Why Use Clearances and Volumes?
A three-compartment model can describe drug in a central compartment and two peripheral compartments. The central compartment is the site of drug entry and elimination; each peripheral compartment exchanges drug with the central one. Writing the model in terms of clearances and volumes gives each process a direct interpretation: clearance describes elimination or exchange capacity, while volume relates an amount to a concentration.
The clearance-volume parameterization does not add new pathways to the mammillary structure. It expresses the same central-to-peripheral exchange and central elimination as rate constants, but in quantities that are often more interpretable and convenient to report. The model below assumes linear, first-order elimination and exchange, constant volumes, and well-mixed compartments.
Amount-Balance Equations
Elimination from the central compartment is proportional to its concentration, with elimination rate \(CL\,C_c\). Exchange with compartment 2 is governed by the concentration difference \(C_c-C_2\), and exchange with compartment 3 by \(C_c-C_3\). A positive difference means net movement out of the central compartment; a negative difference means net movement into it. For an input rate \(R_{\text{in}}(t)\) into the central compartment, the amount balances are:
The same \(Q_i(C_c-C_i)\) term appears with opposite signs in the central and peripheral balances. This represents transfer within the model, not creation or loss of drug. Only the elimination term removes drug from the system. Volumes connect amount and concentration through \(A_i=V_iC_i\). With constant volumes, \(dA_i/dt=V_i\,dC_i/dt\); if volumes varied over time, that simple relationship would need modification.
Connection to First-Order Rate Constants
The clearance-volume form can be translated into the microconstants used to describe first-order rates. Substitute \(C_c=A_c/V_c\), \(C_2=A_2/V_2\), and \(C_3=A_3/V_3\) into the amount equations. For example, the central elimination term becomes \(CL\,A_c/V_c\), so its first-order rate constant is \(CL/V_c\). Applying the same substitution to each exchange term gives:
Here \(k_{10}\) is the central elimination rate constant; \(k_{12}\) and \(k_{13}\) describe outward exchange from the central compartment; and \(k_{21}\) and \(k_{31}\) describe return exchange from the corresponding peripheral compartments. All have units of inverse time. A shared intercompartmental clearance implies the constraints \(k_{12}V_c=k_{21}V_2=Q_2\) and \(k_{13}V_c=k_{31}V_3=Q_3\). These equalities are a useful consistency check when converting between parameterizations.
Worked Example: Converting Clearances and Volumes to Rate Constants
Suppose a hypothetical model has \(CL=5\) L/h, \(Q_2=8\) L/h, \(Q_3=3\) L/h, \(V_c=20\) L, \(V_2=40\) L, and \(V_3=60\) L. Calculate its first-order rate constants.
For central elimination, \(k_{10}=CL/V_c=(5\text{ L/h})/(20\text{ L})=0.25\text{ h}^{-1}\). For exchange with compartment 2, \(k_{12}=Q_2/V_c=8/20=0.20\text{?}\) No: the arithmetic is \(8/20=0.40\), so \(k_{12}=0.40\text{ h}^{-1}\), while \(k_{21}=Q_2/V_2=8/40=0.20\text{ h}^{-1}\).
For exchange with compartment 3, \(k_{13}=Q_3/V_c=3/20=0.15\text{ h}^{-1}\), and \(k_{31}=Q_3/V_3=3/60=0.0500\text{ h}^{-1}\). The shared-clearance checks hold: \(k_{12}V_c=(0.40\text{ h}^{-1})(20\text{ L})=8\text{ L/h}=k_{21}V_2\), and \(k_{13}V_c=(0.15\text{ h}^{-1})(20\text{ L})=3\text{ L/h}=k_{31}V_3\).
Following Drug Movement Through the Model
The equations are also a direct way to determine the instantaneous direction and rate of each process from compartment concentrations. The exchange term is a net flux: it combines movement in both directions and may be positive, zero, or negative. The amount transferred between compartments cancels when their amount balances are added, leaving only input and elimination in the total amount balance.
Worked Example: Calculating Net Exchange and Elimination
In a hypothetical example, let \(A_c=100\) mg, \(A_2=40\) mg, and \(A_3=30\) mg, with \(V_c=20\) L, \(V_2=40\) L, and \(V_3=60\) L. Take \(CL=5\) L/h, \(Q_2=8\) L/h, and \(Q_3=3\) L/h. Assume there is no input at this instant, so \(R_{\text{in}}(t)=0\).
