Difficulty: Medium
Correct Answer: ρ * a * V^2 * (sin α + sin β)
Explanation:
Introduction:
Hydraulic forces on curved vanes are evaluated by resolving momentum change along convenient axes. When a jet meets and leaves a vane tangentially, the tangential component relates to whirl, while the normal component depends on the normal projections of the jet velocity at entry and exit.
Given Data / Assumptions:
Concept / Approach:
By linear momentum, force on the vane equals mass flow rate times the vector change in jet velocity. Resolve along the normal to the vane. For tangential approach/exit, the normal components at inlet and outlet are V sin α and V sin β (both directed into the vane normal on their respective sides). Because the vane is fixed, the normal force component sums these magnitudes.
Step-by-Step Solution:
1) Mass flow rate: m_dot = ρ a V.2) Normal component of inlet velocity: V_n,in = V sin α.3) Normal component of outlet velocity: V_n,out = V sin β (same sense into the vane).4) Force along normal: F_n = m_dot * (V_n,in + V_n,out) = ρ a V^2 (sin α + sin β).
Verification / Alternative check:
Special case β = 0 (perfectly tangential exit) gives F_n = ρ a V^2 sin α, matching flat-plate normal result.
Why Other Options Are Wrong:
Cosine forms correspond to tangential components, not normal.
Difference forms imply opposing normal components, which is incorrect for tangential approach/exit on a fixed vane.
2 cos α is unrelated to normal summation.
Common Pitfalls:
Confusing angle definitions (with tangent vs with vane chord) and mixing tangential/normal projections.
Final Answer:
ρ * a * V^2 * (sin α + sin β)
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