Practical 2
For Practical 2, we have been tasked to complete the air pump challenge at home due to COVID-19.
Background information
An air lift pump is used to lift water from one location to another using compressed air. Liquid enters from one end of the pump (inlet), while a mixture of air and water discharges from the other end (outlet).
The compressed air is mixed with water. Since the density of air is lower than that of water, the air will quickly rise to the top. As the water mixes with air, the density of water decreases, as the density of the air and water mixture is lower than the density of pure water. Thus, as the vapour bubbles rise, they carry along some liquid upward due to fluid pressure.
Constant: Height of U-tube to base of Jug (b = 10cm)
Independent Variable: Length of tubing inside U-tube (a)
Dependent Variable: Flow Rate
Constant: Length of U-tube inside tubing (a = 2cm)
Independent Variable: Height of U-tube from base of Jug (b)
Dependent Variable: Flow Rate
Unfortunately, the water container we used was not big enough, and we were unable to carry out any more runs for Y = 7 cm and Y = 5 cm.
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.
As the distance from the surface of the water to the tip of the air outlet tube increases, the average pump flowrate also 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.
As the distance from the surface of the water to the tip of the U-shape tube that is submerged in water increases, the average flowrate also increases.
3. Summarise the learning, observations and reflection in about 150 to 200 words.
We learnt that the average flow rate increases directly proportionately to the length of the tubing inside the U-tube (a), as shown in Experiment 1. Figure 3 shows this with the gradient of the graph being mostly consistent.
In Experiment 2, we learnt that the average flow rate does not increase consistently, as shown in Figure 4. When Y = 15 cm ,the flowrate was exponentially higher as compared to the previous readings. This is due to the pressure exerted on the water. With the height of the U-tube from the base of the water container (b) being larger, there would be less pressure exerted on the water and the vapour bubbles, and thus an exponentially higher flow rate is expected.
4. Explain how you measure the volume of water accurately for the determination of the flowrate?
The group decided that we will stop the experiment when we collect 50ml of water in the beaker, and measure the amount of time taken for it to reach 50ml. We used a measuring jug with markings at 50ml intervals so that we would be able to see when the water level is at 50ml. We would let the pump run for 10s first into a different container before we use the measuring jug to measure 50ml of water. This ensures that there are no discrepancies in time due to the startup of the pump.
5. How is the liquid flowrate of an air-lift pump related to the air flowrate? Explain your reasoning.
The liquid flowrate is directly proportional to the air flowrate. As the air flowrate increases, the liquid flowrate also increases. This is because as the air bubbles travel up the U-tube, the air bubbles also displace some liquid along with it. As the air bubbles exit the U-tube, the liquid is also pushed out of the U-tube, where it is collected in the measuring jug.
6. Do you think pump cavitation can happen in an air-lift pump? Explain.
Pump cavitation is the formation and collapse of vapour bubbles in the pump. Vapourisation of liquid occurs if the pressure of the liquid at the pump suction is less than the vapour pressure. Bubbles of vapour will form and move towards a region of high pressure inside the pump, where they will collapse. However, since only air is flowing through the air-lift pump, and there is no liquid flowing through the air-lift pump itself, cavitation will not occur.
7. What is the flow regime that is most suitable for lifting water in an air-lift pump? Explain.
The flow regime can be calculated using Reynold’s Number.
Since 826.4482136 < 2100, the most suitable flow regime is laminar flow.
8. What is one assumption about the water level that has to be made? Explain.
The assumption we made is that the water level in the big container remains constant after each run, when water is transferred from the measuring jug back into the big container. This is because there will be some droplets of water left in the measuring jug after each run that is not transferred back into the big container. Furthermore, during the actual experiment, there might be some spillage of water onto the side that is not collected during the experiment. Thus, to simplify the calculation of flow rate, it is assumed that the amount of water collected in the measuring jug is the amount of water transferred back into the big container.
Background information
An air lift pump is used to lift water from one location to another using compressed air. Liquid enters from one end of the pump (inlet), while a mixture of air and water discharges from the other end (outlet).
The compressed air is mixed with water. Since the density of air is lower than that of water, the air will quickly rise to the top. As the water mixes with air, the density of water decreases, as the density of the air and water mixture is lower than the density of pure water. Thus, as the vapour bubbles rise, they carry along some liquid upward due to fluid pressure.
Materials and Experimental Setup
Figure 1. Materials used
1. Water container
2. U-tube
3. Measuring jug
4. Connecting tube
5. Air-lift pump
2. U-tube
3. Measuring jug
4. Connecting tube
5. Air-lift pump
Figure 2. Experimental Setup
Experiment 1
Constant: Height of U-tube to base of Jug (b = 10cm)
Independent Variable: Length of tubing inside U-tube (a)
Dependent Variable: Flow Rate
Table 1. Data for Experiment 1
Experiment 2
Constant: Length of U-tube inside tubing (a = 2cm)
Independent Variable: Height of U-tube from base of Jug (b)
Dependent Variable: Flow Rate
Table 2. Data for Experiment 2
Unfortunately, the water container we used was not big enough, and we were unable to carry out any more runs for Y = 7 cm and Y = 5 cm.
Figure 3: Graph of average flowrate against X for Experiment 1
As the distance from the surface of the water to the tip of the air outlet tube increases, the average pump flowrate also increases.
Figure 4. Graph of Average Flowrate against Y for Experiment 2
In Experiment 2, we learnt that the average flow rate does not increase consistently, as shown in Figure 4. When Y = 15 cm ,the flowrate was exponentially higher as compared to the previous readings. This is due to the pressure exerted on the water. With the height of the U-tube from the base of the water container (b) being larger, there would be less pressure exerted on the water and the vapour bubbles, and thus an exponentially higher flow rate is expected.
The liquid flowrate is directly proportional to the air flowrate. As the air flowrate increases, the liquid flowrate also increases. This is because as the air bubbles travel up the U-tube, the air bubbles also displace some liquid along with it. As the air bubbles exit the U-tube, the liquid is also pushed out of the U-tube, where it is collected in the measuring jug.
6. Do you think pump cavitation can happen in an air-lift pump? Explain.
Pump cavitation is the formation and collapse of vapour bubbles in the pump. Vapourisation of liquid occurs if the pressure of the liquid at the pump suction is less than the vapour pressure. Bubbles of vapour will form and move towards a region of high pressure inside the pump, where they will collapse. However, since only air is flowing through the air-lift pump, and there is no liquid flowing through the air-lift pump itself, cavitation will not occur.
7. What is the flow regime that is most suitable for lifting water in an air-lift pump? Explain.
The flow regime can be calculated using Reynold’s Number.
Figure 5. Calculations for Reynold's Number
Since 826.4482136 < 2100, the most suitable flow regime is laminar flow.
8. What is one assumption about the water level that has to be made? Explain.
The assumption we made is that the water level in the big container remains constant after each run, when water is transferred from the measuring jug back into the big container. This is because there will be some droplets of water left in the measuring jug after each run that is not transferred back into the big container. Furthermore, during the actual experiment, there might be some spillage of water onto the side that is not collected during the experiment. Thus, to simplify the calculation of flow rate, it is assumed that the amount of water collected in the measuring jug is the amount of water transferred back into the big container.
Link to short video of Practical 2: https://www.youtube.com/watch?v=lpyCYnmbixo&ab_channel=NickTay
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