Towards the end of the last lecture that is lecture 33 we were
looking at an example involving the measurement of
flow by use of a venturi. That was example 38 and we will continue with that.
We will also to be comparing as to what will be the performance of
the venturi for the same set of flow conditions, this is of course a
part of example 38.
Image Download From Researchgate
How does it compare with the case of nozzle
and how does it compare with the orifice plate if the venturi is
replaced either nozzle or an orifice plate. Subsequently, we will look
at an important flow meter which is very commonly employed called the drag effect flow meter also named as Rotameter
. It is very common in industrial. practice and in process applications. Let us complete our discussion on the drag effect flow
meter by taking an example and working out the characteristic of
such a meter. Subsequently. we will look at the other types of flow
meters including what we earlier called as the positive displacement type flow meters and some other flow meters
which cannot be classified in one way or the other. The example 38 consisted of a venturi to be used measured the flow of air at 5 bars
and 25°C.
The maximum flow rate is 1kg / s; the minimum flow rate is 0.3. We have to determine the size of the venturi such that the pipe Reynolds number is not less than 100000. We also have calculate the differential pressure developed across the venturi if β = 0.6 for both the
cases that is 0.3kg / s, as well as 1 kg/s. The Reynolds number based on the diameter
is given by the formula 4min/(πDμ), where μ is the dynamic viscosity of the air at
the pressure and temperature which is specified. So all I have to do is substitute and solve
for D because Reynolds number is given as a limiting value of the 105, so I will take
the value of 105 = 4 x 0.3/(πD) where 0.3 is the minimum mass flow rate and D the diameter
which is to be determined multiplied by μ where μ is 18.4 x10-6 .
So I solve for D and I get the value of 0.207 m that is about 20 cm diameter of the pipe.
The value of β which is the ratio of diameter of the venturi at the smallest area region
and the pipe diameter d / D is given as 0.6. Therefore I can calculate d as (0.6 x 0.207)
which is 0.124 m. So the proportions of the venturi required for the particular measurement
is decided based on the Reynolds number which is specified to have a minimum value. And
at Reynolds number at 105 you can you remember that we had given a figure which gave the
value of the coefficient of discharge as a function of Reynolds number and from that
we take the value of C = 0.977 and the value of M the velocity of approach factor is 1/√[1
– β4] that is 1 /√[1 - 0.64 ], it is 1.072.
Now all I have to do is, to use the equation 6 in the previous lecture. With Y = 1, we
are assuming that there are no expansions or waves of compressibility effect. We calculate
the throat area or the area of the smallest section as πd2/4 which comes out to be 0.012
m2 and d is 0.124 which was determined in the previous slide.
So the Δp developed by venturi when the Reynolds number has the minimum value of 105 corresponding
to mass flow rate of 0.3kg /s given by this formula [min/(CMAt)]2/2ρ. This is the formula
and is actually obtained from the characteristic of the venturi. And we already determined
C, M is already determined and all I have to do is to plug all those values so (min
is 0.3, / C is 0.977, M is 1.072 and At is 0.012)2 / (2 x 5.844) this is the density
of the fluid. So we get 47.4 Pascals and that is what we calculated here. This is the pressure
difference across the venturi for the minimum mass flow rate. In fact we can very easily
see that for the maximum mass flow rate, all I have to do is change m. To the value equal
to 1kg / s and I have also got to determine the value of C appropriate to the Reynolds
number corresponding to the maximum flow rate.
And if you remember, the mass flow rate and the Reynolds number are linearly related which
you can see from the formula here. Reynolds number equal to 4 (mass flow rate / π d μ).
So everything else is fixed here and only is now changed from 0.3kg/s to 1kg/s so Reynolds
number will be larger by a ratio of 0.1 to 0.3, 1 / 0.3 of the 105 that is about 3.3x105.
Therefore the Reynolds number increases and correspondingly, the value of the coefficient
also changes, and we can take the value from the same graph so the value of C comes out
to be 0.984 from the graph. This corresponds to Reynolds number 1 / 0.3 x 105 which is
the maximum value in this particular application. This is 3. 3 recurring into 105.
