Harvard-MIT Division of Health Sciences and Technology
HST.542J: Quantitative Physiology: Organ Transport Systems
Instructors: Roger Mark and Jose Venegas
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Departments of Electrical Engineering, Mechanical Engineering, and the Harvard-MIT Division of Health Sciences and Technology
6.022J/2.792J/BEH.371J/HST542J: Quantitative Physiology: Organ Transport Systems
April 1, 2004

Problem 1
A pressure wave, P i
, incident on an arterial (or bronchial) bifurcation will suffer a reflection, P r
, of magnitude
P r
≡ � =
P i
Z
L
− Z o
Z
L
+ Z o where Z o is the upstream arterial impedance, and Z
L is the impedance of the bifurcation. The impedance, Z , of a vessel of radius R may be calculated from the formulæderived in the notes:
Z
2 =
ρ
AC u
C u
≈
2 π R h E
3 where ρ is the fluid density, A is the vessel cross-section, h is the vessel wall thickness, and E is the vessel modulus of elasticity.
Figure 1:
U
1
2 R
0
U
0
2 R
1
U
1
A. Assuming a symmetric bifurcation, ρ = constant, E
0
= E
1
, h
1
R
1
= h
0
R
0
, and no increase in total cross-sectional area across the bifurcation, calculate the reflection coefficient.
C u
≈
A
1
Z
L
ρ
=
=
=
=
2 π R
3 h E
π R
2
1
+
Z l
2
1
Z l
Z l same
6.022j—2004: Solutions to Problem Set 6 2

Z
2
P
P r i
=
=
≡
ρ
AC u
ρ h E
=
2 π 2 R 5
ρ h E
π R 2 2 π R 3
� =
Z
L
− Z o
Z
L
+ Z o
=
� � 1 / 2 � � 1 / 2
1 h
1
E
1
2
R
5
1
− h
0
E
0
R
5
0
� � 1 / 2 � � 1 / 2
1 h
1
E
1
2 R
1
5
+ h
0
R
E
5
0
0
If E
1
= E
0 and h
1
R
1
= h
0
R
0
,
P r
=
P i
1 1
2 R 2
1
1 1
2 R 2
1
−
+
1
R
2
0
1
R
2
0
=
1
�
2
R
0
� 2
R
1
− 1
1
�
R
0
� 2
2 R
1
+ 1
For 2 A
1
= A
0
, R
1
1
= √
2
R
0
.
�
R
0
� 2
R
1
= 2
P r
P i
= 0
B. Suppose the vessels distal to the bifurcation are severely calcified, so that E
1
� E
0
. What will the reflection coeffecient be? Does this suggest a noninvasive method of detecting the presence of severe arterial disease?
Z
L
� Z
0
, so
P
P r i
> 1 and reflection is positive for positive wave in.
2004/24
3 6.022j—2004: Solutions to Problem Set 6

Problem 2
This problem deals with the estimation of the pressure drop to be expected across a vascular steno­ sis. Figure 2 is a sketch of the cross-section of a stenotic artery with a concentric plaque. An idealized model is shown in Figure 3, demonstrating a narrowed region followed by a sudden ex­ pansion where the fluid will generally exhibit turbulent flow. Energy will be lost in two ways: in viscous flow in the narrow region and in turbulent loss in the expansion. In this problem we will concentrate on the latter.
Figure 2: u
1
, A
1
, P
1
Figure 3: u
2
, A
2
, P
2 u
3
, A
3
, P
3
1 2 3
The first task will be to estimate the pressure drop from point 2 to point 3. Position 3 is chosen far enough downstream to be in a region of uniform flow, of velocity u
3
. In order to solve this problem we must make use of the “linear momentum theorem”.
Consider a flow field of fluid with a superimposed control volume (CV). Newton’s second law for the system included within the CV is d
=
�
� system
(1) d t
This equation states that the rate of change of the system momentum equals the sum of forces acting on the mass inside the control volume. Equation 1 may be written as: d dt
�
V
CV
ρ U dV +
�
A
CV
ρ
�
�
U
�
n d A =
�
(2)
The first term is the rate of accumulation of momentum within the control volume; the second term is the net rate of momentum flux out of the CV.
6.022j—2004: Solutions to Problem Set 6 4

