Applied BioFluid Mechanics  Lee Waite and Jerry Fine
.pdf282 Chapter Ten
Again we solve for the derivative dP1/dt 1/CV[PRV/RV PRL1/RL1]. We now have a pair of linear firstorder differential equations which describe the state
of this part of the circuit. Please note that the second equation contains a voltage, PRL1, from the next stage. This is to be expected since the stages are coupled and do not act independently.
Let us now look at this final piece of the circuit. We apply the KCL at the node inside the end load and find,
qRL1 2 qRL2 2 qCL 5 0
P1 2 Pe 
2 
Pe 
2 CL 
dPe 
5 0 

RL1 
RL2 
dt 




Solve for the derivative, and thus obtain,
dPe 
5 
1 
P1 
2 
RL1 
1 RL2 
Pe 


CL RL1RL2 

dt CL RL1 


Finally, we note that PRL1 5 P1 2 Pe , and so we update the last equation in the second stage,
dP1 
5 
1 
PRV 2 
1 
P1 
1 
1 
Pe 





dt CV RV 
CVRL1 

CVRL1 
The differential equations are now presented in their finalized form.


dPRV 
5 2 
RV 
PRV 2 

RV 
P1 1 
RV 
P0 




dt 

LV 

LV 
LV 


















dP1 
5 

1 

PRV 
2 

1 



P1 1 


1 
Pe 

dt 
CV 


CV 

RL1 







RV 






CV RL1 




dPe 

5 


1 

P1 2 

RL1 
1 RL1 
Pe 




dt 






CV RL1RL2 






CLRL1 





For some economy of expression, we switch to the use of matrix notation at this point. Let


2 
RV 



2 
RV 

0 





Lv 


Lv 















[A] 5 

1 

2 

1 



1 



G 
CV RV 
CV RL1 


CV RL1 
W 








0 



1 


2 
RL1 1 RL2 




CL RL1 
CL RL1RL2 

















Lumped Parameter Mathematical Models 
283 
and
5b6 

D 
0 
T 



RV 




LV 


5 

0 

It is convenient to accompany them with a set of relationships for the currents so that these can be output at any solution point. These relationships are as follows.




1 


0 

0 


Y C 





E 


I 
RV 






S 







1 




1 






q1 

0 



2 








RL1 
RL1 




qRL1 















RL1 

RL1RL2 





qC1 5 

1 
2 
1 

1 




P1 

or Q 
5 [R] P 


qRL2 
U 

R 
V 


RL1 


RL1 


Pe 

5 6 
5 6 








1 







qCL 

0 


0 





















RL2 




























0 


1 

2 
RL1 1 RL2 


















The steady state solution of these equations will5 6 now be described. We have calculated the matrix [A] and the vector b describing this linear system. Let the harmonic forcing function be represented by one harmonic
at a time, such as an cos snvtd. (In this example, there is but one harmonic term, so n 1.) The differential equation is
ddt 5P6 5 [A] 5P6 1 5b6a1 cos svtd
A solution of the following form is the particular or steady state solution
to the differential system. 
5C26 cos svtd 


5P6 5 5C16 sin svtd 1 


To find the vectors of unknown constants, 
C1 
and 
C2 , we first differ 

entiate the proposed solution with respect to time,5 6 
and 
5then6 
substitute this 
result into the differential equation. 


v5C16 cos svtd 2 v5C26 sin svtd 
5b6a1 cos svtd 

5 [A] 5C16 sin svtd 1 [A] 5C26 
cos svtd 1 
284 Chapter Ten
Next, simply equate the sine and cosine terms on each side of the equal
sign. 
v5C16 5 [A]5C26 1 
5b6a1 





and 





This leads to 


2v5C26 5 [A] 5C16 






1 
[A]2b5C16 5 5b6a1 







av I 1 


5 

16 



v 




This is a system of linear equations, which is solved for 
C 
, and then 

a simple matrix multiplication yields 5C26. 
21 





1 



















5C16 5 avI 1 

[A]2 b 

5b6a1 






v 







1 












5C26 5 2 


[A] 5C16 






v 




This solution tells us the expected fact that the steady state solution associated with this harmonic will have the same frequency as that harmonic, but will experience a phase difference from it, since both sine and cosine terms are present.
Special case. The data are presented in the following table.
Radius, R 
8 
mm 
Length, L 
10 
cm 
Viscosity, m 
0.004 
Ns/m2 
Density, r 
1,050 
kg/m3 
Mean pressure, a0 
13,300 
Pa 
Fluctuating press, a1 
2,670 
Pa 
Frequency of fluctuation, 
6.28 
rad/s 
first harmonic 





