The mass flow rate, ṁ, is constant before and after the venturi by continuity.
ṁ = ρair A1 V1 (eqn. 3)
where A1 and A2 are the cross-sectional areas of the tube before and after the venturi.
Since the pipe narrows downstream from the flow source, the velocity downstream at a point after the venturi will increase from a point before it in the inverse ratio as that of the cross sectional areas of the pipe before and after the venturi. Hence the velocity after the venturi V2 is given by:
V2 = A1 * V1 (eqn. 4)
A2
If we substitute equation 4 into equation 2 then:
A1 2 – 1 V12 = 2 (P1 – P2)
A2 ρair (eqn. 5)
Since the fluid in the manometer is parrafin and the density of paraffin, ρparrafin = 0.898 g cm-3,
ρair << ρparaffin. Thus the difference in air pressure between two points on the manometer is given by:
ΔP = ρparaffin * g * ΔH (eqn. 6)
Which when substituted into equation 5 gives:
V1 = 2 ρparaffin g ΔH
ρair A1 2 - 1
A2 (eqn. 7)
This value can be used to calculate dm/dt using equation 3.
The Pitot Method
If point 4 can be considered to be the tip of the pitot tube, and point 3 just upstream of it, and the arbitrary datum line the centreline of the pipe (so z can be elimated) the Bernouilli equation can be re-written:
P3 + V32 = P4 + V42
ρair 2 ρair 2 (eqn. 8)
At the tip of the pitot tube the moving air is brought to rest so V4 = 0.
Re-arranging:
V32 = 2 (P4 – P3)
ρair (eqn. 9)
Considering the change in pressure across the manometer tubes in the same way as before:
ΔP = ρparaffin * g * ΔH (eqn. 6)
Substituting equation 6 into equation 9 gives:
V3 = 2 ρparaffin g ΔH
ρair (eqn. 10)
This can be used to calculate the velocity at each radius the pitot is traversed across the tube. This will be a maximum at the tube’s centreline, falling to zero at the boundary of the pipe (the pipe’s wall).
Considering an infinitely small increment of mass flow δṁ through an area of the pipe δA with a width δr at a radius of r from the centreline:
δṁ = ρair V(r) δA (eqn. 11)
where V(r) is some function of r which gives the velocity at radius r from the centreline.
The element being considered is an annulus, and since it is infinitely small it can be considered
δA = 2 Π r * δr (eqn. 12)
so δṁ = ρair V(r) . 2 Π r * δr (eqn. 13)
The total mass flow rate is the sum of these elements from the centreline (r=0) to the edge of the pipe (r=R). Therefore the total mass flow rate ṁ can be approximated by
ṁ -> 2 Π ρair V(r) r dr (eqn. 14)
Plotting a graph of V*r against r gives a velocity profile which allows the calculation of the integral in the above equation.
Apparatus
figure 2
The apparatus were a u-tube paraffin manometer (the manometer had a near-horizontal scale which meant that for a particular change in pressure, the fluid would move a greater distance up the scale than in a vertical manometer giving a higher sensitivity. The scale had been specially calibrated with an adjustment to take in to account the gradient of the tube.) Also a thin transparent pitot tube, a ruler with a millimetre scale for measuring the traverse of the pitot tube from the centreline, a vacuum device to provide an air flow and a hollow cylindrical pipe with a venturi at the upstream end close to the vacuum.
There were a pair of air tubes connected to the upstream and downstream ends of the venturi, and another pair connected to the pitot tube and a point just upstream of the pitot tube. These two pairs of tubes were connected to a pressure tap which allowed the technician to select whether the manometer was reading the pressure difference across the venturi or the pressure difference between the pitot tube and the lower pipe boundary.
Experimental Procedure
First of all the vacuum pump was initiated and set to a flow rate close to its maximum. The pressure tap was set so that the manometer read the pressure difference across the venturi. Once the manometer had settled around a particular value, a reading from the visible leg of the manometer against its scale was taken and recorded. The pressure tap was then set so the manometer was reading the pressure difference between the pitot tube and the pipe boundary. The pitot tube was traversed from the pipe centreline to the pipe boundary at intervals of 2mm. The paraffin level in the manometer was recorded as before for each value. The pitot was then returned to the centreline and the process repeated.
The entire procedure was then repeated when the pump was set at a medium flow rate and once more at a low flow rate. The pump was then turned off and the equipment returned to its original state.
Raw Results
These are the recordings taken from the three experiments.