Let us establish a CV for our model of the sudden expansion in Figure 3. (The volume within the dashed line in Figure 4.)
Figure 4: u
2
, A
2
, P
2 u
3
, A
3
, P
3
The fluid is incompressible, and there is no net acceleration of the mass within the CV, so the first term of Equation 2 disappears. How do we evaluate the second term?
A. Calculate the momentum flux into the CV. (Note that flux is the rate of momentum passing into the CV per unit time.) Do you agree that the answer is:
ρ u
2
( u
2
A
2
) = ρ A
2 u
2
2
Momentum flux into the CV is
ρ u
2
����
× U
2 a
2
� �� � momentum per volume flow unit volume per unit time
= ρ A
2
U
2
2
B. What is the momentum flux out of the CV at point 3?
Momentum flux out of the CV is
ρ A
3 u
2
3
Thus, the net momentum flux out of the CV is
(3)
(4)
ρ A
3 u
2
3
− ρ A
2 u
2
2
C. Use conservation of mass to relate u
3 to u
2 and the cross-sectional areas A
2 and A
3
.
Conservation of mass states that
(5)
5 u
2
A
2
= u
3
A
3 u
3
= u
2
A
2
A
3
(6)
6.022j—2004: Solutions to Problem Set 6

D. Next we need to calculate the right hand side of Equation 2. The forces acting on the CV are pressure forces. Assume that the pressure at the entrance orifice, P
2
, is equal across the entire face of the CV at position 2. What is the net force acting on the CV?
Net force from pressure would be:
P
2
A
3
− P
3
A
3
= A
3
( P
2
− P
3
) (7)
E. What is the pressure drop P
2
− P
3 as a function of the velocity u
2
, the areas A
2 and A
3
, and the fluid density?
We may rewrite Equation 5 as
ρ A
3 u
2
A
2
2
A
2
3
− ρ A
2 u
2
2
= ρ u
2
2
A
2
�
A
2
A
3
− 1
�
(8)
From (8), (7) we have
�
�
ρ u ( u ) d A
( P
2
−
P
2
P
3
) A
3
= ρ u
− P
3
= ρ u
2
2
2
2
A
2
A
2
A
3
�
A
2
�
A
3
A
2
− 1
−
�
1
�
A
3
(9)
F. Now let us focus on the entrance portion of the stenosis shown below, and consider inviscid flow (Figure 5).
Figure 5: u
1
, A
1
, P
1 u
2
, A
2
, P
2
Use the Bernoulli principle and conservation of mass to relate the pressure drop P
1
− P
2 to the characteristics of the fluid, the entrance velocity (u
1
), and the geometry.
Apply Bernoulli principle between (3) and (4):
6.022j—2004: Solutions to Problem Set 6 6

7
P
1
+
P
1
1
2
−
ρ u u
P
2
1
2
2
=
=
=
P
1
− P
2
=
P
2
+ u
1
A
1
A
2
1
2
ρ u
2
2
(continuity)
1
2
�
ρ u
2
2
��
1
ρ u
2
2
1
− u
2
1
�
A
A
1
2
� 2
− 1
�
(10)
G. If we neglect viscous losses in the region of the stenosis, what will be the total pressure drop
P
1
− P
3
? Your derivation should end up with the following:
P
1
− P
3
=
1
2
ρ u
2
1
�
A
1
A
2
− 1
� 2
P
1
− P
3
= ( P
1
− P
2
) + ( P
2
− P
3
)
From (9) and (10) we have
P
1
− P
3
= ρ u
2
2
A
2
A
3
�
A
3
�
A
2
− 1 +
1
2
ρ u
2
1
��
A
A
1
� 2
2
− 1
�
But u
2
= u
1
A
1
A
2
, and A
1
= A
3
; thus
P
1
− P
3
=
P
1
− P
3
=
1
2
ρ u
2
1
�
A
1
�
A
2
A
2
A
1
�
− 1 +
�
A
1
� 2
− 1
A
2
�
1
ρ u
2
2
1
�
A
1
A
2
− 1
� 2
(11)
H. In a vessel where u
1 is 30 cm/sec, and the ratio A
2
/ A
1 is 0.1, estimate the pressure drop in mmHg. (Remember that 1 mmHg=1330 dynes/cm
2
.) What pressure drop would be expected if the lumen diameter is reduced to 25% of its original size?
P
1
− P
3
(mmHg) =
=
1
1330
×
1
2 cm
2
× 1 × 900 sec
2
( 900 )( 81 )
= 27 .
4 mmHg
( 1330 )( 2 )
× ( 10 − 1 )
2
6.022j—2004: Solutions to Problem Set 6

