Torricelli S Law and Draining Pipe Systems

Torricelli’s Law and Draining Pipe Systems

Steven Lansel and Brandon Luders

MATH 2413, Professor McCuan

December 12, 2002 Page 2

I. Abstract

Our goal is to analyze the behavior of a water-filled container with holes for draining in terms of the height of the water column and the exit velocity of the water. Through experimentation we will:

 Derive and verify Torricelli’s Law through a one-hole system with no inflow;  Apply Torricelli’s Law to a system with inflow, and find its equilibrium;  Analyze the effects of multiple holes on draining time;  Consider the shape of a container with minimal draining time; and  Apply Torricelli’s Law to traditional draining problems.

II. Representation

Each system considered in this research is a cylindrical pipe with the top end open and the bottom end closed. Holes were created on the side of the pipe through which water exits the system.

The following definitions will be used throughout the research:

 R, the inner radius of the pipe  r, the radius of all draining holes for that particular pipe  h, the height of the column of water (from the bottom of the lowest hole)  V, the volume of the column of water  v, the velocity of water as it escapes through the holes  f, the distance between the bottom of the hole and the bottom of the tube  A, the cross-sectional area of the pipe  a, the cross-sectional area of each hole in the pipe

III. System 1: One Hole Without Inflow

Apparatus

The pipe used in this system has one hole near the bottom of the pipe through which water exits the system (Figure 1). Four different pipes were used, each with a different combination of values for r and R. With the hole plugged, water is poured into the pipe up to the Page 3 top. The hole is then unplugged, such that water can exit through the hole. We hope to model the resulting behavior of the system and indirectly verify Torricelli’s Law.

Theory

dV Consider , the rate at which the volume of the water column changes. One way we dt can describe this is by considering the amount of water leaving the system in that length of time,

dV av. We thus have that  av . We can also describe this by using the Chain Rule, dt dV dV dh dV dh  . Since for this system dV  Adh , we have that  A . We now set the two dt dh dt dt dt

dV expressions for equal to each other: dt

dV dh  av  A dt dt dh  av  dt A

Since the cross-sections of both the pipe and the hole are circular, we have that a  r 2 and A  R2 , and the problem simplifies to

dh  r 2v  r 2v   . dt R2 R2

We do not know v(t), but we do know from Torricelli’s Law that v(h)  2gh for an ideal draining system, where g is the gravitational constant. However, our system is not ideal. For example, viscosity of the liquid must be considered. Any rotation inside the pipe results in an Page 4 energy loss, as well. Thus, a more general form of the law is appropriate. We thus use v(h)   gh . Past research has shown that water generally takes on the value   0.84.

We now use this function for v to solve for h(t). Note that we use the initial condition h(0) = h0, the initial height of the water column:

dh  r 2v  r 2 gh   dt R2 R2 1 dh  r 2 g  h dt R2  r 2 g 2 h  t  C R2 h(0)  h0 : 2 h0  C  r 2 g 2 h  t  2 h0 R2  r 2 g h  t  h0 2R2 2 2   r 2 g  r 4 2 g r  gh h(t)   t  h   t 2  0 t  h  2 0  4 2 0  2R  4R R

A point of particular significance is the point where h(t) = 0. Beyond that point, the function stays at zero. We can thus make a piecewise graph to model the function, but first we need to find the value of t at which the water column has fully emptied the pipe:

2  r 4 2 g   r  gh  h(t)   t 2   0 t  h  0  4   2  0  4R   R  2 2  2  4 2 2 4 2 4 2 r  gh0 r  gh0  r  g  r  gh  r  gh   r  gh      4 h 0   0    0  2  2   4  0 2  4   4  R  R   4R  R  R   R  t    r 4 2 g  r 4 2 g 2   4  4  4R  2R Page 5

2 r  gh0 2 4 2 2r R  gh 2R2 h t  R  0  0 r 4 2 g r 4R2 2 g r 2 g 2R4

In fact, the point where h(t) = 0 is the vertex of the parabola, as it is at that point where dh/dt = 0.

We also have information to solve for v(t) and V(t):

2   r 2 g    r 2 g   r 2 2 g v(t)   gh(t)   g t  h    g  t  h   t  a gh  2 0   2 0  2 0  2R   2R  2R  4 2   r 2 gh    4 2  2 2  r  g  2  0    r  g  2 2 2 V (t)   R h(t)   R  t   t  h0   t    r  gh0  t   R h0  4   2    2   4R   R    4R 

We now know everything about the behavior of this system. Given r, R, and h0, as well as   0.84 and g = 386 in/s2, we know the following:

2 2R h0 t f  r 2 g  4 2  2   r  g  2 r  gh0  t   t  h ,t  t h(t)   4   2  0 f  4R  R     0,t  t f  r 2 2 g  t  a gh0 ,t  t f v(t)   2R2   0,t  t f   r 4 2 g   t 2    r 2 gh t   R2h ,t  t V (t)   2   0  0 f  4R    0,t  t f

We note that h(t) and V(t) are quadratic, while v(t) is linear.

