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.
Your initial message (if any):
Is flow rate increasing, decreasing, etc.?
I would expect that the rate of flow from the cylinder will decrease as the head pressure in the cylinder decreases.
Is the velocity of the water surface increasing, decreasing, etc.?
I would expect the descending buoy to decrease in speed due to the decreasing liquid column and pressure over time.
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?
The larger the cylinder diameter and the smaller the exiting orifice and the greater the pressure drop across this orifice, the greater the velocity of the water surface. As the pressure decreases inside the cylinder, thus, lowering the pressure drop across the exiting orifice, the lower the velocity of the water surface. Increasing the cylinder diameter and maintaining the same exit opening implies a greater head pressure and pressure drop across the exit orifice, resulting in increased water surface velocity. Decreasing the cylinder diameter, or increasing the exit opening implies less head pressure and less pressure drop across the exit orifice, resulting in decreased water surface velocity.
Explain how we know that a change in velocity implies the action of a force:
The change in velocity is directly related to the pressure in the cylinder. As the force of the liquid column increases or decreases, the velocity is increased or decreased. Therefore, the change in velocity is indicative of force action.
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate
The depth appears to be changing at a slower rate. This is due to the change in the initial picture to the second picture being approximately 1/3 of the column. Then the next picture to the last picture changes what appears to be another 1/3. Therefore, the depth would be changing at a slower and slower rate. If the pictures were clearer, with level marks visible, then the exact change in level over time could be evaluated.
What do you think a graph of depth vs. time would look like?
The graph would be a negative slope from left to right, with the maximum value in the upper left at time zero. As the trend moves to the right, the slope of the line will decrease at a decreasing rate as the time interval increases.
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?
The horizontal distance of the stream decreases as time increases, due top the loss of force on the exit orifice over time.
Does this distance change at an increasing, decreasing or steady rate?
The horizontal distance of the stream appears to be changing at a steady rate as the force changes with changing column.
What do you think a graph of this horizontal distance vs. time would look like?
The horizontal distance vs. time graph would be a negative slope, from left to right, maximum distance at time zero being indicated the far left of the graph. Then the slope of the horizontal distance decreases at a steady rate as the time increases.
The contents of TIMER program as you submitted them:
1 90.90625 90.90625
2 93.28125 2.375
3 95.8125 2.53125
4 98.5 2.6875
5 101.375 2.875
6 104.4063 3.03125
7 107.75 3.34375
8 111.4688 3.71875
9 115.7344 4.265625
10 121.3125 5.578125
11 128.6406 7.328125
12 140.6094 11.96875
The vertical positions of the large marks as you reported them, relative to the center of the outflow hole
1 cm
2.9 cm
4.8 cm
6.7 cm
8.6 cm
10.5 cm
12.4 cm
14.3 cm
16.2 cm
18.1 cm
20.0 cm
21.9 cm
Your table for depth (in cm) vs clock time (in seconds)
0, 21.9
2.375, 20.0
2.53125, 18.1
2.6875, 16.2
2.875, 14.3
3.03125, 12.4
3.34375, 10.5
3.71875, 8.6
4.265625, 6.7
5.578125, 4.8
7.328125, 2.9
11.96875, 1
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
The depth is changing at a slower and slower rate, which is evident by the depth vs. change in time. Initially, the depth changed at a rate of 1.9 cm/2.375 seconds, while at the end of the program the depth was changing at a rate of 1.9 cm/11.96875.
It appears that the laboratory data supports earlier expectations.
Your description of your depth vs. t graph:
The Depth vs. Time has the maximum level (21.9 cm from exit orifice) at time zero, with a negative slope from left to right. As the graph extends to the right, as time increases the slope decreases at a decreasing rate.
