T2a - Fluid Discharge 2023.pptx

Keith Vaugh
Keith VaughSTEM Education & Design à MAGVA Design + Letterpress
KV
FLUID DISCHARGE
{for energy conversion}
KV
Identify the unique vocabulary used in the description and
analysis of fluid flow with an emphasis on fluid discharge
Describe and discuss how fluid flow discharge devices
affects fluid flow.
Derive and apply the governing equations associated with
fluid discharge
Determine the power of flow in a channel or a stream and
how this can be affected by height, pressure, and/or
geometrical properties.
OBJECTIVE
S
KV
FLOW THROUGH
ORIFICES and
MOUTHPIECES
An orifice
Is an opening having a closed perimeter
A mouthpiece
Is a short tube of length not more than two to
three times its diameter
KV
FLOW THROUGH
ORIFICES and
MOUTHPIECES
u1
z1
z2 ②
H
u = u2
①
Orifice
area A
Applying Bernoulli’s equation to
stations ① and ②
putting
z1 - z2 = H, u1 = 0, u2 = u and p1 = p2
KV
TORRICELLI’s
THEOREM
Torricelli’s theorem states that the velocity of the
discharging jet is proportional to the square root of the
head producing flow. This is support by the preceding
derivation;
The discharging flow rate can be determined theoretically
if A is the cross-sectional area of the orifice
Actual discharge
KV
The velocity of the jet is less than that determined by the velocity of jet equ because
there is a loss of energy between stations ① and ② i.e
actual velocity, u = Cu √(2gH)
where Cu is a coefficient of velocity which is determined experimentally and is of the
order 0.97 - 0.98
The paths of the particles of the fluid converge on
the orifice and the area of the discharging jet at B
is less than the area of the orifice A at C
actual area of jet at B = Cc A
where Cc is the coefficient of contraction -
determined experimentally - typically 0.64
Vena contracta
C B
pB
u
uC
pC
KV
Actual discharge = Actual area at B × Actual velocity at B
we see that the relationship between the coefficients is Cd = Cc × Cu
To determine the coefficient of discharge measure the actual discharged volume
from the orifice in a given time and compare with the theoretical discharge.
KV
(a) A jet of water discharges horizontally into the atmosphere from an orifice in
the side of large open topped tank. Derive an expression for the actual
velocity, u of a jet at the vena contracta if the jet falls a distance y vertically for
a horizontal distance x, measured from the vena contracta.
(b) If the head of water above the orifice is H, calculate the coefficient of velocity.
(c) If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a
horizontal distance x of 1.5 m from the vena contracta, calculate the values of
the coefficients of velocity, discharge and contraction, given that the
volumetric flow is 0.117 m3/min and the head H above the orifice is 1.2 m
EXAMPLE 1
KV
Let t be the time taken for a particle of fluid to travel from the vena contracta
A to the point B.
KV
putting x = 1.5m, H = 1.2m and Area, A = 650×10-6m2
KV
THEORY OF
LARGE
ORIFICES
H1
H2
h
δh
D
B
KV
(a) A reservoir discharges through a sluice gate of width B and height D. The
top and bottom openings are a depths of H1 and H2 respectively below
the free surface. Derive a formula for the theoretical discharge through
the opening
(b) If the top of the opening is 0.4 m below the water level and the opening is
0.7 m wide and 1.5 m in height, calculate the theoretical discharge (in
meters per second) assuming that the bottom of the opening is above the
downstream water level.
(c) What would be the percentage error if the opening were to be treated as
a small orifice?
EXAMPLE 2
KV
Given that the velocity of flow will be greater at the bottom than at
the top of the opening, consider a horizontal strip across the
opening of height δh at a depth h below the free surface
For the whole opening, integrating from h = H1 to h = H2
KV
putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 m
KV
NOTCHES
& WEIRS
H
h
δh
H
b
KV
Consider a horizontal strip of width b and height δh at a depth h below the
free surface.