Experiment 1

For all experiments involving this system, we use the four pipes whose measurements are listed in Figure 2. Page 6

For each pipe, the hole was plugged as the pipe was filled with water to the top. Time was kept beginning from the moment the hole was unplugged, making sure that the pipe was always perpendicular to the ground. The time t at which water no longer flowed from the hole was recorded. Five trials were taken for each pipe. All data is listed in Figure 3. Theoretical values were computed through the formula

2 2 2R2 h R h  2  0.121187R h t  0  0    0 f 2 2   2 r  g r  (0.84) 386  r

We note that the percent error for each pipe is very low, less than 20% in all cases and less than 10% in three cases, so this provides evidence to support Torricelli’s Law, as the radii were varied with each pipe. We notice that the percent error generally increases with the time it took to drain the pipe, which may mean that the value given for  may be too low. We rework the formula above to calculate a more accurate value of  :

2R2 h   0 2 g r t f

Pipe BB:   0.91 Pipe BS:   0.99 Pipe SB:   0.81 Pipe SS:   0.93

Analysis 1

In comparing the draining times for each of the four pipes, we do notice several trends.

Looking at the formula for the draining time, we notice that h0, g, and  are all nearly constant if not constant. Thus, if R is doubled, the draining time is quadrupled, and if r is doubled, the draining time is cut by one-fourth. The draining times listed in Figure 3 follow these trends accurately. However, this is all overshadowed by the fact that our value for  might be somewhat off, as the experimental results were inconsistent. It is possible that  is not 0.84, but Page 7 in fact a value unique to our pipes. As all of our pipes were constructed in the same manner, we anticipate that our value of  will be consistent for all pipes. Here, we estimate  to be 0.91.

Experiment 2

In this experiment, each pipe was plugged and filled to the top, then placed on an object of a given height, such as a bucket. The distance from the ground to the bottom of the hole on the pipe was recorded. Time was kept beginning with when the pipe was unplugged, releasing the water. For each trial, when the trajectory of the water landed at a predetermined horizontal distance away from the hole, time was stopped and recorded. This was repeated for various distances for each pipe. For initial conditions, the farthest point the trajectory released was recorded. Furthermore, the results from Experiment 1 were used in the data. The exit velocity v was then derived using projectile motion (Figure 4). Results are listed in Figure 5.

We found, as expected by the function predicted earlier for v(t), that the graph of v(t) is linear. We now seek to compare our graphs for v(t) against those predicted by theory for each pipe. We begin by looking at the equations (Figure 6) and the graphs (Figure 7) for the height of the water column, exit velocity, and volume of the water column for each pipe based on its measurements. We then generate linear regressions based on the data for Experiment 2 for each pipe:

Pipe BB: v(t) = -13.8145t + 188.307 (r = -0.997871) Pipe BS: v(t) = -4.12361t + 214.023 (r = -0.995462) Pipe SB: v(t) = -50.1263t + 194.878 (r = -0.994724) Pipe SS: v(t) = -14.342t + 196.17 (r = -0.996945)

Note that r is the linear correlation factor, measuring how linear the data is. A value of r = -1 would mean that all data is perfectly linear. A comparison of these graphs to the theoretical graphs in shown in Figure 8. Page 8

We now seek to verify the relationships inherent in Torricelli’s Law. Since we have shown that the graph of velocity against time is linear, let v(t) = a – bt.

dh r 2v   dt R2 dh  v(t)r 2  (a  bt)r 2 (bt  a)r 2    dt R2 R2 R2 r 2  bt 2  h(t)    at   h 2   0 R  2 

A comparison of graphs of h(t) here to the theoretical graphs is shown in Figure 9.

a  v v  a  bt  t  b 2 r 2  b  a  v   a  v  r 2  a2  2av  v2 a2  av  h(t)      a   h      h 2  2 b b  0 2  2b b  0 R      R   r 2  a2  2av  v2  2a2  2av  r 2  v2  a2  h(t)     h     h 2   0 2   0 R  2b  R  2b  2 2bR 2 2 h  h0   v  a r 2

We know that when v = 0, h = 0:

2bR2h  a2   0 r 2 2bR2h a2  0 r 2 2bR2h 2bR2h 2bR2h  0  v2  0 r 2 r 2 r 2 2bR2h v2  r 2 R v  2bh   gh r

This proves the square-root relationship of Torricelli’s Law. We now seek to establish a more appropriate value for  : Page 9

R 2b   g r R 2b   r g

We then plug in our values for each pipe:

Pipe BB:   1.07 Pipe BS:   1.17 Pipe SB:   1.02 Pipe SS:   1.09

The average value for  is 1.08.

Analysis 2

We notice that  is not dependent on a, the initial velocity, in the equation above. We also notice that the experimental value for  is much higher than originally predicted. Because

 is unique for each system, we can disregard the old value for  . From now on,  = 1.08 will be used for all experiments.

