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)?

Difficulty: Easy

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

Explanation:


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

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