1.Calculation of the Mean Flow Solution.
For this vessel, we adjust the total fluid resistance in the terminal load (RT RL1 RL2) so, under pressure a0 of 13.3 kPa, we get a steady flow
Lumped Parameter Mathematical Models 
285 
of 4.0 L/min, corresponding to 6.65 10–5 m3/s. The fluid resistance that gives this result is 2 108 Ns/m5. The Reynolds number based on the mean flow across the cross section and also averaged over the time period of 1390 is high, since we have flow through a major vessel, but is not high enough so that the flow begins to leave the laminar range, even at peak velocity.
2.Calculation of the Parameters for the TimeVarying Solution.
We calculate the Fry parameters associated with the first harmonic, as described in Chap. 7. For the resistance per length and inertance per length of the first harmonic, we use the following formulas.
r 
5 c 

8 m 
R 

5 Lr 


vpR4 



V 


v 

V 

lV 5cu 
r 
LV 5 LlV 




pR2 
The results are laid out in the following table.
Angular frequency 

6.28 rad/s 
Womersley number 

10.2 
Resistance coefficient 
cv 
2.22 
Inertance coefficient 
cu 
1.13 
Resistance 
RV 
5.529 105Ns/m5 
Inductance 
LV 
5.927 105 Ns2/m5 



We select a capacitance, CV 2 10–10 m5/N as a typical compliance for a large blood vessel.
3.Calculation of the Parameters for the Terminal Load.
Speaking in the context of the electrical circuit analog, for a blood vessel
having resistance per length RV, inertance per length of LV, and compliance (capacitance) CV, the characteristic impedance is
ZV 5 aRV 1 jvLV b1/2
jvCV
This result is found in the theory of electrical transmission lines. The end load can be shown to have the following impedance.
Ze 5 RL1 1 
RL2 s1 2 jvRL2CLd 

1 1 RL22CL2v2 








286 Chapter Ten
Reflections in a transmission line result from the difference between the impedance in the line at the junction and the connected terminal load. Since reflections, or returning waves, are not pronounced in the circulatory system, we model the interaction between the vessel and the terminal load by matching the two impedances, setting
Ze ZV
Inserting the values previously tabulated, and recalling that RL2 2 108 RL1, the real and imaginary parts of the impedance equality become
1.726 3 108 5 R 

1 
2 3 108 2 R 
L1 


L1 


1 1 39.44s2 3 108 2 RL1d2CL2 , 



(Units will be Ns/m5)
and
21.275 3 107 5 s2 3 108 2 RL1ds21.25587 3 109 1 6.28RL1dCL 1.3944s2 3 108 2 RL1d2C2L
(Units will be Ns/m5)
The physically meaningful solution to the pair of equations is: RL1 = 1.667 108 Ns/m5, RL2 = 3.33 107 Ns/m5, and CL = 1.026 10–8m5/N.
4.Calculation of the Steady State Solution.
For the reader who is following through these calculations, we first give the arrays [A] and {b}.
[A] 
0.9329 
0.9329 
0 
{b} 
0.9329 
9043.2 
29.94 
29.94 


0 


0 
0.5836 
3.511 




0 







We use these to calculate the arrays {C1} and {C2}, leading to the following steadystate solution for the pressures.
5 

6 

C 
22.29 
S sin s6.28td 

C 
8.56 
S 




P 
5 
59.9 
1 
2670 
cos 
s 
6.28t 
d 



192 

102 


















Lumped Parameter Mathematical Models 
287 
From this expression, with the use of the relationship {Q} [R]{P} we can calculate the flow rates. These rates are presented in the plot in Fig. 10.5.
Commentary. We complete the example with some brief comments. First, we note the presence of a flow, qC1, which, in essence, is storage of fluid in the artery segment as it experiences pulsatile flow. A similar flow, qCL, describes the storage and release of fluid as a function of time in the rest of the system. The latter of the flows seems not far out of phase with the flow through the vessel segment, while the former experiences a noticeable phase difference.
This model is intended to illustrate some of the most general procedures that are used in creating the electrical circuit analog. More elaborate models can be generated, for example, by adding multiple segments and branches. More elements per unit length can also be added to increase the spatial resolution of the model. These more complex models can make important contributions to the understanding of the behavior of the cardiovascular system. For example, a clinician may wonder which model parameters have the greatest effect on flow through a given vessel and welldesigned and properly validated models can save enormous time and expense when compared with experiments in animal models.

