A smooth nozzle discharges water under head H with outlet area a and discharge coefficient cd = 1.0. A vertical plate experiences force Fj when held across the free jet, and force Fn when pressed against the nozzle to stop the flow. What is the ratio Fj/Fn?

Introduction / Context:
Fluid jet forces on surfaces are classic momentum problems in hydraulics. Comparing the force on a plate in a free jet versus the force required to arrest the jet at the nozzle exit highlights the difference between momentum flux and stagnation pressure action. Understanding this ratio is useful for designing gates, plates, and nozzle end-stops.

Given Data / Assumptions:

• Nozzle outlet area = a, discharge coefficient cd = 1.0 (ideal discharge).
• Head H gives jet speed v = sqrt(2 g H); thus discharge Q = a * v.
• Case 1: Plate in the free jet (jet is brought to rest at the plate).
• Case 2: Plate held against the nozzle to stop flow right at the exit.

Concept / Approach:

For a free jet arrested by a plate, the force equals the rate of change of momentum: F = rho * Q * v = rho * a * v^2. For a plate closing the nozzle, the jet is not fully developed; the fluid is brought to rest at the exit with stagnation (dynamic) pressure p = 0.5 * rho * v^2 acting over area a, so F = p * a = 0.5 * rho * a * v^2. Therefore, the ratio Fj/Fn equals 2.

Step-by-Step Solution:

Verification / Alternative check:

The result is independent of H and a because both cases scale with rho * a * v^2. Real nozzles with cd ≠ 1.0 only change v slightly; the ratio remains approximately 2 for ideal assumptions.

Why Other Options Are Wrong:

• 1 or 1/2 or 3/2 or 1/4: These contradict the momentum versus stagnation pressure comparison which yields exactly 2 under ideal conditions.

Common Pitfalls:

• Applying pressure times area (0.5 rho v^2 a) to the free jet case; that underestimates the momentum force.
• Using gauge vs. absolute pressure incorrectly—forces depend on differences.

Final Answer:

2