Why Derive Units From the Balance?
A parameter’s name does not determine its units; the balance equation does. In a QSP model, a symbol called a “rate” might mean an amount produced per time, a concentration produced per time, or a first-order constant that multiplies a state. These quantities have different units, even though they may all describe change over time. Deriving units from the equation helps identify what a parameter means and whether the model’s terms can be added.
The focus here is on direct balance equations for production, first-order turnover, and clearance. Assume that amount is measured in a unit such as \(\mathrm{mg}\), concentration in \(\mathrm{mg/L}\), volume in \(\mathrm{L}\), and time in \(\mathrm{h}\). The specific units can change, but the dimensional relationships must remain consistent. Concentration and amount are related by \(C=A/V\); using this conversion to differentiate concentration assumes a fixed compartment volume.
Units of Production and First-Order Turnover
Let \(A\) denote an amount and \(C\) a concentration. The units of a derivative follow from the units of its state divided by time: \(dA/dt\) has units amount/time, while \(dC/dt\) has units concentration/time. Every term added to or subtracted from a derivative must have those same units.
For example, consider an amount balance with a constant production rate \(R_{\mathrm{in}}\) and a first-order loss:
Since \(dA/dt\) has units amount/time, \(R_{\mathrm{in}}\) must also have units amount/time. The product \(k_{\mathrm{out}}A\) must have units amount/time as well. Dividing those units by the units of \(A\) shows that \(k_{\mathrm{out}}\) has units \(1/\mathrm{time}\). It is not an amount-removal rate by itself: the amount removed per time is the product \(k_{\mathrm{out}}A\).
A concentration balance has the corresponding structure. If \(R_C\) is a constant concentration production rate, then \(R_C\) has units concentration/time. In a term \(k_{\mathrm{out}}C\), the same first-order constant has units \(1/\mathrm{time}\), because its product with concentration must have units concentration/time. The numerical value of a rate parameter can depend on the chosen time unit: for example, a value expressed per hour must be converted before it is used in a model whose time unit is minutes.
Worked Example: Checking an Amount-Based Production and Loss Balance
Suppose a hypothetical compartment follows \(dA/dt=R_{\mathrm{in}}-k_{\mathrm{out}}A\), with \(R_{\mathrm{in}}=120\ \mathrm{nmol/h}\), \(k_{\mathrm{out}}=0.30\ \mathrm{h^{-1}}\), and current amount \(A=200\ \mathrm{nmol}\). Check the units and calculate the instantaneous change.
The production term is already in \(\mathrm{nmol/h}\). The loss term is \(k_{\mathrm{out}}A=(0.30\ \mathrm{h^{-1}})(200\ \mathrm{nmol})=60\ \mathrm{nmol/h}\). Thus, both terms have the required units of amount/time, and
At this state, the amount is increasing at \(60\ \mathrm{nmol/h}\). The value \(0.30\ \mathrm{h^{-1}}\) is not itself the amount lost each hour; the amount loss depends on \(A\) through the product \(k_{\mathrm{out}}A\).
Clearance in Amount and Concentration Balances
Clearance is defined through an amount-removal flux. If \(CL\) is clearance and \(C\) is the concentration being cleared, the amount removed per time is \(CL\,C\). Therefore, its units must satisfy \(\mathrm{units}(CL)\times\mathrm{amount/volume}=\mathrm{amount/time}\). It follows that \(CL\) has units volume/time. Clearance is not itself the amount eliminated per time; the product \(CL\,C\) is.
For a fixed-volume compartment, let \(A=VC\). If elimination is represented by \(CL\,C\), the amount balance is \(dA/dt=-CL\,C\). Because \(V\) is constant, \(dA/dt=V\,dC/dt\). Substitution gives
Comparing this with a first-order concentration loss, \(dC/dt=-k_{\mathrm{out}}C\), identifies \(k_{\mathrm{out}}=CL/V\). The units confirm the result: \((\mathrm{volume/time})/\mathrm{volume}=1/\mathrm{time}\). This equivalence depends on the stated fixed-volume setting and on representing the loss as clearance multiplied by concentration.
