#$&*
PHY 202
Your 'flow experiment' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Flow Experiment_labelMessages **
** **
opy this document into a word processor or text editor.
• Follow the instructions, fill in your data and the results of your analysis in the given format.
• Any answer you given should be accompanied by a concise explanation of how it was obtained.
• To avoid losing your work, regularly save your document to your computer.
• When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
The picture below shows a graduated cylinder containing water, with dark coloring (actually a soft drink). Water is flowing out of the cylinder through a short thin tube in the side of the cylinder. The dark stream is not obvious but it can be seen against the brick background.
You will use a similar graduated cylinder, which is included in your lab kit, in this experiment. If you do not yet have the kit, then you may substitute a soft-drink bottle. Click here for instructions for using the soft-drink bottle.
• In this experiment we will observe how the depth of water changes with clock time.
In the three pictures below the stream is shown at approximately equal time intervals. The stream is most easily found by looking for a series of droplets, with the sidewalk as background.
Based on your knowledge of physics, answer the following, and do your best to justify your answers with physical reasoning and insight:
• As water flows from the cylinder, would you expect the rate of flow to increase, decrease or remain the same as water flows from the cylinder?
Your answer (start in the next line):
At first, I would expect the rate of flow to increase because the pressure would decrease as the volume decreased. But this doesn’t sound correct because the cylinder is open to the atmosphere, therefore, it would be a constant pressure. Constant pressure would lead me to believe that the velocity of the liquid leaving the hole would be constant due to the constant pressure.
#$&*
@&
The pressure on the top surface of the water is always the same. The water above the hole, however, contributes to the pressure at the level of the hole.
*@
• As water flows out of the cylinder, an imaginary buoy floating on the water surface in the cylinder would descend.
• Would you expect the velocity of the water surface and hence of the buoy to increase, decrease or remain the same?
Your answer (start in the next line):
I would expect the velocity of the water surface/buoy to decrease because there would be less liquid in the cylinder as time went on, but would almost expect the velocity of increase because of the steady change of volume.
#$&*
• How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated? More specifically how could you determine the velocity of the water surface from the values of the other quantities?
Your answer (start in the next line):
I would think that Bernoulli’s Equation would come into play and that you would have to take into account the pressure head, pressure divided by specific weight of the fluid,; elevation head of fluid,; and the velocity head which is velocity squared and divided by two times gravity. By taking the diameter of the cylinder surface and the exit hole, the total area could be found at these two points and the velocity determined if one other velocity is known.
#$&*
• The water exiting the hole has been accelerated, since its exit velocity is clearly different than the velocity it had in the cylinder.
• Explain how we know that a change in velocity implies the action of a force?
Your answer (start in the next line):
I think its because Force equal mass times acceleration. The fluid has a mass because it does have a specific weight, and the acceleration caused by the change in velocity. Therefore, mass times acceleration equals force.
#$&*
• What do you think is the nature of the force that accelerates the water from inside the cylinder to the outside of the outflow hole?
Your answer (start in the next line):
What we do know is that an increase in velocity is caused by a decrease in pressure.
#$&*
From the pictures, answer the following and justify your answers, or explain in detail how you might answer the questions if the pictures were clearer:
• Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
Flow rate, Q, is equal to the area times the velocity and is expressed in units of m^3./s or ft^3/s. The area is constant and the velocity is changing because it was stated previously that we have acceleration. If we are accelerating, then velocity is assumed to be increasing. If velocity is increasing then the flow rate is increasing. Therefore, I would conclude that it is changing at a faster and faster rate.
@&
Acceleration can be positive or negative.
*@
However, in the picture it appears that the horizontal displacement of the fluid shooting out of the bottom is decreasing as the fluid decreases in the cylinder. This decrease in horizontal displacement leads me to believe there is a decrease in velocity causing a decrease in acceleration. I don’t know if I am seeing this accurately or not due to the pictures.
