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BSL 2B3 - Sarec/Fiq

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<strong>BSL</strong> 2B.3 Laminar flow in a narrow slit<br />

(a) A Newtonian fluid is in laminar flow in a narrow slit formed by two parallel walls a<br />

distance 2B apart. It is understood that B


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p(z). The only nonvanishing components of the stress tensor are τxz = τzx, which depend only<br />

on x.<br />

Consider now a thin rectangular slab (shell) perpendicular to the x−direction extending a<br />

distance W in the y−direction and a distance L in the z−direction. A 'rate of z−momentum'<br />

balance over this thin shell of thickness Δx in the fluid is of the form:<br />

(LW) ( φxz − φ<br />

x<br />

xz<br />

x +Δx ) + (WΔx) ( φzz − φ<br />

z=0<br />

zz<br />

z= L)<br />

+ (ΔxLW)ρg = 0<br />

Dividing the above equation by LWΔx, and the limit taken as Δx approaches zero, we get<br />

∂φxz ∂x − φzz − φ<br />

z=0<br />

zz<br />

z= L − ρg = 0<br />

L<br />

At this point we have to write out explicitly what components φxz and φzz are, making use of<br />

the definition of φ in Eqs. 1.7-1 to 3 and the expressions for τxz and τzz in Table B.1. This<br />

ensures that we do not miss out any of the forms of momentum transport. Hence, we get<br />

φ xz = τ xz + ρv xv z = τ xz<br />

φ zz = p + τ zz + ρv zv z = p<br />

In accordance with the postulates that vz = vz(x), vx = 0, vy = 0, and p = p(z), we see that (i)<br />

since vx = 0, the term ρvxvz is zero; (ii) since vz = vz(x), the term τzz is zero; (iii) since vz =<br />

vz(x), the term ρvzvz is the same at both ends of the slit, so the convective terms are cancel<br />

out.<br />

dτxz dx − p0 − pL − ρg = 0<br />

L<br />

dτxz dx = p0 − pL + ρg<br />

L<br />

The right – hand side may be compactly and conveniently written by introducing the modified<br />

pressure P, which is the sum of the pressure and gravitational terms. The general definition of<br />

the modified pressure is P = p + ρgh, where h is the distance upward (in the direction opposed<br />

to gravity) from a reference plane of choice. Since the z−axis points downward in this<br />

problem, h = − z and therefore P = p − ρgz. Thus, P0 = p0 at z = 0 and PL = pL − ρgL at z = L<br />

giving p0 − pL + ρgL = P0 − PL.<br />

dx = P ⎛⎛<br />

0 − P ⎞⎞<br />

L<br />

⎜⎜ ⎟⎟<br />

⎝⎝ L ⎠⎠<br />

dτ xz<br />

Integration leads to the following expression for the shear stress distribution<br />

τxz = P0 − P ⎛⎛ ⎞⎞<br />

L<br />

⎜⎜ ⎟⎟ x + C1 ⎝⎝ L ⎠⎠<br />

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Copyright © 2005 by Apolinar Picado


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It is worth noting that the two last equations apply to Newtonian and non – Newtonian fluid.<br />

The constant of integration C1 is determined later using boundary conditions.<br />

Substituting Newton's law of viscosity for τxz in above equation gives<br />

−µ dv z<br />

dx = P0 − P ⎛⎛ ⎞⎞<br />

L<br />

⎜⎜ ⎟⎟ x + C1 ⎝⎝ L ⎠⎠<br />

The above first – order differential equation is simply integrated to obtain the following<br />

velocity profile<br />

v z = − P0 − P ⎛⎛ ⎞⎞<br />

L<br />

⎜⎜ ⎟⎟ x<br />

⎝⎝ 2µL ⎠⎠<br />

2 − C1 µ x + C2 Boundary conditions<br />

BC1: at x = B; vz = 0<br />

BC2: at x = − B; vz = 0<br />

Using these, the integration constants may be evaluated as C1 = 0 and C2 = (P0 – PL)B 2 /(2µL).<br />

