Q=5 cm³/min = 5•10⁻⁶/60 m³/s
d=0.5 mm => r=0.25•10⁻³ m,
η=1.5•10⁻³ Pa•s,
L =0.05 m
ρ=1025 kg/m³
Poiseuille’s law
Q=π•P•r⁴/8•η• L.
P = 8• Q•η• L/ π•r⁴=
=8•5•10⁻⁶•1.5•10⁻³•0.05/60•π•(0.25•10⁻³)⁴=
=4074 Pa
P=ρgh -2392
ρgh=P+2392
h=(P+2392)/ ρg=(4074+2392)/1.025•10³•9.8=0.64 m.
If p₀=18 torr ~ 2400 Pa, then
ans. - e
An intravenous blood plasma drip enters a vein in the patient’s arm from a bag raised a height h above the
vein. If the diameter of the 5 cm long needle is 0.5 mm, use Poiseuille's law to find the height h that
results in a 5 cm3/min flow rate.
(Assume the blood pressure in the arm is 18 torr = 2392 Pa and the viscosity of the blood plasma is
1.5*10-3 Pa-s.)
a. 0.17 m
b. 1.55 m
c. 6.46 m
d. 0.52 m
e. 0.66 m
Notes: I tried to solve the problem on my own, but kept coming up with 2.46 m. If someone could just show me how to set it up correctly, that would help greatly.
1 answer
1 answer