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
Equalizing pressures in the two manometer arms as shown in Figure 4:
(6
is the density of the water in the manometer. Since the density of air is very much less than that of water i.e. << then equation 6 can be written as:
(7
Substituting equation 7 into equation 5 and rearranging gives:
(8
The mass flow rate,, can then be found from equation 3.
The Second Experiment: The Pitot tube
Points 3 and 4 are shown widely separated in Figure 4 for clarity. In reality they are very close together. Applying Bernoulli's equation between points 3 and 4 gives:
(9
(Note that z3 = z4 since points 3 and 4 can be ignored as the experiment is taking place over a horizontal system)
The fluid has been brought to rest at the tip of the Pitot tube (the "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 we get,
(11
Thus substituting equation 11 into equation 10 and rearranging:
(12
Application of Mass flow rate: Determining the Mass Flow Rate from the Velocity Traverse
Consider the pipe cross-section shown in Figure 3. 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, A, with thickness,, at distance r from the centre-line, as shown in Figure 3 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 3 – Pipe cross-section showing typical element of area
Apparatus
- A fan attached to a nozzle of decreasing diameter (As seen in figure 1),
- Speed was controlled by a handle attached to the machine
- Series of pipes are connected to the nozzle. The other end is an open face releasing the air into the atmosphere
-
A devise measuring the change in pressure between points 1 and 2 (venturi) and other set of tube measuring the change in pressure between 3 and 4(Pitot tube). A ruler which has a density of 1.88 [mmH2O] will measure the changes in fluid movement.
- A manometer is connected to point four.
Apparatus II... Simplified equipment
- A circular pipe
- Venturi tube
- Water manometer
- Scale
- Pitot tube
- A variable speed pump
Experimental Procedure
- Set up the experiment, as seen above, making sure all rulers are correctly aligned to 0 and that the Pitot tubes at points 1, 2, 3 and 4. The Venturi is between points 1 and 2.
- Turn Control bulge to first speed setting which consists of speeds of 0, 2, 4, 6, 8 and 10.
-
Obtain measurement of ΔHv using the ruler attached to the system. Record in table.
- Move the Pitot tubes at point 1 by 2mm upwards and the point should be moved by 2.8mm to through the entire radius.
- Repeat steps 3 and 4 until a distance of 14.8 mm has been reached. This step should be repeated for speeds of 2, 4, 6, 8 and 10. The pressure point must be set at zero.
-
Take a measurement using the manometer connected to the Pitot tubes at points 1 and 2. This is ΔHp the difference in pressure levels. Record in table the measurement on table.
- Repeat the same steps as taken when measuring repeat step 6 but for the all the speed settings.
Raw Results: Using the Venturi, using points 1 and 2
The Pitot tube: Results obtained at different speeds
Calculation for Venturi, using the raw results:
The diameters of the system are 108.10mm and the narrow tube is 29.60mm.
A1 = 108.10mm
A2 = 29.60mm
Sample Calculation
Calculating A1:
Radius = (108.10/1000)/2 = 0.05405m
So:
Calculating A2:
Radius = (29.60/1000)/2 = 0.0148m
So:
Therefore:
Calculating V1:
Therefore, using V1, we can calculate the Mass Flow rate at the point 1:
Calculating V2:
The Velocities of air pressure at different speeds, the rest of the velocities are worked out in a similar way as above:
Mass flow rate = ρa A1V1 (av)
Mass flow rate = 1.225*0.00917786*0.388
Mass flow rate for the venturi = 0.00436 kg/s
Calculation for Pitot tube, using the raw results: Using the equation and the table below to work out the velocity.
Velocity at speed of 2:
= 7.5 = 1.88 mm H2O, g = 9.81, = 1.225
V1 = [(2*1.88*9.81*7.5)/ (1.225)] 1/2
V1 = 15.02 m/s
V*r = 0.0028*13.99 = 0.0392 [m/s]
Continue similar calculation for all the speeds and take the average velocity to plot a graph against the radius.
Results:
Results Calculated at 2 volts
Results Calculated at 4 volts
Results Calculated at 6 volts
Results Calculated at 8 volts
Results Calculated at 10 volts
This table below is a simplified version results which I have obtained from the table above. This table consists of averages to help me in the further stages.
Now using the trapezium rule I will try to calculate the mass flow rate from the Pitot tube. The trapezium rule is:
A = 0.5h *[Y0+Yn+2(Y1 + Y2 +……Yn-1)]
Adding the value for V*r, in the table above, to use in the final equation to work mass flow rate. The total value of the summation is 1.4005 [m/s].
Mass flow rate equals
r
= ∫ V(r)*r dr
0
Discussion
The analysis of the results shows that the experimental outcome went against the theoretical outcome which I predicted that as the radius decreases the mass flow rate also decreases. Also in my experiment I predicted that the mass flow rate of two the tube would be similar but the final results for the two tubes were far apart. The mass flow increased contradicting my prediction.
After weighing up the possibilities of errors, below I have come up few problems which I think may have affected the results. Any sort of error at start of the calculation process will remain with the results throughout, reducing the accuracy of the results.
The experiment also consisted of error such as our rate of reaction. When trying to read the measuring device weren’t stable and we had to read the value at an angle. The group wasn’t communicating to obtain the best set of results and everyone kept on coming up with different value and the process took really long. We lost time and the last set of results was rushed affecting the overall set off results were rushed contributing to the overall percentage of error. Human error as well as stability level of manometer. The stability of the manometer led to a waiting period but due time we took some of the reading when the water level moving up and down slightly.
Conclusions
- The experiment proved that as we moved away from the surface of the internal diameter the greater the volume of pressure was measured at higher speed and less viscosity.
- I found out that the mass flow rate for two set of tubes was different.
- I found out that at high pressure velocity is low but when velocity is high the level of pressure is low when moving from one to another point on the system.
- There was a large change in velocity and also the mass flow rate from the point 1 to the point 3. As we learnt in aspect of aerospace engineering that decrease in area means increase in velocity.
References
- Fluid Mechanics Book, Frank White, Chapter 2
- Apparatus and Experimental procedure: The online encyclopedia, Wikipedia.org, researches on airflow experiment.
- Basic information about the venturi effect was taken from the Wikipedia page on “Venturi effect”
- P. R Wormleaton’s notes on conservation of mass and continuity)
