The venturi:
The venture is a pipe followed by a short parallel gorge between which the difference that occurs (in pressure) is measured. It is worth nothing that the fluid velocity must increase through the contraction to satisfy the continuity equation.
Applying the Bernoulli equation between points 1 and 2 in Figure 1 gives:
1)
Where,
-
P1 and P2 refer to the pressures at point 1 and 2 respectively
-
V1 and V2 are the mean air velocities at points 1 and 2
-
Z1 and Z2 refer to the heights of points 1 and 2 above the datum
We also have ρa referring to the density of air and g as the acceleration due to gravity.
We take the datum as the venturi pipe centreline, therefore Z1=Z2 = 0 and substituting this into equation one and rearranging gives us:
(2
By continuity the mass flow rate,, is given by:
3)
A1 and A2 are the pipe cross-sectional areas at sections 1 and 2 respectively. Rearranging equation 3 gives:
4)
Substituting equation 4 into equation 2 gives:
5)
Equalising pressures in the two manometer arms as shown in Figure 1 gives:
6)
ρw is the density of the water in the manometer. Since the density of air is very much less than that of water i.e. ρa << ρw the density of air can be ignored and hence equation 6 can be written as:
7)
Substituting equation 7 into equation 5 and rearranging gives:
8)
Where V1 is the velocity and after its calculation, the mass flow rate can be found using equation 3.
The Pitot tube:
Points 3 and 4 are very close together, however in the diagram they are widely separated for clarity. Applying Bernoulli’s equation between points 3 and 4 gives:
9)
(Z3 = Z4 since points3 and 4 are almost conincident)
The fluid has been brought to rest at the tip of the Pitot tube ("stagnation point") so V4 = 0. Thus equation 9 can be written as:
10)
But, using a similar analysis to that for the venturi manometer in equations 6 and 7 gives:
11) (
Thus substituting equation 11 into equation 10 and rearranging:
12)
Determining the Mass Flow Rate from the Velocity Traverse:
Consider the pipe cross-section shown in Figure 2. The velocity will fall from a maximum on the pipe centre-line (r = 0) to zero at the pipe boundary (r = R). Since the flow distribution is symmetrical, the velocity will be the same for any fixed distance, r, from the pipe centre where 0 ≤ r ≤ R.
Thus the increment of mass flow,, flowing through an element of area, δA, with thickness,, at distance r from the centre-line, as shown in Figure 2 is:
13)
V(r) is the velocity at distance r from the centre-line.
The element of area is a thin annulus and so:
14)
Substituting equation 14 into equation 15:
15)
In order to obtain the total mass flow rate the incremental flow rates must be summed from the centre-line to the edge of the pipe, thus:
16)
The value of the integral can be found by plotting a graph of V(r).r against r for the velocity traverse results.
Figure of The Pipe cross section showing typical element of area
Figure 2 – pipe cross-section sowing typical element of area
Source for diagrams and equations: DEN 101 – FLUID MECHANICS handout
Also used for background info: Fluid Mechanics by Martin Widden
Results
Above are the parameters that will be used throughout the whole experiment: room density (and the temperature and pressure of the room used to calculate it). On the day of the experiment the temperature was 21° C (294 K) and the pressure was measured to be 769.26 mm hg (101.9 KPa).
To get the pressure in pascals the following conversion factor was used:
1/0.00755 Pa = 1 mm hg
Thereafter the density was calculated:
Density =
Venturi:
The following parameters were used for calculations involving the venturi.
To calculate the mass flow rate we first had to find the velocities (V1) for each reading. To calculate V1 we used equation 8.
Equation 3 was then used to find the mass flow rate for each voltage. The results are as follows. (The manometer reading is the raw result obtained from the experiment).
Here’s an example (using the reading for 20 volts) of how the mass flow rate was calculated:
First the velocity had to be calculated:
V1 =
V1 = 1.6250 m/s
Then the mass flow rate was calculated:
M = 1.2075 x 0.0092 x 1.6250
= 0.0181 kg/s
Pitot tube:
For the pitot tube we took readings at different heights (starting from the bottom) for each voltage in intervals of three to get six readings overall.
The raw results, obtained at the experiment, for the pitot tube are as follows.
