Why Check an Organ ODE Term by Term?
An organ ODE may look plausible and still be wrong by a factor of 60, 1,000, or an organ volume. A useful check is to write the units of every term before combining them. In a perfused-organ model, the central distinction is between an amount rate, such as milligrams per hour, and a concentration rate, such as milligrams per litre per hour. Both can describe the same organ, but they are not interchangeable.
This tutorial derives a simple perfused-organ balance in both forms. It assumes a well-mixed organ state linked to the outflow concentration by a specified partition relationship. The derivation is then used to check each term, work numerical examples, and identify what changes if organ volume is not constant.
Start With the Amount Balance
Assume that the organ has a fixed volume, receives flow \(Q\), and has equal inflow and outflow, so there is no net accumulation of blood volume in the organ. Assume the concentrations \(C_a\) and \(C_v\) are expressed on the same blood or plasma basis, and that \(Q\) uses the matching volume-per-time unit. The net drug amount entering through perfusion per unit time is therefore \(Q C_a-Q C_v\). Drug is removed within the organ at a first-order rate \(k_{\mathrm{el}}A_T\). Under these assumptions, the organ amount balance is
The units establish whether the terms can be added. If amount is in mg, volume in L, and time in h, then \(Q\) has units L/h, each concentration has units mg/L, and \(k_{\mathrm{el}}\) has units h\(^{-1}\). Thus \(Q(C_a-C_v)\) has units mg/h, as does \(k_{\mathrm{el}}A_T\). The left side, \(dA_T/dt\), also has units mg/h.
To express the outflow concentration using the organ concentration, suppose the model specifies \(C_v=C_T/K\), where \(K=C_T/C_v\) is dimensionless. This relationship is a model assumption: it must be appropriate for the concentration definitions being used. With \(A_T=V_T C_T\), the amount equation becomes
Every term on the right is still an amount rate. In particular, the elimination term is \(k_{\mathrm{el}}V_T C_T\), not just \(k_{\mathrm{el}}C_T\), because the first-order rate constant acts on amount \(A_T=V_T C_T\).
Deriving the Concentration Form
For a fixed \(V_T\), differentiating \(A_T=V_T C_T\) gives \(dA_T/dt=V_T\,dC_T/dt\). Substitute this identity into the amount balance and divide both sides by \(V_T\):
The concentration equation is dimensionally consistent because \(Q/V_T\) has units h\(^{-1}\), the concentration difference has units mg/L, and their product has units mg/L/h. The elimination term \(k_{\mathrm{el}}C_T\) has the same units. The left side is a concentration derivative, also mg/L/h.
The derivation shows exactly why the two equations agree: the concentration equation is the amount equation divided by the fixed organ volume after replacing amount by volume times concentration. It is not valid to divide only one term by volume, or to use an amount rate as though it were a concentration rate.
Worked Example: Checking Every Term in Both Forms
Consider a hypothetical organ with \(V_T=0.750\) L, \(Q=1.20\) L/h, \(C_a=1.50\) mg/L, \(C_T=2.00\) mg/L, \(K=1.50\), and \(k_{\mathrm{el}}=0.0600\) h\(^{-1}\). Assume the stated partition relationship and fixed volume apply. Find the instantaneous amount and concentration rates.
First calculate the outflow concentration without rounding it for the subsequent arithmetic: \(C_v=C_T/K=2.00/1.50=1.333\ldots\) mg/L. The amount in the organ is \(A_T=V_T C_T=(0.750)(2.00)=1.50\) mg.
In the amount equation, the net perfusion rate is \(Q(C_a-C_v)=(1.20)(1.50-2.00/1.50)=0.200\) mg/h. Elimination is \(k_{\mathrm{el}}A_T=(0.0600)(1.50)=0.0900\) mg/h. Therefore, \(dA_T/dt=0.200-0.0900=0.110\) mg/h.
In the concentration equation, the perfusion contribution is \((Q/V_T)(C_a-C_T/K)=(1.20/0.750)(1.50-2.00/1.50)=0.2667\) mg/L/h. The elimination contribution is \(k_{\mathrm{el}}C_T=(0.0600)(2.00)=0.120\) mg/L/h. Thus \(dC_T/dt=0.2667-0.120=0.1467\) mg/L/h. The two rates agree under the fixed-volume conversion: \(V_T(dC_T/dt)=(0.750)(0.1467)=0.110\) mg/h, with the unrounded value giving \(0.110\) mg/h exactly.
Unit Conversions Before Substitution
The units in a formula must match the units of its numerical inputs. For example, if \(Q\) is reported in mL/min while \(V_T\) is in L and time is in hours, convert \(Q\) before calculating \(Q/V_T\). Similarly, expressing concentration in mg/L while using flow in L/h produces mg/h; using flow in mL/min instead produces a different compound unit unless it is converted.
Worked Example: Converting Flow Before Using the Concentration ODE
For a second hypothetical organ, let \(Q=18.0\) mL/min, \(V_T=0.600\) L, \(C_a=0.800\) mg/L, \(C_T=1.20\) mg/L, \(K=2.00\), and \(k_{\mathrm{el}}=0.100\) h\(^{-1}\). Assume constant volume and \(C_v=C_T/K\). Calculate both derivatives using hours as the time unit.
Convert flow: \(18.0\) mL/min \(=0.0180\) L/min, and \((0.0180)(60)=1.08\) L/h. The outflow concentration is \(C_v=1.20/2.00=0.600\) mg/L, and the organ amount is \(A_T=(0.600)(1.20)=0.720\) mg.
