7.10.2 A Two Compartment Model for First Order Absorption

Continuing our discussion of the two compartment model, consider the case now of first

order absorption of the drug. Unsteady mass balance equations may once again be written for each

of the compartments. The equation for the tissue compartment would still be given by Equation

7.66 and the equation that describes the amount of drug yet to be absorbed (A) would still be given

by Equations 7.56 and 7.57. The plasma compartment equation is then given by:

VdC

dtk A k V C k k V Cplasma

plasma

a tp tissue tissue te pt plasma plasma ( ) (7.74)

The initial conditions are that at time equal to zero there is no drug present in either

compartment. These equations may then be solved using Laplace transforms to obtain the

following expressions for the concentration of drug within the plasma and tissue compartments. For

the plasma concentration we have:

C tf D k

V

k A

k A B Ae

k B

k B A Be

k k

A k B keplasma

a

plasma

tp

a

A t tp

a

B t tp a

a a

k ta( )( )

( )( )

( )

( )( )

( )

( )( )

(7.75)

and for the tissue concentration:

C tf D k k

V

e

k A B A

e

k B A B

e

A k B ktissue

a tp

tissue

A t

a

B t

a

k t

a a

a

( )( )( ) ( )( ) ( )( )

(7.76)

The constants A and B have the same dependence on kpt, ktp, and kte as the two compartment IV case and are also given by Equations 7.69. Depending on the type of data one has, Equation 7.75 can be fitted to plasma concentration data to obtain the unknown parameters; f D, Vplasma, ka, ktp, A, and B. In some cases, it may not be possible to resolve the group f D/Vplasma without additional data, perhaps from an IV bolus injection and/or urinary excretion data. Once these parameters have been determined, Equations 7.70 through 7.73 may be used to find the remaining parameters.

Example 7.7 The table below (McNabb et al., 1982) presents plasma nicotine levels in a 49

year old man during and after smoking one cigarette yielding 0.8 mg of nicotine.

Time, minutes Plasma Nicotine Concentration,

ng/ml

0 0

2 5

3 7

4 11

5 10

6 14

8 12

9 13

11 13

12 12

15 12

21 9

25 9

35 8

41 7.5

49 7

59 7

71 6

SOLUTION Figure 7.14 illustrates the solution. We see that the two compartment model

fits this data rather well. The table below summarizes the results of the regression analysis.

Parameter Value

f D/Vplasma 15.4 ng ml-1

A 0.035 min-1

B 0.0013 min-1

ktp 0.013 min-1

kpt 0.02 min-1

kte 0.0037 min-1

Cigarette results show that2 compartment model fits well

Figure 7.14 Example 7.7

Example 7.7 plasma nicotine levels after smoking 1cigarette, analysis using a two compartment model with first order absorption i ..0 17 # of data points Minutes nicotine concentration, ng/ml time

i

0

23

4

56

89

1112

1521

2535

41

4959

71

Ci

0

57

11

1014

1213

1312

129

98

7.5

77

6

0 50 1000

10

20plasma nicotine levels, 1 cig.

Ci

timei

now we perform a nonlinear regression on the 5 unknown parameters of equation 7.75, i.e. (fD/Vp, ka,ktp,A,B)

c ,,,,,time I 0 k a k tp A B ..I 0 k a

+

....k tp A

.k a A ( )B Aexp( ).A time .

k tp B

.k a B ( )A Bexp( ).B time

.k tp k a

.A k a B k aexp .k a time

SSE ,,,,I 0 k a k tp A B

= 0

17

i

Ci

c ,,,,,timei

I 0 k a k tp A B 2

sum of the squares of the error between model and data that is to be minimized TOL .005 I 0 20 k a .4 k tp .01

A .02 B .01 initial guesses of the parameters, may need some fooling around to get reasonable starting values given SSE ,,,,I 0 k a k tp A B 0

I 0

k a

k tp

A

B

minerr ,,,,I 0 k a k tp A B

=I 0 15.3517 ng/ml, the value of D/Vp =k a 0.2851 1/min, the absorption rate constant =k tp 0.0125 1/min, transport constant between tissue and plasma

compartments =A 0.0349 1/min, a mixed constant =B 0.0013 1/min, a mixed constant

0 50 1000

10

20model fit to the data

Ci

c ,,,,,timei

I 0 k a k tp A B

,timei

timei

now we can calculate the other parameters

k te.A B

k tp

=k te 0.0037 1/min, total elimination constant k pt ( )A B k tp k te

=k pt 0.02 1/min, plasma to tissue

Heart Mate Pump 2

https://www.youtube.com/watch?v=TXRfMnG1O_0

LVAD Implant Surgery

LVAD Patient Stories

https://www.youtube.com/watch?v=I4r5_3uS0Nc

The HeartMate II is a high-speed, axial flow, rotary blood pump.

