|Boomer Manual and Download|
|PharmPK Listserv and other PK Resources|
|Previous Page||Previous Chapter||Course Index||Next Chapter||Next Page|
In Figure 15.2.1 we have added an infusion rate constant, k0, to the diagram presented earlier, (Figure 5.4.1). This is a zero order process so the units of k0 are be amount per time, for example 25 mg/min.
Equation 15.2.1 Differential Equation for Drug amount During an IV Infusion
Equation 15.2.1 is the differential equation during the infusion period and it can be integrated to give Equation 15.2.2 using Laplace transforms.
Equation 15.2.2 Integrated Equation for Drug Amount in the body versus Time
and after dividing both sides by the apparent volume of distribution, V.
Equation 15.2.3 Integrated Equation for Drug Concentration versus Time
Calculate Cp Given k0, kel and V at time t
Calculate k0 required to give Cp at time t
You may notice that Equation 15.2.3 for Cp is quite similar to Equation 12.3.4 we had before for the cumulative amount of drug excreted into urine. As you might expect the plot of Cp would be similar in shape.
Figure 15.2.2 Linear Plot of Cp versus Time During a Continuous Infusion
Click on the figure to view the Java Applet window
Java Applet as a Semi-log Plot
This is because the rate of infusion will be constant whereas the rate of elimination will increase as the plasma concentration increases. At steady state the two rates become equal. We can determine the steady state concentration from the differential equation by setting the rate of change of Cp, i.e. dCp/dt = 0.
Equation 15.2.4 Steady State Concentration after Continuous IV Infusion
This could also be calculated from the integrated equation by setting e- kel * t = 0 at t = ∞.
We can now calculate the infusion rate necessary to produce some desired steady state plasma level.
A desired steady state plasma level of theophylline maybe 15 mg/L. The average half-life of theophylline is about 4 hr and the apparent volume of distribution is about 25 liter. What infusion rate is necessary?
First, kel = 0.693/4 = 0.17 hr-1
then k0 = kel * V * Cp = 0.17 * 25 * 15 = 63.8 mg/hr
We would probably use an infusion of 60 mg/hr which would produce a Cpss value given:
Cpss = k0/(kel • V) = 60/(0.17 x 25) = 14.1 mg/L
Calculate Cpss Given k0, kel and V
Calculate k0 required to give Cpss
Copyright 2001-3 David W. A. Bourne (firstname.lastname@example.org)