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BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
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Analysis of Thermal Dilution Method for Cardiac Output Measurement
Last Updated June 19, 2017
One method for measurement of cardiac output is to inject cold saline into the right atrium of the heart and determine how long it takes for the dye to wash out. A thermistor is
placed in the aorta, and temperature is measured as a function of time. The method
can be analyzed through a Laplace transform method.
There are two questions to be answered: 1) How does the temperature change with
time? and 2) How is flow rate related to the temperature measurement?
The following analysis will use concentration instead of temperature because
the nomenclature is simpler. However,
the equations are otherwise the same.
c0(t)
Dye Injected (Q0(t))
Flow In (Q(t))
Flow Out (Q(t) + Q0(t))
c(t)
1) Temperature as a function of time:
The experiment is sketched in Figure 1.
Figure 1: Sketch of the inputs and
"Flow In" is the blood recirculated
outputs for the thermal dilution (dye
through the vena cava. "Flow Out" is the
blood into the aorta. "Dye Injected" is
injection) method. Qt  is from the vethe dye injected through a catheter into
na cava. Q0 t  is from the catheter,
the right atrium. The entire four chambers of the heart are modeled as a single
and Q t   Q0 t  goes to the aorta.
mixing chamber. It is assumed that the
injected dye mixes completely and immediately with the blood from the vena cava. The
conservation of mass states that the difference between mass in and mass out results
in a change of mass within the chamber.
Rate of
mass in
Rate of
mass out
Rate of increase of
mass in the chamber
(i.e. if 1 mg/s of dye flows in and only 0.7 mg/s flows out, then the dye in the chamber
increases at a rate of 0.3 mg/s). Define the following variables:
c(t)
V
c0
Q0(t)
Q(t)
concentration of the dye in the chamber
volume of the chamber
concentration of the dye injected
flow rate of the injected dye into the chamber
flow rate out of the chamber
BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
The rate of mass in is c0Q0 (g/ml x ml/s => g/s).
The rate of mass out is c(Q(t) + Q0(t))
Since the mass in the chamber is cV and V is constant, the rate of change of mass in
the chamber is V dc/dt.
Therefore, the equation governing the concentration is:
c0 Q0 (t )  c(t )Q(t )  Q0 (t )   V dc
dt
Which can be rearranged to a more familiar form:
V dc
dt
 c(t )Q(t )  Q0 (t )   c0 Q0 (t )
Q(t ) is the cardiac output we are trying to measure. It is thus an input to the equation. It
is, of course, a function of time, but we can approximate it as a constant if we decide to
look at time scales larger than one cardiac cycle, in which case we write Qt   Q .
Q0 (t ) is the flow rate of injection, and is also an input. In general, Q0 t   Qt  , so it
can be ignored on the left hand side (but not on the right hand side). The equation then
reduces to.
dct 
Eq. 1
V
 Q c ( t )  c 0 Q0 ( t ) .
dt
The form of this equation is identical to that of a charging RC circuit. Compare it to:
dV t  1
 V t   iin t 
dt
R
Thus, we already know how this equation is going to behave. The solution
has the same form as the RC circuit,
with the RC time constant replaced by
V/Q. Assume that the fluid is injected
from the catheter as a pulse, as shown
in Figure 2. Mathematically this can
be
expressed
as
Q0 t   Qi ut   ut  ti  , where ut  is
the unit step function, Q i is a constant
numerically equal to 1 for the example
in Figure 2 and ti is the total time of
injection, equal to 2 for the example in
Figure 2. To find concentration in the
aorta as a function of time, one must
2
1.5
Q (ml/s)
C
Injection of 2 ml over 2 seconds
1
0.5
0
-1
0
1
2
3
Time (sec)
4
5
Figure 2: Flow rate as a function of time for
injection of 2 ml of fluid into the heart over a
2 second time interval.
BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
solve Equation 1 with the expression for Q0 t  substituted on the right hand side, which
is:
V
dct 
 Q ct   c0Qi u t   u t  t1  .
dt
The equation can be transformed to the Laplace domain to yield:
 1 e  sti
sVC s   Vc0  Q C s   c0Qi  
s
s

 ,

where the second term on the left hand side, which arises from transforming the derivative, is zero because the initial concentration in the aorta is zero. The second term on
the right hand side comes from the shifting theorem. The algebraic manipulation yields:
 1  1 e  sti
C s   c0Qi 
 
s
 sV  Q  s



The inverse transform is more obvious if this equation is rewritten as:


1
e  sti

C s   c0QiV 

 ss  Q V  ss  Q V   .
 sti
The e
factor is a time delay. The inverse transform of
1
s s  a 
can be found in the ta-
bles of Laplace transforms, but for the sake of review it will be obtained from a partial
fraction expansion. Write:
1
A
B
As   
Bs
 


 As  A  Bs  1  A   B and A  1 
ss    s s   ss    ss   
Thus,
 1  1
1
1
1
t


 L1 
  1  e ut 
ss    s  s   
 ss     
So with   Q / V it follows that the inverse transform of Cs  is:
ct  




c0Qi

1  e Qt V ut   1  e Qt ti  V ut  ti 
Q
.
BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
The term modified by ut  looks like the charging of a capacitor in an RC circuit. The
term modified by ut  ti  looks like the discharging of an RC circuit. Therefore, the response to the dye sensor in the aorta (or thermistor in the case of thermal dilution is as
shown in Figure 3. During injection, the concentration in the aorta increases. Once injection stops, the dye is flushed out of the heart in a manner similar to the discharging of
a capacitor. It is instructive to rewrite the equation for ct  for times greater than or
equal to ti , when both of the step functions are active and then reshift the time base
back so that it starts at ti . In this case, the equation becomes:


This form shows that after
t  ti the system is simply
“discharging” all of the dye
that was injected up to that
time. That is, there is an
e  Q t V decaying exponential
with an amplitude factor of
c0Qi
1  e Qti V (which is a
Q
constant, independent of
time).



c0Qi
cQ
 e t ti Q V  e Qt V  0 i 1  e Qti V e Qt V .
Q
Q

2
1.5
Q (ml/s)
ct  
Injection of 2 ml over 2 seconds
1
0.5
0
-1
0
1
2
3
Time (sec)
4
5
Figure 3: Concentration (temperature) in the aorta
as a function of time. The concentration increases
while fluid is being injected and is then flushed out
of the heart after injection stops.
BIEN 435
Louisiana Tech University
Last Updated June 19, 2017
Steven A. Jones
<< Back to BIEN 435 Course Rules