Weighing 12 ounces (about 375 grams) and measuring about 1.5 inches (4 cm) in diameter and 2.5 inches (6 cm) long, it is significantly smaller than other currently approved devices.

The internal pump surfaces are a smooth, polished titanium. Within the pump is a rotor that contains a magnet. The rotor assembly is rotated  by the electromotive force generated by the motor. The rotor propels the blood from the inflow cannula out to the natural circulation. The pump speed can vary from 6,000 rpm to 15,000 rpm, providing blood flow of up to 10 liters per minute.

FDA approval 2008, over 11000 installsEven allows heart to repair itself

https://www.youtube.com/watch?v=ev_01gQVWAM

Now we will take a look at the math model for a rotating cone pump

Process of Hemodialysis

In this chapter we will discuss and analyze several examples of extracorporeal devices.

These devices lie outside (extra) the body (corporeal) and are usually connected to the patient by

an arteriovenous shunt1. In some respects they may be thought of as artificial organs. Their

function is based on the use of physical and chemical processes to replace the function of a failed

organ or to remove an unwanted constituent from the blood. The patient's blood before entering

these devices is infused with anticlotting drugs such as heparin to prevent clotting. A variety of

ancillary equipment may also be present to complete the system. This may include items such as

pumps, flow monitors, bubble and blood detectors, as well as pressure, temperature, and

concentration control systems. It is important to note that these devices do not generally contain

any living cells. Devices containing living cells are called bioartificial organs and these will be

discussed in Chapter 10.

8.1 APPLICATIONS

A variety of extracorporeal devices have been developed. Perhaps the best known are blood

oxygenators that are used in such procedures as open heart surgery and hemodialysis to replace the

function of failed kidneys. Other

examples include hemoperfusion wherein a bed of activated carbon particles are used for cleansing

blood of toxic materials, plasmapheresis2 is used to separate erythrocytes from plasma as a first

step in the subsequent processing of the plasma, immobilized enzyme reactors to rid the body of a

particular substance or to replace lost liver function, and affinity columns are used to remove

materials such as antibodies that have been implicated as the cause of many autoimmune diseases.

1 A shunt is a means of diverting flow, in this case blood from an artery through a device and then back into the body via a vein. 2 See problem 11 in Chapter 5 for a discussion and analysis of a membrane plasmapheresis system.

Dialyzer is where the action is but the restis a complex control system

In many extracorporeal devices a solute must be transported through the blood, perhaps

across a membrane, and then through an exchange fluid. These

transport processes are illustrated in Figure 8.4 for the case of simple diffusion. Concentrations of

the solute at the fluid-membrane interface are assumed to be at equilibrium. The partition

coefficient K, is used to describe the solute equilibrium C KC and C KCmb bs

me es at the

surface of the membrane.

An overall local mass transfer coefficient (K0) for the situation shown in Figure 8.4 may be

written in terms of the individual film mass transfer coefficients on the blood (kb) and exchange

fluid (ke) sides along with the membrane permeability (Pm) and the partition coefficient. Recall that

the transport flux is N = K0(Cb – Ce) = kb(Cb – CbS) = Pm(Cmb – Cme) = ke(Ce

S – Ce) which can be

rearranged and solved for the overall mass transfer coefficient (K0) as

1 1 1 1

0K k K P kb m e

(8.1)

This equation states that the total mass transfer resistance 1

0K

is simply the sum of the individual

mass transfer resistances 1 1 1

k K P kb m e

. The smallest value of kb, KPm, or ke for a given solute

is said to be the controlling resistance.

Recall that we discussed in much detail the membrane permeability in Chapter 5. The

membrane permeability, although best measured experimentally (Dionne et al. 1996; Baker et al.

1997), is defined by Equation 5.90. Also as discussed in Chapter 5, the film mass transfer

coefficients depend on the physical

properties of the fluid and the nature of the flow. They also depend on position due to boundary

layer growth, so we usually use length-averaged values of these coefficients in Equation 8.1.

Techniques for estimating film mass transfer coefficients for a variety of flow conditions

were discussed in section 5.4.8 in Chapter 5. For laminar flow of fluids in a tube, one may use

Equations 5.52 or 5.53 to estimate the film mass transfer coefficient. For other flow situations

Table 5.2 provides a summary of useful mass transfer coefficient correlations.

