Transcript Review Steven A. Jones BIEN 501 Friday, May 14, 2007 Louisiana Tech University Ruston, LA 71272 Slide 1

Review
Steven A. Jones
BIEN 501
Friday, May 14, 2007
Louisiana Tech University
Ruston, LA 71272
Slide 1
Simple Flow Field
What is the pathline?
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Slide 2
Simple Flow Field
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Slide 3
Simple Flow Field
Pathline follows the particle
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Slide 4
Simple Flow Field
What is the streakline?
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Slide 5
What is the Differential Equation
that Describes a Streamline?
Assume we know that:




u x, y, z   f x, y, z i  gx, y, z  j  hx, y, z k
Answer:
Since
So
dx dy dz


u
v
w
x f x, y, z 

,
y g x, y, z 
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dx u
 ,
dy v
dx u

dz w
x f x, y, z 

z hx, y.z 
Slide 6
Continuity
For a two-dimensional flow:
u  cosxe
y
Use the equation of continuity to determine v.
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Slide 7
Answer
u v
u
v
u

0
 v
dx  f  y 
x y
x
y
x

u
  sin x e y
x
v     sinxe
y
 dx  f  y

v  sin x e y  f  y 

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Slide 8
What is the equation for a pathline?
A pathline follows a fluid particle. Assume that you
 
know the entire velocity field: u x, t  and that the
particle passes through the point x0  x0 , y0 , z0 
at time 0.
Answer:



x t  dt  x t   u t dt

dx  
 u x, t 
dt
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Slide 9
Example
Assume that:


 
2
u x, t   x i  2xyj
Is continuity satisfied?
Answer:
Yes
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Slide 10
What is the equation for a pathline?
Assume that:


 
2
u x   x i  2 xyj
What is the equation for the
pathline through (1,2)?
 dy

dx
 u  x ,
 vx 
dt
dt
dx
dy
dy
2
x ,
 2 xy  dt  
dt
dt
2 xy
Answer:
dx
dx
x
2
 2 xy  x 

dy
dy
2y
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Slide 11
What is the equation for a pathline?
dx
x
dx
dy



 lnx    ln2 y   C
dy
2y
x
2y
xe
Write:
 ln2 y C
C  ln 
x
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
2y
Slide 12
What is the equation for a pathline?
x
1 
so

22

2y
  4
4
x
2y
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Slide 13
Answer (Continued)
xt   1  x 2t  f  y 
yt   2  2 xyt  g x 
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Slide 14
Two Compartment Model
Central Compartment
Peripheral Compartment
  k1V1C1
m
qd
C2
C1
V1
  k2V2C2
m
V2
Clearance
Conservation
of Mass
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dC1
V1
 V1k1C1  V2 k 2C2  V1keC1 Central
dt
dC2
Peripheral 
V2
 V1k1C1  V2 k 2C2
dt
Slide 15
Two Compartment Model
V2
In terms of the volume ratio  21 
V1
Conservation
of Mass
dC1
  k1C1   21k 2C2  keC1
dt
dC2
 21
 k1C1   21k 2C2
dt
Central
Peripheral
D
C1 0   C0 , C2 0  0
V1
Initial Conditions
2
Solve the two
ODEs for C1
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d C1
dC1
 k1  k2  k3 
 k2 keC1  0
2
dt
dt
Slide 16
ICs in terms of C1
1  dC1 0

C1 0  C0 , C2 0 
 k1  ke C1 0  0

 21  dt

C1 0   C0
dC1 0 
 k1  ke C0
dt
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Slide 17
Solution
2
dC1
 k1  k2  k3 
 k2 keC1  0
2
dt
dt
D
dC1 0
C1 0  C0 ,
 k1  ke C0
V1
dt
The solution to: d C1
With
Is
Where:

C1  C0 e
1, 2 
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1t
 1   e
2t

k1  k2  ke   k1  k2  ke 2  4k2ke
2
Slide 18
Two Compartment Model

C1  C0 e
Rapid
Release
1t
 1   e
2t

Slow
Release
One
Compartment
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Slide 19
Two Compartment Model
The two-compartment model obeys the same differential
equations as the simple RLC circuit.
It is useful to compare the individual components to the
RLC circuit:
Damping dC
d 2C1
1



k

k

k
 k2 keC1  0
1
2
e
2
dt
dt
Transfer from L to C
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Slide 20
Two Compartment Model
One might expect that overshoot (ringing) could happen.
However, ringing will only happen for imaginary values
of . In our case:
k1  k2  ke   k1  k2  ke 
2
1, 2 
 4k2 ke
2
As you increase k2 or ke, you must also increase (k1+k2+k3).
And for the RLC Circuit:
2
1, 2
R
R
R
  2 4 2
L
LC
 L
2
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Can make the square root
imaginary with small R or
large C.
Slide 21
Two Compartment Model
To see if the square root can become imaginary, minimize
it’s argument w.r.t. ke and see if it can be less than 0.
k1  k2  ke   k1  k2  ke 
2
1, 2 

2
 4k2 ke

d
k1  k2  ke 2  4k2 ke  0
dke
2k1  k2  ke   4k2  0
2k1  2k2  2ke  0
k1  ke  k2
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Slide 22
Two Compartment Model
What value does the argument of the square root take on at
the minimum?
k1  ke  k2
k1  k2  ke 
2
 4k2ke
 k1  ke   k 2   4k 2 ke
2
 k 2  k 2   4k 2 ke  4k 22  4k 2 ke
2
 4k 2 k1  ke   4k 2 ke  4k 2 k1
Since k2 and k1 cannot be negative, the argument of the square
root can never be negative. I.e. no ringing.
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Slide 23
Pharmacokinetic Models
QL
Cv
Q
Vascular
Cp
Js
L
Interstitial
Ci
Ci
Q: Plasma Flow
L: Lymph Flow
q
Js, q: Exchange rates
Cellular
PBPK: Physiologically-Based Pharmocokinetic Model
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Slide 24
Pharmacokinetic Models
dCv
Vv
 QC p  Q  L Cv  J s  Rv
dt
dCi
LCi
Vi
 Js 
 q  Ri
dt
Z
dCc
Vc
 q  Rc
dt
Z: Equilibrium concentration ratio between interstitium
and lymph.
Ci  ZCL
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Slide 25
More Complicated Models
Plasma
Liver

G.I. Track


Kidney
Muscle
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Slide 26
Note on Complexity
• While the equations become more
complicated as more components are
added, the basic concepts remain the
same, and the systems can be analyzed
with the same tools you would use to
analyze a linear system in electrical
engineering (e.g. transfer functions,
Laplace transforms, Mason’s rule).
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Slide 27
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Slide 28
What is the Differential Equation
that Describes a Streamline?
x
f  x, y , z 
dy  az 
g  x, y , z 
x
f  x, y , z 
dz  b y 
hx, y.z 
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Slide 29