IV. System 2: One Hole with Inflow

Apparatus

From now on, only Pipe BS from the experiments above will be used, because it offers the most visible trajectory and the longest draining time. Water from a shower head was allowed to flow into the pipe from the top at a constant rate, with the hole unplugged. Our goal is to analyze this system and determine if there is a height at which the water level stays constant over time. (see Figure 10)

Theory

The differential equation involved is set up as follows, where b is the rate of water entering the pipe: Page 10

dV dV dh  av  b  A dt dt dt dh   r 2v  b   R2 dt dh  R2   r 2 gh  b dt

It is not possible to solve this differential equation by traditional means to find an explicit solution; however, we can analyze it qualitatively, and use numerical methods to graph some solutions.

We first consider the qualitative behavior of the system. There are two ways the height of the water column changes: water comes in at a constant rate and exits according to Torricelli’s

Law. When the water column is near the bottom of the pipe, the height will increase, but if the water column starts near the top of the pipe, the height will decrease. There would thus be an

dh equilibrium point somewhere where  0: dt

R2 (0)  r 2 gh  b b h  r 2 g b2 h   2r 4 2 g

For pipe BS, we know r = 0.125”,  = 0.108, and g = 386. However, we do not know b, and must find it experimentally.

To find b, we took a bucket with a known radius r and height h, and timed how long it took to fill with water to the top. We then performed the following calculations:

V (bucket) r 2h  (5.75)2 (14) b     7.2 in3/s t t (202) Page 11

We can now solve for our theoretical equilibrium water column height h:

b2 (7.2)2 h    47.8 in  2r 4 2 g  2 (0.125)4 (1.08)2 (386)

A slope field modeling the differential equation above is found in Figure 11.

Experiment

We experimentally found the equilibrium height by letting the system run on its own for four minutes, such that the equilibrium height was reached, then determined how long it took the water column to drain. From that the experimental equilibrium height could be determined

(Figure 12). The experimental height of the water column was found to be 53.96 inches.

Analysis

Our percent error here is only 12.9%, so the results are quite accurate. However, they do seem to indicate that a lower value of  might be more appropriate here, not accounting for error. We rework the formula above to calculate the experimental 

b 7.2     1.02 , r 2 gh  (0.125)2 (386)(53.96) which is still in our range of acceptable values.

V. System 3: Two Holes without Inflow

Apparatus

A pipe identical to BS in all but h0 (=56.25”) and f was used which had four holes drilled into one side of the tube. Each hole was spaced 14 inches apart. With the holes plugged, the pipe was filled to the top. The bottom two holes were unplugged while the top two holes remained plugged, thus creating the two-hole system. Let r1 and v1 refer to the upper hole, and let r2 and v2 refer to the lower hole. (see Figure 10)

Theory Page 12

The differential equation is set up as follows:

dV dh   r 2v   r 2v   R2 dt 1 1 2 2 dt dh  r 2v  r 2v  R2 1 1 2 2 dt dh  r 2 gh  r 2 g(h  j)  R2 1 2 dt

Here j represents the distance between the two holes, which in this situation is 14 inches. This is in fact a piecewise function: above the second hole, the above equation is used, while below the second hole, a model similar to the original model is used. The slope field generated by the piecewise differential equation is shown in Figure 13. Numerical analysis of the slope field indicated that the pipe finished draining in approximately 35 seconds.

Experiment

The draining time was solved for experimentally through five trials (Figure 14). This averaged out to be 41.1 seconds, approximately 6 seconds higher than the predicted value.

Analysis

As expected, the pipe drains much faster when there are two holes than when there was just one hole.

VI. System 4: Two Holes with Inflow

Apparatus

The same setup was used as in System 2, but with the two-hole pipe used in System 3.

(see Figure 10)

Theory

This model is essentially a combination of Systems 2 and 3. Its differential equation is set up as follows: Page 13

dV dh  b   r 2v   r 2v   R2 dt 1 1 2 2 dt

dh  R2  r2 gh  r2 g(h  j)  b dt 1 2

As with Systems 2 and 3, this cannot be solved by conventional methods but can be analyzed through slope fields (Figure 15). By setting dh/dt equal to 0, the equilibrium height h can also be solved for. For the system being considered, h = 19.8 inches.

Analysis

A qualitative test of the theory showed that the equilibrium height of the water column fell between the second and third holes of the pipe. Since each hole is spaced 14 inches apart, that means the experimental equilibrium height was between 14 and 28 inches, as predicted by theory. We also notice that this equilibrium height is much lower than the height for the one- hole system.

VII. Conclusions

We developed the following conclusions from the data collected:

 For a draining cylindrical container, o Height and volume decrease quadratically. o Exit velocity decreases linearly. o Torricelli’s Law is obeyed for a non-ideal value of  near 1. o Equilibrium can be achieved if there is a constant steady inflow of water. o The presence of multiple holes reduces draining time and equilibrium height for the system.

Further questions could include looking at influences on  , considering the presence of more holes when draining the system or finding equilibrium, or applying Torricelli’s Law to traditional mixing problems.