Blood flow through extracorporeal devices is typically laminar and in some cases may be

fully developed. The Sherwood number (Sh) for blood for these conditions (kbh/Deffective) is equal to

3.66. In the Sherwood number, h = the flow channel thickness in a flat plate membrane

arrangement or the tube diameter in a hollow fiber. The effective diffusivity of the solute in blood

is given by Deffective and may be quite different from the solute’s bulk diffusivity because of the

presence of the red blood cells. This is discussed in the next section. Recall from Chapter 5 that the

asymptotic limit on the Sh number for fully developed laminar flow is valid when z

hSc 0 05. Re

(concentration field) or z

h 0 05. Re (velocity field). Here, z is defined as the length of the flow

path, h is the characteristic dimension of the flow channel, Re is the Reynolds number, and Sc

the Schmidt number. For solutes in liquids, the Sc is on the order of 1000 so the velocity field will

develop much faster than the concentration field.

The exchange fluid can be either in laminar or turbulent flow. Generally one operates on the

exchange fluid side in such a manner that its contribution to the overall mass transfer resistance

would be negligible. Therefore, ke >> kb and

KPm. If estimates of ke are needed, then one could use Equation 5.52 or 5.53, if the

exchange fluid flow is laminar. For turbulent flow of the exchange fluid, one can consult Table 5.2

Colton and Lowrie (1981) have shown that Equation 8.5 is also equal to the following

expression (actually Equation 5.84 with = H) assuming the red blood cell is impermeable (Dcell =

0).

D

DHK H

H

Heffective

plasmaRBC

( )( )

12 1

2 (8.6)

Equation 8.6 can now be used to estimate the effective diffusivity of a solute in blood. This

equation suggests that for a solute such as urea, its effective diffusivity through blood is about 50 %

of its value in plasma or about 40 % of its aqueous diffusivity.

The basic functional unit of the kidney is the nephron (shown earlier in Figure 7.3). Each

kidney contains about 1 million nephrons. Only about one third of these nephrons are needed to

maintain normal levels of waste products in the blood. If about 90 % of the nephrons lose their

function, then the symptoms of uremia will develop in the patient. Uremia results when waste

products normally removed from the blood by the kidneys start to accumulate in the blood and

other fluid spaces. For example, water generated by normal metabolic processes is no longer

removed and accumulates in the body, leading to edema, and in the absence of additional

electrolytes, almost half of this water enters the cells rather than the extracellular fluid spaces due

to osmotic effects. The normal metabolic processes also produce more acid than base, and this acid

is normally removed by the kidneys. Therefore, in kidney failure, there is a decrease in the pH of

the body's fluids, called acidosis, which can result in uremic coma. The end products of protein

metabolism include such nitrogenous substances as urea, uric acid, and creatinine. These materials

must be removed to ensure continued protein metabolism in the body. The accumulation of urea

and creatinine in the blood, although not life threatening by themselves, is an important marker of

the degree of renal failure. They are also used to measure the effectiveness of dialysis for the

treatment of kidney failure. The kidneys also produce the hormone erythropoietin

that is responsible for regulating the production of red blood cells in the bone marrow. In kidney

failure, this hormone is diminished leading to a condition called anemia and a lowered hematocrit.

If kidney failure is left untreated, death can occur within a few days to several weeks.

Hemodialysis (HD) has been used for nearly 50 years (Kolff 1947) to treat patients with

degenerative kidney failure or end-stage renal disease (ESRD). Hemodialysis has the ability to

replace many functions of the failed kidneys. For example, hemodialysis removes the toxic waste

products from the body, maintains the correct balance of electrolytes, and removes excess fluid

from the body. Hemodialysis can keep patients alive for several years and for many patients, allows

them to survive long enough to receive a kidney transplant. Nearly 500,000 patients are currently

being kept alive by hemodialysis.

Figure 8.5 illustrates the basic operation of hemodialysis. Heparinized blood flows through

a device containing a membrane. The dialysate or exchange fluid flows on the membrane side

opposite the blood. Solutes are exchanged by diffusion between the blood and the dialysate fluid. A

variety of membrane configurations are possible and some of the typical ones were shown earlier in

Figure 8.1. A variety of flow patterns for the blood and dialysate have been used and these are also

summarized in Figure 8.2. The membranes are usually made from such materials as cellulose,

cellulose acetate, polyacrylonitrile, and polycarbonate (Zelman, 1987, Galletti et al. 1995). The

membrane surface area is

on the order of 1 m2. Blood flowrates are in the range of several hundred ml/min and the dialysate

flowrate is about twice that of the blood.

