BSL 2B3

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```BSL 2B.3 Laminar flow in a narrow slit
(a) A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls a
distance 2B apart. It is understood that B &lt;&lt; W, so that “edge effects” are unimportant. Make
a differential momentum balance, and obtain the following expressions for the momentum −
flux and velocity distributions:
⎛ P − P ⎞
τ xz = ⎜ 0 L ⎟x
⎝ L ⎠
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2 ⎤
⎡
(P0 − PL )B2 ⎢ ⎛ x ⎞ ⎥
vz =
1− ⎜ ⎟
2&micro;L
⎢⎣ ⎝ B ⎠ ⎥⎦
In these expressions P = p + ρgh = p − ρgz.
€ average velocity to the maximum velocity for this flow?
(b) What is the ratio of the
(c) Obtain the slit analog of the Hagen − Poiseuille equation.
(d) Draw a meaningful sketch to show why the above analysis is inapplicable if B = W.
(e) How can the result in (b) be obtained from the results of &sect;2.5?
Fig. 2B.3 Flow through a slit, with B &lt;&lt; W &lt;&lt; L.
Solution:
a)
Since the fluid flow is in the z–direction, vx = 0, vy = 0, and only vz exists. Further, vz is
independent of z and it is meaningful to postulate that velocity vz = vz(x) and pressure p =
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p(z). The only nonvanishing components of the stress tensor are τxz = τzx, which depend only
on x.
Consider now a thin rectangular slab (shell) perpendicular to the x−direction extending a
distance W in the y−direction and a distance L in the z−direction. A 'rate of z−momentum'
balance over this thin shell of thickness Δx in the fluid is of the form:
(
)
(
)
(LW) φ xz x − φ xz x +Δx + (WΔx) φ zz z=0 − φ zz z= L + (ΔxLW)ρg = 0
Dividing the above equation by LWΔx, and the limit taken as Δx approaches zero, we get
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∂φ xz φ zz z=0 − φ zz z= L
−
− ρg = 0
∂x
L
At this point we have to write out explicitly what components φ xz and φ zz are, making use of
the definition of φ in Eqs. 1.7-1 to 3 and the expressions for τxz and τzz in Table B.1. This
ensures that we do not miss out any of the forms of momentum transport. Hence, we get
φ xz = τ xz + ρv x v z = τ xz
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φ zz = p + τ zz + ρv z v z = p
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In accordance with the postulates that vz = vz(x), vx = 0, vy = 0, and p = p(z), we see that (i)
since vx = 0, the term ρvxvz is zero; (ii) since vz = vz(x), the term τzz is zero; (iii) since vz =
vz(x), the term ρvzvz is the same at both ends of the slit, so the convective terms are cancel
out.
dτ xz p0 − p L
−
− ρg = 0
dx
L
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dτ xz p0 − p L
=
+ ρg
dx
L
The right – hand side may be compactly and conveniently written by introducing the modified
pressure P, which is the sum of the pressure and gravitational terms. The general definition of
the modified pressure is P = p + ρgh, where h is the distance upward (in the direction opposed
to gravity) from a reference plane of choice. Since the z−axis points downward in this
problem, h = − z and therefore P = p − ρgz. Thus, P0 = p0 at z = 0 and PL = pL − ρgL at z = L
giving p0 − pL + ρgL = P0 − PL.
dτ xz ⎛ P0 − PL ⎞
= ⎜
⎟
dx ⎝ L ⎠
Integration leads to the following expression for the shear stress distribution
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⎛ P − P ⎞
τ xz = ⎜ 0 L ⎟x + C1
⎝ L ⎠
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It is worth noting that the two last equations apply to Newtonian and non – Newtonian fluid.
The constant of integration C1 is determined later using boundary conditions.
Substituting Newton's law of viscosity for τxz in above equation gives
−&micro;
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dv z ⎛ P0 − PL ⎞
= ⎜
⎟ x + C1
dx ⎝ L ⎠
The above first – order differential equation is simply integrated to obtain the following
velocity profile
⎛ P − P ⎞
C
v z = −⎜ 0 L ⎟ x 2 − 1 x + C2
&micro;
⎝ 2&micro;L ⎠
Boundary conditions
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BC1:
at x = B;
vz = 0
BC2:
at x = − B;
vz = 0
Using these, the integration constants may be evaluated as C1 = 0 and C2 = (P0 – PL)B2/(2&micro;L).
Substituting C1 = 0 in the shear stress equation, the final expression is found to be linear as
given by
⎛ P − P ⎞
τ xz = ⎜ 0 L ⎟x
⎝ L ⎠
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Further, substitution of the integration constants gives the final expression for the velocity
profile as
2 ⎤
⎡
(P0 − PL )B2 ⎢ ⎛ x ⎞ ⎥
vz =
1− ⎜ ⎟
2&micro;L
⎢⎣ ⎝ B ⎠ ⎥⎦
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It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian
fluid in a plane narrow slit is parabolic.
b)
The maximum velocity occurs at x = 0 (where dvz/dx = 0). Therefore,
v z, max =
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(P0 − PL )B2
2&micro;L
⎡ ⎛ ⎞2 ⎤
x
v z = v z, max ⎢1− ⎜ ⎟ ⎥
⎢⎣ ⎝ B ⎠ ⎥⎦
Copyright &copy; 2005 by Apolinar Picado
The average velocity is obtained by dividing the volumetric flow rate by the cross – sectional
area as shown below.
B
W
∫ ∫ v (x)dydx
v z = −B B0 Wz
∫ −B ∫ 0 dydx
vz =
v z, max B⎛ x 2 ⎞
∫ ⎜1− ⎟dx
B 0 ⎝ B2 ⎠
vz =
v z, max ⎛ 2 ⎞
⎜ B⎟
B ⎝ 3 ⎠
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vz
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v z, max
=
2
3
Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a
narrow slit is 2/3.
This seems reasonable since v z /vz,max = &frac12; for flow in a circular pipe. For a slit a larger
cross − sectional area carries fluid flowing at the larger velocity than for flow in a circular
pipe. So v z /vz,max for slit &gt; v z /vz,max for a circular pipe.
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c)
€The mass rate of flow
€ is the product of the density ρ, the cross − sectional area (2BW) and the
average velocity v z .
w = ρ (2BW) v z
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w = ρ (2BW)
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2 (P0 − PL )B2
3
2&micro;L
2 (P0 − PL )B3 Wρ
w=
3
&micro;L
The flow rate vs. pressure drop (w vs. ΔP) expression above is the slit analog of the Hagen –
Poiseuille equation (originally for circular tubes). It is a result worth noting because it
provides the starting point for creeping flow in many systems (e.g. radial flow between two
parallel circular disks; and flow between two stationary concentric spheres).
d)
The above analysis is not applicable if B = W, because of the presence of a wall at y = 0 and y
= B would cause vz to vary significantly in y in addition to x, then vz = vz (x, y).
Copyright &copy; 2005 by Apolinar Picado
If W = 2B, then a solution can be obtained for flow in a square duct.
e)
In Eq. 2.5-20, set both viscosities equal to &micro;, p0 − pL = P0 – PL, and set b equal to B.
vz =
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2 (P − P )B2 2 (P0 − PL )B2
vz = ⋅ 0 L
=
2
3&micro;L
3
2&micro;L
2
v z = v z, max
3
vz
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(P0 − PL )B2 ⎛ 8 ⎞ (P0 − PL )B2
⎜ ⎟ =
12&micro;L ⎝ 2 ⎠
3&micro;L
v z, max
=
2
3
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Copyright &copy; 2005 by Apolinar Picado
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