@&
@&
Velocity can decrease at a constant, increasing or decreasing rate, and acceleration can be constant, increasing or decreasing.
*@
*@
#$&*
• What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
The graph of depth vs time would be decreasing and concave upward.
#$&*
• Does the horizontal distance (the distance to the right, ignoring the up and down distance) traveled by the stream increase or decrease as time goes on?
Your answer (start in the next line):
The horizontal distance appears to decrease as time goes on.
#$&*
• Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
The distance is changing at a steady rate.
#$&*
• What do you think a graph of this horizontal distance vs. time would look like? Describe in the language of the Describing Graphs exercise.
Your answer (start in the next line):
I feel as if it would be decreasing at a linear rate because the distance appears to be steadily decreasing.
#$&*
You can easily perform this experiment in a few minutes using the graduated cylinder that came with your kit. If you don't yet have the lab materials, see the end of this document for instructions an alternative setup using a soft-drink bottle instead of the graduated cylinder. If you will be using that alternative, read all the instructions, then at the end you will see instructions for modifying the procedure to use a soft drink bottle.
Setup of the experiment is easy. You will need to set it up near your computer, so you can use a timing program that runs on the computer. The cylinder will be set on the edge of a desk or tabletop, and you will need a container (e.g., a bucket or trash can) to catch the water that flows out of the cylinder. You might also want to use a couple of towels to prevent damage to furniture, because the cylinder will leak a little bit around the holes into which the tubes are inserted.
• Your kit included pieces of 1/4-inch and 1/8-inch tubing. The 1/8-inch tubing fits inside the 1/4-inch tubing, which in turn fits inside the two holes drilled into the sides of the graduated cylinder.
• Fit a short piece of 1/8-inch tubing inside a short piece of 1/4-inch tubing, and insert this combination into the lower of the two holes in the cylinder. If the only pieces of 1/4-inch tubing you have available are sealed, you can cut off a short section of the unsealed part and use it; however don't cut off more than about half of the unsealed part--be sure the sealed piece that remains has enough unsealed length left to insert and securely 'cap off' a piece of 1/4-inch tubing.
• Your kit also includes two pieces of 1/8-inch tubing inside pieces of 1/4-inch tubing, with one end of the 1/8-inch tubing sealed. Place one of these pieces inside the upper hole in the side of the cylinder, to seal it.
• While holding a finger against the lower tube to prevent water from flowing out, fill the cylinder to the top mark (this will be the 250 milliliter mark).
• Remove your thumb from the tube at the same instant you click the mouse to trigger the TIMER program.
• The cylinder is marked at small intervals of 2 milliliters, and also at larger intervals of 20 milliliters. Each time the water surface in the cylinder passes one of the 'large-interval' marks, click the TIMER.
• When the water surface reaches the level of the outflow hole, water will start dripping rather than flowing continuously through the tube. The first time the water drips, click the TIMER. This will be your final clock time.
• We will use 'clock time' to refer to the time since the first click, when you released your thumb from the tube and allowed the water to begin flowing.
• The clock time at which you removed your thumb will therefore be t = 0.
Run the experiment, and copy and paste the contents of the TIMER program below:
Your answer (start in the next line):
1 195.833 195.883
2 198.77 2.937
3 201.254 2.484
4 203.898 2.644
5 206.767 2.778
6 209.328 2.652
7 212.796 3.468
8 216.56 3.764
9 220.454 3.894
10 225.836 5.384
11 231.761 5.923
12 240.584 8.823
13 245.703 5.119
#$&*
Measure the large marks on the side of the cylinder, relative to the height of the outflow tube. Put the vertical distance from the center of the outflow tube to each large mark in the box below, from smallest to largest distance. Put one distance on each line.
Your answer (start in the next line):
Distances are in centimeters.
