Homework Solutions Chapter 5
8. In the figure below a tube with modulus of elasticity of 1 MPa is anchored between two stationary
ports and injected with a volume of water. The wall thickness of the tube is 1 mm. Estimate the vessel
compliance and the volume of water required to raise the pressure by 6 kPa.
3
C=
2πro 3
Eh o
 0.5 
2 

 100 
=
=
1x106 1/1000 
dV
dA
L =
C=
=
dP
dP
> restart;C:= evalf(2*Pi*(0.5/100)^3/1e6/(1/1000),2);
C = 6.28 x 10 -9
> dV:= evalf(C*6e3*0.25/1e-6*ml,2);
DV:= 9.42 ml
>
Answer: C = 0.78 x 10-9 m2/Pa Volume := 1.2 ml
9. Carotid artery modulus of elasticity is reported by Lin et al. as 3.12 x 105 N/m2 and 4.18 x 105 N/m2 in
carotid arterial atherosclerosis, compared to a value of 2.34 x 105 N/m2 in the control group. Estimate
the compliance for both groups. Take the mean diameter of the carotid artery to be 6.6 mm. Take the
wall thickness to be 0.3 mm in the atherosclerotic group and 0.25 mm in the control group.
> restart;C:= evalf(2*Pi*(3.3/1000)^3/4.18e5/(.30/1000),2);
C = 1.8 x 10-9
> restart;C:= evalf(2*Pi*(3.3/1000)^3/2.34e5/(.25/1000),2);
C= 3.86 x 10-9
Answers: Compliance was 1.8 x 10-9 m2/Pa in the atherosclerotic group and 3.86 x 10-9 m2/Pa in the
control group.
10. Estimate the pulse wave velocities in the carotid arteries in problem 9. Take the density of blood to
be 1060 kg/m3.
10. Estimate the pulse wave velocity in the carotid arteries in problem 9. Take the blood density to be
1060 kg/m3.
.
>
E1:=4.18E5;E2:=2.34E5;h1:=0.3/1000;h2:=0.25/1000;r:=3.3/1000;rho:=1060
;
E1 := 418000.
E2 := 234000.
h1 := 0.0003000000000
h2 := 0.0002500000000
r := 0.003300000000
 := 1060
> velocity:=evalf(sqrt(E1*h1/2/r/rho),3);
velocity := 4.23
> velocity2:=evalf(sqrt(E2*h2/2/r/rho),3);
velocity2 := 2.89
Answer: pulse wave velocity for atherosclerotic artery - 4.23 m/s
Pulse wave velocity for control group - 2.89 m/s
11. The diameter of the carotid artery might changes from 6.3 mm to 7.1 mm from diastole to systole.
Find the difference in pulse wave velocity predicted by this change in radius.
> r1:=6.3/2000;r2:=7.1/2000;velocity1:=evalf(sqrt(E2*h2/2/r1/rho),3);
velocity2:=evalf(sqrt(E2*h2/2/r2/rho),3);
r1 := 0.003150000000
r2 := 0.003550000000
velocity1 := 2.96
velocity2 := 2.79
Answer: The difference in velocity between the systolic and diastolic radii is 2.96 - 2.79 m/s = 0.17 m/s
or about 6%.
Alternatively, if you use the non-control group velocity 1 is 3.95 and velocity 2 is 3.71 and the delta is .24 m/s.