Implications of Viscosity

7/30/2019 Implications of Viscosity

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Implications of Viscosity

Viscosity manifests itself through creation of both shear and rotation. One of the most important

characteristics of a real fluid is it satisfies no slip condition on a solid surface. In other words,whenever fluid flows past a solid surface, the layer of fluid in contact with the surface cannot slip

against the surface, with the result that all components of fluid velocity will be zero at the solid,impermeable surface. This condition creates the highest shear on the fluid by a solid non-porous

wall. As fluid particles adjacent to the wall try to stop the next layer of fluid, the shear gradually

loses its strength as we move away from the wall. This is the cause of the boundary layerformation on a solid surface. The internal will be beyond the entrance length, which means that

boundary layer growth from each wall has already met at the center of the channel.

Beyond the entrance length, which is typically 138-140 D for laminar pipe flows and 25-40 D for

turbulent pipe flows, we call the flow fully developed. That is:

Parallel Flows: Instead of assuming fully developed flow, if we assume a parallel flow, itmeans all fluid streamlines are parallel. In such a case of parallel flow along x,

Since all fluid media must satisfy the mass, momentum and energy equation, we find by the

application of continuity equation for incompressible flows,

We therefore find that a parallel flow is indeed fully developed.

x

x

Velocity Profiles dont change with

x fully developed

0wv,0u

0x

u

0x

u,0

z

w

y

v

x

u

0v 0w


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With the introduction of parallel flows, simplification of the governing equations becomes much

simpler. We now develop the applications for some specific types of internal flows.

Plane Poiseuille Flow

This is the case of fully developed incompressible flow between two infinitely large parallel

plates. We seek the velocity profile and shear flow field for such flows. As before, if the flow is

assumed parallel in the x-direction, 0wv,0u . Therefore, the continuity equation reduces

to 0x

u

, which satisfies the fully developed condition. Let us investigate the y- and z-

momentum equations for such a flow. Also, we assume that the body forces are negligible.

Therefore:

0z

p

y

p

from above, which means that pressure is a function of x only. Now we simplify

the x-momentum equation:

(Note thatx

p

was modified to

dx

dpfrom the y and z equation results)

22

2

2

2

2

yz

v

y

v

x

vBy

p

z

vwy

vvx

vut

v:y

2

2

2

2

2

2

zz

w

y

w

x

wB

z

p

z

ww

y

wv

x

wu

t

w:z

0 0 0 0 0 0 0 0

0 0 0 00 0 0 0

2

2

2

2

2

2

xz

u

y

u

x

uB

dx

dp

z

uw

y

uv

x

uu

t

u:x

0 0 0


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u(y)

x

h

hyydx

dp

2

1)y(u 2

Let us further assume that the flow is steady. 0t

u

. Also 0x

u

0x

u

2

2

. Furthermore,

the flow can be assumed to be free from the end conditions since the plates are infinitely long

and deep. 0z

u

, which means 0

z

u

2

2

also. Thus the x-equation simplifies to:

2

2

y

u

dx

dp0

dx

dp1

y

u2

2

We can integrate this equation twice in y to write:

212 CyCy

dx

dp

2

1)y(u

(C1 and C2 = Constants)

Boundary Conditions: Since both plates are stationary, u(0) = 0, u(h) = 0

The velocity profile u(y) may be evaluated with 2C0 , and,

hdx

dp

2

1ChCh

dx

dp

2

10 11

2

To be able to plot this velocity profile, let us assume

[The partial derivatives in velocity are no

longer needed since 0z

u

x

u

t

u

]

x

y

6.25m/sh=1 m

FlowParabolic Velocity Profile


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321 m/N5dx

dp,msec/N10,m1h

Note that the velocity profile starts with a zero value on the wall, reaches a peak value of 6.25

m/s in the middle of the channel (h = 0.5 m) before reducing to zero on the upper wall (h = 1 m)

symmetrically. Also, try to plot the function when 0dxdp (instead of5 N/m3). You will see an

unrealistic curve (showing fluid bulges out along "-" x direction). We can check the volumetricflow rate to claim this point.

h

0yA

dywuAdVQ

, where w = depth of the channel and idywAd

or,

dyhyydx

dp

2

1dyu

w

Q h

0y

2h

0y

3

h

0

23

hdx

dp

12

1

2

y

3

y

dx

dp

2

1

From this expression, it is easy to see that since Q, w, h, and are all positive quantities, Q

cannot be positive unless 0dx

dp . Thus, we make an important discovery for Plane Poiseuille

Flow: A Plane-Poiseuille flow cannot exist if the pressure gradient,dx

dp, is not negative. We also

introduce a new definition of average velocity in this context. Average velocity through any areaA is defined as the volumetric flow rate per unit depth, i.e.

