical distance of the point of intersection of the jets from the water level in the tank. Assume C_{v} = 1 for the orifices.
Let, P be the point of intersection of two jets as shown in Fig. 5.36. If t is the time taken for any liquid particle flowing in the jet from the upper orifice to reach the point P from the plane of the orifice, then,
8=u_{1} t (5.108a)
and (H-3)=\frac{1}{2} g t^{2} (5.108b)
where, u_{1} is the velocity of discharge at the plane of the upper orifice and H is the vertical distance of P from the water level in the tank. Eliminating t from Eqs (5.108a) and (5.108b),
(H-3)=\frac{1}{2} g \frac{64}{u_{1}^{2}}
or \frac{u_{1}^{2}}{g}(H-3)=32 (5.109)
again, applying the Bernoulli′s equation between the top water level and the discharge plane of the upper orifice,
u_{1}^{2}=2 g \times 3=6 g
Substituting this value of u_{1}^{2} in Eq. (5.109), we have
3(H-3)=16
or H = 8.33 m
Similarly, for the jet from the lower orifice,
8=u_{2} t
and, (H-3-h)=\frac{1}{2} g t^{2}
Eliminating t from the above two,
\frac{u_{2}^{2}}{r}(H-3-h)=32
again, u_{2}^{2}=2 g(3+h)
Hence, (3+h)(H-3-h)=16
or 3(H-3)-3 h+h(H-3)-h^{2}=16
Substituting H = 8.33 in the above expression we get
Substituting H = 8.33 in the above expression we get
h^{2}-2.33 h=0
or h(h – 2.33) = 0
which gives h = 0 and h = 2.33 m.
Therefore the distance between the orifices is 2.33 m.