High Flow Rate
Pitot:
figure 3
Venturi: ΔH=0.24 mParaffin
Medium Flow Rate
Pitot:
figure 4
Venturi: ΔH=0.17 mParaffin
Low Flow Rate
Pitot:
figure 5
Venturi: ΔH=0.087 mParaffin
Atmospheric Observations
Atmospheric pressure = 769.25 mmHg
Temperature on the Celcius scale = 20 K
Calculations
Thermodynamic temperature T /K = (273.15 + θ deg C) (eqn. 8)
= (273.15 + 20)
T _ = 293.15 K
Density of Air: where V = Volume
PV = nRT and ρ = n / V
So, ρ = P / RT (reference 2) = 769.25 mmHg / 287.05 * 293.15
= 769.25 * 133.322 / 287.05 * 293.15 (reference 3)
ρair = 1.218770336 kg m-3
Pipe Cross-sectional Areas:
r1 = ½ * 108.3*10-3 = 54.15*10-3 m
A1 = ∏ * (54.15*10-3) = 9.211848665*10-3 m2
r2 = ½ * 29.65*10-3 = 14.825*10-3 m
A2 = ∏ * (14.825*10-3) = 6.904611969*10-4 m2
Sample Calculations
Mass flow rate in the venturi using the results from the experiment with the high flow rate:
“ V1 = 2 ρparaffin g ΔH “
ρair A1 2 - 1
A2 (eqn. 7)
ρparaffin = 898 kg m-3 (reference 1)
ρair = 1.219 kg m-3
A1 = 9.213*10-3 m2
A2 = 6.905*10-4 m2
ΔH = 0.24 mParaffin
V1 = 2 * 898 * 9.81 * 0.24
1.219 9.213*10-3 2 - 1
6.905*10-4
= 4228.5024
1.219 * 177.0224512
= √19.59541771
= 4.426671178
V1= 4.43 m s-1 (3sf)
ṁ = ρair A1 V1 (eqn. 3)
= 1.219 * 9.213*10-3 * 4.43
= 0.04975176621
ṁ _ = 4.98*10-2 kg s-1 (3sf)
Mass flow rate from the pitot traverse, again using the results from the experiment with the high flow rate:
Raw Results
figure 9
Mean height of fluid column in manometer <ΔH> = ΔH1 + ΔH2 (eqn. 9)
2
= (0.223 + 0.223) / 2
<ΔH> = 0.223 mParaffin
Sub. <ΔH> for ΔH in eqn. 10 for each radius, for example:
V = 2 ρparaffin g ΔH (eqn. 10)
ρair
= 2 * 898 * 9.81 * 0.223
1.219
= √3223.120164
= 56.77253001
V = 56.8 m s-1 (3sf)
Since V = 56.8 and r = 0, V*r = 0.
Plotting V*r versus r gives the graph:
figure 10
The trendline has equation y = 52.401x + 0.0087,
So for this flow rate V(r)*r = 52.401r + 0.0087.
Sub. in equation 14.
Taking R as the radius of the pipe = r2 = 14.825*10-3 m (as calculated before), then:
ṁ -> 2 Π ρair V(r) * r dr (eqn. 14)
= 2 Π * 1.219 * ∫ 52.401 r + 0.0087 dr
= 2 Π * 1.219 * [52.401/2 r2 + 0.0087 r]R0
= 2 Π * 1.219 * [52.401/2 * (14.825*10-3)2 + 0.0087 * 14.825*10-3 ]
= 0.04509232974
ṁ = 4.51*10-2 kg s-1 (3sf)
Analysed Results
Using the recordings from the experiment with the high flow rate:
figure 11
Plotting V*r versus r as the theory suggests leads to the following graph:
figure 12
Microsoft Excel reports that the trendline in the graph above has the equation
y= 52.401x + 0.0087, so V(r).r = 52.401r + 0.0087 .
Substituting this in to equation 14 gives a mass flow rate ṁ = 4.51*10-2 kg s-1 for the pitot method.
Substituting ΔH = 0.24 mParaffin in to equation 7 gives a velocity of V1= 4.43 m s-1, which when substituted in to equation 3 gives a mass flow rate of ṁ = 4.97*10-2 kg s-1.
Medium Flow Rate
Using the recordings from the experiment with the medium flow rate:
figure 13
Plotting V*r versus r as before leads to the following graph:
figure 14
The line above has the equation V(r)= 45.213r + 0.0086 .
Substituting this in to equation 14 gives a mass flow rate ṁ = 3.90*10-2 kg s-1 for the pitot method.
