AIR-LIFT PUMP CHALLENGE ⛽
This is our second practical together as a group, unfortunately we have to do it virtually due to the COVID-19 situation😷. In this practical, we are to design an air-lift pump with the given items, and observe the different liquid flowrates.
What is an air-lift pump? How does it work?
An air-lift pump is a device that is used to lift water with the use of air supply. It is a simple device in which liquid enters from the suction of the U-tube submerged in liquid and a mixture of air and liquid discharges from the other end of the U-tube. Air which has a lower density then liquid, it rises quickly, hence the liquid is taken in the ascendant air flow and moves in the same direction of air (up the U-tube and to the discharge).
Making of an air-lift pump
Before making an air-lift pump, we had to make some amendments to the equipment given.
Sketch of our air-lift pump design:
Materials used:
These are the materials used to prepare for our practical air-lift pump experiment.
- U-tube
- Tape
- PVC hose
- Ball valve
- Plastic Connectors
- Pump
- Scissors
Making a mini ring for PVC tube and U-tube
We used a thick magazine and cut it into a strip. Twirling it into a cylindrical shape with a diameter of 2.5cm.
PVC hose:
We used 85.4cm of PVC hose and was cut into 3 parts.
- Connected to the pump - 3.4cm
- Middle section - 38.6cm
- End section towards to U-tube - 43.4cm
1. Team Leader
- Jun Heng 👣
2. Experimenter
- Andrew 🔬
3. Time-keeper
- Pi Ti 🕐
4. Blogger
- Cyane (me😊)
Observing the experiment together: 👀
Photos during the experiment:
We observe the experiment while Andrew does the hands-on. (Our tired faces 🥱)
Experiment Worksheet:
Experimental data:
Experiment 1
b = 10cm
|
a
(cm) |
X
(cm) |
Flowrate
(ml/s) |
Average
pump Flowrate (ml/s) |
||
|
Run
1 |
Run
2 |
Run
3 |
|||
|
2 |
14.0 |
7.000 |
6.833 |
6.833 |
6.890 |
|
4 |
12.0 |
4.666 |
4.333 |
4.500 |
4.500 |
|
6 |
10.0 |
3.000 |
2.833 |
2.75 |
2.861 |
|
8 |
8.0 |
0.750 |
0.500 |
0.833 |
0.694 |
|
10 |
6.0 |
0.000 |
0.000 |
0.000 |
0.000 |
Flowrate is volume of water
collected/transferred divided by time taken
Experiment 2
a = 2cm
|
b
(cm) |
Y
(cm) |
Flowrate
(ml/s) |
Average
pump Flowrate (ml/s) |
||
|
Run
1 |
Run
2 |
Run
3 |
|||
|
10* |
16.0 |
7.000 |
6.833 |
6.833 |
6.890 |
|
12 |
14.0 |
6.000 |
5.833 |
6.000 |
5.944 |
|
14 |
12.0 |
3.333 |
3.333 |
2.667 |
3.111 |
|
16 |
10.0 |
1.333 |
1.500 |
1.167 |
1.333 |
|
18 |
8.0 |
0.000 |
0.000 |
0.000 |
0.000 |
|
20 |
6.0 |
0.000 |
0.000 |
0.000 |
0.000 |
Flowrate is volume of water collected/transferred divided by time taken
Questions:
1. Plot tube length X versus pump flowrate. (X is the distance from the surface of the water to the tip of the air outlet tube). Draw at least one conclusion from the graph.
Figure 1: Graph of Average Flowrate (ml/s) against X (cm).
The average pump flowrate increases with increasing X value. As X increases from 6cm to 14cm, average pump flowrate increases from 0ml/s to 6.89ml/s. There is no flowrate when X is 6cm. Average pump flowrate is observed to follow some linearity as X increases.
Average pump flowrate increases as the distance from the surface of the water to the tip of the air outlet tube increases.
2. Plot tube length Y versus pump flowrate. (Y is the distance from the surface of the water to the tip of the U-shape tube that is submerged in water). Draw at least one conclusion from the graph.