If the diameter were reduced to one-fourth its original size, and area A
2 sixteenth the size of A
1
. Thus would be one-
A
2
A
1
=
1
16
( 6 .
25% )
Using the same formula:
P
1
− P
3
=
=
1
1330
×
1
2
× 1 × 900 × ( 16 − 1 )
2
( 900 )( 225 )
=
( 1330 )( 2 )
76 mmHg
Note therefore the rapid increase in pressure drop across a stenosis as the area ratio drops below 10%. A drop in diameter by a factor of four certainly warrants the term “critical” stenosis.
Realize, of course, that in real life as the stenosis increases, the flow velocity u
1 will in
general not be maintained, but will drop as blood is shunted elsewhere.
P
1
− P
3
(mmHg) =
1
�
1330
×
1
2
��
× 1 × 30
2
�
×
0 .
34
�
A
1
A
2
− 1
� 2
If S ≡ % stenosis =
�
1 − A
2
A
1
�
× 100
Then
�
A
1
A
2
�
− 1 =
S
100 − S
(See Figure 6.)
6.022j—2004: Solutions to Problem Set 6 8

Figure 6:
40
30
20
10
2004/234
9
S (%)
0
10
40
50
60
70
80
90
10 20 30 40 50 60 70
80 90 100
S
100 − S
0
.11
.67
1.0
1.5
2.3
4
9
�
S
100 − S
� 2
0
.012
.45
1.0
2.25
5.29
16
81
� P (mmHg)
�
S
( 0 .
34 )
� 2
100 − S
0
.004
.15
.34
.76
1.8
5.4
27.5
6.022j—2004: Solutions to Problem Set 6

Problem 3
Engineers for a medical products company have developed an inexpensive, hand-held device for monitoring peak expiratory flow rates in asthmatics. The device, shown in Figure 7, consists of a rigid cylinder with a long, narrow slit down the side and a close-fitting, spring-loaded piston diaphragm. Expired air enters the chamber of the peak flow meter from the left through the mouth­ piece, then exits via the slit at a uniform velocity. The length of the slit available for flow, x, is equal to the displacement of the piston. Gas exits from the slit in the form of a jet into the atmosphere.
Assuming the flow to be quasi-steady and inviscid, that the piston moves without friction, and that the displacement is linearly proportional to the force F acting on it:
F = kx answer the following questions. You may take the following as given:
• density of expired air, ρ
• piston diameter, D
• slit height, h ( h ≪ D )
• volume flow rate, Q
• spring constant, k
• exposed slit length, x
Q h
Slit
Figure 7: x
Piston
Spring f = kx
Q
A =
π
D
2
/4
Outlet = hx cm
2
, h << D
P=0
6.022j—2004: Solutions to Problem Set 6 10

Note: The x-direction linear momentum equation cannot be used in this problem because the problem is under-specified. In particular, the restraining force on the cylinder to hold it against the mouth is not given.
A. Given the volume flow rate entering the device, Q, and assuming the piston to be displaced a distance x but non-accelerating, what is the velocity of gas exiting through the slit?
Conservation of mass: area of slit
Q
=
= v slit
= hx v slit
·
Q hx hx
B. What is the pressure inside the chamber of the flowmeter, p o
?
Because we’re told that the piston is not accelerating, we know that the forces on the piston must cancel.
P
0
A kx
P
0
A = kx
π D
2
A =
P
0
=
4 kx
=
A
4kx
π D 2
11
Can also use Bernoulli across slit:
P
0 v
0
P v
1 slit
P
0
+
1
ρv
2
But P
2
0
1
=
=
P
0
=
P
0 and assume
1
2
1
+
ρv
1
2
ρv
2 slit
2 slit
= v slit
1 Q
2
2
ρ h 2 x 2
≫ v
0
6.022j—2004: Solutions to Problem Set 6

