Chapter 16 Page 1 of 11
Problem 2.
∆p=f(L, a, ρ, µ, ω, U)
∆p, L, a, ρ, µ, ω, U;n= 7, dimensional parameters.
Therefore,
L:C1−3C2+C3+C4= 0
M:C2= 0
t:−3C3= 0
aρU C8
2
Chapter 16 Page 2 of 11
Therefore, C10 = 0, C11 =−C12,and C9=−C12. Thus,
Π3≡aω
UC9.(3)
Next,
Therefore, we may write
∆p
ρU2=C1L
aC2
(Re)C3(St)C4.(6)
where, C1, C2, C3, C4are constants.
3
Chapter 16 Page 3 of 11
Problem 3.
Morgan and Young invoke the following assumptions: (1). A collar-like stenosis may be
approximated by a smooth, rigid, axisymmetric constriction in a long straight tube, (2).
The effect of the stenosis geometry is dominant and any influence of wall distensibility is
negligible, (3). Blood can be treated as a Newtonian fluid at the flow rates encountered
in the large arteries where stenosis commonly occur, (4). Blood flow is laminar, and (5).
Steady flow assumption is acceptable although the arterial blood flow is pulsatile.
The stenosis is shown in the following figure: The dimensional variables are designated by
Figure 1: Axisymmetric stenosis.
ˆ. The axial coordinate and velocity are ˆzand ˆu.ˆ
Uis the centerline velocity. The radial
coordinate is ˆrand the corresponding velocity component is ˆv. The dimensionless variables
are: r= ˆr/R0, z = ˆz/R0, R =ˆ
R/R0, u = ˆu/ ¯
U0, v = ˆv/ ¯
U0, U =ˆ
U/ ¯
U0,and p= ˆp/ρ ¯
U0
2,
where, ¯
U0is the average velocity in the unobstructed tube, and ˆpis the pressure and ρis
the density.
4
Chapter 16 Page 4 of 11
This approximation would enable the elimination of the pressure gradient term in both the
integral momentum and integral energy equations. Then it is possible to combine the integral
momentum and energy equations into a single equation in terms of axial velocity:
The equation (2) is subject to (i) u=Uat r= 0; (ii) u= 0 at r=R; (iii) ∂u/∂r = 0 at
r= 0; (iv) ∂3u/∂r3= 0 at r= 0; and, (v) the condition that the net flow through any cross
section must be the same for any incompressible fluid which may be expressed by:
In the above, the condition (iii) is derived from a consideration of the forces on a cylindrical
element having its axis along the tube centerline. If the pressure and the inertial forces
are to be finite as the radius of the element approaches zero, the viscous force, which is
proportional to ∂u/∂r, must approach zero. The condition (iv) is developed by eliminating
the pressure between the axial momentum and radial momentum equations and considering
the resulting equation as rapproaches zero.
Next, it is noted that at high Reynolds numbers, the profile must allow a thin region of
high shear near the wall in the converging section with a relatively flat profile in the core.
To accommodate these requirements, Morgan and Young construct a polynomial fit which
permits the shear near the wall to become large while maintaining a flat core flow. This fit
where a,b, and care unknown coefficients and λis the value of (r/R) at the juncture sepa-
rating the flat and polynomial parts of the profile. The unknown coefficients are determined
from the no slip condition along with two compatibility conditions u=Uand ∂u/∂r = 0
at (r/R) = λ. The constraint (v) enables expressing λin terms of Rand U. Thus the
polynomial fit profile for uis entirely in terms of U,r, and R. The profile is now introduced
into the equation (2). The resulting first order, non-linear ordinary differential equation is
numerically solved by assuming that Poiseuille flow prevails far upstream of the stenosis.
The solution provides the desired velocity profiles. These are plotted by Morgan and Young.
The wall shear stress is evaluated from
∂ˆrˆ
R1 + ˆ
where ˆ
R0is the slope dˆ
R/dˆzof the wall. The results are included in the paper by Morgan
and Young.