Integrating from h = 0 at the free surface to h = H at the bottom of the notch
b must be expressed in terms of h before integrating
KV
For a rectangular notch, put b = constant = B
b = constant
B
H
For a vee notch with an included angle θ, put b
= 2(H - h)tan(θ⁄2)
b = 2(H - h)tan(θ⁄2)
H
h
θ
KV
In the foregoing analysis it has been assumed that
• the velocity of the liquid approaching the notch is very small so that its
kinetic energy can be neglected
• the velocity through any horizontal element across the notch will depend
only on the depth below the free surface
These assumptions are appropriate for flow over a notch or a weir in the side of
a large reservoir
If the notch or weir is located at the end of a narrow channel, the velocity of
approach will be substantial and the head h producing flow will be increased by
the kinetic energy;
where ū is the mean velocity and α is the kinetic energy correction factor to
allow for the non-uniform velocity over the cross section of the channel
KV
Therefore
at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and
x = H + αū2/2g. Integrating between these two limits
For a rectangular notch, putting b = B = constant
KV
Pressure, p, velocity, u, and elevation, z, can cause a stream of fluid to do
work. The total energy per unit weight H of a fluid is given by
If the weight per unit time of fluid is known, the power of the stream can be
calculated;
THE POWER OF
A STREAM OF
FLUID
KV
KV
KV
In a hydroelectric power plant, 100 m3
/s of water flows from an elevation of
12 m to a turbine, where electric power is generated. The total irreversible
heat loss is in the piping system from point 1 to point 2 (excluding the
turbine unit) is determined to be 35 m. If the overall efficiency of the turbine-
generator is 80%, estimate the electric power output.
EXAMPLE 3
KV
Assumptions
1.The flow is steady and incompressible
2.Water levels at the reservoir and the discharge
site remain constant
Properties
We take the density of water to be 1000 kg/m3
The mass flow rate of water through the turbine is
We take point ➁ as the reference level, and thus z2 = 0. Therefore the
energy equation is
(Çengel, et al 2008)
KV
Also, both points ➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the
flow velocities are negligible at both points (V1 = V2 = 0). Then the energy
equation for steady, incompressible flow reduces to
Substituting, the extracted turbine head and the corresponding turbine power
are
Therefore, a perfect turbine-generator would generate 83,400 kW of electricity
from this resource. The electric power generated by the actual unit is
Note that the power generation would increase by almost 1 MW for each
percentage point improvement in the efficiency of the turbine-generator unit.
KV
Flow through orifices and mouthpieces
Theory of small orifice discharge
Torricelli’s theorem
Theory of large orifices
Notches and weirs
The power of a stream of fluid
KV
Andrews, J., Jelley, N., (2007) Energy science: principles, technologies
and impacts, Oxford University Press
Bacon, D., Stephens, R. (1990) Mechanical Technology, second edition,
Butterworth Heinemann
Boyle, G. (2004) Renewable Energy: Power for a sustainable future,
second edition, Oxford University Press
Çengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid
sciences, Third edition, McGraw Hill
Douglas, J.F., Gasoriek, J.M., Swaffield, J., Jack, L. (2011), Fluid
Mechanics, sisth edition, Prentice Hall
Turns, S. (2006) Thermal fluid sciences: An integrated approach,
Cambridge University Press
Young, D., Munson, B., Okiishi, T., Huebsch, W., 2011Introduction to
Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc.
Some illustrations taken from Fundamentals of thermal fluid sciences
1 sur 27

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T2a - Fluid Discharge 2023.pptx

  • 2. KV Identify the unique vocabulary used in the description and analysis of fluid flow with an emphasis on fluid discharge Describe and discuss how fluid flow discharge devices affects fluid flow. Derive and apply the governing equations associated with fluid discharge Determine the power of flow in a channel or a stream and how this can be affected by height, pressure, and/or geometrical properties. OBJECTIVE S
  • 3. KV FLOW THROUGH ORIFICES and MOUTHPIECES An orifice Is an opening having a closed perimeter A mouthpiece Is a short tube of length not more than two to three times its diameter
  • 4. KV FLOW THROUGH ORIFICES and MOUTHPIECES u1 z1 z2 ② H u = u2 ① Orifice area A Applying Bernoulli’s equation to stations ① and ② putting z1 - z2 = H, u1 = 0, u2 = u and p1 = p2
  • 5. KV TORRICELLI’s THEOREM Torricelli’s theorem states that the velocity of the discharging jet is proportional to the square root of the head producing flow. This is support by the preceding derivation; The discharging flow rate can be determined theoretically if A is the cross-sectional area of the orifice Actual discharge
  • 6. KV The velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity, u = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98 The paths of the particles of the fluid converge on the orifice and the area of the discharging jet at B is less than the area of the orifice A at C actual area of jet at B = Cc A where Cc is the coefficient of contraction - determined experimentally - typically 0.64 Vena contracta C B pB u uC pC
  • 7. KV Actual discharge = Actual area at B × Actual velocity at B we see that the relationship between the coefficients is Cd = Cc × Cu To determine the coefficient of discharge measure the actual discharged volume from the orifice in a given time and compare with the theoretical discharge.
