Fluid Mechanics - Lecture 7 - 9 _viscous + NNF Pipe Flowx

1

Fluid Mechanics: OutlineAIM: To be able to predict the behaviour (processing implications) of fluids in simple flows

under conditions relevant to many situations encountered by process engineers

Module 1 : Properties of Fluids and Fluid Statics

Lecture 1. Fluid Properties

Lecture 2. Dimensional Analysis

Lecture 3. Non-Newtonian Fluids

Lecture 4. Fluid Statics: Fluids at Rest

Module 2: Fluid Dynamics: Fluids in Motion

PART A – Mass (continuity), Momentum and Energy Conservation

Lecture 5 Regimes, Continuity, Momentum (Bernoulli)

Lecture 6 Bernoulli, Flow measurement

Lecture 7 Mechanical Form of Energy Equation (MFEE)

PART B – Fluid Flow in Pipes

Lecture 8-9 Viscous Pipe Flow

Lecture 10 Non-Newtonian Pipe Flow

Lecture 11 Pumping

Lecture 12 Compressibility

2

Module 2 Fluid Dynamics Part B:

Viscous Pipe Flow

Velocity distributions for laminar flow & Pipe flow

(Lecture 8 – 9 )

3

ReadingWhite, Chapter 6 or Potter, …(PWR) Chapter 7

Most of these chapters is covered.

Main issues

- Pipe flow regimes (White 6.1, 6.2)

- Velocity distribution in pipe flow (calculate v profiles in pipe) (W6.4)

-major losses in pipes: Moody Chart (W6.3-6.4)

- Minor Losses in pipes (W6.6-6.7)

- Non-circular ducts (W6.8)

-Typical pipe flow problems (W6.6)

4

µρ d v

Re=

Pipe Flow: Reynolds Number

....... forces

....... forces

Laminar Flow Turbulent Flow

http://video.google.com/videoplay?docid=1827702182265329855#

v

t

v

t

5

µρ d v

Re=

Pipe Flow: Reynolds Number

Symbol note: µµµµ = viscosity µµµµ used for Newtonian fluids (no shear dependence)

ηηηη used for non-Newtonain fludis (shear dependence)v = velocity V, U, u also commonly usedρρρρ = densityd = diameter D also used. Characteristic dimension for flow field

…so you may see this:

Sometimes engineers use kinematic viscosity (ν = µ/ρ)

If in doubt, look at the dimensions. When considering viscosity, remember µ (water) ~ 1 mPas

ηρ D U

Re=

υD v

Re=

6

∆P

vP ∝∆

)275.1( −=∝∆

m

vP m

v

Log

(∆P

)

vP ∝∆

)275.1( −=∝∆

m

vP m

)log(v

Pipe Flow: Hagen (1839)

Hagen first observed the transition from laminar to turbulence.Didn’t realise it was dependent on fluid viscosity

7

Pipe Flow: Hagen & ReynoldsHagen (1839) discovered that there were 2 flow regimes, but Reynolds (1883)

showed that this depended on Re. They carried out a number of experiments

measuring the relationship between pressure drop and water velocity.

As the velocity increases, ∆P/L increases

linearly (slope = 1). This is the laminar

region.

At point B the line becomes irregular

(scattered data points) until point C when it

becomes straight with slope = 1.7-2.0

(depending on the roughness of the pipe

walls). This is the turbulent region.

ReA ~ 2300

log u

log

∆P/L

If the velocity decreases from high to low, the laminar region is re-entered at a lower velocity at point A. From

Hagens data for pipe flow, point A occurs at 2100 (Re) and point C occurs at 4000 (Re). A – B - C is the

Transistional region. ReA~ 2300 is now an accepted design value for pipe flow transition.

8

Upon entry, viscous effects grow from the wall as fluid flows down tube.Viscous boundary layers grown downstream to retard flow. This accelerates central core flow to maintain constant Q (continuity)

“Inviscid” region

Growing boundary layers

Developed velocity profilev =f(r)

Viscous Pipe Flow

P

xLe

9

Fully Developed Flow in a PipeWhen a fluid with uniform velocity enters a pipe, the layers of fluid adjacent to

the walls are slowed down, forming a boundary layer where viscous forces

dominate over inertial forces.

The thickness of the boundary layer increases downstream from the mouth to a

maximum, after which flow in the pipe is fully developed. If the two boundary

layers meet, laminar flow exists in the pipe. If the boundary layers don’t meet,

inertial forces are dominating and the flow is turbulent.

