Why Equilibration Has a Characteristic Time
A perfusion-limited tissue is assumed to exchange drug rapidly enough that tissue concentration remains in equilibrium with the blood leaving the tissue. The tissue does not necessarily reach its equilibrium concentration immediately: the rate of approach depends on how quickly blood delivers drug, how much tissue must equilibrate, and how much drug partitions into the tissue. A characteristic time scale makes these effects explicit.
The half-time discussed here is an equilibration half-time, not an elimination half-life. It describes how long it takes for the difference between the current tissue concentration and its equilibrium value to be reduced by one-half under a specified, constant arterial concentration.
From Tissue Mass Balance to the Time Constant
Assume a well-mixed tissue with fixed volume \(V_T\), constant blood flow \(Q_T\), and constant partition coefficient \(K_{T:B}\). Let \(A_T\) denote the amount of drug in the tissue, and let \(C_T\), \(C_A\), and \(C_V\) denote tissue, arterial-blood, and venous-blood concentrations. Use a consistent blood concentration basis throughout. Under the perfusion-limited assumption, venous blood is in equilibrium with the tissue, so \(C_V=C_T/K_{T:B}\). With no local input or elimination, the tissue mass balance is \(dA_T/dt=Q_T(C_A-C_V)\), and \(A_T=V_TC_T\).
If \(C_A\) is held constant, substitution gives the concentration equation below. The constant target concentration is \(K_{T:B}C_A\); the tissue concentration approaches this target exponentially.
The solution follows by subtracting the target concentration from both sides. Define the concentration gap \(D(t)=K_{T:B}C_A-C_T(t)\). Then \(dD/dt=-D/\tau_T\), whose solution is \(D(t)=D(0)e^{-t/\tau_T}\). Thus:
At \(t=\tau_T\), the remaining gap is \(e^{-1}\), or about 36.8%, of its initial value. To find the half-time, set the remaining gap equal to one-half its initial value: \(e^{-t_{1/2,T}/\tau_T}=1/2\). Taking logarithms gives the stated result.
This definition concerns the gap to equilibrium, whether tissue concentration is rising or falling. It does not mean the tissue concentration itself is halved. For example, a tissue starting above its target loses half its excess above the target after one half-time.
Calculating a Half-Time and a Tissue Profile
Worked Example: Time Constant and Half-Time
Consider a hypothetical tissue with \(Q_T=0.90\) L/h, \(V_T=0.30\) L, and \(K_{T:B}=4.0\). Calculate its time constant and equilibration half-time. The units are compatible: volume times a dimensionless partition coefficient divided by volume per time gives time.
Time constant.
Half-time.
The equilibration half-time is approximately 0.924 hours, or \(0.9241\times60=55.45\) minutes. Under the stated assumptions, after that time the tissue-to-target concentration gap is half the initial gap.
Worked Example: Time to Reach a Target Fraction of Equilibrium
For a hypothetical tissue, let \(Q_T=1.20\) L/h, \(V_T=0.45\) L, and \(K_{T:B}=2.5\). Arterial concentration is held at \(C_A=1.60\) mg/L, and the initial tissue concentration is \(C_T(0)=0.40\) mg/L. Find the time at which the tissue has completed 90% of its change toward equilibrium.
The equilibrium tissue concentration under this constant arterial concentration is \(K_{T:B}C_A=(2.5)(1.60)=4.00\) mg/L. Completing 90% of the change means 10% of the initial gap remains. First calculate the time constant:
The gap declines in proportion to \(e^{-t/\tau_T}\). Set that fraction to 0.10 and solve:
As a concentration check, 90% of the initial 3.60 mg/L gap is 3.24 mg/L. The tissue concentration at that point is \(4.00-3.24=0.760\) mg/L. It has moved 90% of the way from 0.40 mg/L toward 4.00 mg/L, as required. The time is longer than one half-time because reducing the remaining gap to 10% takes \(\ln(10)/\ln(2)\), or about 3.322, half-times.
How Flow, Volume, and Partitioning Affect the Time Scale
The formula \(\tau_T=V_TK_{T:B}/Q_T\) gives a direct comparison when the other quantities are held fixed. Increasing flow shortens the time constant because more blood passes through the tissue per unit time. Increasing tissue volume lengthens it because more tissue capacity must equilibrate. Increasing partitioning also lengthens it: at equilibrium, more drug is present in tissue relative to blood, so a given flow takes longer to change the tissue concentration by the same fraction of its gap. The half-time changes in the same proportions because it is a fixed multiple of \(\tau_T\).
