Why Units Are a Structural Check
A compartment model describes amounts moving between spaces and being eliminated from the body. Its parameters do not all have the same units: an amount, a concentration, a volume, a clearance, and a rate constant represent different kinds of quantities. Checking their units is a quick way to catch mistakes in equations and calculations before interpreting a concentration-time profile.
Dimensional consistency is necessary, but it is not sufficient to establish that a model or parameter value is correct. Two expressions can have matching units and still represent different biological processes. Here the focus is narrower: verify that the quantities in one- and two-compartment calculations can be combined as written.
In a well-mixed compartment, concentration is amount divided by volume. For the central compartment, \(C_c=A_c/V_c\). Multiplying a clearance by concentration gives an amount-per-time rate: \((\mathrm{L/h})(\mathrm{mg/L})=\mathrm{mg/h}\). This is why elimination clearance is often used in an amount-balance equation as \(CL\,C_c\).
From Clearances and Volumes to Rate Constants
A rate constant describes the fraction of an amount transferred or eliminated per unit time. In the standard linear two-compartment model, \(k_{10}\) is elimination from the central compartment, \(k_{12}\) is transfer from central to peripheral, and \(k_{21}\) is transfer from peripheral to central. The subscript identifies the direction and destination convention; the defining relationships below specify the denominators.
The denominator rule follows from the amount flux. Elimination from the central compartment is \(CL\,C_c=CL\,A_c/V_c\), so its coefficient multiplying \(A_c\) is \(CL/V_c\). Central-to-peripheral transfer is \(Q\,C_c=Q\,A_c/V_c\), giving \(Q/V_c\). Peripheral-to-central transfer is \(Q\,C_p=Q\,A_p/V_p\), giving \(Q/V_p\). Since \((\mathrm{L/h})/\mathrm{L}=\mathrm{h}^{-1}\), all three coefficients are rates per unit time.
Here \(R_{\mathrm{in}}\) is an input rate, with units of amount per time. Every term on the right side of either equation must have units of amount per time, matching the derivative on the left. For example, \((k_{10}+k_{12})A_c\) has units \((1/\mathrm{h})(\mathrm{mg})=\mathrm{mg/h}\). If an amount rather than a rate is entered as \(R_{\mathrm{in}}\), the equation is dimensionally inconsistent.
Worked Unit Checks
Worked Example: Converting Clearance Before Calculating a Rate
Suppose a hypothetical one-compartment calculation uses \(CL=0.12\ \mathrm{L/min}\) and \(V=18\ \mathrm{L}\). Find the elimination rate constant in \(\mathrm{h}^{-1}\).
The units must use a common time scale. Convert clearance to liters per hour:
Now divide clearance by volume:
The liters cancel and the remaining unit is inverse hours. Dividing \(0.12\ \mathrm{L/min}\) by \(18\ \mathrm{L}\) without converting would instead give \(0.00667\ \mathrm{min}^{-1}\), which is equivalent: \(0.00667\times60=0.400\ \mathrm{h}^{-1}\). The numerical values differ because the time unit differs, not because the underlying rate differs.
Worked Example: Checking a Two-Compartment Amount Balance
Consider hypothetical parameters \(CL=4\ \mathrm{L/h}\), \(Q=6\ \mathrm{L/h}\), \(V_c=20\ \mathrm{L}\), and \(V_p=30\ \mathrm{L}\). At a particular time, let \(A_c=80\ \mathrm{mg}\) and \(A_p=30\ \mathrm{mg}\), with no input. Calculate the compartment rates of change and check their units.
First calculate the microconstants:
Substitute into the amount-balance equations. With \(R_{\mathrm{in}}=0\):
Each product is a rate constant in \(\mathrm{h}^{-1}\) times an amount in mg, so each contribution is in \(\mathrm{mg/h}\). The negative central rate means the central amount is decreasing at that instant; the positive peripheral rate means the peripheral amount is increasing. Adding the rates gives \(-34+18=-16\ \mathrm{mg/h}\). Internal transfer cancels in the total, leaving elimination alone: \(C_c=80/20=4\ \mathrm{mg/L}\), and \(CL\,C_c=(4\ \mathrm{L/h})(4\ \mathrm{mg/L})=16\ \mathrm{mg/h}\).
Worked Example: Correcting a Wrong Denominator
Suppose hypothetical values are \(Q=10\ \mathrm{L/h}\), \(V_c=25\ \mathrm{L}\), and \(V_p=40\ \mathrm{L}\). A calculation reports \(k_{12}=Q/V_p=0.250\ \mathrm{h}^{-1}\) and \(k_{21}=Q/V_c=0.400\ \mathrm{h}^{-1}\). Are these assignments correct?
