Control volumes: what holds the nozzle on
引入| English | 释义 | 拼音 |
|---|---|---|
| control volume | 控制体积 | kòng zhì tǐ jī |
| continuity/kɒntɪˈnjuːɪti/ | 连续性方程 | lián xù xìng fāng chéng |
| Bernoulli | 伯努利方程 | bó nǔ lì fāng chéng |
The bolts feel the push
- A wash-down nozzle must be bolted on, and the bolt load comes from momentum and pressure together. The clean way to find it is a control volume 控制体积: draw a boundary and account for everything that crosses it.
- Cut where pressure and velocity are known, and through the bolts whose force you want.
Two balances
- Mass (continuity 连续性方程): $Q = V_1A_1 = V_2A_2$ for a liquid.
- Momentum: $\sum F = \dot m(V_2 - V_1)$, where $\sum F$ includes the pressure forces on the cut faces as well as the force from the walls.
Which term is it?
Sort each quantity into its place in the control-volume balance.
12 L/s through a 30 mm nozzle: jet speed (one decimal, m/s)?
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Q over the area 7.069e-4 m2.
Put the steps of a control-volume force calculation in order.
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The boundary comes first; the sign reversal comes last.
The nozzle bolts
- $Q = 12$ L/s enters an 80 mm pipe and leaves a 30 mm nozzle: $V_1 = \mathbf{2.39}$ m/s and $V_2 = \mathbf{17.0}$ m/s.
- Bernoulli 伯努利方程 to the jet ($p_2 = 0$ gauge) gives $p_1 = \tfrac12\rho(V_2^2 - V_1^2) = \mathbf{141}$ kPa.
- $p_1A_1 + F = \rho Q(V_2 - V_1)$ gives $F = 174.7 - 708.6 = -534$ N: the fluid pushes the nozzle downstream with 534 N, and the bolts carry it in tension.
Inlet gauge pressure upstream of that nozzle, V1 = 2.39 m/s (kPa, whole number)?
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Half rho times (V2 squared minus V1 squared).
The bolt force (N, whole number, magnitude)?
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p1 A1 minus rho Q (V2 - V1) = 708.6 - 174.7.
At the nozzle, which term is larger?
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708.6 N of pressure against 174.7 N of momentum gain.
Turning the flow: a 90° bend
- The same 12 L/s turns through $90°$ at 150 kPa gauge. Each direction carries $pA + \rho QV = 754.0 + 28.6 = 782.6$ N, so the anchor holds $\sqrt2 \times 782.6 = \mathbf{1107}$ N at $45°$.
- At this low speed the pressure term is 26 times the momentum term, which is why low-velocity, high-pressure lines still need thrust blocks.
Anchor force on the 90 degree bend, 782.6 N per direction (N, whole number)?
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Root two times the per-direction force.
A jet on a plate
- Turn the nozzle's jet onto a flat plate square to it. The plate turns the flow through 90°, so all the jet's momentum flux is removed: $F = \rho QV = 998 \times 0.012 \times 17.0 = \mathbf{203}$ N.
- The jet is at atmospheric pressure, so there is no pressure term here, only momentum. That is the force a wash-down operator feels holding the plate.
The 12 L/s, 17.0 m/s jet on a fixed plate normal to it: force (N, nearest whole)?
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$\rho QV = 998 \times 0.012 \times 16.98 = 203$ N.
Worked example: a plate that moves away
- Let the plate recede at $u = 5$ m/s. In the plate's frame the jet arrives at $17.0 - 5 = 12.0$ m/s, and only $\rho A(V - u)$ kg/s reaches it.
- $F = \rho A(V - u)^2 = 998 \times 7.069 \times 10^{-4} \times 12.0^2 = \mathbf{101}$ N, and the power delivered is $Fu = 506$ W. Turbines exploit exactly this trade between force and speed.
Same jet (30 mm nozzle) on a plate receding at 5 m/s: force (N, nearest whole)?
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$998 \times 7.069 \times 10^{-4} \times (16.98 - 5)^2 = 101$ N.
A plate moving away from the jet feels the same force as a fixed one.
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The relative speed falls and so does the mass flow reaching it: 101 N against 203 N.
Three control-volume slips
- Leaving out the pressure on the cut faces. It is often the biggest term.
- Gauge at one face, absolute at the other. Use one reference throughout.
- Force on the fluid reported as force on the pipe. They are equal and opposite.
The force on the pipe is equal and opposite to the force on the fluid.
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Newton's third law across the wall.
Recap
- $Q = VA$ is conserved: 2.39 m/s in the pipe becomes 17.0 m/s in the jet.
- $\sum F = \dot m(V_2 - V_1)$, component by component, pressure forces included.
- Nozzle bolts: 534 N in tension; bend anchor: 1107 N at 45°.