Topic: Fluid Mechanics
A constant-heat water tank has, on one of its vertical sides two identical small orifices issuing two horizontal jets in the same vertical plane. The vertical distance between the centers of orifices is 1.5 m and the jet trajectories intersect at a point 0.5 m below the lower orifice. What is the approximate height of water level in the tank above the point of intersection trajectories?
- A. 1.0 m
- B. 2.5 m
- C. 0.5 m
- D. 2.0 m
Answer: B. 2.5 m
Explanation: Let the lower orifice be 2 and the upper orifice be 1. Vertical coordinates: $y_{2}=0.5m$, $y_{1}=1.5+0.5=2m$. Orifice 1 and 2 are identical, so their coefficient of velocity $C_{v}$ is the same. $C_{v} = \frac{x}{\sqrt{4yH}} \Rightarrow x^{2} = 4yH \cdot C_{v}^{2}$. At the intersection, x is same. So $y_{1}H_{1} = y_{2}H_{2}$. If H is head for orifice 2, $H_{2}=H$, and $H_{1}=H-1.5$. Thus, $2 \times (H-1.5) = 0.5 \times H \Rightarrow 1.5H = 3 \Rightarrow H = 2m$. The height of the water level above the intersection point is $H + y_{2} = 2 + 0.5 = 2.5~m$.