Corresponding to that this is the value of C which I am going to get. And all I have
to do is now substitute this value of C and the corresponding value of the mass flow rate
in the formula which we used here earlier, all I have to do is replace by this one so
we will say instead of minimum I am going to replace it by maximum value and the value
of C by the value which I just now wrote down; C = 984 corresponding to the new value of
the C and the max = 1kg / s. So all I have to do is to substitute these values and the
value of Δ p will be obtained, this we can call as a maximum Δ p obtained using this
particular venturi and this will be [1kg/s / (0.984 x 1.072 the velocity of approach
factor x 0.012)]2 is the area of the venturi at the smallest section x 1 / (2 x 5.844)
which is the density of the fluid. This is the pressure difference which is developed
equal to 519.2 Pascals.
Actually I can convert it to a more useful unit in terms of millimeters of water. All
I have to do is divide it by assuming the water density as 1000 which is good enough
of the approximation for this problem; you can simply divide this by g which is 9.81
so this can be converted into millimeters of water which will be 52.9 mm of water and
divided by 9.81 that comes here. So this is the pressure drop or pressure difference between
the inlet and the smallest diameter section of the venturi which is what is measured by
the manometer. So 52.9 mm of water we are going to get.
Now, let us look at what is going to happen if I do the following? I am having the venturi
and now I want to compare it with the case of nozzle, this is the venturi, this is the
nozzle and I have an orifice plate all having the same value of β in each case. So what
I have to do is to workout the appropriate values of C for the case of nozzle and C for
the orifice plate. That is the only thing which is going to change because all other
conditions are given to be the same. That is the same mass flow rate and the same diameter
of the pipe because we already determined the diameter of the pipe such that the pipe
Reynolds number is not less than 105. So we are going to specify this as these conditions
have been fixed and I want to compare the venturi, the nozzle and the orifice plate.
Therefore the mass flow of air is 1kg / s and 5 bars and 20°C, there is a inlet pressure
and temperature, the pipe diameter is 0.207 as determined in the earlier case and the
β value is 0.6 these are held fixed. The corresponding velocity of approach factor
is also 1.072. All these are fixed for all the three cases. So, the performance of a
venturi, long nozzle and an orifice plate with corner taps are compared in the table
given in the slide.
What is going to happen is that the value of C is different for the three cases and
therefore to that extent there will be changes in the performance. So, if I look at the comparison
of the venturi, nozzle and orifice, the values of C are determined and in fact you remember
in the example 38, I had 0.984, and for the nozzle, I use the
appropriate expression which was given earlier. So, this works out to be 0.988. For the orifice
it comes to 0.605. These three cases correspond to the Reynolds number equal to 3.33 recurring
into 105. This is for the highest mass flow rate case. The corresponding values of K the
flow coefficient can also be calculated nothing but C x m, m value is same for these three
cases is equal to 1.072 so this comes to 1.055, 059 and 0.649.
We see that the value of K is the smallest for the orifice and for the venturi and the
nozzle they are of more or less the same order of magnitude, very close to each other. Therefore
we expect the delta p generated by both venturi and nozzle to be close to each other. You
can see that 519.2 worked out in the example 38 and 515.2 for the nozzle slightly lower
because the value of K is slightly higher and for the orifice you get the highest value
of 1372 Pascals. In fact, we convert these 2 mm of water as indicated. Simply you have
to divide by 9.81 and you have 52.9 for the case of venturi, 52.5 in the case of nozzle
and 139.8 for the orifice plate. So the orifice plate indeed gives you the biggest measurable
effect of delta p. That means it is the most sensitive in terms of the amount of pressure
drop you are going to get for a given condition. But, the penalty will be that we are going
to lose most of the 139.8 mm of water pressure drop which will not be recovered and most
of it is going to be dissipated. In the case of venturi of course, 52.9 only a few percentage
of this will be lost and most of it will be recovered. That means the pressure at the
end of the expansion portion where the diameter changes back from the smaller diameter to
the pipe diameter, the value will be more or less equal to the pressure earlier. That
means the pressure drop is equal to penalty which we are going to pay in this case. The
venturi requires the longest length for the installation, nozzle is next to that and the
orifice requires the least amount of length for the installation. That means for the point
of view of installation, all we need is the two flanges and we can put the orifice plate
whereas, in the case of venturi and nozzle we require some more sophisticated arrangement,
certain length has been introduced and so on.
