1. Torricelli’s Law and Draining Pipes
MATH 2413
Professor McCuan
Steven Lansel, Brandon Luders
December 10, 2002

2. Question
How is the rate at which water exits a draining container affected by various factors?
The only force at work is gravity.
Water exits faster with more water in the container.
Exit velocity, height, and volume are all functions of time.
What are these functions?

3. The System Parameters r: radius of the exit hole
R: inner radius of the pipe
h0: height of the tube (from bottom of exit hole)
f: distance from bottom of exit hole to ground
Functions
h(t): height of the water column
V(t): volume of water column
v(t): exit velocity

4. The System (continued)
Initial conditions: h0, V0, v0 when t=0
Cross-sectional area:
Exit hole: a = pr2
Pipe: A = pR2
Four pipes:
Two values for r
Two values for R
Each pipe has a different set of initial conditions

5. The Pipes Pipe R r f h0 BB 1” 0.25” 4” 55.25” BS 0.125” 3.3” 58.6” SB
0.5”
57.4” SS
3.25”
55.7”

6. Finding v(t) For each pipe:
Fill with water to the top with the hole plugged.
Elevate the pipe while keeping it vertical.
Let the water drain. Start keeping time.
When the trajectory hits a predetermined horizontal distance, stop timing.
For initial velocity, measure the farthest point the trajectory reaches.
For draining time, see how long it takes to drain the pipe completely.

7. Data Tables Pipe BB Distance Time Trial 1 0” 13.30 s Trial 2 8”
16”
10.37 s
Trial 4
28”
6.65 s
Trial 5
40”
4.04 s
Trial 6
58”
0 s
Pipe BS
Distance
Time
Trial 1 0”
50.40 s
Trial 2 8”
45.54 s
Trial 3
16”
39.12 s
Trial 4
28”
28.93 s
Trial 5
40”
22.70 s
Trial 6
62”
0 s
Pipe SB
Distance
Time
Trial 1 0”
3.80 s
Trial 2 8”
3.31 s
Trial 3
16”
3.04 s
Trial 4
28”
2.04 s
Trial 5
40”
1.01 s
Trial 6
60”
0 s
Pipe SS
Distance
Time
Trial 1 0”
13.1 s
Trial 2 8”
12.1 s
Trial 3
16”
10.4 s
Trial 4
28”
6.89 s
Trial 5
40”
4.59 s
Trial 6
58”
0 s

8. Projectile Motion
Projectile motion can be used to convert the trajectory of the water to the initial velocity, by assuming that horizontal velocity is constant.
f includes the height of the bucket used (14 inches) and the distance from the hole to the bottom of the tube.
g is in inches per second squared. (g = 386)

9. Velocity Plots Pipe BB Velocity Time Trial 1 0 in/s 13.30 s Trial 2
Pipe BS
Velocity
Time
Trial 1
0 in/s
50.40 s
Trial 2
26.7 in/s
45.54 s
Trial 3
53.4 in/s
39.12 s
Trial 4
93.5 in/s
28.93 s
Trial 5
133.6 in/s
22.70 s
Trial 6
207.1 in/s
0 s
Pipe SB
Velocity
Time
Trial 1
0 in/s
3.80 s
Trial 2
26.7 in/s
3.31 s
Trial 3
53.4 in/s
3.04 s
Trial 4
93.5 in/s
2.04 s
Trial 5
133.6 in/s
1.01 s
Trial 6
200.4 in/s
0 s
Pipe SS
Velocity
Time
Trial 1
0 in/s
13.1 s
Trial 2
26.8 in/s
12.1 s
Trial 3
53.5 in/s
10.4 s
Trial 4
93.7 in/s
6.89 s
Trial 5
133.8 in/s
4.59 s
Trial 6
194.0 in/s
0 s

10. Linear Regressions Pipe BB: v(t) = -13.8145t + 188.307 (r = -0.997871)
Pipe BS: v(t) = t (r = )

11. Linear Regressions Pipe SB: v(t) = -50.1263t + 194.878 (r = -0.994724)
Pipe SS: v(t) = t (r = )

12. Analysis
All of our data was strongly linear (linear correlation factors were all between and -1).
v(t) is most likely a linear function.
Let v(t) = a – bt.

13. Deriving h(t) Overview: Find two descriptions for dV/dt.
Set them equal to each other.
Find a formula for dh/dt.
Plug in v(t).
Solve for h(t).

