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So the Reynolds number at position 1 is 5,000 and the Reynolds number at position 2 is 6,244. The
conclusion still holds.
Since the flow is turbulent, the wall shear stress is
4.11. For this problem, Bernoulli’s equation is used to relate the pressure and velocity. Choosing a
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Since the area drops by a factor of 5, the average velocity increases by a factor of 5. The radius
decrease by 5.5361/2 = 2.35. Consequently, the Reynolds number increases by 2.35 times to 2348
which places this just above the laminar limit. Assuming laminar flow through is acceptable but
there may be bursts of turbulence. If the flow were turbulent, <v2> would equal the centerline
velocity.
4.12. The integral momentum balance is
The z component of the force is the only non-zero component of the integral form of the
conservation of linear momentum. At 1 and 2 the result is:
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4.13. Assume that the flow is laminar. The centerline is horizontal so that gravity does not influence
flow and flow is steady. Equation (4.3.8) reduces to:
The control volume is applied to a region over which fully developed flow as shown in the figure
Rearranging results in a relation for the average velocity at location 2 in terms of the average
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The shear stress is zero over surfaces 1 and 2 and the pressure is zero over surface 3. Because the
detailed velocity field through the taper is not known, the shear stress cannot be computed.
4.14. Neglect any frictional resistance acting on the plunger. For this problem use Bernoulli’s
equation along the centerline and assume that flow is steady.
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4.15. The expression for the pressure drop is (from Equation 4.4.28)
The units are (kg/m3)(m/s)(1/s)m = N/m2 = Pa. To convert to mm Hg, divide by 133.3 Pa/mm Hg.
The acceleration term is then (4.12 mm Hg)cos(2πt).
The pressure drop, determined from the modified Bernoulli equation (4.4.28), neglecting viscous
losses:
The convective term cannot be neglected because of its magnitude and phase lag.
4.16. (a) Defining the control volume as:
From the integral form of the conservation of momentum,
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(b) The cardiac output is the flow rate from the heart
Human
Mouse
Cardiac output (L min–1)
5.0
0.012
Hear rate (beats per minute)
60
600
Both Re and τw are very different for the mouse and human. Thus, blood flow in the human does not
represent a dynamically scaled model.
4.18. (a) From the boundary condition at y = 0, a0 = 0. From the other two conditions:
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4.19. The flow is steady and two-dimensional. Gravity forces can be neglected. The characteristic
length in the x-direction is L and the characteristic length in the y-direction is h. The characteristic
velocity in the x-direction is U. The characteristic velocity in the y-direction can be determined from
the dimensionless form of the conservation of mass. Define the following dimensionless variables as
follows.
As a result, the conservation of mass (Table 3.1) for an incompressible fluid in terms of the
dimensionless variables is:
In order for the scaling to be appropriate, U/L and V/h must be of the same magnitude. Since
h/L<<1, equation (S4.19.1) indicates that V~Uh/L<<U. The quantity Uh/L represents the
characteristic velocity in the y-direction.
With this information about the scaling of velocity in the y-direction, the x-component of the Navier-
Stokes equation (equation (3.3.26))
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The first term in parentheses on the left hand side of equation (S4.19.4) is the Reynolds number.
Since we assume that the Reynolds number is small (order 1 or less), then the left hand side can be
neglected because these terms are multiplied by h/L. Further the second derivative of vx with respect
to x can be neglected because the terms are multiplied by (h/L)2. As a result, the dimensional form
of the x-component of the Navier-Stokes equation simplifies to
For the y-component of the Navier-Stokes equation, we have
The let-hand side of equation (S4.19.7) can be neglected because the terms are multiplied by (h/L)3.
Likewise, viscous stresses due to vy can be neglected because the terms are multiplied by (h/L)2. As
a result, the y-component of the Navier-Stokes equation reduces to:
4.20.
FN=1
1+
α
2
( )
1
2
6
µ
UW
α
2
⎛
⎝
⎜ ⎞
⎠
⎟ ln
β
−2
β
−1
β
+1
⎛
⎝
⎜
⎞
⎠
⎟
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Note that the ratio value given at the end of Example 4.6 is incorrect. It should be 0.0243.
To support a net force of 44.48 N, the velocity U is
4.21. Beginning with Equation (4.7.25)
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The pressure gradient is:
The pressure is reduced by the second term on the right hand side of equation p4.12.6. The product
(vw/vxo)(L/h) is of order unity or smaller. The pressure gradient is smaller than the value for a solid
wall for 0<x<L/2 and larger for L/2<x<L.
4.22. This problem is similar to the problem for peristaltic pumping in a cylindrical tube, which was
presented in section 4.8 of the text. Here, Cartesian coordinates (x,y) are used, with x-velocities vx
and Vx in the wave (moving) and fixed frames of reference, respectively. The equation of motion in
the moving frame is
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We now take the time averages of equations (S4.22.7) and (S4.22.8) over the period T of one
peristaltic contraction. The involves the integrals
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Thus there are positive and negative values of vx, and flow reflux occurs. After considerable
manipulation, we find a second condition in which reflux occurs,
which has velocity profile