20.0, 18.7, 17.0, 15.35, 13.65, 11.9, 10.1, 8.3, 6.4, 4.5, 2.6, 0.7, 0
#$&*
Now make a table of the position of the water surface vs. clock time. The water surface positions will be the positions of the large marks on the cylinder relative to the outflow position (i.e., the distances you measured in the preceding question) and the clock times will as specified above (the clock time at the first position will be 0). line Enter 1for each event, and put clock time first, position second, with a comma between.
For example, if the first mark is 25.4 cm above the outflow position and the second is 22.1 cm above that position, and water reached the second mark 2.45 seconds after release, then the first two lines of your data table will be
0, 25.4
2.45, 22.1
If it took another 3.05 seconds to reach the third mark at 19.0 cm then the third line of your data table would be
5.50, 19.0
Note that it would NOT be 3.05, 19.0. 3.05 seconds is a time interval, not a clock time. Again, be sure that you understand that clock times represent the times that would show on a running clock.
The second column of your TIMER output gives clock times (though that clock probably doesn't read zero on your first click), the third column gives time intervals. The clock times requested here are those for a clock which starts at 0 at the instant the water begins to flow; this requires an easy and obvious modification of your TIMER's clock times.
For example if your TIMER reported clock times of 223, 225.45, 228.50 these would be converted to 0, 2.45 and 5.50 (just subtract the initial 223 from each), and these would be the times on a clock which reads 0 at the instant of the first event.
Do not make the common error of reporting the time intervals (third column of the TIMER output) as clock times. Time intervals are the intervals between clicks; these are not clock times.
Your answer (start in the next line):
20.0, 18.7, 17.0, 15.35, 13.65, 11.9, 10.1, 8.3, 6.4, 4.5, 2.6, 0.7, 0
0, 49.87
0.7, 44.751
2.6, 35.928
4.5, 30.005
6.4, 24.621
8.3, 20.727
10.1, 16.963
11.9, 13.495
13.65, 10.843
15.35, 8.065
17.0. 5.421
18.7, 2.937
20.0, 0.00
#$&*
You data could be put into the following format:
clock time (in seconds, measured from first reading) Depth of water (in centimeters, measured from the hole)
0 14
10 10
20 7
etc. etc.
Your numbers will of course differ from those on the table.
The following questions were posed above. Do your data support or contradict the answers you gave above?
• Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
Yes it does. The water seems to be decreasing at a steady state and then at a decreasing rate.
#$&*
• Sketch a graph of depth vs. clock time (remember that the convention is y vs. x; the quantity in front of the 'vs.' goes on the vertical axis, the quantity after the 'vs.' on the horizontal axis). You may if you wish print out and use the grid below.
Describe your graph in the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph begins to decrease at a steady rate and then begins to decrease at a decreasing rate. The curve is concave upward.
#$&*
caution: Be sure you didn't make the common mistake of putting time intervals into the first column; you should put in clock times. If you made that error you still have time to correct it. If you aren't sure you are welcome to submit your work to this point in order to verify that you really have clock times and not time intervals
Now analyze the motion of the water surface:
• For each time interval, find the average velocity of the water surface.
Explain how you obtained your average velocities, and list them:
Your answer (start in the next line):
6.81 mm/s, 8.05 mm/s, 7.56 mm/s, 7.20 mm/s, 7.54 mm/s, 5.77 mm/s, 5.31 mm/s, 5.14 mm/s, 3.42 mm/s, 3.38 mm/s, 2.27 mm/s, 0.78 mm/s.
Average Velocity of entire experiment = (0.78 + 6.81) / 2 = 3.80 mm/s
For each interval, I took the displacement of water and divided it by the time interval.
#$&*
• Assume that this average velocity occurs at the midpoint of the corresponding time interval.
What are the clock times at the midpoints of your time intervals, and how did you obtain them? (Give one midpoint for each time interval; note that it is midpoint clock time that is being requested, not just half of the time interval. The midpoint clock time is what the clock would read halfway through the interval. Again be sure you haven't confused clock times with time intervals. Do not make the common mistake of reporting half of the time interval, i.e., half the number in the third column of the TIMER's output):
Your answer (start in the next line):
1.4865, 4.179, 6.743, 9.454, 12.169, 15.229, 18.845, 22.674, 27.313, 32.9665, 40.3395, 47.3105
#$&*
• Make a table of average velocity vs. clock time. The clock time on your table should be the midpoint clock time calculated above.
