SME
Biotransport · Problem type
A constant infusion
Use this for “What infusion rate holds 10 mg/L when clearance is 5 L/h, and how long until the level is 90% of the way there?”
Chart
C(12) ≈ 6.988
This is an example. Type your own values to start a new problem.
Cₛₛ (mg/L)
=
R (mg/h)
CL (L/h)
k (h⁻¹)
=
CL (L/h)
V (L)
t₉₀ (h)
=
ln
10
k (h⁻¹)
R is the infusion rate and V the volume of distribution; t₉₀ is how long the level takes to get 90% of the way to Cₛₛ.
t½
Half-life
calculated
t
Time since the start
example
C
Concentration at t
calculated
Formulas
Cₛₛ
=
R
CL
10
=
50
5
k
=
CL
V
0.1
=
5
50
t
1
2
=
ln(2)
k
6.931
=
ln(2)
0.1
t₉₀
=
ln(10)
k
23.03
=
ln(10)
0.1
C
=
Cₛₛ
×
(1
−
e
−
k
×
t
)
6.988
=
10
×
(1
−
e
−0.1 × 12
)
Assumptions
•
The drug runs at a steady rate into one well-mixed compartment and is cleared at first order.
•
The level rises as C = Cₛₛ(1 − e−kt) toward the steady level, where infusion in equals clearance out.
•
Each half-life closes half the gap that is left, so 90% takes ln 10 ÷ ln 2 = 3.32 half-lives.
Step-by-step
WE KNOW
R = 50 mg/h, CL = 5 L/h, V = 50 L, t = 12 h
FIND
steady level (Cₛₛ), elimination rate constant (k), half-life (t½), time to 90% (t₉₀), concentration at t (C)
Step 1 · Find steady level (Cₛₛ)
Use Cₛₛ = R ÷ CL
The level stops rising when clearance takes out the drug as fast as it runs in.
Cₛₛ
=
50
5
Cₛₛ = 10 mg/L
Step 2 · 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 3 · Find half-life (t½)
Use t½ = ln(2) ÷ k
Half
the
gap
is
closed
when
e
−
kt
=
1
2,
so
kt
=
ln
2.
t
1
2
=
ln(2)
0.1
t
1
2
=
0.6931
0.1
t
1
2
=
6.931
h
Step 4 · Find time to 90% (t₉₀)
Use t₉₀ = ln(10) ÷ k
At
90%
a
tenth
of
the
gap
is
left:
e
−
kt
=
1
10,
so
kt
=
ln
10.
t₉₀
=
ln(10)
0.1
t₉₀
=
2.303
0.1
t₉₀
=
23.03
h
Step 5 · Find concentration at t (C)
Use C = Cₛₛ × (1 − e−k × t)
The
share
of
the
way
to
the
steady
level
grows
as
1
−
e
−
kt
.
C
=
10
×
(1
−
e
−0.1 × 12
)
C
=
10
×
(1
−
e
−1.2
)
C
=
10
×
(1
−
0.3012)
C
=
10
×
0.6988
C
=
6.988
mg/L
Check
Put the numbers back in. Both sides should match.
10
=
50
5
✓
0.1
=
5
50
✓
6.931
=
ln(2)
0.1
✓
23.03
=
ln(10)
0.1
✓
6.988
=
10
×
(1
−
e
−0.1 × 12
)
✓
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Repeated doses and the steady state
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An oral dose: peak time and peak level
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