MODELING BLOOD FLOW IN
THE CARDIOVASCULAR
SYSTEM
MA432 – Spring 2015
Mette S Olufsen
Department of Mathematics,
North Carolina State University
NC STATE University
Overview
§  Introduction to cardiovascular dynamics
§  Algebraic model
§  Differential equations model
§  Data
§  Subset selection and parameter estimation
§  Numerical solution with personal data
§  Summary
Before 1628
Pliny the Elder (Roman), Galen (Greek),…;
Two distinct types of blood were thought to exist:
§  “Nutritive blood” was thought to be made by the liver
and carried through veins to the organs, where it was
consumed
§  “Vital blood” was thought to be made at the heart and
pumped through arteries to carry the “vital spirits”
§  Blood was thought to be produced and consumed
at the ends of a transport system whereas idea of a
circulatory blood system was unthinkable
It was believed that the heart acted not to pump blood, but to suck it
in from the veins and that blood flowed through the septum of the
heart from one ventricle to the other through a system of tiny pores
Ancient View
William Harvey’s 1615 modeling:
Stroke volume is 70 milliliters/beat
Heart beats 72 times/minute
Body tissue
Cardiac output of
7,258 liters/day
Brain
Vital spirits flow
Heart
Nutritive blood flow
Air flow
Liver
Nutrition
Lungs
Stomach
This seemed absurd when compared to the amount of blood in a human
body which in most humans is around 5 liters and even more absurd
compared to the amount of liquid intake per day (in food and in drinking)
which is less than 5 liter per day
Modeling
§ 
Modeling gave birth to the view that blood must circulate.
Harvey successfully announced the discovery of the
circulatory cardiovasclar system
§ 
Harvey discovered the circulation of blood 46 year before
the discovery of the light microscope
§ 
Consequently Harvey changed the view of the world
using a simple mathematical model making the
inaccessible accessible
1. 
2. 
3. 
In 1615 Harvey wrote (hand-written notes) that he was convinced that
blood circulated
Anton Van Leeuwenhoek's microscope from 1674 (the first functioning
light microscope) was sufficiently strong to make the capillaries visible
Marcello Malpighi, made the discovery simultaneously and
independently, published his discovery of the capillaries in 1675. He is
often credited the discovery of the capillaries by use of the microscope
Right side
Sys veins (blue)
Pulm veins (red)
Pulmonary
circulation
CV System Head
Left side
Sys Arteries (red)
Pulm Arteries (blue)
Exchange of O2
(in) and CO2 (out)
Lungs
Heart
Systemic
circulation
Body
Exchange of O2
(out) and CO2 (in)
Systemic Arteries
§ 
§ 
§ 
§ 
Large Arteries (cm)
Small Arteries (mm)
Arterioles (100 µm)
Capillaries (50 µm)
Arterioles and Capillaries
Pressure and
Area
Pressure
Pulse
pressure
Vessel area
Total vessel
area (cm2)
Systolic
pressure
Diastolic
pressure
L ventricle
Aorta
Mean
pressure
Arteries
Arterioles
Capil- Ven
laries oules
Veins
Cardiovascular Modeling §  Blood pressure p (mmHg)
§  Force per unit area
§  mmHg relates to the height of a column of mercury
that can be supported by the pressure in question
§  Blood flow q (ml/min) or (l/min)
§  Cardiac output is the flow pumped by the
heart
§  Blood volume V (ml) or (l)
§  Stressed volume and unstressed volume
Compartment Models
In-flow
Volume, Concentration,
Amount
dV
= flow in − flow out
dt
= qin − qout
Out-flow
Why mmHg ?
The Reverend Hales demonstrating blood pressure in a horse.
Reproduced from: History of hypertension series. Sandwich, Kent; Pfizer Ltd., 1980.
