SME
Biotransport · Problem type
Repeated doses and the steady state
Use this for “A 500 mg dose every 8 h, clearance 5 L/h. What is the average steady-state level?”
Chart
One dose every 8, 6 doses, each adding a copy of the first curve: the troughs rise toward 8.16, the steady average 12.5 is dashed
This is an example. Type your own values to start a new problem.
Cₛₛ (mg/L)
=
D (mg)
CL (L/h)
×
τ (h)
V
Volume of distribution
example · L
k
Elimination rate constant
calculated · h⁻¹
R
Accumulation factor
worked out (not typed)
Formulas
k
=
CL
V
0.1
=
5
50
Cₛₛ
=
D
CL ×
τ
12.5
=
500
5 × 8
R
=
1
1 −
e
−
k
×
τ
1.816
=
1
1 −
e
−0.1 × 8
Assumptions
•
Each dose enters one well-mixed compartment at once and is cleared at first order.
•
Doses add: the level is the sum of every dose so far, each fading as e−kt.
•
At steady state each interval clears one whole dose, so the average level stops rising.
Step-by-step
WE KNOW
D = 500 mg, τ = 8 h, CL = 5 L/h, V = 50 L
FIND
elimination rate constant (k), average steady level (Cₛₛ), accumulation factor (R)
Step 1 · Find elimination rate constant (k)
Use k = CL ÷ V
Clearance over volume is the share cleared each hour.
k
=
5
50
k
=
0.1
h⁻¹
Step 2 · Find average steady level (Cₛₛ)
Use Cₛₛ = D ÷ (CL × τ)
At
steady
state
each
interval
clears
one
dose:
the
average
level
is
D
CLτ
.
Cₛₛ
=
500
5 × 8
Cₛₛ
=
500
40
Cₛₛ = 12.5 mg/L
Step 3 · Find accumulation factor (R)
Use R = 1 ÷ (1 − e−k × τ)
Each
dose
leaves
e
−
kτ
of
the
last
behind;
the
doses
sum
to
1
1 −
e
−
kτ
of
one.
R
=
1
1 −
e
−0.1 × 8
R
=
1
1 −
e
−0.8
R
=
1
1 − 0.4493
R
=
1
0.5507
R
=
1.816
Check
Put the numbers back in. Both sides should match.
0.1
=
5
50
✓
12.5
=
500
5 × 8
✓
1.816
=
1
1 −
e
−0.1 × 8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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