Substituting ΔH = 0.17 m in to equation 7 gives a velocity of V1= 3.73 m s-1, which when substituted in to equation 3 gives a mass flow rate of ṁ = 4.18*10-2 kg s-1.
Low Flow Rate
Finally, using the recordings from the experiment with the low flow rate:
figure 15
Plotting V*r versus r as before leads to the following graph:
figure 16
The line above has the equation V(r)= 32.935r + 0.0009 .
Substituting this in to equation 14 gives a mass flow rate ṁ = 2.78*10-2 kg s-1 for the pitot method.
Substituting ΔH = 0.087 m in to equation 7 gives a velocity of V1= 2.67 m s-1, which when substituted in to equation 3 gives a mass flow rate of ṁ = 2.99*10-2 kg s-1.
Discussion
Theory suggests that the mass flow rate ṁ should have the same value all along the air pipe, since mass is conserved and none is stored in the pipe.
As might be expected, the mass flow rate calculated using the two different methods is of the same order of magnitude in all three experiments. Taking the case of the high flow rate experiment, the mass flow rate calculated using the pitot traverse method was 4.51*10-2 kg s-1, whereas using the venturi method it was 4.97*10-2 kg s-1. This represents an actual difference of just 4.6*10-3 m, which is a percentage difference of 9.7% . Once a reasonable margin for error is taken in to account it is clear that the two methods are equivalent in terms of the answer produced. However in terms of accuracy and the amount of experimental time required the two methods differ considerably.
Using the pitot method relies on 7 individual manometer readings, whereas the venturi method relies on only 1 – each measurement increases the opportunity for errors to be made, so one of the reasons for the difference in the mass flow rates calculated by the two methods could be experimental error. The pitot method in particular requires minute readings to be taken for each traversal of the tube in the flow, which read by eye could be large sources of error. The proximity of the points to the trendlines on the graphs show that the accuracy of the experimental method could be improved. With the pitot method it is necessary to plot a graph, estimate the trend and then perform an integration. If the graph is plotted inaccurately then there will be a larger error in the results. The graph itself means that a computer is almost certainly necessary in order to use the pitot method and the integration requires the use of more advanced mathematical skills and a much larger number of calculations than the venturi method. This leaves the calculation more open to errors and takes more of the engineer’s valuable time and financial resources.
Potential sources of error in the readings are:
- Reading the manometer scale or ruler incorrectly – for example, a parallax error if the scale is not read at eye level
- Positioning the pitot tube incorrectly – not parallel with the air flow both horizontally and vertically, causing the fluid level on the manometer to change
- The inaccuracy as a result of the difficulty of traversing the pitot over minute distances and holding it steady in place while readings are taken
- The inability of the electric pump to maintain a constant flow rate and constant operating temperature (which will affect the air pressure and therefore the values used in the calculations)
- The fact that the use of a pitot tube in the air flow causes turbulence which can affect the velocity
- The fact that the pitot tube itself has dimensions and so it can never measure the air pressure at the pipe’s wall
- Varying atmospheric conditions in the laboratory
- Plotting the graph incorrectly or errors in estimating the trend of the graph for the pitot method
Conclusions
If it is assumed that both methods lead to the correct value of the mass flow rate, then it must be concluded from this experiment that the venturi method is more accurate than the pitot method because it relies on fewer measurements and resources and the method does not involve any estimation.
The venturi method is the better of the two because:
- The venturi method requires fewer readings to be taken and calculations to be made than the pitot method leaving less scope for error and taking less time
- The calculations required for the venturi method rely on fewer mathematical skills than those for the pitot method, which requires an understanding of differential calculus. This means that the venturi method leaves less room for mistakes to be made
- Measuring the traversal of the pitot by eye could be inaccurate since the distance moved is minute
- A venturi can be left permanently as a physical feature of a pipe since it requires no later user intervention – a pitot tube needs to be adjusted as measurements are taken.
Performing these experiments has allowed me to conclude that:
-
Air velocity increases with distance from the pipe walls.
This shows the existence of some type of turbulence at the wall which interferes with the fluid flow.
-
Air velocity decreases with the radius of the pipe.
At the same flow rate, air moving through a wide pipe will have a lower velocity than air moving through a narrower one.
References
1. Marcus Materials Co.
“Densities of Various Materials”
2. Richard Shelquist, 18 Oct 2005
“Air Density and Density Altitude”
3. Departmental Guidline ENG/GUI/1 “Physical Quantities and Units, Departmental Policy”