Let us look at how a typical installation will be. The typical installation of any of
these types of flow meters requires, for example, if you have any of the flow meters, it requires
a straight length of tube upstream of the flow, it requires a certain length of the
pipe which should be striped. I cannot have a corner or bend in the pipe because, when
there is a bend the flow becomes some what disturbed from its parallel type of flow arrangement.
For example, if I have a bend like this, there will be a secondary flow and the flow will
be disturbed like this. Therefore we should avoid putting the flow meter right next to
a bend that is very close to a bend. So we require may be 20 to 50 diameters of the pipe
or may be more depending on the installation. Similarly, we require a certain straight length
of tube downstream of the meter. If you do not want have any effect of the down stream
and the upstream then we need some straight length both before and after the flow meter.
The flow meter itself is now either one of the three we have. It may be an orifice plate
which of course will be the smallest in terms of the length, the nozzle will be slightly
longer and venturi of course will be a very long installation. The length of the venturi
is the longest of the three.
The other point is, if you take the venturi for example, this is typically the venture,
so you are going to measure the pressure difference between these two points, this is delta p
what I am going to measure.
So normally what is done is, you have several taps on the peripheries so you have the pipe
here, I have taps, this is the pipe thickness and I have several taps that means they are
all communicated to the same, and I am going to measure the average of the pressure in
the 4 taps, in this case it is either 4 or more than 4. So, if we look at the cross section
of the pipe the pressure taps will be there like that, and on the wall here, on the periphery
all of them are connected to the same hole which is going to be connected to one arm
of the pressure difference measuring devices. So here also, we will have several holes on
the periphery all of them connected to this tap, and here, I have several holes which
are connected to this tap. So it is the average pressure around the periphery which is normally
measured in the case of an installation like this. With this we have finished looking at
the variable area devices. Just to recapitulate the variable area devices we are going to
deliberately introduce a change in the area of cross section which will affect the velocity
of the flow and the kinetic energy is converted to potential energy in the form of pressure
head and what is measured is the pressure difference between the two chosen stations
or sections along the length of the installation.
Let us now look at what are the other kinds of instruments which can be used for measuring
the flow rate. One of them is called the Rotameter or it is also called the drag effect meter.
We will look at the principle of operation of a Rotameter, how it is constructed, what
are its characteristic etc. we will derive what is called the Rota meter equation and
then we will look at the design considerations and take up example 39.
This is the schematic of a Rotameter. Essentially what it has is a vertical tapered tube and
the direction of gravity is from top to bottom. It has to be aligned vertically with respect
to the local plane. For example, if you are having a table top experiment, the table top
is supposed to be horizontal and the Rotameter must be exactly vertical. To the horizontal
plane at the local place for making measurements the Rotameter must be exactly vertical. In
the case of the three measuring meters the venture, nozzle and the orifice plate normally
we use the horizontal orientation. The meter will be located horizontally in the pipeline.
And many times the pipeline is actually horizontal and therefore it does not involve in any problem.
But if we are going to install a Rotameter you have to make an arrangement such that
locally the flow direction is exactly vertical from bottom to top, it should be from bottom
to top. The flow direction is fixed from the bottom to top against the gravity. Let us
look at the schematic of the Rotameter.
Essentially, it consists of what is called Bob, Bob is this piece of material which is
shown in the form of a cylinder with a small cap on to it, the diameter of the Bob is this
d, and it is normally equal to the diameter at the bottom of the tapered tube. Of course
here I shown the tapered tube as very high but actually the tapered tube is very small,
and by looking at it we may not be able to say that it is a tapered tube. It is a very
small taper we are talking about.
How does it work? Just imagine that the Bob is in some location
and this location is measured from the bottom here. So Y is measured in this direction from
here. So the Bob is in some location so you can take this position as the corresponding
Y. Now suppose, there is a flow of fluid it may be either gas or liquid flowing like this,
it has to negotiate the Bob go around the Bob like this, in the space available between
the Bob and the outer tube. The area available for flow is actually the annular area between
this d and the local diameter of the outer tube, there is an annulus which is formed.