Give your table below, giving one average velocity and one clock time in each line. You will have a line for each time interval, with clock time first, followed by a comma, then the average velocity.
Your answer (start in the next line):
(s, mm/s)
1.4865, 6.81
4.179, 8.05
6.743, 7.56
9.454, 7.20
12.169, 7.54
15.229, 5.77
18.845, 5.31
22.674, 5.14
27.313, 3.42
32.9665, 3.38
40.3395, 2.27
47.3105, 0.78
#$&*
• Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph increases linearly between 1.5 and 4.2 s, then becomes concave downward and decreases steadily to 12.2 seconds and becomes concave upward till time 15.2 seconds and begins to be concave downward and decrease at an increasing rate till time 18.8 s. Then it begins to decrease at a decreasing rate till time 22.7 seconds. Then decreases at an increasing rate till 27.3 s. It begins to move upward till point 32.9 seconds and then begins to decrease at a decreasing rate till time 47.3 s.
#$&*
• For each time interval of your average velocity vs. clock time table determine the average acceleration of the water surface. Explain how you obtained your acceleration values.
Your answer (start in the next line):
I took the interval between the midpoint clock times and divided the velocity at each of these clock times to get the accerleration.
aAve = vAve / ‘dt
#$&*
@&
The apparent concavities may be simply the result of uncertainties in measurement. Ideally we would expect the graph to in fact be linear, and this system should behave in a nearly ideal manner.
*@
• Make a table of average acceleration vs. clock time, using the clock time at the midpoint of each time interval with the corresponding acceleration.
Give your table in the box below, giving on each line a midpoint clock time followed by a comma followed by acceleration.
Your answer (start in the next line):
1.486, 4.58
4.179, 2.98
6.743, 2.94
9.454, 2.65
12.16, 2.78
15.22, 1.88
18.84, 1.46
22.67, 1.34
27.31, 0.73
32.9, 0.60
40.33, 0.30
47.31, 0.11
#$&*
Answer two questions below:
• Do your data indicate that the acceleration of the water surface is constant, increasing or decreasing, or are your results inconclusive on this question?
• Do you think the acceleration of the water surface is actually constant, increasing or decreasing?
Your answer (start in the next line):
The data indicates that the acceleration is decreasing at a decreasing rate.
@&
According to your velocity vs. midpoint clock time results, the average rate of change of velocity with respect to clock time for from 4 sec to 22 sec is about -.15 cm/s^2, and from 22 sec to 40 sec is about the same.
This appears to be consistent with a constant average acceleration.
It isn't clear how you got the accelerations you listed, but they do not seem consistent with your table. It appears that you might have divided velocities by clock times, rather than dividing changes in velocity by changes in clock time.
*@
#$&*
Go back to your graph of average velocity vs. midpoint clock time. Fit the best straight line you can to your data.
• What is the slope of your straight line, and what does this slope represent? Give the slope in the first line, your interpretation of the slope in the second.
• How well do you think your straight line represents the actual behavior of the system? Answer this question and explain your answer.
• Is your average velocity vs. midpoint clock time graph more consistent with constant, increasing or decreasing acceleration? Answer this question and explain your answer.
Your answer (start in the next line):
Slope of the graph is approximately -0.145
The straight line does not represent the behavior. There are too many peaks going on throughout the graph.
The average velocity vs midpoint clock time is more consistent with decreasing acceleration. The graphs match up.
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
2 hrs
#$&*
You may add any further comments, questions, etc. below:
Your answer (start in the next line):
#$&*
*#&!
@&
Good, except that your calculations of acceleration do not appear to be correct.
I will ask you to revise those calculations.
Also see my other notes.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
&#
*@