Transcript Torricelli's Law and Draining Pipes

Torricelli’s Law
and
Draining Pipes
MATH 2413
Professor McCuan
Steven Lansel, Brandon Luders
December 10, 2002
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?
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
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
The Pipes
Pipe
R
r
f
h0
BB
1”
0.25”
4”
55.25”
BS
1”
0.125”
3.3”
58.6”
SB
0.5”
0.25”
3.3”
57.4”
SS
0.5”
0.125”
3.25”
55.7”
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.
Data Tables
Pipe BB
Distance
Time
Pipe BS
Distance
Time
Trial 1
0”
13.30 s
Trial 1
0”
50.40 s
Trial 2
8”
11.87 s
Trial 2
8”
45.54 s
Trial 3
16”
10.37 s
Trial 3
16”
39.12 s
Trial 4
28”
6.65 s
Trial 4
28”
28.93 s
Trial 5
40”
4.04 s
Trial 5
40”
22.70 s
Trial 6
58”
0s
Trial 6
62”
0s
Pipe SB
Distance
Time
Pipe SS
Distance
Time
Trial 1
0”
3.80 s
Trial 1
0”
13.1 s
Trial 2
8”
3.31 s
Trial 2
8”
12.1 s
Trial 3
16”
3.04 s
Trial 3
16”
10.4 s
Trial 4
28”
2.04 s
Trial 4
28”
6.89 s
Trial 5
40”
1.01 s
Trial 5
40”
4.59 s
Trial 6
60”
0s
Trial 6
58”
0s
Projectile Motion
1
2f
 f   gt 2  t 2 
t 
2
g
d  vt  v 
vd
d
g
d
t
2f
g
2f
•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)
2f
g
Velocity Plots
Pipe BB
Velocity
Time
Pipe BS
Velocity
Time
Trial 1
0 in/s
13.30 s
Trial 1
0 in/s
50.40 s
Trial 2
26.2 in/s
11.87 s
Trial 2
26.7 in/s
45.54 s
Trial 3
52.4 in/s
10.37 s
Trial 3
53.4 in/s
39.12 s
Trial 4
91.7 in/s
6.65 s
Trial 4
93.5 in/s
28.93 s
Trial 5
131.0 in/s
4.04 s
Trial 5
133.6 in/s
22.70 s
Trial 6
189.9 in/s
0s
Trial 6
207.1 in/s
0s
Pipe SB
Velocity
Time
Pipe SS
Velocity
Time
Trial 1
0 in/s
3.80 s
Trial 1
0 in/s
13.1 s
Trial 2
26.7 in/s
3.31 s
Trial 2
26.8 in/s
12.1 s
Trial 3
53.4 in/s
3.04 s
Trial 3
53.5 in/s
10.4 s
Trial 4
93.5 in/s
2.04 s
Trial 4
93.7 in/s
6.89 s
Trial 5
133.6 in/s
1.01 s
Trial 5
133.8 in/s
4.59 s
Trial 6
200.4 in/s
0s
Trial 6
194.0 in/s
0s
Linear Regressions
Pipe BB: v(t) = -13.8145t + 188.307 (r = -0.997871)
Pipe BS: v(t) = -4.12361t + 214.023 (r = -0.995462)
Linear Regressions
Pipe SB: v(t) = -50.1263t + 194.878 (r = -0.994724)
Pipe SS: v(t) = -14.342t + 196.17 (r = -0.996945)
Analysis
All of our data was strongly linear (linear
correlation factors were all between -0.99
and -1).
v(t) is most likely a linear function.
Let v(t) = a – bt.
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).
What is dh/dt?
dV
  av
dt
Outflow from the pipe:
dV dV dh
dh


A
Chain rule: dt dh dt
dt
Set them equal!
dh  av

dt
A
dV
dh
 av  A
dt
dt
dh  pr v  r v


dt
pR 2
R2
2
2
Deriving h(t), Part 2
v(t )  a  bt
dh  v(t )r 2  (a  bt)r 2 (bt  a)r 2



dt
R2
R2
R2

r 2  bt 2
h(t ) 
 at   h0

R 2  2

Height Slope Fields
Height Graphs
2
BB := tpiecewise( t11.3129, .4317031250t 9.767619494t55.25)
2
BS := tpiecewise( t42.6496, .03221570312t 2.747973947t58.6)
2
SB := tpiecewise( t3.02669, 6.265787500t 37.92920788t57.4)
2
SS := tpiecewise( t11.148, .4481875000t 9.992806162t55.7)
Deriving Torricelli’s Law
Overview:
– Reinsert v into the equation, eliminating t.
– Solve for a.
– Express v in terms of h.
av
v  a  bt  t 
b
2
r 2  b  a  v 
r 2  a 2  2av  v 2 a 2  av 
 a  v  
h(t ) 
 a
 h0 

 h0



2b
b 
 b  
R 2  2  b 
R 2 
r 2  a 2  2av  v 2  2a 2  2av 
r 2  v 2  a 2 
h(t ) 
 h0 
 h0
2


2b
2
b
R 2 
R



2bR2
r2
h  h0   v 2  a 2
When v = 0, h = 0:
2
a 
2bR 2 h0
r2
a2 
2
2bR h
r
v2 
2bR2 h
r2
2
2bR 2 h0
r2