Therefore if we say that the value of the velocity is um in that annular region, the
Bob is going to experience three forces. One of course, is the weight of the Bob which
is vertically downwards then there is the force of the buoyancy because it is displaying
a certain volume of the material which is flowing, either gas or a liquid which is upwards,
by Archimedes principle, and, thirdly, there is a drag force which is because of the flowing
fluid negotiating over that and flowing and is joining that. It flows with a larger velocity
um in the annular region and goes back and joins there. So because of the form of this
particular Bob material and because of the flow which negotiates that and joins at the
back there is something called the form drag. The form drag is because of the subtle changes
in the pressure acting on the surface of the Bob and this usually is written in the form
of a drag coefficient multiplied by the kinetic energy of the fluid per unit mass.
These 3 forces are going to now be in balance for a particular value of the flow rate which
is in this direction if the value of um is going to be a constant value because the drag
force, the weight of the Bob and the force due to buoyancy must be always in balance,
and this will be for a given value for um, the um gets fixed because of that and what
is changing here is only the area available for flow that is the annular area surrounding
the Bob, that is area available for the flow.
Actually what is happening in this case is that, the area is changing. More or less it
is a linear change in area. The taper is small, the change in area is linear and therefore
the volumetric flow rate which we infer from the position of the Bob is a linear function
of the position of the Bob from the datum which is here. Let us now look at the way
this turns out. We want to look at the Rotameter equation. We will actually derive this equation.
Here is a simple sketch. This is the tapered tube, and this is the axis and the gravity
force is acting in the downward direction like this, the flow is taking place from below
and here is the Bob and ρb is density and volume is Vb. This is the density and this
is the volume of the Bob.
In fact the weight of the Bob will be nothing but (ρb Vb g) the acceleration due to gravity
at the location where we are doing the experiment or where we have set up the Rotameter. This
is one force which is to be taken into account. The force of buoyancy will be equal to (ρf
Vb g), it is the weight of the displaced fluid whose volume is equal to the volume of the
Bob. So, if the density ρf is the fluid density, Vb is of course the volume of the Bob multiplied
by g gives you the second force. This is actually upwards and this is downwards. There is the
third force which we call as the drag force and mostly it is not due to viscous effects,
but it is only due to the way the flow negotiates the body. Therefore it is mostly due to the
form of the object. Therefore it is called the form drag. This drag force we will call
it as Fd equal to some coefficient, Cd times the kinetic energy of the fluid, which is
given by density into um2 / 2 Ab, the frontal area of the Bob material, all for unit volume.
This force is also in the upper direction.
So what we need to do now? The force acting in the downward direction
must be exactly balanced by the sum of the forces acting in the upward direction. That
means, I have to equate one to the sum of the two forces which are given by two end
three. So you can say, for equilibrium, the downward force this is the weight must be
the sum of drag and the buoyancy forces, the two forces which are acting in the upper direction
must be equal to W.
Therefore we can just do that. So (ρb Vb g) is the weight of the Bob must be balanced
by the buoyancy force (ρf Vb g) plus the drag force which is now given equal to (Cd
Ab ρf um2/2). In the previous slide here this is proportional to Cd multiplied by the
area which is intercepted by the flow. That will be the frontal area of the Bob material.
So Ab is the frontal area of the Bob. This is the equation, and I can solve for um and
you can rearrange like this. So um equal to this minus this and then you appropriately
divide and um square is here, therefore I am going to get a square root and Vb is common
for both the cases multiplied by (ρb - ρf )/ (ρf Ab). Therefore the value of the velocity
um is equal to this. So what does it mean? It means that the velocity cannot change.
Whatever may be the position of the Bob equilibrium requires that um is a cost. That means if
I am allowing certain amount of fluid to flow through the system it will decide the position
along the length of the tapered tube where the value of the velocity in the annular region
between the Bob and the outer tube will be exactly equal to um. So the equilibrium position
will be determined by the requirement that the velocity um be a constant given by this
particular quantity assuming that Cd is a constant. So normally what is done is that
the Cd is dependent on the shape of the Bob and so on.
