torricelli’s law and draining pipes math 2413 professor mccuan steven lansel, brandon luders...

Torricelli’s Law Torricelli’s Law andand

Draining PipesDraining PipesMATH 2413MATH 2413

Professor McCuanProfessor McCuanSteven Lansel, Brandon LudersSteven Lansel, Brandon Luders

December 10, 2002December 10, 2002

QuestionQuestion

How is the rate at which water exits How is the rate at which water exits a draining container affected by a draining container affected by various factors?various factors?– The only force at work is gravity.The only force at work is gravity.– Water exits faster with Water exits faster with

more water in the container.more water in the container.– Exit velocity, height, andExit velocity, height, and

volume are all volume are all functions of time.functions of time.

– What are these What are these functions?functions?

The SystemThe 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

The System (continued)The System (continued)

Initial conditions: Initial conditions: hh00, , VV00, , vv00 when t=0 when t=0Cross-sectional area:Cross-sectional area:– Exit hole: Exit hole: aa = = rr22

– Pipe: Pipe: AA = = RR22

Four pipes:Four pipes:– Two values for Two values for rr– Two values for Two values for RR– Each pipe has a different set of initial Each pipe has a different set of initial

conditionsconditions

The PipesThe Pipes

PipePipe RR rr ff hh00

BBBB 1”1” 0.25”0.25” 4”4” 55.25”55.25”

BSBS 1”1” 0.125”0.125” 3.3”3.3” 58.6”58.6”

SBSB 0.5”0.5” 0.25”0.25” 3.3”3.3” 57.4”57.4”

SSSS 0.5”0.5” 0.125”0.125” 3.25”3.25” 55.7”55.7”

Finding Finding vv((tt))For each pipe:For each pipe:– Fill with water to the top Fill with water to the top

with the hole plugged.with the hole plugged.– Elevate the pipe while Elevate the pipe while

keeping it vertical.keeping it vertical.– Let the water drain. Start Let the water drain. Start

keeping time.keeping time.– When the trajectory hits a When the trajectory hits a

predetermined horizontal predetermined horizontal distance, stop timing.distance, stop timing.

– For initial velocity, For initial velocity, measure the farthest measure the farthest point the trajectory point the trajectory reaches.reaches.

– For draining time, see For draining time, see how long it takes to drain how long it takes to drain the pipe completely.the pipe completely.

Data TablesData TablesPipe BB Distance Time

Trial 1 0” 13.30 s

Trial 2 8” 11.87 s

Trial 3 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

Projectile MotionProjectile Motion

f

gdv

f

gd

t

dvvtd

g

ft

g

ftgtf

2

2

22

2

1 22

•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)

Velocity PlotsVelocity PlotsPipe BB Velocity Time

Trial 1 0 in/s 13.30 s

Trial 2 26.2 in/s 11.87 s

Trial 3 52.4 in/s 10.37 s

Trial 4 91.7 in/s 6.65 s

Trial 5 131.0 in/s 4.04 s

Trial 6 189.9 in/s 0 s

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

Linear RegressionsLinear RegressionsPipe BB: v(t) = -13.8145t + 188.307 (r = -0.997871)

Pipe BS: v(t) = -4.12361t + 214.023 (r = -0.995462)

Linear RegressionsLinear RegressionsPipe SB: v(t) = -50.1263t + 194.878 (r = -0.994724)

Pipe SS: v(t) = -14.342t + 196.17 (r = -0.996945)

AnalysisAnalysis

All of our data was All of our data was stronglystrongly linear (linear linear (linear correlation factors were all between -0.99 correlation factors were all between -0.99 and -1).and -1).

vv((tt) is most likely a linear function.) is most likely a linear function.

Let Let vv((tt) = ) = aa – – btbt..

Deriving Deriving hh((tt))

Overview:Overview:– Find two descriptions for Find two descriptions for dV/dtdV/dt..– Set them equal to each other.Set them equal to each other.– Find a formula for Find a formula for dh/dtdh/dt..– Plug in Plug in vv((tt).).– Solve for Solve for hh((tt).).

