Welcome to Week 5 Blog (Practical)
23 May 2021
This week we would be sharing more on the experiment that we did which is on the Air Lift Pump Challenge, due to the tightened SMM restrictions we would be bringing you this practical from the comfort of our home (NOT really). Well, let us begin!
1. Making of Air Lift Pump
Figure 1: Material used for making of a small hole
Then, we measure 2cm from the bottom of the pipe, before we proceed to heat up the small metal needle to poke through the pipe. However, due to lack of manpower, we are not able to film the process where the making of the small hole is made thus no videos is being shown for that.
Figure 2: Measurement of hole
Lastly, after putting all of the different components together, this is how the setup looks like.
(A)
Experiment
Worksheet
Experiment 1
b = 10cm
|
a
(cm) |
X
(cm) |
Flowrate
(ml/s) |
Average
Flowrate (ml/s) |
||
|
Run
1 |
Run
2 |
Run
3 |
|||
|
2 |
15 |
12.22 |
12.00 |
12.33
|
12.18 |
|
4 |
13 |
11.00 |
11.17 |
11.00 |
11.06 |
|
6 |
11 |
7.67 |
7.83 |
7.50 |
7.67 |
|
8 |
9 |
3.33 |
3.00 |
3.67 |
3.33 |
|
10 |
7 |
2.33 |
2.00 |
2.33 |
2.22 |
Flowrate
is volume of water collected/transferred divided by time taken
Experiment 2
a = 2cm
|
b
(cm) |
Y
(cm) |
Flowrate
(ml/s) |
Average
Flowrate (ml/s) |
||
|
Run
1 |
Run
2 |
Run
3 |
|||
|
10* |
17 |
12.22 |
12.00 |
12.33 |
12.18 |
|
12 |
15 |
8.17 |
8.00 |
8.17 |
8.11 |
|
14 |
13 |
7.17 |
7.00 |
7.33 |
7.17 |
|
16 |
11 |
3.00 |
3.17 |
3.33 |
3.17 |
|
18 |
9 |
2.50 |
2.67 |
2.50 |
2.56 |
|
20 |
7 |
1.17 |
1.00 |
1.33 |
1.17 |
Flowrate
is volume of water collected/transferred divided by time taken
*This is the same setting as the first run in experiment1. You do not
need to repeat it. Just record the results will do.
(B)
Questions
& Tasks
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.
[Answer]
Figure 4: Graph of Tube length, X,
versus pump flowrate
From the graph, it can be seen that when the
distance from the surface of the water to the tip of the air outlet tube
increases, the average pump flowrate increases. Thus, in order to achieve a
higher average flowrate, the distance from the surface of the water to the tip
of the air outlet must be larger.
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.
[Answer]
Figure 5: Graph of Tube length, Y, versus Pump Flowrate.
From the graph, it can be seen that when the
distance from the surface of the water to the tip of the U-shape tube is larger,
the higher the pump flow rate. Thus, to achieve a higher flowrate, submerge the pipe
nearer to the bottom of the pail.
3.
Summarise
the learning, observations, and reflection in about 150 to 200 words.
[Answer]
The more
the pipe is submerged in the pail, more water can be pumped out, resulting in a
larger flowrate. This can be seen from our results and graph as when Y/X
increases, the average pump flow rate also increases. However, the pump flowrate
was not constant, but rather a pulsating flow.
Due
to the COVID-19 situation, only 1 of us was the experimenter, this increases
the experimental errors present as we were not there to assist the experimenter
other than giving comments. For example, making sure that a/b were constant for
the different experiments. We were also not given lab apparatus as well such as
a retort stand to eliminate the errors said.
This
made it challenging for us and thus the results obtained would not be as
accurate. However, despite these challenges, the results obtained coincides
with our hypothesis which is the more the pipe is submerged, the larger the
amount of water collected, which would yield a larger pump flowrate.
4.
Explain how you measure the volume of water
accurately for the determination of the flowrate?
[Answer]
A measuring cup with increments of 10ml was used to
measure the amount of water pumped out from the pail. The volume of water
pumped out was then divided over the time taken in seconds to get the flowrate of
water displaced by the pump.
5.
How is the liquid flowrate of an air-lift pump
related to the air flowrate? Explain your reasoning.
[Answer]
The higher the air flowrate the higher the liquid
flowrate. A higher airflow rate produced by the pump would suggest that more
pressure is being produced and is acting on the liquid. This would mean that
more pressure energy can be converted into kinetic energy will result in an
increase in the liquid flowrate.
6. Do you think pump cavitation can happen in an air-lift pump? Explain
[Answer]
No.
As the air-lift pump does not have any priming requirements, it would not cavitate
as pump cavitation only happens when priming requirements are needed to prevent
pump from undergoing cavitation through air-bound/vapour lock.
7. What is the flow regime that is most suitable for lifting water in an air-lift pump? Explain.
[Answer]
The slug flow regime is the most suitable for lifting water in an air-lift pump. The air-lift pump is exposed to two-phase flow conditions as the air is being pushed upwards and lifting the water to a higher level.
By using slug flow regime,
it is able to achieve the highest efficiency with an air-lift pump
8. What is one assumption about the water level that has to be made? Explain.
[Answer]
The water level in the pail remains constant throughout the experiment. As water is pumped out of the pail, the water level in the pail will gradually drop.
Thus, by assuming water level to be constant, ensures that x and y values remain constant as water is pumped out of the pail.
Reflection:
Overall, despite it being an online experiment and only one person was allowed to operate it, the team had fun doing this experiment and enjoyed it.
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