5
Chapter 16 Page 5 of 11
Problem 4.
In a pure pressure-gravity flow, the effects of friction are negligible compared with the effects
due to changes of the external pressure and the elevation in the gravity field. Lengthwise al-
terations of speed, pressure, area, etc., are brought about by changes in peand z. Changes in
pemay be brought about by : (i) active muscle tone; (ii) elastic constrictions, or sphincters,
as where veins pass from the abdominal cavity to the thoracic cavity, and in the pulmonary
system; (iii) weights, (iv) pressurizing cuffs, and (v) clamps etc.,. Changes in zare important
because (i) in the vertical position of a human being, the hydrostatic pressure exceeds the
venous and arterial pressure levels;(ii) during aircraft maneuvers; and, (iiI) during certain
phases of space flight when effective gmay be greatly increased.
Retaining only the terms in d(pe+ρgz) in the table of influence coefficients for the tube law,
the governing equations of a pure pressure-gravity flow are written as,
1−S21
α
dα
dx =−2
3α3
2dΠ
dx ,(2)
3α∗3
6
Chapter 16 Page 6 of 11
Integration of equation (9) between limits Π and Π∗corresponding to Sand 1, gives
Shapiro provides graphs of equation(10). The graphs show that increasing values of Π drive
Stowards unity and decreasing values of Π drive Saway from unity. From these, it is
concluded that: (i) Choking occurs when the value of Π continually increases with either
S < 1 or S > 1; (ii) Continuous transition through S= 1 may be achieved by means of an
increase in Π until S= 1 is reached, with dΠ/dx becoming zero exactly at S= 1 followed
then by a decrease of Π.
7
Chapter 16 Page 7 of 11
Problem 5.
To determine the volume flux for the flow with the Power-law model, consider an annu-
lar volume element of length Land thickness dr in the flow, as shown in the figure below:
r
dr
a
L
Figure 2: Representative volume element.
Let there be a constant pressure gradient, −∆p/L, across the element of length L.
Force due to pressure gradient on the annular element = +∆p
L(2πrL)dr. (1)
For a power-law fluid,
τ=µ˙γn.(6)
Therefore ˙γ=∆p
L
r
2µ1
n
.(7)
8
Chapter 16 Page 8 of 11
where u(r) is velocity component in rdirection.
From equations 7 and 8,
du
dr =−∆p
L
r
2µ1
n
.(9)
With equation 11,
Q=πZa
0
r2∆p
L
r
2µ1/n
dr (14)
=∆p
2µL1/n nπ
3n+ 1a3n+1
n(15)
9
Chapter 16 Page 9 of 11
Problem 6.
To determine the volume flux for the flow with Herschel-Bulkley model, we first note that
τ=µ˙γn+τ0, τ ≥τ0
and ˙γ= 0 , τ < τ0(1)
There is yield stress and as a consequence the flow region includes plug flow in the core as
shown in figure Let the radius of the plug flow region be rp.
Core region
a
r
p
Velocity profile
Figure 3: Representative volume element.
For r≤rp, τ(r) = τ0= constant (2)
Therefore in the core,˙γ= 0.(3)
Therefore in the core,du
dr = 0.(4)
Therefore,
u= constant (5)
=up(say) (6)
(7)
For a constant pressure gradient, −(∆p/L), across an element of length Lin the core of
region, from a force balance,
2πrpLτ0=∆p
Lπrp
2L(8)
dr =−∆p
2µL1
(r−rp)1
10
Chapter 16 Page 10 of 11
By integration,
u=−∆p
2µL1/n n
n+ 1 (r−rp)n+1
n+C. (12)
where Cis a constant of integration.
Now, at r=a, u = 0 (no-slip condition). Therefore,
Then, after integration and considerable algebra,
Q=∆p
2µL1/n nπ
n+ 1a3n+1
nhξ2(1 −ξ)n+1
n+ (1 + ξ)(1 −ξ)2n+1
n
11
Chapter 16 Page 11 of 11