What is What is dh/dtdh/dt??

Outflow from the pipe:Outflow from the pipe:

Chain rule:Chain rule:

Set them equal!Set them equal!

avdt

dV

dt

dhAav

dt

dV

A

av

dt

dh

2

2

2

2

R

vr

R

vr

dt

dh

dt

dhA

dt

dh

dh

dV

dt

dV

Deriving Deriving hh((tt), Part 2), Part 2btatv )(

2

2

2

2

2

2 )()()(

R

rabt

R

rbta

R

rtv

dt

dh

0

2

2

2

2)( hat

bt

R

rth

Height Slope FieldsHeight Slope Fields

Height GraphsHeight Graphs

:= BB t ( )piecewise ,t 11.3129 .4317031250 t2

9.767619494 t 55.25

:= BS t ( )piecewise ,t 42.6496 .03221570312 t2

2.747973947 t 58.6

:= SB t ( )piecewise ,t 3.02669 6.265787500 t2

37.92920788 t 57.4

:= SS t ( )piecewise ,t 11.148 .4481875000 t2

9.992806162 t 55.7

Deriving Torricelli’s LawDeriving Torricelli’s Law

Overview:Overview:– Reinsert Reinsert vv into the equation, eliminating into the equation, eliminating tt..– Solve for Solve for aa..– Express Express vv in terms of in terms of hh..

b

vatbtav

0

222

2

2

0

2

2

2

2

2

2)( h

b

ava

b

vava

R

rh

b

vaa

b

vab

R

rth

0

22

2

2

0

222

2

2

22

222)( h

b

av

R

rh

b

avavava

R

rth

2202

22avhh

r

bR

When v = 0, h = 0:

20

22 2

r

hbRa

20

22 2

r

hbRa

20

22

20

2

2

2 222

r

hbRv

r

hbR

r

hbR

2

22 2

r

hbRv ghbh

r

Rv 2

Torricelli’s LawTorricelli’s Law

Ideal law:Ideal law:Experimental factors cause decrease Experimental factors cause decrease in effectivenessin effectiveness– Rotational motionRotational motion– ViscosityViscosity

More appropriate law: More appropriate law: Would a better value for alpha work?Would a better value for alpha work?We can use this to theoretically We can use this to theoretically describe the motion of the pipes!describe the motion of the pipes!

ghv 2

84.0, ghv

Physical Proof of Torricelli’s Physical Proof of Torricelli’s LawLaw

Bernoulli’s equation for ideal fluid:Bernoulli’s equation for ideal fluid:

Let point Let point aa be at the top of the container, be at the top of the container, and point and point bb at the hole at the hole

constant2

ρvρgyP

2

2

ρvρgyP0ρgyP

2b

bbaa atmba PPP

2

ρvρgyP0ρgyP

2b

batmaatm 2gh)y2g(yv ba

2b

2ghvb

What is alpha?What is alpha?gb

r

R 2g

b

r

R 2

Pipe BBPipe BB 1.071.07

Pipe BSPipe BS 1.171.17

Pipe SBPipe SB 1.021.02

Pipe SSPipe SS 1.091.09

AverageAverage 1.081.08

Theoretical Theoretical hh((tt))

Overview:Overview:– Plug in Torricelli for Plug in Torricelli for vv((tt), not ), not aa – – btbt..– Integrate with respect to Integrate with respect to dtdt..– Solve for Solve for hh((tt).).

