Forces on Pipe Bends & Nozzles: Fluid Mechanics
Applying ΣF = ṁ(V₂ − V₁) to a stationary vane — force components vs deflection angle, and the 2ṁV limit at 180°.
The last lesson gave you the engine: <em>ΣF = ṁ(V₂ − V₁)</em>, the steady momentum balance for a control volume. This lesson drives it on its cleanest, most instructive track — a free jet of water striking a stationary curved vane. Because the jet is in the open air, every gauge pressure is zero and the pressure terms drop out entirely, leaving nothing but the momentum-flux change. That strips the problem down to its pure geometry: how hard does the vane get pushed, and in which direction, as you bend the jet through a larger and larger angle? The answer is a single pair of equations, and the interactive below lets you feel them move. The destination is a famous limit — turn the jet completely around (180°) and the force on the vane doubles to <em>2ṁV</em>.
A free jet: pressure drops out, geometry is all that's left
A water jet issuing into the open air is, at every point along its length, at atmospheric pressure. In the momentum balance that means every gauge pressure is zero, and the pressure-force terms that dominated the reducing-bend problem of Lesson 1 simply vanish. What remains is the purest form of the momentum equation: the force on the fluid equals the change in its momentum flux, and the change comes entirely from the vane bending the jet's direction.
Assume the vane is stationary and frictionless. 'Frictionless' means the vane does no work on the fluid, so the fluid's speed is the same leaving as arriving — only its direction changes. (A real vane with friction would slow the jet slightly; we ignore that here to see the geometry cleanly.) So the incoming velocity is V along +x, and the outgoing velocity is V at angle θ above +x: V₂ = (V cosθ, V sinθ).
Plugging into ΣF = ṁ(V₂ − V₁) and remembering the result is the force on the fluid, then flipping the sign for the force on the vane, gives the two equations that govern every vane problem. They are short enough to carry in your head, and the simulator below lets you watch them move as you drag the deflection angle.
Reading the force vs the deflection angle
Those two equations repay a slow read. Start with the x-component of the force on the vane, ṁV(1 − cosθ). At θ = 0 — no deflection, the vane is parallel to the jet — cos0 = 1 and the force is zero: a vane that does not turn the jet feels no push. As θ grows, 1 − cosθ grows, and the x-force rises. At θ = 90°, cos90° = 0 and the x-force is ṁV. At θ = 180°, cos180° = −1 and the x-force reaches ṁV(1 − (−1)) = 2ṁV — the famous result for a jet turned completely back on itself.
That doubling is not a coincidence; it is the momentum equation doing arithmetic. To reverse the jet, the vane must remove the incoming +x momentum (a change of ṁV) and supply an equal −x momentum to send it back the other way (another ṁV). The two add: a total swing of 2ṁV in the x-momentum, and so a 2ṁV x-force on the vane. A curved blade that turns a jet through 180° — the working principle of a Pelton wheel bucket — extracts twice the momentum of a flat plate that merely stops it.
The y-component, −ṁV sinθ, is the sideways push. It is zero at θ = 0 and θ = 180° (both pure x-flow in or out) and largest in magnitude at θ = 90°. Its sign tells you the deflected jet pushes the vane opposite to the deflection: a jet turned upward pushes the vane down. Drag the angle slider in the simulator and watch both components and the resultant force vector respond.
A jet striking a flat plate held normal to the flow is a special case you already have the answer to. The plate turns the axial jet into a radial sheet, destroying all the +x momentum — which is exactly the θ = 90° case with no outgoing x-component. So the force on a flat plate is ṁV, the same as any 90° deflection. This is why a flat plate is the reference: it is the vane problem with the geometry set to its simplest non-trivial value.
- Momentum flux of the jet: ṁ = ρAV = 1000(0.002)(20) = 40 kg/s. So ṁV = 800 N.
- Incoming: V₁ = (20, 0). Outgoing at θ = 90°: V₂ = (0, 20). (Speed unchanged; direction turned 90° upward.)
- Force on the fluid by the vane = ṁ(V₂ − V₁) = 40((0, 20) − (20, 0)) = 40(−20, 20) = (−800, 800) N.
- Force on the vane by the fluid = −(−800, 800) = (800, −800) N.
- So Fx on the vane = +800 N, Fy on the vane = −800 N (the deflected jet pushes the vane forward and down); |F| = √(800² + 800²) = 800√2 ≈ 1131 N.
- Check with the formula: Fx = ṁV(1 − cos90°) = 800(1 − 0) = 800 N ✓; Fy = −ṁV sin90° = −800(1) = −800 N ✓.
- ṁ = ρAV = 40 kg/s, so ṁV = 800 N.
- θ = 180°: cos180° = −1. Formula: Fx on vane = ṁV(1 − cosθ) = 800(1 − (−1)) = 800(2) = 1600 N.
- This is the classic 2ṁV = 2(800) = 1600 N result: reversing the jet removes its +x momentum and supplies equal −x momentum, doubling the force on the vane.
- ṁ = ρAV = 1000(0.003)(10) = 30 kg/s, so ṁV = 300 N.
- Fx on vane = ṁV(1 − cos60°) = 300(1 − 0.5) = 300(0.5) = 150 N.
- At 60° the vane removes half the x-momentum, giving half the flat-plate force.
Check your understanding
- For a free jet (gauge pressure zero everywhere) on a stationary frictionless vane deflected through θ, the speed is unchanged and only the direction changes.
- Force on the vane: Fx = ṁV(1 − cosθ), Fy = −ṁV sinθ — the reaction to the momentum change on the fluid.
- The x-force grows with deflection: 0 at θ = 0, ṁV at θ = 90° (also the flat-plate result), and 2ṁV at θ = 180° (jet reversed).
- Pressure plays no role because the jet is at atmospheric pressure throughout; the problem is purely geometric.
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