Pressure - Flow - Volume
q
Blood Flow In (q1)
System
ppout
pin
R1
Blood Flow Out (q2)
pext
Psv=0
C
pin − pout
q=
R
V = C ( p − pext )
R2
Normal Values Variables
Systemic
P
V
Pulmonary
Unit
Psa
100
Ppa
15
mmHg
Psv
2
Ppv
5
mmHg
Vsa
1
Vpa
0.1
liter
Vsv
3.5
Vpv
0.4
liter
V0
5
liter
§  Stroke volume 70
ml/beat
§  Heart rate
beats/min
§  Cardiac output
80
5.6 l/min
liter
Large Arteries and Veins 1.5 . Ma the ma tic al Mo de l of the Un co ntr oll ed Cir cu lat ion
Rn Qp =P pu -P nu
Pulmonary
(1 '4 '8 )
(1 .4 .e )
rmed that the Pressure
nen the distending Presbut P.t - 4ho.t* and
"
I chest. In fact, Rho,,* ls
l. and this contributes to
*toti. volume VBn' This
r the chest is oPened durssume for simPlicitY that
19) without modification'
r i.t E* .rc i." t 1'8 thr ou gh
V = CP
i.
Vp u = Cp uP pu
Vp u = Cp uP pu
= KsP*
Right q*
heart
Left heart
q" = KlPpv
Vru = Cru Pru
Vru = Cru P.u
Systemic
i
hn co nt ro lle d
rave been develoPed above
RrQr=Pru-Pru
Fig ure 1.6 . Eq ua tio ns of the cir cu lat ion ' On e eq ua tio n is sh ow n'f or ea ch o
For .dditiotta equation that r
eight principal elements oiiir"
Systemic and Pulmonary Capillaries 1.5 . Ma the ma tic al Mo de l of the Un co ntr oll ed Cir cu lat ion
Rn Qp =P pu -P nu
(1 '4 '8 )
(1 .4 .e )
rmed that the Pressure
nen the distending Presbut P.t - 4ho.t* andin
"
I chest. In fact, Rho,,* ls
l. and this contributes to
*toti. volume VBn' This
in
r the chest is oPened durssume for simPlicitY that
19) without modification'
r i.t E* .rc i." t 1'8 thr ou gh
p − pout
Q=
R
RQ = p − pout
i.
P
Pulmonary
Vp u = Cp uP
papu
= KsP*
Right q*
heart
Left heart
q" = KlPpv
Vru = Cru Pru
Vru = Cru P.u
Psv
P
= Cp uP pu
Vp upv
Systemic
Psa
i
hn co nt ro lle d
rave been develoPed above
RrQr=Pru-Pru
Fig ure 1.6 . Eq ua tio ns of the cir cu lat ion ' On e eq ua tio n is sh ow n'f or ea ch o
For .dditiotta equation that r
eight principal elements oiiir"
The Heart
1.4. The Heart as a Pair of Pumps
l1
and veins are priressuredifferences
essels,while their
be of resistanceis
smallest arteries,
rt but where large
Lflow and pressure
Lowfor changes in
incewe can always
- P)lQ.In fact,
conditions where
/hen R changes, a
on usually involves
re smooth muscles
generated by the
stancescirculating
Lctsof metabolism.
lutoregulation.
rmpliancerelations
rn to say that the
lssure,but the relavely lesscompliant
nplicity.
DIASTOLE
SYSTOLE
Figure 1.3. Cross section of the left side of the heart with the left ventricle in
DIASTOLtr (relaxed) and SYSTOLE (contracted). LA : left atrium, LV : left
ventricle, Ao : Aorta. The numbers in the chambers are pressures in millimeters
of mercury (mmHg). Note that the inflow (mitral) valve is open in diastole and
closed in systole, whereas the outflow (aortic) valve is closed in diastole and open
in systole. Pressures are approximately equal across an open valve, but can be
very unequal, with the downstream pressure higher, across a closed valve. The
valves ensure the unidirectional flow of blood from left atrium to left ventricle to
aorta.