So normally, the design of the Bob is such that the Cd is more or less a constant for
the range of operation of a particular Rotameter. Secondly, the density of the Bob and the density
of the fluid both are going to come into the picture. So the density of the Bob does not
change when you use the Rotameter for measuring the flow rate of one fluid and another fluid.
So ρf is going to change. That means that the calibration is dependent on the fluid
which is being used.
Normally what is done is that it is calibrated for a particular fluid and if the density
of the fluid is different we have to use the correction factor for that. For example, if
we use air at one particular pressure and temperature and if the temperature changes
and density of air will be changed therefore ρf will be different so you have to correct
that.
That means the Rotameter calibration is dependent on the fluid and its temperature
in the case of gases so that you have take in to account some of these thing during the
operation of the thing. Now that we have got the um let us look at what the flow rate is.
So, volume flow rate is given by the velocity times the area available for the flow so um
x A available for the flow. The um is already found which is a constant given by some quantity
dependent on the Bob density, fluid density, Cd and Ab and so on which are all fixed. Therefore
only the Af is going to vary. Suppose you have a tapered tube here, and I have a Bob
like this, this diameter is d the area available between here and here, so I can say if the
local diameter is d which is a function of y, y is measured from the bottom from some
datum, d at y I will say d at y equal to d plus some ay where a is called the tapered
parameter, so I can say that Af area for flow is (π / 4 d)2 – d2. That is just the area
available here around the periphery of that circle, that is the annular area.
Now I can square this so [d(y)]2 is nothing but d2 + 2 ady + a2 y2 and, I am assuming
that a is small which is small taper, I can ignore this factor as being too small. If
we ignore that you see that d(y2 d2 + 2ady) and this [d(y)]2 – d2 and now this d2 will
cancel so this whole thing can be rewritten by taking the difference between that and
that as 2ad(yπ/4) = Af. Therefore the volumetric flow
rate Q is given by um and for um the expression is already available 2g Vb(ρb - ρf) / (ρf
Cd Ab) this is your umπ/(4x2ady) so this will cancel off this 2 so π ady / 2 multiplied
by this is actually the area.
This is the area factor, the velocity factor, this is constant, and therefore, you can see
this is also a constant, is the constant, pi is the constant, 2 is the constant so this
is nothing but some C times y, C is the gage constant. So basically the Rotameter is a
linear device where the volumetric flow rate is given by some constant times the position
of the Bob which has noted down. Actually what you have done in practice is, if you
go back to the sketch here, along this tube a scale will be marked which will give you
the position of the Bob.
For example, you can look at this edge here and find out where it is and that will be
your position of the Bob, position of the Bob is given by the edge of this thing here
and corresponding division on the outside will give you the height above the data, above
this particular position. So, that gives the y so you actually measure only the y that
is the scale reading on the periphery of the tapered tube and from that you multiply by
a constant C which is called the gage constant for the Rotameter and you have the volumetric
flow rate. Rotameters are made for application with gases and with liquids; they can span
very large range of values for the volumetric as well as the mass flow rate. We have very
small Rotameters which uses a Bob made of cork or pith some material with very low density
so that it can be used for gases and it could be only a few millimeters diameters and few
millimeters tall. So the Rotameters may be a few centimeters in diameter and may be almost
half meter in length.
Therefore, depending on the flow rate depending on the fluid we can have different Bob materials,
the Bob can be either hollow or solid to adjust the density to require the extent because
when we talk about Bob density, it is the average density of the Bob, and if we have
a hole inside the density comes down. Therefore, by having a Bob with a hollow inside you can
get a density which you want. and you will also see that, if I go back to that equation
you have Q = √[2gVb (ρb - ρf)/ρf] and imagine ρb = 2(ρm). If ρb = 2(ρf) you
have (ρb - ρf) / ρf that will become one, then this becomes equal to 1.
So the density and the Bob, density of the fluid both drop off and it becomes a very
simple expression so this will be one so it will be (2g Vb) / (Cd Ab) and that will be
the value of C. This is a specific case where the density is taken equal to twice the density
of the fluid then the density drops off from the equation. In fact it is found that, even
if the density of the fluid varies a few percent may be 5% or 10% this is not going to alter
the thing too much, the error is not very significant.
