发表于:2007-04-25 17:04:00
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Bernoulli's TheoremMany flowmeters are based on the work of Daniel Bernoulli in the 1700s. Bernoulli's theorem relates to the Steady Flow Energy Equation (SFEE), and states that the sum of:
Pressure energy,
Kinetic energy and
Potential energy
will be constant at any point within a piping system (ignoring the overall effects of friction). This is shown below, mathematically in Equation 4.2.2 for a unit mass flow:
Equation 4.2.2
Where:
P1 and P2 = Pressure at points within a system (Pa)
u1 and u2 = Velocities at corresponding points within a system (m/s)
h1 and h2 = Relative vertical heights within a system (m)
ρ = Density (kg/m3)
g = Gravitational constant (9.81 m/s2)
Bernoulli's equation ignores the effects of friction and can be simplified as follows:
Pressure energy + Potential energy + Kinetic energy = Constant
Equation 4.2.3 can be developed from Equation 4.2.2 by multiplying throughout by 'ρg'.
Equation 4.2.3
Friction is ignored in Equations 4.2.2 and 4.2.3, due to the fact that it can be considered negligible across the region concerned. Friction becomes more significant over longer pipe lengths. Equation 4.2.3 can be further developed by removing the 2nd term on either side when there is no change in reference height (h). This is shown in Equation 4.2.4:
Equation 4.2.4
Example 4.2.2
Determine P2 for the system shown in Figure 4.2.4, where water flows through a diverging section of pipe at a volumetric rate of 0.1 m3/s at 10°C.
The water has a density of 998.84 kg/m3 at 10°C and 2 bar g.
Fig. 4.2.4 System described in Example 4.2.2
From Equation 4.1.4:
Equation 4.1.4
Where:
qv = Volumetric flowrate (m/s)
A = Cross-sectional area (m2))
u = Velocity (m/s)
By transposing the Equation 4.1.4, a figure for velocity can be calculated:
Equation 4.2.4 is a development of Equation 4.2.3 as described previously, and can be used to predict the downstream pressure in this example.
Equation 4.2.4
Example 4.2.2 highlights the implications of Bernoulli's theorem. It is shown that, in a diverging pipe, the downstream pressure will be higher than the upstream pressure. This may seem odd at first glance; it would normally be expected that the downstream pressure in a pipe is less than the upstream pressure for flow to occur in that direction. It is worth remembering that Bernoulli states, the sum of the energy at any point along a length of pipe is constant.
In Example 4.2.2, the increased pipe bore has caused the velocity to fall and hence the pressure to rise. In reality, friction cannot be ignored, as it is impossible for any fluid to flow along a pipe unless a pressure drop exists to overcome the friction created by the movement of the fluid itself. In longer pipes, the effect of friction is usually important, as it may be relatively large. A term, hf, can be added to Equation 4.2.4 to account for the pressure drop due to friction, and is shown in Equation 4.2.5.
Equation 4.2.5
With an incompressible fluid such as water flowing through the same size pipe, the density and velocity of the fluid can be regarded as constant and Equation 4.2.6 can be developed from Equation 4.2.5 (P1 = P2 + hf).
Equation 4.2.6
Equation 4.2.6 shows (for a constant fluid density) that the pressure drop along a length of the same size pipe is caused by the static head loss (hf) due to friction from the relative movement between the fluid and the pipe. In a short length of pipe, or equally, a flowmetering device, the frictional forces are extremely small and in practice can be ignored. For compressible fluids like steam, the density will change along a relatively long piece of pipe. For a relatively short equivalent length of pipe (or a flowmeter using a relatively small pressure differential), changes in density and frictional forces will be negligible and can be ignored for practical purposes. This means that the pressure drop through a flowmeter can be attributed to the effects of the known resistance of the flowmeter rather than to friction.
Some flowmeters take advantage of the Bernoulli effect to be able to measure fluid flow, an example being the simple orifice plate flowmeter. Such flowmeters offer a resistance to the flowing fluid such that a pressure drop occurs over the flowmeter. If a relationship exists between the flow and this contrived pressure drop, and if the pressure drop can be measured, then it becomes possible to measure the flow.
Quantifying the relationship between flow and pressure drop
Consider the simple analogy of a tank filled to some level with water, and a hole at the side of the tank somewhere near the bottom which, initially, is plugged to stop the water from flowing out (see Figure 4.2.5). It is possible to consider a single molecule of water at the top of the tank (molecule 1) and a single molecule below at the same level as the hole (molecule 2).
With the hole plugged, the height of water (or head) above the hole creates a potential to force the molecules directly below molecule 1 through the hole. The potential energy of molecule 1 relative to molecule 2 would depend upon the height of molecule 1 above molecule 2, the mass of molecule 1, and the effect that gravitational force has on molecule 1's mass. The potential energy of all the water molecules directly between molecule 1 and molecule 2 is shown by Equation 4.2.7.
Equation 4.2.7
Where:
mv = Mass of the molecule
g = Gravitational constant (9.81 m/s2)
h = Height of molecule above the hole
Fig. 4.2.5 A tank of water with a plugged hole near the bottom of the tank
Molecule 1 has no pressure energy (the nett effect of the air pressure is zero, because the plug at the bottom of the tank is also subjected to the same pressure), or kinetic energy (as the fluid in which it is placed is not moving). The only energy it possesses relative to the hole in the tank is potential energy.