  • 8. KV (a) A jet of water discharges horizontally into the atmosphere from an orifice in the side of large open topped tank. Derive an expression for the actual velocity, u of a jet at the vena contracta if the jet falls a distance y vertically for a horizontal distance x, measured from the vena contracta. (b) If the head of water above the orifice is H, calculate the coefficient of velocity. (c) If the orifice has an area of 650 mm2 and the jet falls a distance y of 0.5 m in a horizontal distance x of 1.5 m from the vena contracta, calculate the values of the coefficients of velocity, discharge and contraction, given that the volumetric flow is 0.117 m3/min and the head H above the orifice is 1.2 m EXAMPLE 1
  • 9. KV Let t be the time taken for a particle of fluid to travel from the vena contracta A to the point B.
  • 10. KV putting x = 1.5m, H = 1.2m and Area, A = 650×10-6m2
  • 12. KV (a) A reservoir discharges through a sluice gate of width B and height D. The top and bottom openings are a depths of H1 and H2 respectively below the free surface. Derive a formula for the theoretical discharge through the opening (b) If the top of the opening is 0.4 m below the water level and the opening is 0.7 m wide and 1.5 m in height, calculate the theoretical discharge (in meters per second) assuming that the bottom of the opening is above the downstream water level. (c) What would be the percentage error if the opening were to be treated as a small orifice? EXAMPLE 2
  • 13. KV Given that the velocity of flow will be greater at the bottom than at the top of the opening, consider a horizontal strip across the opening of height δh at a depth h below the free surface For the whole opening, integrating from h = H1 to h = H2
  • 14. KV putting B = 0.7 m, H1 = 0.4 m and H2 = 1.9 m
  • 16. KV Consider a horizontal strip of width b and height δh at a depth h below the free surface. Integrating from h = 0 at the free surface to h = H at the bottom of the notch b must be expressed in terms of h before integrating
  • 17. KV For a rectangular notch, put b = constant = B b = constant B H For a vee notch with an included angle θ, put b = 2(H - h)tan(θ⁄2) b = 2(H - h)tan(θ⁄2) H h θ
  • 18. KV In the foregoing analysis it has been assumed that • the velocity of the liquid approaching the notch is very small so that its kinetic energy can be neglected • the velocity through any horizontal element across the notch will depend only on the depth below the free surface These assumptions are appropriate for flow over a notch or a weir in the side of a large reservoir If the notch or weir is located at the end of a narrow channel, the velocity of approach will be substantial and the head h producing flow will be increased by the kinetic energy; where ū is the mean velocity and α is the kinetic energy correction factor to allow for the non-uniform velocity over the cross section of the channel
  • 19. KV Therefore at the free surface, h = 0 and x = αū2/2g, while at the sill , h = H and x = H + αū2/2g. Integrating between these two limits For a rectangular notch, putting b = B = constant
  • 20. KV Pressure, p, velocity, u, and elevation, z, can cause a stream of fluid to do work. The total energy per unit weight H of a fluid is given by If the weight per unit time of fluid is known, the power of the stream can be calculated; THE POWER OF A STREAM OF FLUID
  • 21. KV
  • 22. KV
  • 23. KV In a hydroelectric power plant, 100 m3 /s of water flows from an elevation of 12 m to a turbine, where electric power is generated. The total irreversible heat loss is in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to be 35 m. If the overall efficiency of the turbine- generator is 80%, estimate the electric power output. EXAMPLE 3
  • 24. KV Assumptions 1.The flow is steady and incompressible 2.Water levels at the reservoir and the discharge site remain constant Properties We take the density of water to be 1000 kg/m3 The mass flow rate of water through the turbine is We take point ➁ as the reference level, and thus z2 = 0. Therefore the energy equation is (Çengel, et al 2008)
  • 25. KV Also, both points ➀ and ➁ are open to the atmosphere (P1 = P2 = Patm) and the flow velocities are negligible at both points (V1 = V2 = 0). Then the energy equation for steady, incompressible flow reduces to Substituting, the extracted turbine head and the corresponding turbine power are Therefore, a perfect turbine-generator would generate 83,400 kW of electricity from this resource. The electric power generated by the actual unit is Note that the power generation would increase by almost 1 MW for each percentage point improvement in the efficiency of the turbine-generator unit.