The second point is that the tapered parameter. If the tapered parameter is not small then
a straight non linearity will be there and of course this can be taken care of by adjusting
the scale on the side of the Rotameter to include that non linearity in it. Therefore
when you take the reading automatically it has been corrected for the slight nonlinearity
if there is any non linearity in it. Here is example 39.
A Rotameter is to be designed for the particular condition as given here. To measure maximum
flow rate of 200 lpm of water at 27˚C this is the fluid whose volumetric flow rate I
want to determine. The 200 lpm is maximum. The Bob has a diameter of 15 mm and a volume
of 3 mm that is 3 cm3. The effective density of the Bob is twice the density of water.
If we take the density equal to twice the density of the fluid, then you do not have
worry about the density. It will not come into the picture.
The total length of the Rotameter tube is 150 mm. That means ymax the maximum possible
value of y is 150 mm and this must correspond to this 200 mm. And y = 0 corresponds to 0
flow and if there is no flow the Bob will be at the bottom at its datum level and when
the full volume flow rate is there in the Rota meter it will go to the 15 cm or 150
mm level which will be the maximum level possible or in this particular case. And the diameter
of tube at inlet is also equal to 15 mm so the Bob will be the just sitting close at
the opening at the bottom and as soon as you start the flow it will float up and it will
take a position which corresponds to the volumetric flow rate which we are measuring.
So we want to determine the tapered parameter, tube paper is what is to be determined for
a drag coefficient of 0.8. The drag coefficient is given to have value of 0.8 so we want to
find the value of tube tapered of this particular case. This is example 39. The Rotameter problem:
The volume of the Bob is given as 3ml and a milliliter is 10-6 m3. I will convert everything
to SI units.
We will assume g is 9.81 standard values, Vb the volume of the Bob and then the diameter
of the tube at the bottom is 15 mm so it will be 0.015 m and the diameter of the Bob is
also equal to that this is also equal to Bob diameter. So we can determine the frontal
area Ab which is the frontal area of the Bob which comes in the calculation of the drag
force, this frontal area will be ( π d2) / 4 so (π / 4) (0.015)2 m2 and this comes
to 1.767 x 10-4 m2. So, the frontal area of the Bob is fixed. We are also given the maximum
flow rate Qmax. It is given by 200 lpm and 1lt is again 10-3 factor will come in terms
of cubic meters divided by 60 because it is given in liters per minute and I am calculating
in seconds. So this will be so many cubic meter per second and this comes to 3.33 x
10-3 m3 / s. This is the maximum volumetric flow rate.
We are also given the value of Cd to be 0.8 and we also know that y maximum equal to 150mm
which is 0.15m. And if I go now to the equation which we derived earlier, we see that the
taper parameter, a can be solved for, the taper parameter is what we want to determine,
this will be (2 K) which is the gage constant by π√[ 2g Vo / Cd Ab]. This can be done
easily by going back to the Rota metric equation and reworking this thing. And I am assuming
ρb = 2ρf and therefore the density factor is this thing and what is K? K y = Q is the
formula. Therefore K ymax = Qmax. Therefore K is nothing but Qmax / ymax. Both of them
are given, and the value of Qmax /ymax comes out 0.022.
We put the Qmax value and the ymax value and we just work it out, and it will come to 0.022.
So, that value is known and all we have to do substitute all the values which will be
2 x0.022 / π √[2 x 9.81 x 3 x 10-6 is the volume / Cd is 0.8, the frontal area is 1.767
x 10-4 ] and this works out to be 0.022 the taper parameter comes exactly equal to 0.022.
In fact the taper parameter if you remember d at any y value equal to d + a y. So we can
say dmax = d + a (ymax) and all I have to do is to substitute, d = 0.015 + 0.022 x0.15
ymax, this comes to 0.0183 m. So the diameter at entry if you look at the Rotameter tube
the diameter here is, 15mm and here 18.3mm and this length is 15cm. So this is designed
for the Rotameter. The entry diameter is 15mm, exit diameter is 18.3mm, the length of the
taper tube is 15cm and the taper parameter is the value which we just described.