Figure 2: Graph of Average Flowrate (ml/s) against Y (cm)
The average pump flowrate increases with increasing Y value. Average pump flowrate increases from 0ml/s to 6.89ml/s as Y increases from 6cm to 16cm. There is no flowrate when Y has a range of value from 6 cm to 8 cm. Average flowrate is observed to be slightly exponential with increasing Y.
Average pump flowrate increases as distance from the surface of the water to the tip of the U shape tube that is submerged in water increases.
3. Summarise the learning, observations and reflection in about 150 to 200 words.
Learning: The larger the distance of the surface of the water to the outlet of the airflow, the greater the volume of water pumped into the other side of the U-bend.
Observation: In experiment 1, as the length of a increases, the average flowrate of the experiment decreases exponentially. In experiment 2, as length of b increases, the average flowrate of the experiment decreases gradually.
Reflection: My experiments can be improved on, in terms of the setting up I think that we should have tightly secured the pipe for the air outlet so that the pipe of the air outlet does not move about or fall off the u bend, this would improve on my efficiency and reduce the work time. We should have used the same water level from the start of the experiment, this would reduce the amount of time spent on the experiments as we had to redo the experiments to only have one changing variable, to keep the experiment fair.
4. Explain how you measure the volume of water accurately for the determination of the flowrate?
We used a jug with an accuracy to 5.00ml. We estimated the value to the nearest tens-place. We also used a timer to help time the duration of flow, which is kept constant at 60 seconds.
5. How is the liquid flowrate of an air-lift pump related to the air flowrate? Explain your reasoning.
The faster the air flowrate, the faster the liquid flowrate of the air-lift pump. As air flowrate increases, the amount of air mixing with the liquid increases. There are more lighter air bubbles carrying up heavier water volumes, decreasing the density per volume of mixture. Hence, larger air flowrate allows an increase in liquid flowrate of an airlift pump. However, when air flowrates become significantly high, the bubbles formed become too large and unstable, eventually resulting in an annular flow whereby most fluid coming out is gas instead of liquid. This would in turn cause a decrease in the liquid flowrate of an air-lift pump.
6. Do you think pump cavitation can happen in an air-lift pump? Explain.
Cavitation occurs when the liquid in the pump turns into vapour at low pressure and air bubbles are created. An air-lift pump requires an air supply allowing liquid to mix with air, causing the mixture to be less dense than the rest of the liquid creating the slug flow thus displacing upwards to the discharge by the surrounding liquid of higher density. In an air-lift pump, there is no rapid creation and subsequent collapse of air bubbles in the fluid. Therefore, I think that pump cavitation will not happen in an air-lift pump.
7. What is the flow regime that is most suitable for lifting water in an air-lift pump? Explain.
Air lift pumps are most suitable in turbulent slug-flow regime. Airlift pumps are simple devices using injected gas flow as the driving force for pumping liquids or mixtures to higher levels. The air is injected in the lower part of the pipe that carries the fluid that is to be transported. Thereafter, air that has a lower density than that of the fluid rises, causing the fluid to move in the same direction as the air bubbles due to fluid pressure.
Slug flow is a two-phase flow of both liquid and gas, where intermittent sequence of heavy liquid is followed by light air bubbles flowing through the tube.
Due to the working principles of the airlift pump, the flow regime has to be in two-phases, gas and liquid, in order for the airlift pump to move liquid vertically upwards. Hence, the difference in densities between the gas and the liquid is required for the airlift pump to work. Other laminar flow regimes are unable to allow efficient lifting of water in the airlift pump as the driving force for the airlift pump to work (gas bubbles) is eliminated.
8. What is one assumption about the water level that has to be made? Explain.
The water level height remains constant. The turbulent flow of the water is observed to have made splashes, however it is assumed that no water is lost during this process. Recollection and the reusing of the water after measuring each flowrate is also assumed to have no loss of water outside the pail. Water level for all experimental values across the same experiment is the same.
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