Question

A water jet (density = 1000 kg/m$^3$) is approaching a vertical plate, having an orifice at the center, as shown in the figure. While a part of the jet passes through the orifice, remainder flows along the plate. Neglect friction and assume both the inlet and exit jets to have circular cross-sections. If $V = 5$ m/s, $D = 100$ mm and $d = 25$ mm, magnitude of the horizontal force (in N, rounded off to one decimal place) required to hold the plate in its position is $________________$.

Hint:

When solving jet impact problems, always separate the flow into two components: through the orifice (retains velocity direction) and deflected flow (loses horizontal component). Apply momentum conservation carefully.

Answer 1

Correct Answer: 183

Step 1: Understand the problem.
A water jet with velocity $V$ and diameter $D$ strikes a plate with an orifice of diameter $d$. Some water passes through the orifice, while the rest is deflected. The required horizontal force is obtained from momentum change in the jet.
Step 2: Write given data.
Density of water: $\rho = 1000 \ \text{kg/m}^3$
Velocity: $V = 5 \ \text{m/s}$
Inlet diameter: $D = 0.1 \ \text{m}$
Orifice diameter: $d = 0.025 \ \text{m}$
Step 3: Compute areas.
Inlet jet area: \[ A_1 = \frac{\pi}{4} D^2 = \frac{\pi}{4} (0.1)^2 = 7.854 \times 10^{-3} \ \text{m}^2 \] Orifice area: \[ A_2 = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.025)^2 = 4.909 \times 10^{-4} \ \text{m}^2 \]
Step 4: Compute mass flow rates.
Inlet mass flow: \[ \dot{m}_1 = \rho A_1 V = 1000 \times 7.854 \times 10^{-3} \times 5 = 39.27 \ \text{kg/s} \] Through orifice: \[ \dot{m}_2 = \rho A_2 V = 1000 \times 4.909 \times 10^{-4} \times 5 = 2.455 \ \text{kg/s} \] Deflected flow: \[ \dot{m}_3 = \dot{m}_1 - \dot{m}_2 = 39.27 - 2.455 = 36.815 \ \text{kg/s} \]

Step 5: Apply momentum principle.
- Inlet momentum flux (horizontal): \[ M_{in} = \dot{m}_1 V = 39.27 \times 5 = 196.35 \ \text{N} \] - Outlet momentum flux (horizontal): Only orifice contributes, since deflected flow has zero horizontal momentum. \[ M_{out} = \dot{m}_2 V = 2.455 \times 5 = 12.275 \ \text{N} \] - Net force on plate: \[ F = M_{in} - M_{out} = 196.35 - 12.275 = 184.075 \ \text{N} \]
Step 6: Round off.
\[ F \approx 184.1 \ \text{N} \] Final Answer: \[ \boxed{184.1 \ \text{N}} \]