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Mass Balance for a Perfused Tissue · Tutorial 46 of 799

Converting Between Tissue Amounts and Concentrations

Learn to convert a perfused-tissue amount balance into a concentration balance, account for changing tissue volume, and check the units of every term.

Intermediate 10 min read

What You'll Learn

  • Define tissue concentration as amount divided by the volume used to represent the tissue.
  • Derive the concentration-rate equation from the amount balance.
  • Identify the dilution or concentration effect caused by changing tissue volume.
  • Convert amount, volume, concentration, and time units without changing the underlying balance.
  • Diagnose common errors in volume-change terms and unit conversions.

From Tissue Amount to Tissue Concentration

A mass balance tracks how much drug is present in a tissue. Many model outputs and measurements, however, are expressed as concentration. Converting between these descriptions requires more than dividing a rate by a volume whenever the volume can change: the amount and volume may both vary over time.

Let \(A_T(t)\) denote the drug amount in a tissue at time \(t\), and let \(V_T(t)\) be the tissue volume used to define its concentration. The average tissue concentration is the amount per unit of that volume. This is a bookkeeping definition; by itself, it does not assert that the drug is spatially uniform or that any particular concentration is the one governing exchange with blood.

Definition: Tissue concentration is \(C_T(t)=A_T(t)/V_T(t)\), where amount and volume refer to the same tissue and time. Equivalently, \(A_T(t)=V_T(t)C_T(t)\). The amount balance describes change in \(A_T\); the concentration balance describes change in \(C_T\).

Write the tissue amount balance in generic form as \(dA_T/dt=J_{\mathrm{in}}-J_{\mathrm{out}}+R_T\). Here \(J_{\mathrm{in}}\) and \(J_{\mathrm{out}}\) are sums of amount rates entering and leaving the tissue, respectively, and \(R_T\) is any additional net amount rate, such as a local source minus a local loss. Each term on the right has units of amount per time. The conversion below does not require a particular expression for these rates.

Deriving the Concentration Balance

Since \(A_T=V_TC_T\), differentiating both sides with respect to time gives the product rule. This is the key step: if the tissue volume changes, the amount can change because concentration changes, because volume changes, or both.

$$ \frac{dA_T}{dt} = V_T\frac{dC_T}{dt} + C_T\frac{dV_T}{dt} $$

Substitute the amount balance for \(dA_T/dt\), then isolate \(dC_T/dt\). For positive tissue volume, the result is

$$ \frac{dC_T}{dt} = \frac{J_{\mathrm{in}}-J_{\mathrm{out}}+R_T}{V_T} - \frac{C_T}{V_T}\frac{dV_T}{dt} $$

The first term is the net amount rate expressed per unit tissue volume. The second term accounts for the volume change. If tissue volume increases while the amount is momentarily held fixed, concentration falls; if volume decreases with amount held fixed, concentration rises. Those effects are represented by the minus sign and the factor \(C_T(dV_T/dt)/V_T\).

For constant volume, \(dV_T/dt=0\), so the conversion simplifies to \(dC_T/dt=(dA_T/dt)/V_T\). This is the familiar “amount rate divided by volume” form. It is valid as written only when the volume is constant over the interval being modeled, or when volume change is otherwise negligible for the intended calculation.

Worked Example: Constant-Volume Tissue

Consider a hypothetical tissue with constant volume \(V_T=2.0\ \mathrm{L}\). At a particular time, its total input rate is \(12\ \mathrm{mg/h}\), its total output rate is \(7\ \mathrm{mg/h}\), and there is no additional local source or loss. Find the amount rate and concentration rate.

The amount balance gives

$$ \frac{dA_T}{dt} = 12\ \mathrm{mg/h}-7\ \mathrm{mg/h} = 5\ \mathrm{mg/h}. $$

Because the volume is constant, the volume-change term is zero. Thus

$$ \frac{dC_T}{dt} = \frac{5\ \mathrm{mg/h}}{2.0\ \mathrm{L}} = 2.5\ \mathrm{mg/(L\,h)}. $$

The positive rate means the tissue concentration is increasing at that instant, at \(2.5\ \mathrm{mg/L}\) per hour. The amount and concentration rates describe the same net accumulation on different scales.