  • 26. KV Flow through orifices and mouthpieces Theory of small orifice discharge Torricelli’s theorem Theory of large orifices Notches and weirs The power of a stream of fluid
  • 27. KV Andrews, J., Jelley, N., (2007) Energy science: principles, technologies and impacts, Oxford University Press Bacon, D., Stephens, R. (1990) Mechanical Technology, second edition, Butterworth Heinemann Boyle, G. (2004) Renewable Energy: Power for a sustainable future, second edition, Oxford University Press Çengel, Y., Turner, R., Cimbala, J. (2008) Fundamentals of thermal fluid sciences, Third edition, McGraw Hill Douglas, J.F., Gasoriek, J.M., Swaffield, J., Jack, L. (2011), Fluid Mechanics, sisth edition, Prentice Hall Turns, S. (2006) Thermal fluid sciences: An integrated approach, Cambridge University Press Young, D., Munson, B., Okiishi, T., Huebsch, W., 2011Introduction to Fluid Mechanics, Fifth edition, John Wiley & Sons, Inc. Some illustrations taken from Fundamentals of thermal fluid sciences

Notes de l'éditeur

  1. To appreciate energy conversion such as hydro, wave, tidal and wind power a detailed knowledge of fluid mechanics is essential. During the course of this lecture, a brief summary of the basic physical properties of fluids is provided and the conservation laws of mass and energy for an ideal (or inviscid) fluid are derived. The application of the conservation laws to situations of practical interest are also explored to illustrate how useful information about the flow can be derived. Finally, the effect of viscosity on the motion of a fluid around an immersed body (such as a turbine blade) and how the flow determines the forces acting on the body of interest.
  2. Orifices and mouthpieces can be used to measure flow rate. An orifice is an opening which has a closed perimeter. They are generally made in the walls or the bottom of a tank containing fluid and as a consequence the contents can flow through opening thereby discharging. Orifice’s may be classified by size, shape, shape of the upstream edges and the conditions of discharge. A mouthpiece is a tube not more than two to three times its diameter. Such would be fitted to a circular opening or orifice of the same diameter of a tank thereby creating an extension of the orifice. The contents of the tank could be discharged through this.
  3. An orifice is an opening found in the base or side of a tank. The pressure acting on the fluid surface forces a discharge through this opening, therefore the volumetric flow rate discharged will depend on the the head of fluid above the level of the opening. The term “small orifice” is used when the opening is small compared to the head producing flow, i.e. it can be assume that this head does not vary appreciably from point to point across the orifice. A small orifice in the side of a large tank is with a free surface open to the atmosphere is illustrated. At point on the free surface ① the pressure is atmospheric pressure p1. As the large, the velocity u1 can be considered to be negligible. The conditions in the region of the Orifice are uncertain, but at a point ② in the jet, the pressure p2 will again be atmospheric pressure with a velocity u2 which equals the jet velocity u. Establishing the datum for potential energy at the centre of the orifice and applying Bernoulli’s equ (assuming no loss in energy)
  4. u = √(2gH) can be applied to compressible or non-compressible fluids. H is expressed as the head of the fluid flowing through the orifice (H = p/ρg) The actual discharge is considerably less than the theoretical discharge therefore a coefficient of discharge must be introduced Cd There are two reasons for the difference between the theoretical and actual. What are these? The velocity of the jet is less than that determined by the velocity of jet equ because there is a loss of energy between stations ① and ② i.e actual velocity = Cu √(2gH) where Cu is a coefficient of velocity which is determined experimentally and is of the order 0.97 - 0.98
  5. In the plane of the of the orifice the streamlines have a component of velocity towards the centre and the pressure at C is greater than atmospheric pressure. At B, a small distance outside the orifice, the streamlines have become parallel. This section through B is referred to as the vena contracta
  6. The values of the coefficient of discharge, coefficient of velocity and the coefficient of contraction are determined experimentally. If the orifice is not in the bottom of the tank, then to determine measure the actual velocity, the jet’s profile must be measured.
  7. If orifice has a large opening compared to the head producing flow, then the discharge calculated using the formula for a small orifice and the head measured from the centre line of the orifice will yield an inaccurate result, i.e. the velocity will vary substantially from top to bottom. There for a large orifice the preferred method to is calculate the flow through a thin horizontal strip across the orifice opening an integrate from top to bottom in order to determine the theoretical discharge. From this the actual discharge can be determined providing the coefficient of discharge is known.
  8. An opening in the side of a tank or a reservoir which extends above the free surface is referred to as a Notch. It is essentially a large orifice which does not have an upper edge and therefore has a variable area depending upon the level of the free surface. A weir is a notch on much larger scale i.e. it may have a large width in the direction of flow. The method developed for determining the theoretical flow through a large orifice is also used to determine the theoretical flow for a notch.
  9. This theory applies to a notch of any shape.
  10. NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch).
  11. NOTE - the value of ū is obtained by dividing the discharge by the full cross-sectional area of the channel (not the notch).