2

3

hdx

dp

12

1

wh

hdx

dp

12

1

A

QV

If we evaluate the maximum velocity in this flow,

0hy2dx

dp

2

10

dy

du

2

hy , which occurs at the center of the channel.


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2/hy

2max2/hy

hyydx

dp

2

1u)y(u

dx

dp

8

1

Therefore we notice that the maximum velocity

2

3

dx

dp

12

1

dx

dp

8

1

V

umax

or, V23umax for this flow.

Shear Stress Distribution:

hy2dx

dp

2

1

x

v

y

uyx

2

hy

dx

dp

2

1

If we plot this function along with the velocity profile, we notice a linear variation of shear stressand shear force as follows:

These plots were made with 0dxdp as stated before.

x

y

u(y)

h=1 m umaxx F x


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Couette Flow

This type of flow is also between infinite parallel plates. However, the boundary conditions are a

little different from Plane Poiseuille Flows. Here one of the plates remains stationary, whereasthe other moves with a constant velocity, U. For visualization, we assume the bottom plate

stationary and the top plate moving.

All the assumptions applicable to the derivation of Plane Poiseuille flows hold in the case ofCouette flows. Thus, we may skip part of the derivation and start with the velocity profile.

212 CyCy

dx

dp

2

1)y(u

Now, 0C0)0(u 2

hChdx

dp

2

1UU)y(u 1

2

hdx

dp

2

1

h

UC1

h

Uyhyy

dx

dp

2

1)y(u 2

If we compare the above velocity profile with that obtained for Plane Poiseuille flows, we find

xu=0

hFlow

U

For 0dx

dp

h

For 0dx

dp

For 0dx

dp

Couette Flow Velocity Profiles


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the right hand side has an additional term,h

Uy. The plot of just this term is a linear velocity

profile from y = 0, u = 0 to y = h, u = U. Thus, the Couette flow velocity profile may be thought

of as the superposition of the Plane Poiseuille flows velocity profile and this additional linear

profile. Because of this additional fluid momentum, Couette flows can exist even with mild

adverse pressure gradient (i.e., 0dxdp ). Recall that the existence of Q > 0 makes the flow

possible.

Since we know the velocity profile h

Uyhyy

dx

dp

2

1)y(u 2

, all the flow quantities such

as volumetric flow rate, average velocity, maximum velocity, shear stress and shear force

distributions can be computed as before using their respective formulae.

Hagen Poiseuille Flow (or, Pipe Flow)

Now we come to derive the most popular application of the internal flows, commonly known as

Hagen Poiseuille Flow or, simply pipe flows. Since pipes have cylindrical geometry, we use the

cylindrical form of the momentum equations. Let us assume an incompressible, steady flowthrough a circular pipe without any appreciable body forces. Assuming a parallel flow in the z-

direction, 0Vz , but 0VVr .

Continuity equation 0z

VV

r

1Vr

r

zr

0zVz

As in the case of Plane Poiseuille flow, writing out the momentum equations in and r direction

will simply result in 0p

r

p

. Therefore, let us focus on z-direction.

z

VV

V

r

V

r

VV

t

V:z zz

zzr

z

2

z2

2

z2

2

z

2

z2

zz

VV

r

1

r

V

r

1

r

VB

dz

dp

We can further assume 0Vz

because of the cylindrical symmetry.

0 0

0 0 0 0

0


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r

z

r

V

r

1

r

V

dz

dp0 z

2

z2

r

V

r

1

r

V

dz

dp z2

z2

[ 0V

z

V

t

V zzz

]

r

Vr

rrdz

dp z

dzdpr

drdVr

drd z

or, integrating twice over r, we get

21

2

z CrlnCdz

dp

4

r)r(V

(C1, C2 = Constants)

Since the pipe radius is R, the boundary conditionsmay be written as 0)Rr(Vr

and 0)0r(dr

dVz .

The second boundary condition is due to flow symmetry at r = 0, whereas the first one is due to

no-slip condition. Solving the constants C1 and C2 we get

2

22

zRr1

dzdp

4R)r(V

As in the case of Plane Poiseuille flows, 0dz

dp for this flow to exist (i.e., Q > 0).

Some additional results are:


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x

ydr

dz

dp

8

RQ

4

,

dz

dp

8

RV

2

, V2VmaxZ

, and

dz

dp

2

r

dr

dVzzr

[Note: You must use an annular area element zedrr2Ad

to derive V and Q results.]

Conclusion

Poiseuille flow is the pure pressure-driven fluid motion in channels with fixed walls, whileCouette flow is the pure shear-driven motion of a fluid between walls which are moving

relative to each other.

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Implications of Viscosity