Both reported numbers have the right dimensions: clearance divided by volume gives inverse time. But the compartment assignments are reversed. Central-to-peripheral transfer leaves the central compartment, so its flux is \(Q\,C_c=Q\,A_c/V_c\). Therefore:
Peripheral-to-central transfer leaves the peripheral compartment, so its flux is \(Q\,C_p=Q\,A_p/V_p\). Thus:
This example shows why dimensional analysis alone cannot catch every error: the incorrect assignments still carry units of inverse time. Check the units and also trace each flux back to the concentration, and hence volume, of its donor compartment.
Concentrations, Doses, and Time Units
For an instantaneous intravenous bolus into the central compartment, the initial central amount is the dose \(D\), and the initial central concentration is \(C_c(0)=D/V_c\). The dose must be an amount and the volume must be in matching volume units. For example, mg divided by L gives mg/L. If a model output is requested in \(\mathrm{\mu g/mL}\), convert the concentration after calculating it or convert all inputs consistently; do not silently treat mg/L as \(\mathrm{\mu g/mL}\) without noting that their numerical values are equal.
Time units also carry through exponential expressions. If an amount or concentration profile contains \(\exp(-kt)\), then \(kt\) must be dimensionless. A rate in \(\mathrm{h}^{-1}\) therefore requires time in hours. Using \(k=0.4\ \mathrm{h}^{-1}\) with \(t=30\) minutes as though \(t=30\) hours changes the exponent by a factor of 60. Convert the time to \(0.5\) hours, or convert the rate to \(\mathrm{min}^{-1}\), before evaluating the expression.
Practical Checks and Common Failure Modes
A reliable check starts by writing the units next to every supplied value, rather than only checking the final answer. Reduce each expression to its basic dimensions and confirm that additions combine like units. Then check whether the numerical value is sensible for the stated time scale. A rate of \(0.4\ \mathrm{h}^{-1}\) is not numerically the same as \(0.4\ \mathrm{min}^{-1}\), even though both are rates.
- Mixing time units. Dividing a clearance in L/min by a volume in L gives a rate in min\(^{-1}\), not h\(^{-1}\). Convert one of the time units before reporting a result in the other.
- Using the wrong compartment volume. \(k_{12}\) uses \(V_c\), whereas \(k_{21}\) uses \(V_p\). Follow the source of the transferred amount.
- Confusing amount and concentration. A dose in mg is not a concentration in mg/L. Divide by a volume to obtain concentration.
- Adding unlike quantities. Clearance and volume cannot be added, nor can an amount be added directly to an amount-per-time rate.
- Assuming correct units prove a correct model. A rate constant can have valid units and still be assigned to the wrong pathway or calculated with the wrong compartment volume.
Unit checking complements, rather than replaces, the usual approaches to compartment calculations. Directly substituting parameters into the amount-balance equations makes flux directions explicit; dimensional analysis is a quick audit of those substitutions. Converting units before calculation is usually safer than trying to repair a result afterward, particularly when concentrations, doses, and times use different scales. Neither a correct unit check nor a plausible numerical magnitude establishes that parameters can be uniquely estimated or that the structural model is appropriate.
Check Your Understanding
Answer each question, then compare with the worked solutions below.
- If \(CL=3\ \mathrm{L/h}\) and \(V_c=15\ \mathrm{L}\), what is \(k_{10}\), including units?
- What units must \(R_{\mathrm{in}}\) have in an amount-balance equation?
- For \(Q=8\ \mathrm{L/h}\), \(V_c=16\ \mathrm{L}\), and \(V_p=32\ \mathrm{L}\), calculate \(k_{12}\) and \(k_{21}\).
- A dose of \(120\ \mathrm{mg}\) is placed into a central compartment with \(V_c=24\ \mathrm{L}\). What is the initial concentration?
- Why is a dimensionally correct value for \(k_{12}\) not enough to confirm that it was calculated correctly?
Solutions. 1. \(k_{10}=CL/V_c=(3\ \mathrm{L/h})/(15\ \mathrm{L})=0.200\ \mathrm{h}^{-1}\). 2. \(R_{\mathrm{in}}\) must be amount per time, such as mg/h, because it is added to terms in \(dA_c/dt\). 3. \(k_{12}=Q/V_c=8/16=0.500\ \mathrm{h}^{-1}\), and \(k_{21}=Q/V_p=8/32=0.250\ \mathrm{h}^{-1}\). 4. \(C_c(0)=D/V_c=(120\ \mathrm{mg})/(24\ \mathrm{L})=5.00\ \mathrm{mg/L}\). 5. Both \(Q/V_c\) and \(Q/V_p\) have units of inverse time; to verify \(k_{12}\), also confirm that central is the donor compartment and therefore \(V_c\) is the denominator.