We regard the ventricle as a compliance vessel whose
his collaborators:
compliance changes with time. Thus, the ventricle is described by
1.4 . Th e He art as a Pa ir of Pu m
Pressure Volume Loop
A: Inflow valve closes
B: Outflow valve opens
C: Outflow valve closes
D: Inflow valve opens
V = Va + Cry rtot eP
V Inf low val veclo se
Ou tflo w val veoP e
Ou tflo w val veclo s
Inf low val veoP en
t
rr of time. During diastole'
[igh value C6i6s1.1. 8s t.rre
s the ventricle contracts'
i
V = Va + Cd ias tolcP
.:0 . so tha t the sY sto )ic
rolume
f.
(1 .4 .5 )
lsh all us e. Fin all Y, if F
SetcardiacoutPut
vd
ve s
V=V +C
vE p
P
Figure 1.5' Pressure-volume diagram for either ventricle' Landmarks on
d
diastole
ume axis (V) are Va : "deaJv
i'e'' the volume at zero tra
olume"'
pressure, Ves : end-systolic volume, Vno : end-diastolic volume' La
on the pressure (P; a*it are the venous (inflow) pressure P-.'' u.od the
(outflow) pressure P"' Since the venous pressure is essentially the sam
atr ial pre ssu re, itm ayb eco nsi
d
Le: and Right Heart 1.5 . Ma the ma tic al Mo de l of the Un co ntr oll ed Cir cu lat ion
V = Vd + C(t)P(t)
(1 '4 '8 )
VES = Vd + Csystole Pa (Min)
V
V
(1 .4 .e )
P
=V +C
P (Max)
= V − V = Cdiastole Pv − Csystole Ps
EDthat the
d
diastole v
rmed
Pressure
nen the distending PresED
ES
bustroke
t P.t - 4ho.t* and
"
I chest. In fact, Rho,,* ls
ibutes tov
l. anstroke
d this contrdiastole
*toti. volume VBn' This
r the chest is oPened durssume for simPlicitY that
19) without modification'
r i.t E* .rc i." t 1'8 thr ou gh
V
Rn Qp =P pu
-P nu
Pulmonary
≈C
P
= Cp uP pu
Vp upv
Vp u = Cp uP pu
q* = KsP*
q" = KlPpv
Right heart
Left heart
Vru = Cru Pru
Vru = Cru P.u
Psv
i.
i
hn co nt ro lle d
rave been develoPed above
Systemic
RrQr=Pru-P
ru
Fig ure 1.6 . Eq ua tio ns of the cir cu lat ion ' On e eq ua tio n is sh ow n'f or ea ch
For .dditiotta equation that
eight principal elements oiiir"
Le: and Right Heart 1.5 . Ma the ma tic al Mo de l of the Un co ntr oll ed Cir cu lat ion
Rn Qp =P pu -P nu
Pulmonary
(1 '4 '8 )
(1 .4 .e )
rmed that the Pressure
nen the distending Presstroke
diastole v
but P.t - 4ho.t* and
"
I chest. In fact, Rho,,* ls
stroke
diastole v
l. and this contributes to
*toti. volume VBn' This
v
r the chest is oPened durssume for simPlicitY that
19) without modification'
r i.t E* .rc i." t 1'8 thr ou gh
V
≈C
Q = FV
P
= FC
Q = KP
P
= Cp uP pu
Vp upv
Vp u = Cp uP pu
P
q* = KsP*
q" = KlPpv
Right heart
Left heart
Vru = Cru Pru
Vru = Cru P.u
Psv
i.
