Solution for Homework 1
1.
2. P gh , is the density of mercury
In SI unit:
120mmHg=13.5x103x9.8x0.12=1.59x104 N/m2
80mmHg=13.5x103x9.8x0.08=1.06x104 N/m2
In cm-g-s system:
120mmHg=13.5x980x12=1.59x105 dyn/cm2
80mmHg=13.5x980x8=1.06x105 dyn/cm2
3.
( )
( )

Solution for Homework 3
1.
According to Moens-Kortewegs formula on wave speed,
Eh
c=
2!R
!
D
2 !10 3
2
!
R
2 = 4.5 !10 5 Pa
E = c2
= 1.52
h
0.05D
The increased wave speed reflects the increased Youngs modulus in the artery, indicates the
hardening of the

Solution for homework #5
1. From Gorlins equation, Q = EOA 51.6 !p = 2.6 * 51.6 * 4 = 268ml/s
The performance index=EOA/Actual Cross-section area=
2.6
= 0.34
! 3.12
4
To calculate the Reynolds number, we need to use the effective orifice area and the effe

Biomedical Fluid Mechanics
Homework
Page 1 of 4
_
Homework #2
Due 9/24/10 Friday
1. Blood is ejected from left ventricle to aorta. For the aorta with a diameter of 2.5cm and an
average velocity of ~0.25m/s, what is the Reynolds number? Is the flow turbu

Solution for homework #4
1.
We can estimate the work done by the left ventricle of the heart by the area of the P-V loop.
W ! 105mmHg 100ml = 292
lb
0.0035 ft 3 = 1.03lb " ft
ft 2
Assume that heart beat 60 times per minute.
Wday ! 86400 "1.03lb # ft = 88

Solution for Midterm Exam
1.
(a), (b), (c) and (d)
In ascending aorta, the relation between pressure gradient and flow rate does not satisfy the
Hagen-Poiseuilles equation. There are many factors that can change this relation, including,
(1) Flow in aorta

Solution to Midterm Exam II
1. Impedance mismatch means that the characteristic impedances of two adjacent blood vessels
do not match each other. There will be wave reflection from the point where there is impedance
mismatch.
Characteristic impedance of a

Solution for Homework 2
1.
Re
VD
In cm-g-s system, 1 g / cm3 , V 25 cm / s , D 2.5 cm , 0.04
Re = 1563, the flow is laminar.
Entrance length, Le 0.06 d Re 234 cm
The entrance length is longer than human overall height. The flow is not fully developed in

Solution for Homework 1
1. (10 points)
End of systole:
P = 120mmHg-(1050kg/m3)(9.81m/s2)(30cm)(1m/100cm)(1mmHg/133.32Pa) = 96.82mmHg
End of Diastole:
P = 80mmHg-(1050kg/m3)(9.81m/s2)(30cm)(1m/100cm)(1mmHg/133.32Pa)=56.82mmHg