2

2

2

2

R

ghr

R

vr

dt

dh

CtR

grh

2

2

2

:)0( 0hh

Ch 02

02

2

22 htR

grh

02

2

2ht

R

grh

020

22

4

242

02

2

42)( ht

R

ghrt

R

grht

R

grth

Theoretical Theoretical hh((tt) Equations) Equations

hBBhBB = 0.26598 = 0.26598tt^2 - 7.66689^2 - 7.66689tt + 55.25, + 55.25, tt < 14.4126 < 14.4126

hBShBS = 0.01662 = 0.01662tt^2 - 1.97398^2 - 1.97398tt + 58.6, + 58.6, tt < 59.3726 < 59.3726

hSBhSB = 4.25565t^2 - 31.2586 = 4.25565t^2 - 31.2586tt + 57.4, + 57.4, tt < 3.67259 < 3.67259

hSShSS = 0.26598 = 0.26598tt^2 - 7.69805^2 - 7.69805tt + 55.7, + 55.7, tt < 14.4712 < 14.4712

Comparing Comparing hh((tt) Graphs) Graphs

Special Case: Draining TimesSpecial Case: Draining Times

How long does it take to drain each pipe?How long does it take to drain each pipe?PipePipe IdealIdeal ActualActual Percent ErrorPercent ErrorBBBB 14.4 s14.4 s 13.3 s13.3 s 7.6%7.6%BSBS 59.4 s59.4 s 50.4 s50.4 s 15.2%15.2%SBSB 3.7 s3.7 s 3.8 s3.8 s 2.7%2.7%SSSS 14.5 s14.5 s 13.1 s13.1 s 9.7%9.7%This is based on alpha being 0.84. It is This is based on alpha being 0.84. It is too low.too low.How did we derive ideal draining times?How did we derive ideal draining times?

Deriving Draining TimeDeriving Draining Time

Solve for when Solve for when hh((tt) = 0.) = 0.

04

)( 020

22

4

24

ht

R

ghrt

R

grth

4

24

40

24

40

24

20

2

4

24

04

242

20

2

20

2

242

44

R

gr

R

ghr

R

ghr

R

ghr

R

gr

hR

gr

R

ghr

R

ghr

t

g

h

r

R

gRr

ghRr

R

gr

R

ghr

t 02

2

2240

42

4

24

20

2

22

2

Alpha is 0.84.

R, r, and h0 vary with each pipe.

g is 386.

Modeling Other FunctionsModeling Other Functions

We can use this to also model the velocity We can use this to also model the velocity and height:and height:

f

f

f

f

f

f

f

tt

tthRtghrtR

grtV

tt

ttghatR

grtv

tt

tthtR

ghrt

R

grth

g

h

r

Rt

,0

, 4

)(

,0

,2)(

,0

,4)(

2

02

022

2

24

02

22

020

22

4

24

02

2

ExtensionsExtensions

More complicated systemsMore complicated systems

Equilibrium PointsEquilibrium Points

The height of the water The height of the water column is affected by two column is affected by two factors:factors:– Water leaving through hole Water leaving through hole

(variable rate)(variable rate)– Water entering through top Water entering through top

(constant rate)(constant rate)

Equilibrium when those two Equilibrium when those two are equalare equal

Finding Equilibrium (Experimental)Finding Equilibrium (Experimental)

Keep the pipe (pipe BS) unplugged and fill Keep the pipe (pipe BS) unplugged and fill it with water coming from a constant it with water coming from a constant source of water (for example, a source of water (for example, a showerhead).showerhead).

After about four minutes, plug the hole.After about four minutes, plug the hole.

Measure the time it takes for the Measure the time it takes for the equilibrium water column to drain. Use equilibrium water column to drain. Use this to find the height of the water column.this to find the height of the water column.

Experimental ResultsExperimental ResultsTrial 1 Trial 2 Trial 3 Trial 4 Trial 5 Average

42.9 s 39.1 s 43.5 s 43.7 s 44.1 s 42.66 s

097398.101662.0 02 htthBS

t = 42.66 s

h0 = 53.96 in

Differential EquationDifferential Equation

This is not This is not solvable by typical solvable by typical ODE methods.ODE methods.

Slope fields and Slope fields and Euler’s method Euler’s method can be used to can be used to numerically numerically interpret this ODE.interpret this ODE.

bghrdt

dhR

dt

dhRbvr

dt

dhA

dt

dVbav

dt

dV

22

22

Equilibrium Slope FieldEquilibrium Slope Field

There is an equilibrium point near h ~ 47 inches.

