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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