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Phy 232
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.
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I am working on all my labs now; although I am still quite behind, please consider re-instating me as I need this course to complete my electrical engineering degree. I will complete this course!!
** #$&* Is flow rate increasing, decreasing, etc.? **
Flow Experiment
In this experiment you will use the 250-ml graduated cylinder from your lab kit.
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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):
A time measurement was taken as the liquid was drained after every 20 mL which left the apparatus. Plotting the time vs. volume, it can be shown that the rate of flow decreased as the cylinder drained. Physically, this can be described: as the liquid drains, less force (mass of the volume of liquid * acceleration) is applied to the opening hold of the cylinder, and less water leaves during each time interval. Eventually the flow stops. This is when the water level is below the hole in the cylinder.
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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):
If the rate of flow is decreasing as the liquid drains, then the buoy velocity, i.e. dy/dt, would decrease as well. 20 mL on the cylinder corresponds to ~3 cm in the y direction.
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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):
One could calculate the velocity of water passing through the cross section of a cylinder by measuring the distance between each volume measurement of the cylinder as was done in the previous solution. The velocity of the water surface would then be dy/dt instead of dV/dt. The water surface velocity would then be inversely proportional to the diameter of the cylinder, i.e. the larger the diameter of the cylinder, the slower the surface velocity.
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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):
dV/dt = acceleration, and mass * acceleration = force; a force is implied due to the known existence of acceleration associated with the flow of mass (in this case, expressed as a volume).
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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):
If there were no opening in the cylinder, then the force at a specific point on the cylinder would be equal to the sum of all forces at that point. Neglecting the motion of the water, then the forces would be zero. When the point is opened, i.e. a hole is formed, then the opposing force which once kept the water inside the cylinder would no longer exist and the water molecules at the point would move through the hole in the direction of the applied force.
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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):
The depth appears to be moving at a slower and slower rate.
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
It would look like a y = k/x graph (k is a constant).
The general shape of the graph is similar when you aren't too close to the x or y axis, but the y = k / x graph has a vertical and a horizontal asymptote, neither of which corresponds to the behavior of this system. For example the flow actually stops.
The function is actually quadratic; the graph corresponds to a section of a parabola. The point corresponding to the flow stopping is the vertex.
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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):
Decrease.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
Inreasing.
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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):
It would look like y = k/x (k is a constant).
The horizontal range of the outflowing water is in fact linear.
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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.
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):
t=0, V=25
deltaTIME1=1.7 s (V=210 mL)
deltaTIME2=2.3 s (V=170 mL)
deltaTIME3=2.3 s (V=130 mL)
deltaTIME4=2.9 s (V=90 mL)
deltaTIME5=4.0 s (V=50mL)
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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):
y0=25.5 cm
y1=21 cm
y2=16.5 cm
y3=12.0 cm
y4=7.5 cm
y5=3.0 cm
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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). Enter 1 line for 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):
25.5, 0.0
21.0, 1.7
16.5, 4.0
12.0, 6.3
7.5, 9.2
3.0, 13.2
0, 19.1
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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):
The data supports the answers; however, this is due to my running a similar experiment before answering the questions. Woops! The depths is changing at a slower and slower rate.
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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 looks like a y=k/x graph (k is a constant).
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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):
v1=-2.6 cm/s
v2=-2.0 cm/s
v3=-2.0 cm/s
v4=-1.6 cm/s
v5=-1.1 cm/s
v6=-0.8 cm/s
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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):
Clock times would be the averages between the two measured clock times.
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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):
-2.6, .85
-2.0, 2.85
-2.0, 5.15
-1.6, 7.75
-1.1, 11.2
-0.8, 16.15
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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):
It looks fairly linear.
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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):
Average acceleration is change avg velocity / change in time.
0.6
0.0
0.4
0.5
0.3
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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):
0.6, 1.85
0.0, 4
0.4, 6.45
0.5, 9.48
0.3, 14.18
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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 results are noisy; however when neglecting the a=0 m/(s*s) measurement, it is shown that the acceleration is approximately constant. The acceleration should be constant in an ideal measurement.
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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):
The slope is ~.5 m/s/s. The slope represents the system fairly well; however, there is error in human measurement that dilutes the reliability of the data. The clock time graph is more consistent with constant acceleration, because a line of best fist can be more readily interpolated as apposed to doing the same with the acceleration points. Furthermore, the acceleration points are estimates of acceleration at the midpoints of data points of data that are also estimates at midpoints.
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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):
1 - 1.5 hours.
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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Copy your document into the box below, be sure you have filled in your name and other identifying information at the top of this form, and submit:
Soft-drink bottle alternative
For students who use the soft-drink bottle as an alternative to the graduated cylinder:
Use a bottle which has a uniform cylindrical section. Most bottles are tapered at the top and the bottom, with a uniform cylindrical section in between.
Remove the label so you can easily observe the water level in the cylindrical section.
If you have a drill and a small drill bit (between about 1/16 inch and 1/8 inch), drill a hole near the bottom of the cylindrical section. If not, use something sharp and try to create a hole about 1/8 inch on a side; it's generally not difficult to make a triangular hole using a sharp blade. You can practice on a spare bottle, or if you prefer near the very top of the cylindrical section.
Instead of the milliliter markings, use a ruler or a ruler copy (the latter is a printable copy of a number of rulers; you may cut out a section and tape it to your bottle, which gives you a convenient scale to use for the experiment).
If you use this alternative, you should include the shape and dimensions of the hole, as best you can determine them.
Author information goes here.
Copyright © 1999 [OrganizationName]. All rights reserved.
Revised: 07/02/10
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