Removal of waste products from the plasma must also be carefully controlled in order to

avoid the disequilibrium syndrome. If waste products are removed from the blood too fast by

dialysis, then the osmolarity of the plasma becomes less than that of the cerebrospinal fluid

resulting in a flow of water from the plasma and into the spaces occupied by the

brain and spinal cord. This increases the local pressure in these areas and can lead to serious side

effects.

Decreases acidity

Has a lower osmoticpressure

The rate at which water is removed (ultrafiltration) from the plasma by the pressure gradient

across the dialysis membrane can be estimated using Equation 5.95. These equations can be

simplified by assuming only the plasma proteins are impermeable. Hence the following equation is

obtained.

plasmaiccotonmeanP PSLQ (8.7)

oncotic plasma represents the oncotic pressure of the plasma proteins, about 28 mm Hg, and Pmean is

the average transmembrane pressure difference which is given by the next equation.

PP P P P

meanb b d d

,in ,out ,in ,out

2 2 (8.8)

Note that to avoid problems with the calculations it is best to simply use absolute pressures

in Equation 8.8, recognizing that since the dialysate is subatmospheric its reported gauge pressure

will be negative.

The pressure drop on the blood side

is typically on the order of 20 mmHg and that on the dialysate side is about 50 mmHg. The mean

transmembrane pressure drop is on the order of a few hundred mmHg. The hydraulic conductance,

LP, is typically about 3 ml/hr/m2/mmHg. High flux membranes can have values of the hydraulic

conductance as high as 20 ml/hr/m2/mmHg, however protein deposition on the plasma side of the

membrane can reduce this in a linear manner by about 6 %/hr (Zelman, 1987). Example 8.1

illustrates the calculation of the ultrafiltration rate in a typical membrane dialyzer.

Example 8.1 Using the membrane properties listed above, calculate the ultrafiltration rate

of water assuming a membrane surface area of 1 m2. How much water would be removed

after 6 hours of dialysis ? Assume blood enters the device at 120 mmHg (gauge) and leaves

at 100 mmHg (gauge). The dialysate fluid enters at -150 mmHg (gauge) and leaves at - 200

mmHg (gauge).

SOLUTION

mmHg282

)200760()150760(

2

)760100()760120(x

mmHghrm

ml3xlmQ

22

= 771 ml/hr (13ml/min)

After 6 hours of dialysis, about 10 pounds of water will have been removed.

Solute transfer in the dialyzer can be analyzed with the help of the model shown in Figure

8.6 showing cocurrent flow of the blood and dialysate. The Q's represent the volumetric flowrates

(ml/min) of the blood (b) and the dialysate (d) and the C's represent the species concentration

(usually in mg/dL). The mass transfer rate of species i across the dialysis membrane is given by the

following equation:

M Q C Q C Q C Q Ci bin

bin

bout

bout

dout

dout

din

din

i i i i (8.9)

The clearance of the dialyzer for a particular solute is defined as the volumetric flowrate of blood

(ml/min) entering the dialyzer that is completely cleared of the solute. The dialyzer clearance (CLD)

for solute i is then given by the next equation.

CLM

CDi

bin

i

(8.10)

The clearance for such solutes as urea, creatinine, and uric acid is on the order of 100 ml/min.

Higher molecular weight solutes show a reduced clearance.

The term dialysance is also used to describe the solute removal characteristics of a dialyzer.

Dialysance (DB) is defined by the next equation. We see that the mass transfer rate of species i (Mi)

is now divided by the maximum

solute concentration difference, ie. indi

inbi CC , between the blood and the dialysate.

DM

C CBi

bin

din

i i

(8.11)

For a single pass dialyzer, Cdin

i0, and the dialysance is then the same as the clearance.

Because of water removal, the inlet and outlet flowrates of the blood and dialysate streams

are not equal. Rather the following equation holds between the ultrafiltration rate and the flowrates

of these streams.

ind

outd

outb

inb QQQQQ (8.12)

If ultrafiltration is important, then we can use this relationship in combination with Equations 6.9

and 6.10 to derive an expression for the clearance that allows for examination of the importance of

ultrafiltration on the clearance of a given solute. Explicitly including ultrafiltration, the solute

clearance is given by the following equation.

ini

outi

ini

outiini

b

b

b

bbinbD C

CQ

C

CCQCL (8.13)