Worked Example: Expanding Tissue Volume

Suppose that at one instant a hypothetical tissue contains \(12\ \mathrm{mg}\) of drug in \(3.0\ \mathrm{L}\), so \(C_T=4.0\ \mathrm{mg/L}\). Its amount is increasing at \(6.0\ \mathrm{mg/min}\), while its volume is increasing at \(0.10\ \mathrm{L/min}\). Calculate the concentration rate.

Use the general conversion, rather than dividing the amount rate by volume alone:

$$ \frac{dC_T}{dt} = \frac{1}{V_T}\frac{dA_T}{dt} - \frac{C_T}{V_T}\frac{dV_T}{dt}. $$

Substitute the values, keeping the units explicit:

$$ \frac{dC_T}{dt} = \frac{6.0\ \mathrm{mg/min}}{3.0\ \mathrm{L}} - \frac{4.0\ \mathrm{mg/L}}{3.0\ \mathrm{L}} (0.10\ \mathrm{L/min}) = 2.0-0.1333 = 1.867\ \mathrm{mg/(L\,min)}. $$

The tissue concentration is increasing, but more slowly than the amount-rate-per-volume value of \(2.0\ \mathrm{mg/(L\,min)}\). The increasing volume reduces concentration growth by \(0.1333\ \mathrm{mg/(L\,min)}\) at this instant.

Units Must Balance in Every Term

The equation is independent of the particular units chosen, but its numerical values depend on those units. A reliable check is to verify dimensions term by term. If amount is in mg, volume in L, and time in min, then \(dA_T/dt\) has units \(\mathrm{mg/min}\); dividing by \(V_T\) gives \(\mathrm{mg/(L\,min)}\). Also, \(dV_T/dt\) has units \(\mathrm{L/min}\), so \((C_T/V_T)(dV_T/dt)\) has units \((\mathrm{mg/L})/\mathrm{L}\times\mathrm{L/min}=\mathrm{mg/(L\,min)}\). Both terms can therefore be subtracted.

Changing the concentration unit changes the numerical values of the concentration and its rate, unless the new unit is exactly equivalent on the same scale. For example, \(1\ \mathrm{mg/L}=1\ \mathrm{\mu g/mL}\), so reporting concentration in either of these equivalent units gives the same numerical value. With the time unit unchanged, the corresponding concentration-rate numbers are also the same. By contrast, \(1\ \mathrm{mg/mL}=1000\ \mathrm{mg/L}\), so a rate reported in \(\mathrm{mg/(mL\,min)}\) has a numerical value one-thousandth of the same rate reported in \(\mathrm{mg/(L\,min)}\).

Volume units must be converted consistently too. If concentration is in \(\mathrm{mg/mL}\), then an amount in mg divided by a volume in mL gives that concentration. Pairing a numerical volume in mL with a concentration in \(\mathrm{mg/L}\) without converting one of them creates a factor-of-1000 error. Likewise, changing time from hours to minutes changes rate values: a rate per minute is one-sixtieth of the corresponding rate per hour.

Worked Example: Flow Rates and Equivalent Concentration Units

For a hypothetical tissue, an input stream has flow \(0.25\ \mathrm{L/min}\) and concentration \(8\ \mathrm{\mu g/mL}\). An output stream has flow \(0.20\ \mathrm{L/min}\) and concentration \(5\ \mathrm{mg/L}\). Let the tissue volume be constant at \(0.50\ \mathrm{L}\). Find the net amount rate and the tissue concentration rate, assuming these are the only amount fluxes.

First express both stream concentrations in \(\mathrm{mg/L}\). Since \(8\ \mathrm{\mu g/mL}=8\ \mathrm{mg/L}\), the input amount rate is

$$ J_{\mathrm{in}} = (0.25\ \mathrm{L/min})(8\ \mathrm{mg/L}) = 2.0\ \mathrm{mg/min}. $$

The output amount rate is

$$ J_{\mathrm{out}} = (0.20\ \mathrm{L/min})(5\ \mathrm{mg/L}) = 1.0\ \mathrm{mg/min}. $$

Therefore \(dA_T/dt=2.0-1.0=1.0\ \mathrm{mg/min}\). With fixed tissue volume,

$$ \frac{dC_T}{dt} = \frac{1.0\ \mathrm{mg/min}}{0.50\ \mathrm{L}} = 2.0\ \mathrm{mg/(L\,min)} = 0.0020\ \mathrm{mg/(mL\,min)}. $$

The final equality follows because \(1\ \mathrm{L}=1000\ \mathrm{mL}\). It illustrates a genuine change in numerical value when converting from mg/L to mg/mL. It does not apply to mg/L versus \(\mathrm{\mu g/mL}\), which are equivalent concentration units with the same numerical scale.