Finding Equilibrium (Theory)Finding Equilibrium (Theory)

0dt

dh

bghrR 22 )0(

gr

bh

2

gr

bh

242

2

bghrdt

dhR 22

What is What is b b (or not 2(or not 2bb)?)?

bb is the rate at which water is the rate at which water enters the pipe and can be enters the pipe and can be determined experimentally.determined experimentally.A container (bucket) with A container (bucket) with known volume was filled by known volume was filled by the water source. The time the water source. The time it took to fill the source was it took to fill the source was recorded.recorded.

Bucket:CylindricalV = r2h

r = 5.75 inches

h = 14 inches

t = 202 seconds

/sin 2.7)202(

)14()75.5()bucket( 322

t

hr

t

Vb

Finding Equilibrium (Theory, Part 2)Finding Equilibrium (Theory, Part 2)

in 8.47)386()08.1()125.0(

)2.7(242

2

242

2

gr

bh

•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

Two-Hole System (no inflow)Two-Hole System (no inflow)

dt

dhRjhgrghr

dt

dhRvrvr

dt

dhRvrvr

dt

dV

222

21

22

221

21

22

221

21

)(

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.

Two-Hole Slope FieldTwo-Hole Slope Field

Experimental CalculationsExperimental Calculations

Trial 1Trial 1 Trial 2Trial 2 Trial 3Trial 3 Trial 4Trial 4 Trial 5Trial 5 AverageAverage

41.6 s41.6 s 40.9 s40.9 s 41.2 s41.2 s 40.9 s40.9 s 41.0 s41.0 s 41.1 s41.1 s

The numerical theoretical solution to the system was calculated to be 35 seconds.

Two-Hole System (inflow)Two-Hole System (inflow)

bjhgrghrdt

dhR

dt

dhRvrvrb

dt

dV

)(

22

21

2

22

221

21

Two-Hole Inflow Slope FieldTwo-Hole Inflow Slope Field

Qualitative AnalysisQualitative Analysis

A qualitative view of the system showed A qualitative view of the system showed that the equilibrium point was between the that the equilibrium point was between the second and third holes of the four-hole second and third holes of the four-hole system (between 14 and 28 inches).system (between 14 and 28 inches).

The numerical solution was 19.8 inches.The numerical solution was 19.8 inches.

dt

dhRjhgrghr 22

22

1 )(

Applying Torricelli’s LawApplying 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.

:= h1 t .01662 t2

1.97398 t 58.6

212

2

2 ' hhgR

rh

Applications with InflowApplications with Inflow

Suppose water flows into Tank 1 at rate Suppose water flows into Tank 1 at rate b.b. := ODE1 3.141592654 ( )( )D hh1 t 1.048453910 ( )hh1 t 7.2

:= ODE2 ( )( )D hh2 t .016984375 386 ( )( )hh1 t ( )hh2 t

Calculus of VariationsCalculus of Variations

Given a constant volume, what is the Given a constant volume, what is the shape of the container that drains it in the shape of the container that drains it in the shortest amount of time?shortest amount of time?

Two scenarios:Two scenarios:– Actual shape?Actual shape?– Degenerative case?Degenerative case?

ConclusionsConclusions

For a draining cylindrical container,For a draining cylindrical container,– Height and volume decrease quadratically.Height and volume decrease quadratically.– Exit velocity decreases linearly.Exit velocity decreases linearly.– The container drains in finite time.The container drains in finite time.– Torricelli’s Law is obeyed for a non-ideal Torricelli’s Law is obeyed for a non-ideal

value of alpha near 1.value of alpha near 1.– Equilibrium can be achieved if there is a Equilibrium can be achieved if there is a

constant inflow.constant inflow.– Multiple holes increases the rate of decrease Multiple holes increases the rate of decrease

and decreases emptying time.and decreases emptying time.

ghv ghv 2