Meanwhile, at the position opposite the hole, molecule 2 has a potential energy of zero as it has no height relative to the hole. However, the pressure at any point in a fluid must balance the weight of all the fluid above, plus any additional vertical force acting above the point of consideration. In this instance, the additional force is due to the atmospheric air pressure above the water surface, which can be thought of as zero gauge pressure. The pressure to which molecule 2 is subjected is therefore related purely to the weight of molecules above it.
Weight is actually a force applied to a mass due to the effect of gravity, and is defined as mass x acceleration. The weight being supported by molecule 2 is the mass of water (m) in a line of molecules directly above it multiplied by the constant of gravitational acceleration, (g). Therefore, molecule 2 is subjected to a pressure force m g.
But what is the energy contained in molecule 2? As discussed above, it has no potential energy; neither does it have kinetic energy, as, like molecule 1, it is not moving. It can only therefore possess pressure energy.
Mechanical energy is clearly defined as Force x Distance,
so the pressure energy held in molecule 2 = Force (m g) x Distance (h) = m g h, where:
m = Mass of all the molecules directly between and including molecule 1 and molecule 2
g = Gravitational acceleration 9.81 m/s2
h = Cumulative height of molecules above the hole
It can therefore be seen that:
Potential energy in molecule 1 = m g h = Pressure energy in molecule 2.
This agrees with the principle of conservation of energy (which is related to the First Law of Thermodynamics) which states that energy cannot be created or destroyed, but it can change from one form to another. This essentially means that the loss in potential energy means an equal gain in pressure energy.
Consider now, that the plug is removed from the hole, as shown in Figure 4.2.6. It seems intuitive that water will pour out of the hole due to the head of water in the tank.
In fact, the rate at which water will flow through the hole is related to the difference in pressure energy between the molecules of water opposite the hole, inside and immediately outside the tank. As the pressure outside the tank is atmospheric, the pressure energy at any point outside the hole can be taken as zero (in the same way as the pressure applied to molecule 1 was zero). Therefore the difference in pressure energy across the hole can be taken as the pressure energy contained in molecule 2, and therefore, the rate at which water will flow through the hole is related to the pressure energy of molecule 2.
In Figure 4.2.6, consider molecule 2 with pressure energy of m g h, and consider molecule 3 having just passed through the hole in the tank, and contained in the issuing jet of water.
Fig. 4.2.6 The plug is removed from the tank
Molecule 3 has no pressure energy for the reasons described above, or potential energy (as the fluid in which it is placed is at the same height as the hole). The only energy it has can only be kinetic energy.
At some point in the water jet immediately after passing through the hole, molecule 3 is to be found in the jet and will have a certain velocity and therefore a certain kinetic energy. As energy cannot be created, it follows that the kinetic energy in molecule 3 is formed from that pressure energy held in molecule 2 immediately before the plug was removed from the hole.
It can therefore be concluded that the whole of the kinetic energy held in molecule 3 equals the pressure energy to which molecule 2 is subjected, which, in turn, equals the potential energy held in molecule 1.
The basic equation for kinetic energy is shown in Equation 4.2.8:
Equation 4.2.8
Where:
mv = Mass of the object (kg)
u = Velocity of the object at any point (m/s)
If all the initial potential energy has changed into kinetic energy, it must be true that the potential energy at the start of the process equals the kinetic energy at the end of the process. To this end, it can be deduced that:
Equation 4.2.9
Equation 4.2.10
Equation 4.2.10 shows that the velocity of water passing through the hole is proportional to the square root of the height of water or pressure head (h) above the reference point, (the hole). The head 'h' can be thought of as a difference in pressure, also referred to as pressure drop or 'differential pressure'.
Equally, the same concept would apply to a fluid passing through an orifice that has been placed in a pipe. One simple method of metering fluid flow is by introducing an orifice plate flowmeter into a pipe, thereby creating a pressure drop relative to the flowing fluid. Measuring the differential pressure and applying the necessary square-root factor can determine the velocity of the fluid passing through the orifice.
The graph (Figure 4.2.7) shows how the flowrate changes relative to the pressure drop across an orifice plate flowmeter. It can be seen that, with a pressure drop of 25 kPa, the flowrate is the square root of 25, which is 5 units. Equally, the flowrate with a pressure drop of 16 kPa is 4 units, at 9 kPa is 3 units and so on.
Fig. 4.2.7 The square-root relationship of an orifice plate flowmeter
Knowing the velocity through the orifice is of little use in itself. The prime objective of any flowmeter is to measure flowrate in terms of volume or mass. However, if the size of the hole is known, the volumetric flowrate can be determined by multiplying the velocity by the area of the hole. However, this is not as straightforward as it first seems.
It is a phenomenon of any orifice fitted in a pipe that the fluid, after passing through the orifice, will continue to constrict, due mainly to the momentum of the fluid itself. This effectively means that the fluid passes through a narrower aperture than the orifice. This aperture is called the 'vena contracta' and represents that part in the system of maximum constriction, minimum pressure, and maximum velocity for the fluid. The area of the vena contracta depends upon the physical shape of the hole, but can be predicted for standard sharp edged orifice plates used for such purposes. The ratio of the area of the vena contracta to the area of the orifice is usually in the region of 0.65 to 0.7; consequently if the orifice area is known, the area of the vena contracta can be established. The subject is discussed in further detail in the next Section.
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