Real FE Civil Exam Q part 45: Fluid Mechanics 5: Water Jet Problem
Автор: Average Civil Engineer, PE
Загружено: 2025-09-25
Просмотров: 106
Описание:
"A water jet strikes a chute, changes direction by 90°, and suddenly forces act on the system! 🚀 This is a classic FE Civil Exam Fluid Mechanics problem where we apply the momentum equation step by step to find the jet force. Perfect for anyone preparing for the FE Civil or PE Civil Exam."
Key Formulas (Simple)
Momentum Equation:
Force = ṁ × ΔV
Mass Flow Rate:
ṁ = ρ × Q
Volumetric Flow Rate:
Q = A × V
Area of Jet:
A = π d² / 4
Velocity Components:
Vx = V cos(θ), Vy = V sin(θ)
Force Components:
Fx = ṁ (ΔVx), Fy = ṁ (ΔVy)
Important Concepts
Momentum principle: The force on a system equals the change in momentum of the fluid.
Control volume analysis: Always consider inlets, outlets, and the direction of velocity vectors.
Vector breakdown: Forces and velocities must be split into x and y components.
Symmetry check: Sometimes components cancel out (like Fx in this case).
Newton’s 3rd Law: The force on the chute is equal and opposite to the force on the fluid.
Final result: For this problem, the chute feels ~38 lbf downward, no force in x.
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