SME
Biotransport
Blood flow
Use this for “How much blood flows through a 2 mm-radius artery 10 cm long under a 100 Pa pressure drop?”
Diagram
vₘₐₓ = 2 vavg = 2 × 0.1429 = 0.2857 m/s at the centre, 0 at the wall.
Not to scale: the vessel is far longer than it is wide.
This is an example. Type your own values to start a new problem.
Q (mL/min)
=
6π
×
ΔP (Pa)
×
(
r (mm)
)
4
8
×
μ (mPa·s)
×
L (cm)
R (Pa·s/m³)
=
ΔP (Pa)
Q (mL/min)
Formulas
Q
=
6π × ΔP ×
r
4
8 ×
μ
×
L
108
=
6π × 100 ×
2
4
8 × 3.5 × 10
R
=
ΔP
Q
÷ (6 ×
10
7
)
5.57 ×
10
7
=
100
108 ÷ (6 ×
10
7
)
Assumptions
•
Steady laminar flow of a Newtonian fluid through a rigid, straight tube (Poiseuille’s law).
•
Blood is close to Newtonian in vessels wider than about 0.5 mm; μ = 3.5 mPa·s.
•
Q goes as r⁴: halving the radius cuts the flow to 1/16 at the same ΔP.
•
In SI, Q = πΔPr⁴ ÷ (8μL). With r in mm, L in cm and μ in mPa·s it comes out in 10⁻⁷ m³/s, and 10⁻⁷ m³/s is 6 mL/min: hence 6π.
Step-by-step
WE KNOW
ΔP = 100 Pa, r = 2 mm, L = 10 cm, μ = 3.5 mPa·s
FIND
blood flow (Q), resistance (R)
Step 1 · Find blood flow (Q)
Use Q = (6π × ΔP × r⁴) ÷ (8 × μ × L)
Poiseuille’s
law,
πΔPr⁴
8
μL
;
the
6
turns
mm,
cm
and
mPa·s
into
mL/min
(
10
-7
m
3
/s
is
6
mL/min).
Q
=
6π × 100 ×
2
4
8 × 3.5 × 10
Q
=
107.7
mL/min
Step 2 · Find resistance (R)
Use R = ΔP ÷ (Q ÷ (6 × 10⁷))
Resistance
is
the
pressure
drop
per
unit
of
flow,
with
Q
turned
into
m
3
/s.
R
=
100
107.7 ÷ (6 ×
10
7
)
R
=
100
1.795 ×
10
-6
R
=
5.57 ×
10
7
Pa·s/
m
3
Check
Put the numbers back in. Both sides should match.
107.7
=
6π × 100 ×
2
4
8 × 3.5 × 10
✓
5.57 ×
10
7
=
100
107.7 ÷ (6 ×
10
7
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.