BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
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.
1) Temperature as a function of time:
Flow In (Q(t)) c
0
(t) c(t)
Dye Injected (Q
0
(t))
Flow Out (Q(t) + Q
0
(t))
The experiment is sketched in Figure 1.
"Flow In" is the blood recirculated through the vena cava. "Flow Out" is the blood into the aorta. "Dye Injected" is the dye injected through a catheter into the right atrium. The entire four chambers of the heart are modeled as a single
Figure 1: Sketch of the inputs and outputs for the thermal dilution (dye injection) method.
Q
0
 
Q
 
is from the vena cava. and
Q
 

Q
0
 
is from the catheter,
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) concentration of the dye in the chamber
V volume of the chamber c
0 concentration of the dye injected
Q
0
(t) flow rate of the injected dye into the chamber
Q(t) flow rate out of the chamber

BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
The rate of mass in is c
0
Q
0
(g/ml x ml/s => g/s).
The rate of mass out is c(Q(t) + Q
0
(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: c
0
Q
0
( t )
 c ( t )

Q ( t )

Q
0
( t )


V dc dt
Which can be rearranged to a more familiar form:
V dc dt
 c ( t )

Q ( t )

Q
0
( t )

 c
0
Q
0
( 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
0
( t )
is the flow rate of injection, and is also an input. In general,
Q
 

Q
0
 

Q
 
Q .
, so it can be ignored on the left hand side (but not on the right hand side). The equation then reduces to.
V dc

Q c ( t )
 c
0
Q
0
( t ) .
Eq. 1 dt
The form of this equation is identical to that of a charging RC circuit. Compare it to:
C dV dt

1
R
V
 
 i in
 
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
Q
0
 

Q i
 u t
 expressed t
 t i
 
, where u as
is
2
1.5
1
0.5
0
Injection of 2 ml over 2 seconds the unit step function,
Q i is a constant numerically equal to 1 for the example
-1 0 1 2
Time (sec)
3 4 5 in Figure 2 and t i 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
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 is:
Q
0 substituted on the right hand side, which
V dc

Q c
 c
0
Q i
 u
   t
 t
1
 
. dt
The equation can be transformed to the Laplace domain to yield: sVC
 

Vc
 

Q C
 
 c
0
Q i


1 s
 e
 st i 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:
C
 c
0
Q i

 sV
1

Q




1 s
 e
 st i s


The inverse transform is more obvious if this equation is rewritten as:
C
 c
0
Q i
V

 s
 s

1
Q V
  s e
 st i

Q V
 


.
The e
 st i factor is a time delay. The inverse transform of s
 s
1
 a

can be found in the tables of Laplace transforms, but for the sake of review it will be obtained from a partial fraction expansion. Write: s
 s
1
  

A s
 s
B
 
 s
A

 s s



 

 s
 s
Bs
  

As

A
 
Bs

1

A
 
B and A

1

Thus,
1 s
 s
  

1
 s


1
 s
  

L

1 s
1
 s
  

1


1
 e
  t
 u
 
So with
 c
   c
0
Q i
Q


Q / V

1

it follows that the inverse transform of
C
 
is: e

Q t V
 u
  

1
 e

Q
 t
 t i

V
 u
 t
 t i


.

BIEN 435
Steven A. Jones
Thermal Dilution Method
January 8, 2004
The term modified by term modified by u
 t
 t u i
 

looks like the charging of a capacitor in an RC circuit. The
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
 
for times greater than or equal to t , when both of the step functions are active and then reshift the time base i back so that it starts at t . In this case, the equation becomes: i c
 
 c
0
Q i
Q

 e
  t
 t i

Q V  e

Q t V

 c
0
Q i
Q

1
 e

Q t i
V
 e

Q t V .
This form shows that after t
 t i
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 c
0
Q i
Q

1
 e

Qt i
V

(which is a
2
1.5
1
0.5
Injection of 2 ml over 2 seconds constant, independent of time).
0
-1
Figure 3:
0 1 2
Time (sec)
3 4 5
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 April 13, 2020
Steven A. Jones