- 1x U-tube
- 1x PVC Tube
- 1x Air-lift pump
- 1x Big pail
- 1x Measuring cup
- 1x Pair of scissors
- Lots of tape
- Permanent Markers
- PVC tube is inserted into the long end of the U-tube, ensuring that the distance between the inserted end of the PVC tube to the long end of the U-tube is 2cm (this will be called distance a).Check that there are no kinks on the PVC tube and tape the remaining PVC tube and U-tube together.
- Measure 10cm from the bottom of the large pail and place a mark. This will be the distance of the long end of the U-tube to the bottom of the large pail (which will be called distance b).
- Place the new contraption into the pail, where the U-tube would comfortably lie on the side.
- Fill the large pail with water, until it reaches near the brim. Record the water level.
- Move the long end of the U-tube until it is near or at the 10cm mark on the pail. Check that there are no kinks on the PVC tube.
- Tape the U-tube onto the pail, ensuring that it does not move.
- Place the measuring cup on the short end of the U-tube, ensuring that the short end must be inside the measuring cup to prevent splashing.
- Do a few test runs with the air-lift pump to see if there is adjustments needed.
- Start stopwatch when the air-lift pump is turned on.
- Stop the stopwatch once the experimenter sees that there is 300ml of water inside the measuring cup. Record the time taken. Return the water collected into the pail.
- Repeat steps 1-10 for 3 runs, and then using different values of a and b.
Water level = 28cm = a + b + X/Y
Volume of collected water = 300ml for X/Y ≥ 14cm, 100ml for X/Y <14cm
Experiment 1
b = 10cm
Experiment 2
a = 2cm
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.
From the graph, we can see that X has a logarithmic relationship with average flow rate, whereby higher X means higher flow rate.
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.
From the graph, we can see that Y has a logarithmic relationship with average flow rate, whereby higher Y means higher discharge flow rate.
Summarise the learning, observations and reflection in about 150 to 200 words.
From this experiment, we learnt that the distance from the water surface where air was introduced into the water affected the liquid flow rate at the outlet of the U-tube. The farther the distance, the higher the discharge flow rate. This is because the mass of water above is higher, causing more water to be pushed into the u-tube. This was observed in both the experiments and the graphs, when X/Y which is the distance from the water surface increases, the discharge flow rate will increase.
We also learnt that there is a minimum distance from water surface/Y to have a discharge flow rate, e.g exp 2 when Y=8cm, there was no discharge flow rate.
We also learnt that the collection of water and the whole set up of the experiment is quite challenging without the school’s practical lab as some materials were hard to obtain like the plastic container film or the exact 5L water jug. The water collection is also really difficult as the water splashes everywhere and when the water is collected there are a lot of water droplets around the side of the container while tons of bubbles in the water itself.
Explain how you measure the volume of water accurately for the determination of the flowrate?
We measured the volume of water collected by using a bottle with measurements.
Each measurement is either 300ml for when X and Y is more than 14cm, or 100ml for when X and Y is less than 14cm. The difference in measurement is due to the significantly lower flow rate and thus to conserve time, 100ml was used.
How is the liquid flow rate of an air-lift pump related to the air flowrate? Explain your reasoning.
The higher the air flow rate, the higher the water flow rate. As the air flow rate increases, more air is introduced into the water which causes the density of water and air mixture to decrease even further. Due to the increase of the difference of density, more of the water and air mixture is pushed out, causing the liquid flow rate out to increase.
Do you think pump cavitation can happen in an air-lift pump? Explain.
Cavitation cannot occur as an airlift pump does not require the water to be vaporized and thus no water can enter the pump.
What is the flow regime that is most suitable for lifting water in an air-lift pump? Explain.
Turbulent flow is more suitable as the increase in speed will disrupt the flow, causing the flow to become homogeneous. This allows the water and air that are both in the flow to mix better, allowing density to decrease even lower which will result in a higher liquid flow rate.
What is one assumption about the water level that has to be made? Explain. The water level would have stayed the same throughout the experiment. This is because the volume may change due to a lot of factors such as evaporation and amount of water returned to the pail after a run. Due to this, X and Y must be considered constant. To allow both variables to be constant, water level must be assumed constant so that the reference point is the same.
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