Your explanation and list of average average velocities:
The average velocity is found by taking the total distance traveled in one interval divided by the total time of that interval:
Int(1) = 1.9 cm/2.3750 sec = 0.8 cm/sec
Int(2) = 1.9 cm/2.53125 sec = 0.75 cm/sec
Int(3) = 1.9 cm/2.6875 sec = 0.71 cm/sec
Int(4) = 1.9 cm/2.875 sec = 0.66 cm/sec
Int(5) = 1.9 cm/3.03125 sec = 0.63 cm/sec
Int(6) = 1.9 cm/3.34375 sec = 0.57 cm/sec
Int(7) = 1.9 cm/3.71875 sec = 0.51 cm/sec
Int(8) = 1.9 cm/4.265625 sec = 0.45 cm/sec
Int(9) = 1.9 cm/5.578125 sec = 0.34 cm/sec
Int(10) = 1.9 cm/7.328125 sec = 0.26 cm/sec
Int(11) = 1.9 cm/11.96875 sec = 0.16 cm/sec
The midpoints of your time intervals and how you obtained them:
Divide each time interval by 2 to get the midpoint for that particular time interval. Then, that midpoint is added to previous total time intervals to get the midpoint clock time for that particular interval. The following list contains the midpoints for each time interval (all in seconds):
1.1875
3.640625
6.25
9.03125
11.984375
15.171875
18.703125
22.6953125
27.6171875
34.0703125
43.71875
Your table of average velocity of water surface vs. clock time:
1.1875, 0.8
3.640625, 0.75
6.25, 0.71
9.03125, 0.66
11.984375, 0.63
15.171875, 0.57
18.703125, 0.51
22.6953125, 0.45
27.6171875, 0.34
34.0703125, 0.26
43.71875, 0.16
Your description of your graph of average velocity vs clock time:
The graph is a linear graph with a negative slope with the maximum value of 0.8 cm per time unit in the upper left of the graph, and a near linear decreasing slope as the time interval increases to the right. The linearity of the indicated slope could result from human error while taking the indicated time snapshots.
Your explanation of how acceleration values were obtained:
The average acceleration is obtained by subtracting the subtracting the previous average velocity from the next average velocity, this will provide the change in velocity, positive or negative, between the intervals. After that value is obtained it is divided by time, to give the average acceleration by delta(V)/delta(T). The following is the average acceleration values for the time intervals:
Int(0) = 0.8 cm/s / 1.1875 sec = 0.67 cm/s^2
Int(1) = -0.05 cm/s / 2.453125 sec = -0.02 cm/s^2
Int(2) = -0.04 cm/sec 2.609375 sec = -0.015 cm/s^2
Int(3) = -0.05 cm/s / 2.78125 sec = -0.018 cm/s^2
Int(4) = -0.03 cm/s /2.953125 sec = -0.01 cm/s^2
Int(5) = -0.06 cm/s / 3.1875 sec = -0.019 cm/s^2
Int(6) = -0.06 cm/s / 3.53125 sec = -0.017 cm/s^2
Int(7) = -0.06 cm/s / 3.9921875 sec = -0.015 cm/s^2
Int(8) = -0.11 cm/s / 4.9240625 sec = -0.022 cm/s^2
Int(9) = -0.08 cm/s / 6.453125 sec = -0.012 cm/s^2
Int(10) = -0.10 cm/s / 9.6484375 sec = -0.01 cm/s^2
Your acceleration vs clock time table:
1.1875, 0.67
3.640625, -0.02
6.25, -0.015
9.03125, -0.018
11.984375, -0.01
15.171875, -0.019
18.703125, -0.017
22.6953125, -0.015
27.6171875, -0.022
34.0703125, -0.012
43.71875, -0.01
According to the evidence here, is acceleration increasing, decreasing, staying the same or is in not possible to tell?
My data indicates the acceleration of the water surface is inconclusive, due to the variations of increasing and decreasing values of acceleration. However, I beleive that the acceleration of the water surface is constant due to the unchanging gravity upon the liquid column. With f=ma, the force changes as a result of the changing mass, not the acceleration.
For reasons a bit beyond the scope of this course the acceleration of the water surface should indeed be constant. Due to the magnification of the effect of timing uncertainties as you first calculate average velocities then calculate average accelerations, it takes very good data (and of course a correct analysis) to ascertain this.
The reason for the uniformity of acceleration is that water pressure is proportional to depth, so by the work-energy theorem outflow velocity is proportional to the square root of depth. Run this through a fairly simple differential equation and a quadratic depth vs. t pops out. A quadratic depth vs. t function entails uniform acceleration; this fact and the solution of the differential equation require calculus.