Jet on Fixed Plate — Normal Impact What is the force exerted by a water jet (normal impact) on a fixed flat plate in terms of specific weight w (N/m^3), jet area a (m^2), and jet velocity V (m/s)?

Introduction:The impact force of a jet on surfaces follows directly from Newton's second law applied to a control volume. Using specific weight instead of density is common in hydraulic machinery problems, necessitating careful unit conversions.

Given Data / Assumptions:

• Jet strikes the plate normally and comes to rest tangentially (no rebound).
• Steady incompressible flow, neglecting air effects and splashing losses.
• Specific weight w = rho * g.

Concept / Approach:Force equals rate of change of momentum. For normal impact on a fixed plate, the outgoing normal velocity is approximately zero, so the momentum reduction equals rho * a * V * V. Express density as rho = w / g to obtain the force in terms of w.

Step-by-Step Solution:Mass flow rate: m_dot = rho * a * V = (w / g) * a * V.Momentum change per second: m_dot * V = (w / g) * a * V^2.Hence, force on plate: F = (w / g) * a * V^2.

Verification / Alternative check:Dimensional check: w/g has units of kg/m^3; multiplied by a * V^2 yields N.

Why Other Options Are Wrong:w * a * V: missing one factor of V and g; incorrect units.(w * a * V^2) / (2 * g): the 1/2 factor applies to energy, not momentum force in this case.(w * V) / a: dimensionally inconsistent.

Common Pitfalls:Mixing energy head expressions (V^2 / 2g) with momentum-based force calculations.

Final Answer:F = (w / g) * a * V^2