Library / Hydraulics, open channels and fluid power / HYD-34
Hydraulics, open channels and fluid power
Fluid force on a right-angle elbow (x-component)
Horizontal component of the force of the fluid on an elbow that turns the flow from the +x direction to +y, from the inlet gauge pressure, area and speed.
- Ref
- HYD-34
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- Fx [N]
A steady flow entering an elbow along +x and leaving along +y changes its momentum, so the fluid pushes on the elbow. Using gauge pressures, the x-component of the force of the fluid on the elbow is Fx = p₁A₁ + ρ Q U₁, and the y-component is Fy = −p₂A₂ − ρ Q U₂. The weight of the fluid and the elbow are excluded, so thrust blocks must also carry them. Here Q = A U at each section.
Horizontal component of the force of the fluid on an elbow that turns the flow from the +x direction to +y, from the inlet gauge pressure, area and speed. The vertical component, Fy = −p₂A₂ − ρQU₂, and the resultant are in the derived rows.
Calculator
- Force y-component, −p₂A₂ − ρ Q U₂−1,540 N
- Resultant force magnitude2,556 N
- Direction of the resultant, from +x−37.05 °
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The fluid pushes the elbow with 2,040 N in x and −1,540 N in y, a resultant of 2,556 N.
- Most of the force is the pressure; the momentum part is only 40 N in x, which is why thrust blocks depend mainly on line pressure.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| p1 Inlet gauge pressure | 1,000 to 1e8 Pa | 50 | 1,000,000 | 1,000,000 | +10% | +9.8 % |
| A1 Inlet area | 1e-4 to 1 m² | 60 | 200,000 | 200,000 | +10% | +9.8 % |
| Q Discharge | 1e-4 to 10 m³/s | 2,000 | 22,000 | 20,000 | +10% | +0.196 % |
| U1 Inlet mean speed | 0.01 to 50 m/s | 2,000 | 3,000 | 999.8 | +10% | +0.196 % |
| ρ Liquid density | 500 to 2,000 kg/m³ | 2,020 | 2,080 | 60 | +10% | +0.196 % |
| p2 Outlet gauge pressure | 1,000 to 1e8 Pa | 2,040 | 2,040 | 0 | +10% | 0 % |
| A2 Outlet area | 1e-4 to 1 m² | 2,040 | 2,040 | 0 | +10% | 0 % |
| U2 Outlet mean speed | 0.01 to 50 m/s | 2,040 | 2,040 | 0 | +10% | 0 % |
Sweep of Inlet gauge pressure
| Inlet gauge pressure (Pa) | Force on the elbow, x-component (N) |
|---|---|
| 1,000 | 50 |
| 4,217 | 82.17 |
| 17,780 | 217.8 |
| 74,990 | 789.9 |
| 316,200 | 3,202 |
| 1,334,000 | 13,380 |
| 5,623,000 | 56,270 |
| 2.371e7 | 237,200 |
| 1e8 | 1,000,000 |
Sweep one variable
Variables and how each one contributes
Force on the elbow, x-component
What the equation gives you. Everything else feeds it.
Inlet gauge pressure
The gauge pressure at the inlet, in Pa. Raises the x-force in direct proportion to the inlet area. Raising it by 10 % (to 220,000) moves the result by +9.8 %, from 2,040 to 2,240 N.
Outlet gauge pressure
The gauge pressure at the outlet, in Pa. Changes only the y-component, so the x-force is unchanged. Raising it by 10 % (to 165,000) moves the result by +0 %, from 2,040 to 2,040 N.
Inlet area
The inlet area, in m². Raises the pressure force on the inlet in direct proportion. Raising it by 10 % (to 0.011) moves the result by +9.8 %, from 2,040 to 2,240 N.
Outlet area
The outlet area, in m². Changes only the y-component, so the x-force is unchanged. Raising it by 10 % (to 0.011) moves the result by +0 %, from 2,040 to 2,040 N.
Liquid density
The liquid density, in kg/m³. Raises the momentum part in direct proportion. Raising it by 10 % (to 1,100) moves the result by +0.196 %, from 2,040 to 2,044 N.
Discharge
The discharge, in m³/s. Raises the momentum part in direct proportion. Raising it by 10 % (to 0.022) moves the result by +0.196 %, from 2,040 to 2,044 N.
Inlet mean speed
The mean speed at the inlet, in m/s. Raises the momentum part in direct proportion. Raising it by 10 % (to 2.2) moves the result by +0.196 %, from 2,040 to 2,044 N.
Outlet mean speed
The mean speed at the outlet, in m/s. Changes only the y-component, so the x-force is unchanged. Raising it by 10 % (to 2.2) moves the result by +0 %, from 2,040 to 2,040 N.
Worked example with the values above
- Pressure force at the inlet: p₁A₁ = 200,000 × 0.01 = 2,000 N.
- Momentum flow: ρ Q U₁ = 1,000 × 0.02 × 2 = 40 N.
- Fx = 2,040 N; the y-component is −1,540 N.
Assumptions and limits
- Steady, incompressible, inviscid flow outside the boundary layer, at atmospheric pressure where stated.
- Uniform velocity profiles at the sections used.
Common mistakes
- Using the area of the pipe where the jet area is needed.
- Mixing gauge and absolute pressures in a force balance.
- Forgetting that the force on the support is the reaction to the force on the fluid.
Statistics and uncertainty
Discharge and velocity coefficients vary with the Reynolds number and with the geometry, and measured jet distances carry the reading error of the ruler, so the results are no better than 2 % to 5 %.
Where it is used
- Timing tank drainage and calibrating nozzles and orifices.
- Sizing thrust blocks, deflectors and jet impact plates.
Maths you need
- Torricelli’s law: jet speed √(2gh) from a head h
- Momentum equation: force = rate of change of momentum
- Continuity: Q = A U