The concentrations are \(C_c=100/20=5\) mg/L, \(C_2=40/40=1\) mg/L, and \(C_3=30/60=0.5\) mg/L. Central elimination is \(CL\,C_c=(5\text{ L/h})(5\text{ mg/L})=25\) mg/h. The net exchange rates from the central compartment are \(Q_2(C_c-C_2)=(8\text{ L/h})(4\text{ mg/L})=32\) mg/h and \(Q_3(C_c-C_3)=(3\text{ L/h})(4.5\text{ mg/L})=13.5\) mg/h.
Thus \(dA_c/dt=0-25-32-13.5=-70.5\) mg/h, \(dA_2/dt=32\) mg/h, and \(dA_3/dt=13.5\) mg/h. Adding these rates gives \(-70.5+32+13.5=-25\) mg/h, exactly the negative elimination rate. Drug is leaving the central compartment faster than it is entering it, while both peripheral amounts are increasing at this instant.
Assumptions, Checks, and Limitations
This parameterization relies on a specific structure. Each compartment is well mixed, so its concentration is represented by one value. Volumes are constant, and elimination is linear in central concentration. Each peripheral compartment connects directly to the central compartment, with no direct exchange between peripheral compartments and no elimination from them. The model treats each connection as a shared clearance, so exchange depends on the difference between the connected concentrations.
When using or fitting this model, check that units match, volumes are positive, and clearances are nonnegative. If rate constants and volumes are supplied instead, check that both directions of each connection imply the same \(Q_i\). A mismatch can indicate inconsistent inputs, a unit conversion error, or a model whose parameter definitions do not use a shared intercompartmental clearance. Also examine whether the proposed structure is suitable: systematic disagreement between observed concentrations and model predictions may reflect an inadequate compartment structure or assumptions, rather than a simple arithmetic error.
Worked Example: Checking Whether Rate Constants Imply One Shared Clearance
Suppose hypothetical values are \(V_c=20\) L, \(V_2=40\) L, \(k_{12}=0.40\text{ h}^{-1}\), and \(k_{21}=0.25\text{ h}^{-1}\). Can these two rate constants represent exchange through a single \(Q_2\)?
The outward rate implies \(Q_2=k_{12}V_c=(0.40\text{ h}^{-1})(20\text{ L})=8\) L/h. The return rate implies \(Q_2=k_{21}V_2=(0.25\text{ h}^{-1})(40\text{ L})=10\) L/h. Since \(8\ne10\), the values are inconsistent with one shared \(Q_2\) under this parameterization. For example, if \(k_{12}=0.40\text{ h}^{-1}\) and the volumes are retained, consistency requires \(k_{21}=8/40=0.20\text{ h}^{-1}\).
Clearance-volume parameters and microconstants are alternative descriptions of the same linear compartment model when the shared-clearance constraints hold. Clearances and volumes can make the model's quantities easier to interpret, while rate constants express fractional rates of change. Neither parameterization removes the structural assumptions: a model with peripheral elimination, time-varying volumes, nonlinear processes, or different exchange pathways would require different equations. A numerical conversion alone cannot establish that the selected structure adequately describes a particular data set.
Common Mistakes
- Using concentration in place of amount. The balance equations describe amount changes; exchange and elimination terms use concentration multiplied by clearance to produce amount per time.
- Assigning a separate clearance to each direction. In this parameterization, a connection has one shared \(Q_i\). Its two rate constants differ because they divide by different compartment volumes.
- Assuming exchange always moves outward from the central compartment. The sign of \(C_c-C_i\) determines the net direction; if it is negative, the net flow is toward the central compartment.
- Adding exchange as a system-wide loss. Exchange appears with opposite signs in the connected compartment balances. Only elimination, and any input, changes the total amount in the modeled system.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- What physical process is represented by \(Q_3(C_c-C_3)\), and what are its units?
- If \(Q_2=6\) L/h, \(V_c=30\) L, and \(V_2=60\) L, calculate \(k_{12}\) and \(k_{21}\).
- If \(C_c=2\) mg/L and \(C_2=5\) mg/L, what is the direction of net exchange between the central compartment and compartment 2?
- Why do exchange terms cancel when the three amount-balance equations are added?
Solutions. (1) It is the net exchange rate from the central compartment toward compartment 3, with its sign indicating net direction; its units are \((\text{L/h})(\text{mg/L})=\text{mg/h}\). (2) \(k_{12}=Q_2/V_c=6/30=0.20\text{ h}^{-1}\), and \(k_{21}=Q_2/V_2=6/60=0.10\text{ h}^{-1}\). (3) Since \(C_c-C_2=2-5=-3\) mg/L, net exchange is from compartment 2 toward the central compartment. (4) Each exchange term enters one compartment's balance positively and the connected compartment's balance negatively, so each transfer sums to zero across the system; elimination remains as a loss.