i
hn co nt ro lle d
rave been develoPed above
Systemic
RrQr=Pru-Pru
Fig ure 1.6 . Eq ua tio ns of the cir cu lat ion ' On e eq ua tio n is sh ow n'f or ea ch
For .dditiotta equation that
eight principal elements oiiir"
Compartment Model
1.5 . Ma the ma tica l Mo de l of the Un con tro lled Cir cul ati on
ti)
Rn Qp =P pu -Pn u
(1 '4 '8 )
(1 .4 .e )
he Pressure
ending Pres4ho.t* and
ct, Rho,,* ls
ontributes to
e VBn' This
oPened durmPlicitY that
modification'
1'8 thr ou gh
lle d
eloPed above
Vpu = Cpu Ppu
Vpu = Cpu Ppu
q* = KsP*
q" = KlPpv
Vru = Cru Pru
Vru = Cru P.u
RrQr=Pru-Pru
Fig ure 1.6 . Eq ua tio ns of the cir cul ati on ' On e eq ua tio n is sho wn 'fo r ea ch of the
For .dditiottal equations that relate
eight principal elements oiiir"
Variables and Parameters
§  Variables
§  Pressure (P)
§  Flow (Q)
§  Volume (V)
§  Parameters
§  Resistance (R)
§  Compliance (C)
§  Heart pumping (K)
Solve Equations
Find variables (parameters)
1.5 . Ma the ma tica l Mo de l of the Un con tro lled Cir cul ati on
§  Summary equa?on ti)
Rn Qp =P pu -Pn u
(1 '4 '8 )
Vi = (1
Ci.4Pi.e )
V = ∑V
rmed that the Pressure
0
i
nen the distending Presbut P.t - 4ho.t* and
"
I chest. In fact, Rho,,* ls
i ibutes to i−1
l. and this contr
i is
*toti. volume VBn' Th
r the chest is oPened durh citY tha
L/R
h
t
ssume for simPli
19) without modification'
i gh
r i.t E* .rc i."t 1'8 thr ou
1
Q = ( P − Pi+1 )
R
Q =K
Vpu = Cpu Ppu
Vpu = Cpu Ppu
q* = KsP*
q" = KlPpv
P
Q=Q
Vru = Cru Pru
Vru = Cru P.u
i.
Variables: V, P, Q hn co nt ro lle d
Parameters: K, C, R i
rave been develoPed above
RrQr=Pru-Pru
Fig ure 1.6 . Eq ua tio ns of the cir cul ati on ' On e eq ua tio n is sho wn 'fo r ea ch of the
For .dditiottal equations that relate
eight principal elements oiiir"
Solving the Equations
§  The heart
Q = K R Psv & Q = K L Ppv
Q
Psv =
KR
Q
& Ppv =
KL
§  Capillaries
(
1
( Psa − Psv )
Rs
Q=
1⎛
Q⎞
1 ⎛
Q⎞
P
−
&
Q
=
P
−
sa
pa
Rs ⎜⎝
K R ⎟⎠
Rp ⎜⎝
K L ⎟⎠
& Q=
1
Ppa − Ppv
Rp
)
Q=
⎛
⎛
1 ⎞
1 ⎞
Psa = Q ⎜ Rs +
&
P
=
Q
R
+
pa
⎜⎝ p K ⎟⎠
K R ⎟⎠
⎝
L
Solving the Equations §  Volume equations V=CP
Csv
Vsv = Csv Psv =
Q
KR
Vi i = pv, sa, pa
§  Define Ti
Csv
Tsv =
⇒
KR
Ti i = pv, sa, pa
Vsv = TsvQ
Solving the Equations
§  Volume and flow equations revisited
Vi = TiQ
V0 = Vsa + Vsv + Vpv + Vpa
V0 = TsaQ + TsvQ + TpvQ + TpaQ
V0
Q=
Tsa + Tsv + Tpv + Tpa
Function of parameters!
Solving the Equations
§  Use same approach to predict
V0
Q=
Tsa + Tsv + T pv + T pa
Vi = TiQ = ...
Vi
Pi =
= ...
Ci
Normal Parameter Values
Systemic
Pulmonary
Units
R
Rs
17.5
Rp
1.79
mmHg min/liter
C
Csa
0.01
Cpa
.00667
liter/mmHg
Csv
1.75
Cpv
0.08
Right
Left
K
KR
2.8
V
V0
5.0
KL
Unit
1.2
liter/min/mmHg
liter
Parameter values can be calculated from solution formulas
Homework •  Solve all equa?ons for V, P and Q as a func?on of the model parameters (R, C, K) •  V: ( Vsa =, Vsv =, Vpa=, Vpv =) •  P: (Psa =, Psv =, Ppa =, Ppv =) •  Q: (Ql = Qr = Qs = Qp = Q) •  You are expected to write equa?ons and get values using parameters from the table