Worked Example: Converting Clearance to a Concentration Loss
In a hypothetical fixed-volume compartment, let \(V=2.5\ \mathrm{L}\), \(CL=0.75\ \mathrm{L/h}\), and \(C=4.0\ \mathrm{mg/L}\). Calculate the amount eliminated per time and the concentration change due to elimination.
The amount-elimination flux is \(CL\,C=(0.75\ \mathrm{L/h})(4.0\ \mathrm{mg/L})=3.0\ \mathrm{mg/h}\); the litres cancel, leaving amount/time. The corresponding first-order constant is
The concentration loss is therefore \(\frac{dC}{dt}=-(0.30\ \mathrm{h^{-1}})(4.0\ \mathrm{mg/L})=-1.2\ \mathrm{mg/(L\,h)}\). As a consistency check, multiplying this derivative by the fixed volume gives \((2.5\ \mathrm{L})(-1.2\ \mathrm{mg/(L\,h)})=-3.0\ \mathrm{mg/h}\), matching the amount balance. The negative sign denotes loss; the elimination flux itself is conventionally reported as a positive magnitude.
Turnover at Steady State
A simple turnover balance combines constant production with first-order loss. In a concentration formulation, write \(dC/dt=R_C-k_{\mathrm{out}}C\), where \(R_C\) is a concentration production rate. Under constant parameters, a steady state is a concentration \(C_0\) for which the derivative is zero. Setting the balance to zero gives \(R_C=k_{\mathrm{out}}C_0\), and hence \(C_0=R_C/k_{\mathrm{out}}\). The units are concentration/time divided by \(1/\mathrm{time}\), which yields concentration.
The same reasoning applies to an amount balance, \(dA/dt=R_{\mathrm{in}}-k_{\mathrm{out}}A\), giving \(A_0=R_{\mathrm{in}}/k_{\mathrm{out}}\). The units are amount/time divided by \(1/\mathrm{time}\), which yields amount. These steady-state relationships are consequences of the balance and the assumption that the derivative is zero; they do not change the units of the production rate or turnover constant.
Worked Example: Finding Production and Clearance From a Baseline
Suppose a hypothetical fixed-volume compartment has baseline concentration \(C_0=8.0\ \mathrm{ng/mL}\), first-order loss constant \(k_{\mathrm{out}}=0.25\ \mathrm{h^{-1}}\), and volume \(V=1.5\ \mathrm{L}\). At steady state, calculate the concentration production rate, the equivalent amount production rate, and the clearance corresponding to this loss constant.
At steady state, \(R_C=k_{\mathrm{out}}C_0\). Thus \(R_C=(0.25\ \mathrm{h^{-1}})(8.0\ \mathrm{ng/mL})=2.0\ \mathrm{ng/(mL\,h)}\). The result has concentration/time units, as required. Since \(1.5\ \mathrm{L}=1500\ \mathrm{mL}\), the equivalent amount production rate is
For this fixed volume, \(CL=k_{\mathrm{out}}V\), so \(CL=(0.25\ \mathrm{h^{-1}})(1.5\ \mathrm{L})=0.375\ \mathrm{L/h}\). Checking the baseline elimination flux gives \((0.375\ \mathrm{L/h})(8.0\ \mathrm{ng/mL})=3.0\ \mathrm{\mu g/h}\), after converting \(0.375\ \mathrm{L}\) to \(375\ \mathrm{mL}\). Production and elimination therefore match at baseline, as required for steady state.
A Short Derivation and a Practical Audit
The unit relationships can be derived directly by treating the state units as dimensions. Let \([A]\) denote amount, \([V]\) volume, and \([t]\) time. Then \([C]=[A]/[V]\). For the amount balance \(dA/dt=R_{\mathrm{in}}-k_{\mathrm{out}}A\), the left side has units \([A]/[t]\). Equality of units for each term requires \([R_{\mathrm{in}}]=[A]/[t]\) and \([k_{\mathrm{out}}][A]=[A]/[t]\), so \([k_{\mathrm{out}}]=1/[t]\). For clearance, \([CL][C]=[A]/[t]\); substituting \([C]=[A]/[V]\) gives \([CL]=[V]/[t]\). Finally, \([CL/V]=1/[t]\), consistent with its equivalence to a first-order constant for a fixed volume.