The amount-form perfusion rate is \((1.08)(0.800-0.600)=0.216\) mg/h. The elimination rate is \((0.100)(0.720)=0.0720\) mg/h. Hence \(dA_T/dt=0.216-0.0720=0.144\) mg/h.
For the concentration form, \(Q/V_T=1.08/0.600=1.80\) h\(^{-1}\). The perfusion contribution is \((1.80)(0.800-0.600)=0.360\) mg/L/h, and elimination contributes \((0.100)(1.20)=0.120\) mg/L/h. Therefore \(dC_T/dt=0.240\) mg/L/h. Multiplying by volume gives \((0.600)(0.240)=0.144\) mg/h, matching the amount rate.
When Organ Volume Changes
The simple conversion \(dA_T/dt=V_T\,dC_T/dt\) requires constant volume. If \(V_T\) varies with time, the product rule instead gives \(dA_T/dt=V_T\,dC_T/dt+C_T\,dV_T/dt\). The extra term has units \((\mathrm{mg/L})(\mathrm{L/h})=\mathrm{mg/h}\), so it is an amount rate. Omitting it would generally make the amount and concentration descriptions disagree.
Worked Example: Accounting for a Changing Organ Volume
Suppose a hypothetical organ has \(V_T=0.800\) L, \(C_T=1.50\) mg/L, and its volume is increasing at \(dV_T/dt=0.0100\) L/h. At this instant, let the net perfusion contribution be \(0.600\) mg/h and elimination be \(0.250\) mg/h. These amounts imply \(dA_T/dt=0.600-0.250=0.350\) mg/h. Find the concentration derivative.
Use the product-rule form and solve for \(dC_T/dt\): \[ \frac{dC_T}{dt} =\frac{1}{V_T}\left(\frac{dA_T}{dt}-C_T\frac{dV_T}{dt}\right). \] The volume-change term is \(C_T(dV_T/dt)=(1.50)(0.0100)=0.0150\) mg/h. Therefore, \(dC_T/dt=(0.350-0.0150)/0.800=0.41875\) mg/L/h, or \(0.4188\) mg/L/h to four significant figures.
Checking back, \((0.800)(0.41875)+(1.50)(0.0100)=0.335+0.0150=0.350\) mg/h. Thus the concentration derivative and the amount rate are consistent when the volume-change contribution is included.
Diagnostics, Alternatives, and Common Errors
An amount-based equation and a concentration-based equation are alternative state-variable descriptions of the same organ balance when their assumptions and transformations agree. Amount states make dose and transfer rates easy to express directly as amount per time. Concentration states can be convenient when observations or partition relationships are stated as concentrations. Neither form is automatically safer: the conversion between them must be applied to every term.
| Check | Consistent form | Warning sign |
|---|---|---|
| Perfusion in amount form | \(Q(C_a-C_v)\), with units amount/time | Flow multiplied by an amount rather than a concentration |
| First-order elimination in amount form | \(k_{\mathrm{el}}A_T\), with units amount/time | A concentration used where the rate law specifies amount |
| Perfusion in concentration form | \((Q/V_T)(C_a-C_v)\), with units concentration/time | Using \(Q(C_a-C_v)\) directly in an equation for \(dC_T/dt\) |
| Amount-to-concentration conversion | Divide the entire fixed-volume amount balance by \(V_T\) | Dividing some terms by volume but not others |
| Changing volume | Include \(C_T\,dV_T/dt\) in the product rule | Using \(dA_T/dt=V_T\,dC_T/dt\) despite changing volume |
A practical audit is to annotate each term with units, convert all inputs to a common time and volume basis, and then verify that the sum on each side has the units of the state derivative. A second check is to convert the computed concentration rate back to an amount rate by multiplying by \(V_T\) when volume is constant. If this does not reproduce the amount balance, look for a missing volume factor, an inconsistent concentration basis, or an unconverted flow unit.
Common mistake. A modeler writes \(dC_T/dt=Q(C_a-C_v)-k_{\mathrm{el}}C_T\) and argues that both terms describe drug entering or leaving the organ. That reasoning overlooks the units: the perfusion term is amount/time, while the elimination term is concentration/time. For a concentration-state equation with constant volume, the correction is to divide the amount balance by \(V_T\), giving \((Q/V_T)(C_a-C_v)-k_{\mathrm{el}}C_T\). Conversely, in the amount equation, use \(k_{\mathrm{el}}A_T\), not \(k_{\mathrm{el}}C_T\).
Check Your Understanding
Answer each question, then compare your work with the solutions below.
- If \(Q\) is in L/h and concentration is in mg/L, what are the units of \(Q(C_a-C_v)\)?
- For a fixed-volume concentration equation, why does the perfusion coefficient appear as \(Q/V_T\) rather than \(Q\)?
- A model uses \(k_{\mathrm{el}}=0.20\) h\(^{-1}\) and \(A_T=3.0\) mg. What is the elimination amount rate, and what are its units?
- If \(V_T\) changes with time, what additional term appears when differentiating \(A_T=V_T C_T\)?
Solutions. (1) \(Q(C_a-C_v)\) has units \((\mathrm{L/h})(\mathrm{mg/L})=\mathrm{mg/h}\), an amount rate. (2) The amount balance is divided in full by the fixed volume to obtain \(dC_T/dt\); therefore \(Q/V_T\) has units h\(^{-1}\), and multiplying by a concentration gives concentration/time. (3) \(k_{\mathrm{el}}A_T=(0.20)(3.0)=0.60\) mg/h. (4) The product rule gives \(dA_T/dt=V_T\,dC_T/dt+C_T\,dV_T/dt\); the additional term is \(C_T\,dV_T/dt\), with units amount/time.