× 10–5 
Flow rate in electrical analog model 

10 


8 

/s) 
6 




3 


(m 

q1 
rate 
4 


qRL1 

Flow 

qC1 




2 
qRL2 

qCL 





0 

–2 










0 
0.2 
0.4 
0.6 
0.8 
1 
1.2 
1.4 
1.6 
1.8 
2 





Time (s) 





Figure 10.5 Flow rate results for example problem.
288 Chapter Ten
10.2.3 Summary of the lumped parameter electrical analog model
Many scientists have used similar models to mimic blood flow and pressure conditions in animal circulation as well as in the human circulatory system, with the goal of better understanding of the relationship between the two.
For example, Bauernschmidt et al. (1999a) simulated flow hemodynamics during pulsatile extracorporeal perfusion. Control of perfusion is achieved by surgeons, anesthesiologists, and perfusionists making realtime decisions. This leads to variations of the perfusion regimens between different hospitals and between different surgical teams in the same hospital. To develop a computercontrolled scheme for the integrated control of extracorporeal circulation, they developed a mathematical model for simulating hemodynamics during pulsatile perfusion. The model was constructed in Matlab using Simulink. They divided the human arterial tree into a 128segment multibranch structure. Peripheral branches were terminated by resistance terms representing smaller arterioles and capillary beds. Bauernschmidt studied the effects of different perfusion regimens with differing amounts of flow and pulsatility and found the model to be useful for a realistic simulation of different perfusion regimens.
In 2004, L. R. John developed a mathematical model of the human arterial system, based on an electrical transmission line analogy. The authors concluded that quantitative variations of blood pressure and flow waveforms along the arterial tree from their model followed clinical trends.
10.3 Modeling of Flow through the Mitral Valve
Assessment of ventricular function and quantification of valve stenosis and mitral regurgitation are important in clinical practice as well as in physiological research (Thomas and Weyman, 1989, 1991; Takeuchi et al., 1991; Thomas et al., 1997; Scalia et al., 1997; Rich et al., 1999; Firstenberg et al., 2001; DeMey et al., 2001; Garcia et al., 2001). Approximately 400,000 patients are diagnosed with congestive heart failure in the United States each year. Elevated diastolic filling pressure in these patients leads to the development of congestive heart failure symptoms (Jae et al., 1997). Noninvasive assessment of diastolic function that does not require the use of intracardiac pressure has been an important goal, and in recent years, Doppler electrocardiography has become the “diagnostic modality of choice” (Garcia et al., 2001) to assess diastolic function.
The goal of the research that created this model was development and validation of a mathematical model of flow through the mitral valve
Lumped Parameter Mathematical Models 
291 
the current, which flows through it. Convective resistance will be discussed in more detail in Sec. 10.3.3. The term V dAdt represents the time rate of change of flow rate due to the closing or opening of the mitral valve. If the valve were held completely open, this term would be zero. Section 10.3.4 describes, in greater detail, the variablearea mitral valve that is used in this model.
The second differential equation is written as follows:
(10.12)
where Ca represents atrial compliance in m5/N QN/mm3 2R. The pressure inside the atrium is related to its volume through atrial compliance, Ca d(volume)/dt. The volume is also related to flow rate. In one sense, the flow into the atrium determines its final volume and therefore the final atrial pressure.
Similarly, the third equation describing the system is the equation for change in ventricular pressure with time. The ventricular pressure is similarly related to the ventricular volume and the ventricular compliance, Cv. The change in pressure of the ventricle is also related to “active relaxation.” The active relaxation term, resulting from ventricular geometry, is described in more detail in Sec. 10.3.2.
dPv 5 
q 2 active relaxation term 
(10.13) 
dt 
Cv 

In all three equations, q represents flow rate in m3/s, dq/dt represents the time rate of change of flow rate in m3/s2, Pa represents left atrial pressure in N/m2, Pv represents left ventricular pressure in N/m2, and M represents the inertance term, which is analogous to inductance in the electrical circuit and has the units of kg/m4.
M in Eq. (10.14) can be calculated as
M 5 rl/A 
(10.14) 
where r represents blood density in kg/m3,l represents the blood column length through the mitral valve with units of meters, and A represents the mitral valve effective area in m2.
Rv in Eq. (10.15) represents viscous resistance, which is analogous to resistance in the electrical circuit and has the units kg/m4s. Rv can be calculated as
Rv 5 8 rlm/A2 
(10.15) 
where m represents viscosity with units of Ns/m2.