The taper parameter will also mean this angle and that angle turns out to be a very small
angle actually that will be 0.62˚. So, that angle we call it as α and α is 0.63˚, a
small angle for the taper. And in fact, if you were to calculate including the quadratic
term, we can calculate the error in line arising for this particular instrument that is given
by [a2 y2 max / d2 max ] 100 by s. As a percent of the area, I can find this out and a2 y2
will give you about 3.23%. So the non linearity from the instrument is about 3% which may
be quite small for most applications we may have in mind.
Now let us look at some other type of flow meters. There are the positive displacement
meters then I have got the vortex shedding type flow meter and lastly the turbine flow
meter.
The positive displacement meters are characterized by, you have a device which is going to measure
or which is going to send the measured volume of the material from one side to the other
side so what I mean by that? For example, if we take a positive displacement type with
lobed rotors then for each rotation of this lobe a certain volume of the material is taken
from the inlet, and it is going to be sent to the outlet. A certain volume of the material
is taken from the inlet and it is going to be sent to the outlet. So this is rotating
like this, and this is rotating like this and each time it rotates a certain volume
of the material is taken from the inlet side and it is dumped to the outlet side.
So in this case, because we know the volume between the lobed rotor and the casing, if
this volume is exactly known or measured accurately, then each rotation will correspond to, one
volume being taken from this side and taken to the outside by this lobe and this lobe
rotor will exactly take the same amount in the rotation. Therefore by knowing the volume
and the number of rotations performed per second all I have to do is measure the rotation
speed of the rotor and I will be able to find out the volumetric flow rate dividing the
volume by the time it takes for this to be taken from this side to this side. So, this
is positive displacement. That means the amount of material coming from here is taken by the
volume here and sent to the other side by what is called positive displacement of the
material.
Another example is a syringe which has got a very accurately bold cylinder a piston and
if you advance the piston from one location to another location, an equivalent volume
of the material will be going out of the syringe. A syringe can be coupled to a device which
will push it by a push rode and you can measure exactly how much volume has been displaced
by looking at the reading on the cylinder of the syringe. You can actually look at the
divisions and then find out what is the amount of material which has been dumped or which
has been delivered. So syringe type of device is very often used in chemical instrumentation.
In instruments where you want to deliver a certain amount of fluid per second you can
use a syringe pump.
Of course syringe pump cannot operate continuously. When the piston position has gone from one
extreme to another extreme, you will have to stop and refill the syringe and again restart
the thing and so on, the syringe is a positive displacement type of meter. The second one
is the vortex shedding flow meter. It is noticed that if you have an obstruction like this
in a flow, for example, I can have different shapes for the obstruction, the flow which
comes from here gives raise to a vortex which is shed at a known frequency at a certain
rate. That means if you look at the frequency at which the vortex is shed, it follows the
relation which is given here, this is the Strouhal number which is nothing but the (f
D) / V where f is the frequency of the vortex shedding, D is the frontal diameter of the
body and V is the velocity.
So the Strouhal number has a constant value in this range from here to here, and it means
that the velocity here is given by the frequency of shedding of the carbon vortices which are
developed at the periphery here at this corner and this corner and that is given by the frequency
of shedding multiplied by the diameter divided by Strouhal number. The Strouhal number is
a constant in that range, and therefore V is proportional to the frequency of the vortex
shedding.
If you are able to measure the frequency of the vortex shedding it is just by monitoring
the pressure at the back and each time in the vortex shedding the pressure will go through
a pulse so there will be a change in pressure and you can measure the number of such events
and then find out how much time it took and immediately I can find out what is the velocity.
So, vortex shedding meter is one of them which can be used for measuring the velocity and
this velocity is nothing but the average velocity of the flow. Turbine type flow meter: This
is another meter which can be used. It is essentially like a turbine.
So, if a certain volumetric flow rate takes place from the left to right the turbine is
nothing but a device with blades on the periphery of the rotor and the number of times it is
going to rotate is proportional to the velocity of the fluid which passes through from this
side to this side. It consists of a rotor then the bearing to support and also straightening
waves so that the velocity is in the direction parallel to the axis and C is the reluctance
pick up. Each time it rotates the reluctance pick up will register by producing a pulse
and you can again count the number of pulses per second and determine the velocity of the
fluid to the turbine type flow meter. Thank you.