Choosing and Checking a Formulation

The amount formulation and concentration formulation are alternative state descriptions, not different mass-balance principles. The amount equation is often the clearest starting point for bookkeeping: inputs, outputs, and local processes are all written as amount rates. The concentration equation is useful when concentration is the desired state or output, but requires the tissue volume and its time dependence to be specified. If \(V_T\) is known and positive, either formulation can be converted into the other using \(A_T=V_TC_T\).

The concentration equation also requires care about how each flux is expressed. An amount rate can be divided by volume to give its contribution to concentration change. If a term is already a concentration rate, dividing it by volume again would be dimensionally incorrect. When an amount rate is represented as a volume flow multiplied by a concentration, the flow and concentration units must combine to amount per time before the result is used in the amount balance.

Useful diagnostics are straightforward. Check that the volume in the concentration definition is the same volume used in the balance; check the dimensions of both concentration-rate terms; and verify the sign of the volume correction. With zero net amount rate and increasing volume, the formula must give a decreasing concentration. With constant volume, it must reduce to amount rate divided by volume. If either check fails, revisit the algebra, sign convention, or unit conversion.

A common mistake is to use \(dC_T/dt=(dA_T/dt)/V_T\) even when \(V_T\) changes. That omits the term \(-C_T(dV_T/dt)/V_T\), incorrectly attributing all amount change to concentration change. Another error is to treat equivalent units as if they rescale a value: \(1\ \mathrm{mg/L}\) and \(1\ \mathrm{\mu g/mL}\) have the same numerical value. Unit conversions should be applied to quantities and rates consistently, not inferred from the unit labels alone.

The conversion has limits as a modeling description. It assumes \(V_T(t)\) is defined and sufficiently regular to differentiate, and that the amount \(A_T\) and concentration \(C_T\) refer to the same drug pool. An average concentration \(A_T/V_T\) does not by itself specify spatial gradients, binding, or which concentration drives a particular transport process. Those require additional model structure; they do not alter this amount-to-concentration identity.

Check Your Understanding

Answer each question, then compare with the concise solutions below.

  1. A tissue has fixed volume \(4.0\ \mathrm{L}\) and net amount rate \(-8.0\ \mathrm{mg/h}\). What is its concentration rate?
  2. At an instant, \(C_T=3.0\ \mathrm{mg/L}\), \(V_T=2.0\ \mathrm{L}\), \(dA_T/dt=2.0\ \mathrm{mg/min}\), and \(dV_T/dt=0.20\ \mathrm{L/min}\). Find \(dC_T/dt\).
  3. Explain why the numerical value of \(dC_T/dt\) does not differ when concentration is reported in mg/L rather than \(\mathrm{\mu g/mL}\), provided the time unit is unchanged.
  4. A concentration rate is \(0.006\ \mathrm{mg/(mL\,h)}\). Express it in \(\mathrm{mg/(L\,h)}\).

Solutions. (1) Constant volume gives \((-8.0\ \mathrm{mg/h})/(4.0\ \mathrm{L})=-2.0\ \mathrm{mg/(L\,h)}\). (2) The amount contribution is \(2.0/2.0=1.0\ \mathrm{mg/(L\,min)}\); the volume correction is \((3.0/2.0)(0.20)=0.30\ \mathrm{mg/(L\,min)}\). Thus \(dC_T/dt=1.0-0.30=0.70\ \mathrm{mg/(L\,min)}\). (3) Since \(1\ \mathrm{mg/L}=1\ \mathrm{\mu g/mL}\), the same concentration and its rate have the same numerical value in these equivalent units when time is unchanged. (4) Since \(1\ \mathrm{mg/mL}=1000\ \mathrm{mg/L}\), the rate is \(6.0\ \mathrm{mg/(L\,h)}\).