2
2bR h0
r
2
2
v 
2
2bR h0
r
2
R
v
2bh   gh
r
Torricelli’s Law
Ideal law: v  2gh
Experimental factors cause decrease in
effectiveness
– Rotational motion
– Viscosity
More appropriate law: v   gh,  0.84
Would a better value for alpha work?
We can use this to theoretically describe the
motion of the pipes!
Physical Proof of Torricelli’s Law
Bernoulli’s equation for ideal fluid:
ρv
P  ρgy 
 constant
2
2
Let point a be at the top of the container,
and point b at the hole
ρv
P  ρgy  0  P  ρgy 
Pa  Pb  Patm
2
2
b
a
a
b
b
Patm  ρgy a  0  Patm  ρgy b 
ρv
2
2
b
v2b  2g(ya  yb )  2gh
vb  2gh
What is alpha?
R
2b   g
r
R 2b

r g
Pipe BB
1.07
Pipe BS
1.17
Pipe SB
1.02
Pipe SS
1.09
Average
1.08
Theoretical h(t)
Overview:
– Plug in Torricelli for v(t), not a – bt.
– Integrate with respect to dt.
– Solve for h(t).
h(0)  h0 :
2
dh  r 2 v  r  gh


dt
R2
R2
2 h0  C
2 h
 r 2 g
R
2 h
r  g
2
R2
t C
h
 r 2 g
2R
2
2
2
t  2 h0
t  h0
2
4 2
  r 2 g

r
 gh0
r

g
2


h(t ) 
t  h0

t 
t  h0
4
2
 2R 2

4R
R


Theoretical h(t) Equations
hBB = 0.26598t^2 - 7.66689t + 55.25,
hBS = 0.01662t^2 - 1.97398t + 58.6,
hSB = 4.25565t^2 - 31.2586t + 57.4,
hSS = 0.26598t^2 - 7.69805t + 55.7,
t < 14.4126
t < 59.3726
t < 3.67259
t < 14.4712
Comparing h(t) Graphs
Special Case: Draining Times
How long does it take to drain each pipe?
Pipe
Ideal
Actual
Percent Error
BB
14.4 s
13.3 s
7.6%
BS
59.4 s
50.4 s
15.2%
SB
3.7 s
3.8 s
2.7%
SS
14.5 s
13.1 s
9.7%
This is based on alpha being 0.84. It is too
low.
How did we derive ideal draining times?
Deriving Draining Time
Solve for when h(t) = 0.
 r 4 2 g  2  r 2 gh0
t   
h(t )  
2
 4R 4 

R




t  h  0
0


2
2
2


 r 4 2 g 
 r 4 2 gh0   r 4 2 gh0 
r  gh0
r

gh
r
 gh0
0






 
4
h
 
2
2
4  0



2
4
4
 R
  R

R
R
R
 4R 



 

t

4 2
 r 4 2 g 
r
 g

2
 4R 4 
2R 4


2
r 2 gh0
t
R
2
r 4 2 g

2r 2 R 4 gh0
r 4 R 2 2 g
2R 4
Alpha is 0.84.
R, r, and h0 vary with each pipe.
g is 386.

2R 2
r 2
h0
g
Modeling Other Functions
We can use this to also model the velocity
and height:
2R 2 h
tf 
0
r 2
g
 r 4 2 g 
 r 2 gh0
2
t   


h(t )   4 R 4 
R2




0, t  t f


t  h , t  t
0
f


  r 2 2 g

t  a gh0 , t  t f
v(t )   2 R 2

0, t  t f



 p r 4 2 g 
t 2   p r 2 gh0 t  p R 2 h0 , t  t f

V (t )   4 R 2 



0, t  t f

Extensions
More complicated systems
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
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.
Experimental Results
Trial 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
hBS  0.01662 t 2  1.97398 t  h0  0
t = 42.66 s
h0 = 53.96 in
Differential Equation
dV
 av  b
dt
dV
dh
A
dt
dt
2
2 dh
p r v  b  p R
dt
2 dh
pR
 p r 2 gh  b
dt
This is not
solvable by typical
ODE methods.
Slope fields and
Euler’s method
can be used to
numerically
interpret this ODE.
Equilibrium Slope Field
There is an equilibrium point near h ~ 47 inches.
Finding Equilibrium (Theory)
dh
0
dt
pR
2 dh
dt
 p r 2 gh  b
pR 2 (0)  pr 2 gh  b
h
b
pr 2 g
h
b2
p 2 r 4 2 g
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
V (bucket) pr 2h p (5.75)2 (14)
b


 7.2 in3/s
t
t
(202)
Finding Equilibrium (Theory, Part 2)
h
b
2
p r  g
2 4 2

(7.2)
2
p (0.125) (1.08) (386)
2
4
2
 47.8 in
•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)
dV
2
2
2 dh
 p r1 v1  p r2 v2  p R
dt
dt
2
2
2 dh
 r1 v1  r2 v2  R
dt
2
2
2 dh
 r1  gh  r2  g (h  j )  R
dt
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 Field
Experimental Calculations
Trial 1
41.6 s
Trial 2
40.9 s
Trial 3
41.2 s
Trial 4
40.9 s
Trial 5
41.0 s
Average
41.1 s
The numerical theoretical solution to the
system was calculated to be 35 seconds.
Two-Hole System (inflow)
dV
2
2
2 dh
 b  p r1 v1  p r2 v2  p R
dt
dt
2 dh
pR
 p r12 gh  p r22 g (h  j )  b
dt
Two-Hole Inflow Slope Field
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.
dh
 r  gh  r  g (h  j )  R
dt
2
1
2
2
2
Applying Torricelli’s Law
2
h1 := t.01662t 1.97398t58.6
r2
h2 '  2  g
R

h1  h2

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.
Applications with Inflow
Suppose water flows into Tank 1 at rate b.
ODE1 := 3.141592654D( hh1)( t )1.048453910 hh1( t ) 7.2
ODE2 := D( hh2)( t ).016984375 386 ( hh1( t )  hh2( t ) )
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?
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.
v  gh v  2gh
– Equilibrium can be achieved if there is a
constant inflow.
– Multiple holes increases the rate of decrease
and decreases emptying time.