Worked Example: Changing One Parameter at a Time
Use the hypothetical baseline values \(Q_T=0.60\) L/h, \(V_T=0.24\) L, and \(K_{T:B}=3.0\). Calculate the baseline half-time, then examine what happens when each of the three parameters is doubled separately, with the others unchanged.
The baseline time constant and half-time are:
If flow is doubled to 1.20 L/h, then \(\tau_T=(0.24)(3.0)/1.20=0.60\) h and \(t_{1/2,T}=0.4159\) h. Doubling flow halves the half-time. If volume is doubled to 0.48 L, then \(\tau_T=(0.48)(3.0)/0.60=2.40\) h and \(t_{1/2,T}=1.663\) h. If partitioning is doubled to 6.0, then \(\tau_T=(0.24)(6.0)/0.60=2.40\) h and \(t_{1/2,T}=1.663\) h. Doubling either volume or partitioning doubles the half-time in this model.
These calculations isolate one parameter at a time; they do not imply that the parameters are physiologically independent in every application. They show how each enters the model’s characteristic time when the remaining quantities are fixed.
Assumptions, Diagnostics, and Alternatives
The half-time is useful only to the extent that the model assumptions describe the situation being interpreted. The calculation assumes constant \(Q_T\), \(V_T\), and \(K_{T:B}\); a well-mixed tissue; rapid tissue–blood equilibration; and no local formation, elimination, or additional input. The constant-target exponential solution also assumes \(C_A\) is constant during the interval. In a whole-body model, arterial concentration often changes as other tissues distribute drug or elimination occurs. The tissue equation can still be used, but its target \(K_{T:B}C_A(t)\) is then moving, so the concentration profile need not be a single exponential approaching a fixed value.
Useful checks include verifying that \(V_TK_{T:B}/Q_T\) has units of time, confirming that predicted concentrations approach \(K_{T:B}C_A\) under constant \(C_A\), and checking that the distance to that target shrinks exponentially. A fitted profile that has multiple phases, persistent curvature on a plot of the log concentration gap against time, or a target that shifts as arterial concentration changes may not support one meaningful half-time. Such patterns can reflect changing inputs or parameters, or failure of the well-mixed, perfusion-limited representation. A single tissue half-time should not be forced onto a profile whose driving concentration is changing substantially.
The time constant and half-time are two ways of reporting the same exponential response: \(\tau_T\) is the time for the gap to fall to about 36.8%, while \(t_{1/2,T}\) is the time for it to fall to 50%. Neither specifies the equilibrium concentration; that is determined by partitioning and arterial concentration. When arterial concentration varies, solving the coupled tissue-and-blood differential equations is generally more appropriate than applying the fixed-target half-time formula to the observed profile. If rapid tissue–blood equilibration is not credible, a model with additional tissue exchange structure may be needed rather than merely changing the half-time.
Check Your Understanding
Try each question, then compare your reasoning with the solutions below.
- A tissue has \(V_T=0.20\) L, \(K_{T:B}=5.0\), and \(Q_T=1.0\) L/h. Calculate \(\tau_T\) and \(t_{1/2,T}\).
- After one time constant, what fraction of the original concentration gap remains? What fraction remains after two half-times?
- With \(V_T\) and \(K_{T:B}\) fixed, what happens to the half-time if \(Q_T\) is tripled?
- Why is it generally incorrect to interpret \(t_{1/2,T}\) as the time for tissue concentration itself to fall by half?
- What feature of the arterial concentration makes the fixed-target exponential solution unsuitable as a complete description of the tissue profile?
Solutions. (1) \(\tau_T=(0.20\ \mathrm{L})(5.0)/(1.0\ \mathrm{L/h})=1.00\) h, so \(t_{1/2,T}=(1.00\ \mathrm{h})(0.6931)=0.6931\) h. (2) After one time constant, the gap fraction is \(e^{-1}=0.3679\). After two half-times it is \((1/2)^2=0.25\). (3) Since \(\tau_T=V_TK_{T:B}/Q_T\), tripling flow reduces the half-time to one-third its original value. (4) The half-time applies to the difference between tissue concentration and its target, not necessarily to tissue concentration itself; the target may be nonzero and the tissue may start above or below it. (5) If \(C_A\) changes over time, the target \(K_{T:B}C_A(t)\) also moves. The concentration is then driven by a changing target, rather than approaching one fixed value according to a single exponential.