Substituting C1 = 0 in the shear stress equation, the final expression is found to be linear as<br />

given by<br />

τxz = P0 − P ⎛⎛ ⎞⎞<br />

L<br />

⎜⎜ ⎟⎟ x<br />

⎝⎝ L ⎠⎠<br />

Further, substitution of the integration constants gives the final expression for the velocity<br />

profile as<br />

v z = (P 0 − P L )B 2<br />

2µL<br />

1− x<br />

⎡⎡ 2<br />

⎛⎛ ⎞⎞ ⎤⎤<br />

⎢⎢ ⎜⎜ ⎟⎟ ⎥⎥<br />

⎣⎣ ⎢⎢ ⎝⎝ B⎠⎠<br />

⎦⎦ ⎥⎥<br />

It is observed that the velocity distribution for laminar, incompressible flow of a Newtonian<br />

fluid in a plane narrow slit is parabolic.<br />

b)<br />

The maximum velocity occurs at x = 0 (where dvz/dx = 0). Therefore,<br />

v z,max = (P 0 − P L )B2<br />

2µL<br />

v z =v z,max 1− x ⎛⎛<br />

⎜⎜<br />

⎝⎝ B<br />

⎞⎞ ⎡⎡ 2⎤⎤<br />

⎢⎢ ⎟⎟ ⎥⎥<br />

⎣⎣ ⎢⎢ ⎠⎠ ⎦⎦ ⎥⎥<br />

Copyright © 2005 by Apolinar Picado


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The average velocity is obtained by dividing the volumetric flow rate by the cross – sectional<br />

area as shown below.<br />

v z =<br />

∫<br />

B<br />

∫<br />

∫<br />

W<br />

−B 0<br />

B<br />

−B<br />

v z = v z,max<br />

B<br />

v z = v z,max<br />

B<br />

v z<br />

v z,max<br />

= 2<br />

3<br />

∫<br />

v z(x)dydx<br />

∫<br />

0<br />

W<br />

⎛⎛<br />

dydx<br />

1− x2<br />

B 2<br />

B<br />

0 ⎜⎜<br />

⎝⎝<br />

2<br />

3 B<br />

⎛⎛ ⎞⎞<br />

⎜⎜ ⎟⎟<br />

⎝⎝ ⎠⎠<br />

⎞⎞<br />

⎟⎟ dx<br />

⎠⎠<br />

Thus, the ratio of the average velocity to the maximum velocity for Newtonian fluid flow in a<br />

narrow slit is 2/3.<br />

This seems reasonable since<br />

€<br />

v z /vz,max = ½ for flow in a circular pipe. For a slit a larger<br />

cross − sectional area carries fluid flowing at the larger velocity than for flow in a circular<br />

pipe. So v z /vz,max for slit > v z /vz,max for a circular pipe.<br />

c)<br />

€ The mass rate of flow € is the product of the density ρ, the cross − sectional area (2BW) and the<br />

average velocity v z .<br />

w = ρ (2BW) v z<br />

€<br />

w = ρ (2BW) 2 (P0 − PL )B<br />

3<br />

2<br />

2µL<br />

w = 2 (P0 − PL )B<br />

3<br />

3 Wρ<br />

µL<br />

The flow rate vs. pressure drop (w vs. ΔP) expression above is the slit analog of the Hagen –<br />

Poiseuille equation (originally for circular tubes). It is a result worth noting because it<br />

provides the starting point for creeping flow in many systems (e.g. radial flow between two<br />

parallel circular disks; and flow between two stationary concentric spheres).<br />

d)<br />

The above analysis is not applicable if B = W, because of the presence of a wall at y = 0 and y<br />

= B would cause vz to vary significantly in y in addition to x, then vz = vz (x, y).<br />

Copyright © 2005 by Apolinar Picado


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If W = 2B, then a solution can be obtained for flow in a square duct.<br />

e)<br />

In Eq. 2.5-20, set both viscosities equal to µ, p0 − pL = P0 – PL, and set b equal to B.<br />

v z = (P 0 − P L )B2<br />

12µL<br />

⎛⎛ 8⎞⎞<br />

⎜⎜ ⎟⎟ =<br />

⎝⎝ 2⎠⎠<br />

(P0 − PL )B2<br />

3µL<br />

v z = 2<br />

2 ⋅ (P0 − PL )B2<br />

=<br />

3µL<br />

2 (P0 − PL )B<br />

3<br />

2<br />

2µL<br />

v z = 2<br />

3 v z,max<br />

v z<br />

v z,max<br />

= 2<br />

3<br />

Copyright © 2005 by Apolinar Picado

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