In practice, this audit is most useful when performed term by term before interpreting parameter values. It checks the balance structure, not whether the proposed biology is correct. A dimensionally consistent model can still have an unsuitable mechanism or parameter value.
Record whether the balance is for amount or concentration, and state the time unit and, when needed, volume unit.
Use amount/time for an amount derivative or concentration/time for a concentration derivative.
Check that each production or loss term has the same units as the derivative; use products such as \(k_{\mathrm{out}}A\) or \(CL\,C\) rather than assigning the product’s units to its parameter.
Use \(A=VC\) to move between amount and concentration formulations, and verify that a fixed-volume conversion is appropriate for the balance being used.
Common Mistakes, Diagnostics, and Limitations
A frequent error is to label \(k_{\mathrm{out}}\) as an amount/time rate because it appears in a loss term. The term that removes amount is \(k_{\mathrm{out}}A\); the constant alone has units \(1/\mathrm{time}\). Similarly, \(CL\) is not an amount/time rate. The elimination flux is \(CL\,C\), while \(CL\) has units volume/time. Multiplying clearance by a concentration without matching volume units can produce a numerical result that looks plausible but is wrong by a conversion factor.
Useful warning signs include an equation that adds terms with different units, a purported first-order constant reported in amount/time, or a clearance reported in amount/time without reference to concentration. A concentration-based production rate and an amount-based production rate also cannot be exchanged numerically without accounting for volume. If the volume changes, differentiating \(A=VC\) involves both \(dC/dt\) and the change in \(V\); the fixed-volume conversion above should not be applied unchanged.
Amount-based balances make production and removal fluxes explicit in amount/time. Concentration-based balances express change per volume per time and can be convenient when observations and states are concentrations. Clearance parameterization describes elimination as a volume processed per time, whereas first-order parameterization describes fractional loss per time. They are interchangeable through \(k_{\mathrm{out}}=CL/V\) only under the fixed-volume and balance assumptions stated here. Neither form is universally preferable; choose one that matches the model’s state representation and define its units clearly.
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- An amount balance includes \(R-kA\), where \(A\) is measured in \(\mathrm{mg}\) and time in days. What units must \(R\) and \(k\) have?
- A concentration is \(6\ \mathrm{mg/L}\), and clearance is \(0.4\ \mathrm{L/h}\). What is the amount-elimination flux, including units?
- A concentration turnover model has \(R_C=1.5\ \mathrm{\mu g/(mL\,h)}\) and \(k_{\mathrm{out}}=0.30\ \mathrm{h^{-1}}\). What is its steady-state concentration?
- For a fixed \(2.0\ \mathrm{L}\) compartment with \(k_{\mathrm{out}}=0.15\ \mathrm{h^{-1}}\), calculate the equivalent clearance and state its units.
Solutions. (1) \(R\) must be \(\mathrm{mg/day}\), because it is added to an amount derivative; \(k\) must be \(\mathrm{day^{-1}}\), because \(kA\) must be \(\mathrm{mg/day}\). (2) \(CL\,C=(0.4\ \mathrm{L/h})(6\ \mathrm{mg/L})=2.4\ \mathrm{mg/h}\). (3) \(C_0=R_C/k_{\mathrm{out}}=(1.5\ \mathrm{\mu g/(mL\,h)})/(0.30\ \mathrm{h^{-1}})=5.0\ \mathrm{\mu g/mL}\). (4) \(CL=k_{\mathrm{out}}V=(0.15\ \mathrm{h^{-1}})(2.0\ \mathrm{L})=0.30\ \mathrm{L/h}\).