C. What is the relationship between the displacement of the piston, x, and the other given parameters?
Use Bernoulli across slit:
P
0
+
1
2
ρv
2
0
But P out
P
0
=
=
P out
+
1
2
ρv
2 slit
0 and assume v slit
≫ v
0
=
1
2
ρv
2 slit
=
1
2
ρ
Q
2 h 2 x 2
=
4kx
π D 2
So x
3 x
=
=
π ρ D
2
Q
2
8 kh 2
�
π ρ D
2
Q
2 � 1 / 3
8 kh 2
(from B)
If Bernoulli was used in part B, then here we must balance the forces:
1 Q
ρ
2 h 2 x
2
2
·
P
0
A
π D
2
4
=
= kx kx x
3 = x =
π ρ D
2
Q
2
8 kh 2
�
π ρ D
2
Q
2 � 1 / 3
8 kh
2
2004/448-solution
6.022j—2004: Solutions to Problem Set 6 12

Problem 4
You are conducting experiments in your laboratory on a new artificial lung which consists of two parallel membranes separated by a small distance d. (See Figure 8.) Blood flows through the gap under steady laminar conditions due to a pressure gradient, − d p / dx, and oxygen transport occurs across the membranes. Assume blood to be a Newtonian fluid of viscosity µ .
Figure 8:
O
2 direction of flow, Q y
+d /2 d z d x d y x
–d /2
O
2
A. Using the coordinate system shown in Figure 8, what are the boundary conditions on the fluid velocity v x
( y ) and the shear rate γ ˙ =
∂v x
( y )
?
∂ y
1: No-slip condition: V x
= 0 at y = ± d / 2
2: Shear rate zero at axis of symmetry: dV x
/ dy = 0 at y = 0
B. Set up the differential equation of motion for the control volume dx dy dz.
For steady laminar flow, the rate of change of momentum is zero, hence the sum of all forces acting on the control volume must be zero.
13
Force due to pressure gradient: F p
Force due to viscous shear: F
−
F p
− d τ ( y ) − d d v
+ F v
P ( x ) x or µ d
2
V x
( y ) dy 2
= −
= d d d x
( P ( x ) dydz ) dx = dy
[ τ ( y ) dxdz] dy =
− d P ( x ) d x
− d τ ( y ) dy dxdydz dxdydz
= 0
= 0 where
− d P d x
( x )
= constant and τ ( y ) = µ dV x
( y ) d y
= d P ( x ) dx
6.022j—2004: Solutions to Problem Set 6

C. Solve the equation (subject to the boundary conditions) for the blood velocity in the x-direction as a function of y. Sketch the result.
Solving the differential equation by integrating once: dV x dy
1 d P
= ·
µ dx
· y + c
1
Boundary condition 2 requires c
1
= 0.
Integrating again:
V x
( y ) =
1
2 µ
· d P d x
· y
2 + c
2
Using boundary condition 1 y
+d /2
V x
( y ) =
1
�
− d P
�
��
·
2 µ dx d
� 2
− y
2
2
� x
–d /2
D. Using the result of (C), determine the flow rate per unit width, Q / dz, in terms of the pressure gradient − d p / dx, the gap width d, and the blood viscosity µ .
+d
–d
/2
/2 d z d y
Q
Q dz
Q dz
= 2
� d / 2
V x
( y ) dydz
0
= 2
� d / 2
0
1
2 µ
�
− d P
� �� dx d
� 2
2
�
− y
2 dy
= d
3
12 µ
�
− d P
� dx
E. What is the resistance, R, per unit width of the structure?
Resistance =
� P 12 µ� x 12 µ L
Q
= d
3
= d
3 dz where � x = L
F. What is the shear rate at the membrane surface in terms of the average flow velocity, V ?
The shear rate at the membrane surface is given by
6.022j—2004: Solutions to Problem Set 6 14

γ ˙ | wall
=
=
∂ V x �
�
∂ y �
� y d / 2
1 ∂ P d
µ ∂ x 2
We can obtain the average velocity from total flow and cross-sectional area:
Q Q
A
= d
= d
2
12 µ
�
− d P
� dx
Thus
6
γ V d
2004/83
15 6.022j—2004: Solutions to Problem Set 6