14. What is dh/dt?
Outflow from the pipe:
Chain rule:
Set them equal!

15. Deriving h(t), Part 2

16. Height Slope Fields BB BS SB SS

17. Height Graphs BB BS SB SS

19. Deriving Torricelli’s Law
Overview:
Reinsert v into the equation, eliminating t.
Solve for a.
Express v in terms of h.

21. When v = 0, h = 0:

22. Torricelli’s Law Ideal law:
Experimental factors cause decrease in effectiveness
Rotational motion
Viscosity
More appropriate law:
Would a better value for alpha work?
We can use this to theoretically describe the motion of the pipes!

23. Physical Proof of Torricelli’s Law
Bernoulli’s equation for ideal fluid:
Let point a be at the top of the container, and point b at the hole

24. What is alpha? Pipe BB 1.07 Pipe BS 1.17 Pipe SB 1.02 Pipe SS 1.09
Average
1.08

25. Theoretical h(t) Overview: Plug in Torricelli for v(t), not a – bt.
Integrate with respect to dt.
Solve for h(t).

27. Theoretical h(t) Equations
hBB = t^ t , t <
hBS = t^ t , t <
hSB = t^ t , t <
hSS = t^ t , t <

28. Comparing h(t) Graphs BB BS SB SS

29. Special Case: Draining Times
How long does it take to drain each pipe?
Pipe Ideal Actual Percent Error BB s 13.3 s 7.6% BS s 50.4 s 15.2% SB 3.7 s 3.8 s 2.7% SS s 13.1 s 9.7%
This is based on alpha being It is too low.
How did we derive ideal draining times?

30. Deriving Draining Time
Solve for when h(t) = 0.

31. Alpha is 0.84.
R, r, and h0 vary with each pipe.
g is 386.

32. Modeling Other Functions
We can use this to also model the velocity and height:

33. Extensions
More complicated systems

34. Equilibrium Points
The height of the water column is affected by two factors:
Water leaving through hole (variable rate)
Water entering through top (constant rate)
Equilibrium when those two are equal

35. Finding Equilibrium (Experimental)
Keep the pipe (pipe BS) unplugged and fill it with water coming from a constant source of water (for example, a showerhead).
After about four minutes, plug the hole.
Measure the time it takes for the equilibrium water column to drain. Use this to find the height of the water column.

36. Experimental Results Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Average
h0 = in

37. Differential Equation
This is not solvable by typical ODE methods.
Slope fields and Euler’s method can be used to numerically interpret this ODE.

38. Equilibrium Slope Field
There is an equilibrium point near h ~ 47 inches.

39. Finding Equilibrium (Theory)

40. What is b (or not 2b)?
b is the rate at which water enters the pipe and can be determined experimentally.
A container (bucket) with known volume was filled by the water source. The time it took to fill the source was recorded.
Bucket: Cylindrical V = pr2h
r = 5.75 inches
h = 14 inches
t = 202 seconds

41. Finding Equilibrium (Theory, Part 2)
Our experimental height was approximately 6 inches too high. (12.9 % error)
Sources of potential error:
Value for alpha
Experimental errors (timing, synchronization, etc.)
Theoretical conversion vs. actual conversion

42. Two-Hole System (no inflow)
The system acts in a piecewise fashion:
-Above the second hole, water is draining out of both holes.
-Below the second hole, the system acts just as the original system did.

43. Two-Hole Slope Field

44. Experimental Calculations
Trial 1
Trial 2
Trial 3
Trial 4
Trial 5
Average
41.6 s
40.9 s
41.2 s
41.0 s
41.1 s
The numerical theoretical solution to the system was calculated to be 35 seconds.

45. Two-Hole System (inflow)

46. Two-Hole Inflow Slope Field

47. Qualitative Analysis
A qualitative view of the system showed that the equilibrium point was between the second and third holes of the four-hole system (between 14 and 28 inches).
The numerical solution was 19.8 inches.

48. Applying Torricelli’s Law
Suppose two tanks are arranged such that one tank (Tank A) empties into a lower tank (Tank B) through which water can also leave. The tanks and holes are identical, and there is no inflow. A B

49. Applications with Inflow
Suppose water flows into Tank 1 at rate b.

50. Calculus of Variations
Given a constant volume, what is the shape of the container that drains it in the shortest amount of time?
Two scenarios:
Actual shape?
Degenerative case?

51. Conclusions For a draining cylindrical container,
Height and volume decrease quadratically.
Exit velocity decreases linearly.
The container drains in finite time.
Torricelli’s Law is obeyed for a non-ideal value of alpha near 1.
Equilibrium can be achieved if there is a constant inflow.
Multiple holes increases the rate of decrease and decreases emptying time.