As you can see we have insufficient readings for 90 volts. This is because we were unable to attain any further results at this level and therefore we cannot calculate the mass flow rate for this voltage. However, for the sake of completeness, we have included it in our tables.
Again to calculate the mass flow rate for each voltage we first had to find the velocity of each reading. Equation 12 was used to calculate this. However we couldn’t use equation 3 this time to find the mass flow rate. Instead we had to use equation 16.
Here is an example (using 20 volts and r equals 15 mm) for calculating the velocity:
V3 =
= 18.3 m/s
An integral was introduced in equations 16 which could not be carried out due to inadequate information. Therefore graphs had to be plotted and the regions to the axes calculated.
The graph of V3.r against r was required (were r is the distance from the centre line of the measurements). The following table shows the data calculated in order to plot the graphs.
Here’s an example of how the first figure was obtained:
V(r)r = V3 x r = 18.3 x 15
= 275 m2/103s
The points were plotted and in order to approximate the area under the line each graph occupies a line of best fit was added.
The following table shows the maximum, for v(r)r, reached by each trend.
The shape of each graph to the axis was treated as a triangle in order to simplify the calculations necessary.
The following table shows the mass flow rate attained for each voltage. This was calculated using equation 16.
Here’s an example (using the first result for 20 volts) of how the mass flow rate was calculated:
First the area under the graph had to be calculated:
Area = (294.4/1000 x 15/1000) = 0.0022 m2
Calculation of mass flow rate:
M = 2 x ∏ x 1.2075 x 0.0022
= 0.0167 kg/s
Discussion
Since we couldn’t get a reading for 90 volts for the pitot tube we cannot compare the flow rates for this value. However for the sake of completeness the venturi flow rate for 90 volts has been included.
Comparing the venturi results with the pitot tube’s we can see that they are somewhat similar (all accurate to 1 d.p.). In fact all but one of the results (when the voltage is 20) differ only by a level of 4% or less. However similar, differences in mass flow rates did occur, therefore we shall examine the possible reasons behind this.
We will first consider the sources of error in our experiment and any other discrepancies that may have occurred.
- The fluid in the manometer was fluctuating in sum circumstances and therefore an precise reading could not be taken,
- Even when no fluctuations occurred a human error could have occurred when reading the results (there were differences in opinion on occasions as to the actual results). The measurement of pressure in the room could also have been affected by this factor,
- The pitot tube was not always dead centre (i.e. was not perfectly aligned to the direction of flow) all the time. In fact we cannot be sure it ever was. A pitot tube can achieve an accuracy level of better than 1% when aligned correctly,
- When calculating the velocities and the areas under the graphs – for the calculation of mass flow rate for the pitot tube – the maximum value of r (refer to equation 16) was approximated to 15 rather than the more accurate 14.825: since the diameter of the tube was 29.65 mm. This was done to simplify the calculations and save time where possible. This shouldn’t have a significant impact on the final result.
- The value of the temperature was taken to be 294 K rather than the more accurate 294.25 K. Again, this was done to keep the procedure simple. Also the temperature was assumed constant for the duration of the experiment. In real life this can be unrealistic,
- The integral required to carry equation 16 could not be found by means of integration due to inadequate information (if the integral was applied the results obtained would be more concrete). When we plotted the graphs there were positive correlations and hence a line of best fit was used and this may have been less accurate.
Although both the venturi and pitot tube are suppose to provide good results, it is said the readings obtained from the pitot tube are more accurate. The pitot tube gives an estimate of the flow rate through a tube where the average velocity occurs. Since this point was unknown to us many readings were taken at different points and the average calculated. The venture however simply takes its measurement from the centreline.
Conclusion
The aim of the experiment was to measure the flow of air using two methods and the results compared. Differences (which were expected) have occurred however the sources of error discussed should be enough to bridge the gap. Overall the experiment was a success. Should this experiment be repeated my recommendation would be to take more samples of pitot tube readings to get a better approximation.
References
The following references were used for the background theory and application of theory in order to complete this report:
- DEN 101 – FLUID MECHANICS handout
- Fluid mechanics by Martin Widden
- http://www.flowmeterdirectory.com/flowmeter_artc/flowmeter_artc_02111201.html
