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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Faculty of Engineering andTechnical Studies
4.1 Application of Bernoulli Equation
X0
Figure 4.1: Streamlines passing a non-rotating obstacle
4.1.1 Pitot tube A point in a fluid stream where the
velocity is reduced to zero is known as a stagnation point.
Any non-rotating obstacle placed in the stream produces a stagnation point next to its upstream surface.
The velocity at X is zero: X is a stagnation point.
By Bernoulli's equation the quantity p + ½V2 + gz is constant along a streamline for the steady frictionless flow of a fluid of constant density.
If the velocity V at a particular point is brought to zero the pressure there is increased from p to p + ½V2.
For a constant-density fluid the quantity p + ½V2 is therefore known as the stagnation pressure of that streamline while ½V2 – that part of the stagnation pressure due to the motion – is termed the dynamic pressure.
A manometer connected to the point X would record the stagnation pressure, and if the static pressure p were also known ½V2 could be obtained by subtraction, and hence V calculated.
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
A right-angled glass tube, large enough for capillary effects to be negligible, has one end (A) facing the flow. When equilibrium is attained the fluid at A is stationary and the pressure in the tube exceeds that of the surrounding stream by ½V2. The liquid is forced up the vertical part of the tube to a height :
h = p/g = ½V2/g = V2/2g
above the surrounding free surface. Measurement of h therefore enables V to be calculated.
(4.3)ghV 2
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Measurement of the static pressure may be made at the boundary of the flow, as illustrated in (a), provided that the axis of the piezometer is perpendicular to the boundary and the connection is smooth and that the streamlines adjacent to it are not curved
A tube projecting into the flow (Tube c) does not give a satisfactory reading because the fluid is accelerating round the end of the tube.
Figure 4.3: Piezometers connected to a pipe
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
4.1.2 Pitot static tube The tubes recording static pressure
and stagnation pressure are frequently combined into one instrument known as a Pitot-static tube
The ‘static’ tube surrounds the ‘total head’ tube and two or more small holes are drilled radially through the outer wall into the annular space.
The position of these ‘static holes’ is important. This instrument, when connected to a suitable manometer, may be used to measure point velocities in pipes, channels and wind tunnels.
Figure 4.5: Pitot static tube
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
4.1.3 Venturi meter The Venturi meter is a device for
measuring discharge in a pipe.
It consists of a rapidly converging section, which increases the velocity of flow and hence reduces the pressure.
It then returns to the original dimensions of the pipe by a gently diverging ‘diffuser’ section. By measuring the pressure differences the discharge can be calculated.
This is a particularly accurate method of flow measurement as energy losses are very small.
Figure 4.6: A Venturi meter
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Example 4.1 A Venturi meter with an entrance diameter of 0.3
m and a throat diameter of 0.2 m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96. Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the volume flowing when the pressure difference between the entrance and the throat is measured as 0.06 m on a water U-tube manometer.
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Consider a large tank, containing an ideal fluid, having a small sharp-edged circular orifice in one side.
If the head, h, causing flow through the orifice of diameter d is constant (h>>d), Bernoulli equation may be applied between two points, (1) on the surface of the fluid in the tank and (2) in the jet of fluid just outside the orifice. Hence :
lossesg
V
g
Ph
g
V
g
P 0
22
222
211
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Now P1 = Patm and as the jet in unconfined, P2 = Patm. If the flow is steady, the surface in the tank remains stationary and V1 0 (z2=0, z1=h) and ignoring losses we get :
or the velocity through the orifice,
(4.7)
This result is known as Toricelli’s equation.Assuming no loses, ideal fluid, V constant across jet at (2), thedischarge through the orifice is
hzzg
V 21
22
2
ghV 22
20 VAQ
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
For the flow of a real fluid, the velocity is less than that given by eq. 4.7 because of frictional effects and so the actual velocity V2a, is obtained by introducing a modifying coefficient, Cv, the coefficient of velocity:
Velocity, (4.8)
or (typically about 0.97)
gh2AQ 0
ghCV va 22
velocityltheoretica
velocityactualCv
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
As a real fluid cannot turn round a sharp bend, the jet continues to contract for a short distance downstream (about one half of the orifice diameter) and the flow becomes parallel at a point known as the vena contracta (Latin : contracting vein).
d0
approx. d0/2
Vena contracta
P = Patm
Figure 4.8: The formation of vena contracta
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
In a nozzle, the flow contracts gradually to the outlet and hence the area of the jet is equal to the outlet area of the nozzle.
i.e. Cc = 1.0
therefore Cd = Cv
Taking a datum at the nozzle, Torricelli’s equation gives the total energy head in the system as it assumes an ideal fluid and hence no loss of energy, i.e. theoretical head :
(4.11)
dnozzle
contraction within nozzle
Figure 4.9: Contraction within a nozzle
g
Vht 2
22
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Weir Assumptions assume that the velocity of the fluid approaching the
weir is small so that kinetic energy can be neglected.
assume that the velocity through any elemental strip depends only on the depth below the free surface.
These are acceptable assumptions for tanks with notches or reservoirs with weirs, but for flows where the velocity approaching the weir is substantial the kinetic energy must be taken into account (e.g. a fast moving river).
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
A General Weir EquationTo determine an expression for the theoretical flow through a notch we will consider a horizontal strip of width b and depth h below the free surface, as shown in the figurevelocity through the strip
V = discharge through the strip,
Integrating from the free surface, h = 0, to the weir crest, h = H gives the expression for the total theoretical discharge,
Qtheoretical =
gh2ghhbAVQ 2
dhbhgH
0
21
2
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Rectangular WeirFor a rectangular weir the width does not change with depth so there is no relationship between b and depth h. We have the equation, b = constant = B.
Substituting this with the general weir equation gives:
(4.14)
Figure 4.11 : A rectangular weir
dhhgBQH
O
ltheoretica 21
2
23
23
2HgB
To calculate the actual discharge we introduce a coefficient of discharge, Cd, which accounts for losses at the edges of the weir and contractions in the area of flow, giving :
(4.15)23
23
2HgBCQ dactual
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
Example 4.2 Water flows over a sharp-crested weir 600 mm wide. The
measured head (relative to the crest) is 155 mm at a point where the cross-sectional area of the stream is 0.26 m2. Calculate the discharge, assuming that Cd = 0.61.
As first approximation,
H = 155 mm
Cross sectionalArea = 0.26 m2
23
23
2HgBCQ dactual
23
)155.0(/62.196.03
261.0 2 msmm
= 0.0660 m3/s
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)
• Chapter 4 emphasized basically on the application of Bernoulli equation in order to solve problems related to fluid mechanics and the application of momentum equation to solve type of flows problem.
• Students should concentrate more on the examples given in chapter 4 and try to relate the concept in the real scenario.
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Subject Matter Expert/Author: Assoc. Prof. Dr Othman A. Karim (OUM)