Fully developed flow occurs at Le distance; Le/d ~f (Re)

Laminar: Le /d ~ 0.06 Re (calculate max Le ~ )

Turbulent: Le /d ~ 4.4 Re1/6 (smooth walls)

Le Fully developed flow

10

Laminar Flow in a PipeLaminar flow occurs when:

Viscosity is high/low ? Pipe diameter is large/small ? velocity is high/low?

Laminar flow can be pictured as layers of fluid which do not mix.

Consider a fluid flowing from left to right through a horizontal pipe.

Pressure causes the fluid to flow (i.e. higher pressure at z=0 than at z=L )

P0 PL

Rr

τ =τw = max

z=0 z=L

z τ = 0

How do we determine the velocity profile ?

11

Rate of mtm in =

Setting up momentum balance over a thin shell of fluid.

Recall for steady state:

Rate of mtm out- Sum of forces

acting on system(surface and body)

+ 0

This is obtained by considering the shear stress across shell

E.g. pressure forces (acting on surfaces) and gravity forces (acting on the volume as a whole)

Pressure, P0 Pressure, PL

r

z

Mtm flow in & out by viscous transfer

Laminar Flow in a Pipe: MTM balance

12

Laminar Flow in a Pipe: recap


r

zflow

r

∆rP0 (2 π r ∆r) PL (2 π r ∆r)

Forces must balance in steady state:

(2 π r L )τ|r+∆r - (2 π r L) τ |r + (2 π r ∆r)(P0- PL) = 0

Pressure Force on annulus at z=0

Pressure Force on annulus at z=L

Consider force balance on annulus of fluid, and integrate across all rz

(2 π r L)τ|rShear Force at r

(2 π r L)τ|r+∆rShear Force at r+∆r

13

Laminar Flow in a Pipe: recap


r

zflow

(2 π r L)τ|r+∆r

(2 π r L)τ|r

r

∆rP0 (2 π r ∆r) PL (2 π r ∆r)

Forces must balance in steady state:

(2 π r L )τ|r+∆r - (2 π r L) τ |r + (2 π r ∆r)(P0- PL) = 0

Consider force balance on annulus of fluid, and integrate across all rz

14

Laminar Flow in a PipeVery long tube (no end effects; fully developed flow), Length L, radius R.


Shear Force at r: (2 π r L)τ|rShear Force at (r+∆r) (2 π r L )τ|r+∆r

Horizontal cylinder: gravity force neglected (can also be ‘adsorbed’ into pressure terms if vertical)

Pressure force acting on annular surface at z= 0 P0 (2 π r ∆r)

Pressure force acting on annular surface at z= L -PL (2 π r ∆r)

r

z

Mtm flow in & out by viscous transfer:

flow

15

Laminar Flow in a PipeAdd up the contributions to the momentum balance (In – Out + ΣF = 0):

⇒ (2 π r L )τ|r+∆r - (2 π r L)τ|r + (2 π r ∆r)(P0- PL) = 0

⇒ Divide by 2 π L∆r

c = 0 b/c at r = 0, τ ~ finite! = 0

rL

PP

∆r

rrL0rx∆rr

−=−+ ττ

rL

PP

dr

)d(r L0

−=⇒τ

Cr2L

PPr 2L0 +

−=τr

Cr

2L

PP L0 +

−=⇒τ

rL

PP

∆r

rr

0L0rx∆rr

−=−

→∆+⇒ ττ

rLim

r2L

PP L0

−=ττ=0

R2L

PP L0max

−== wττL

PRw 2

∆=τ

16

Laminar Flow in a PipeSubstitute fluid constitutive model, i.e. Newtonians Law of viscosity:

⇒

Integrate:

At r = R, vz = 0 (no slip) ⇒

Parabolic velocity profile

r2L

PPL0

−=τ r2L

PP

dr

dv L0z

−=− µ

Cr L 4

PPv 2L0

z +

−−=µ

2L0 R L 4

PPC

−=µ

( )22L0z rR

L 4

PPv −

−=µ

dr

dvzµτ −=

vz = 0

vz =max

( )

L 4

PP,v

2L0

,maxz µR

v CLz

−==

−

−=2

2L0z

r1 R

L 4

PPv

Rµ

−

−=2

2L0z

r1 R

L 4

PPv

Rµ

17

Laminar Flow in a Pipe

Calculate Average Velocity…..

Calculate Volume rate of flow ,

Hagen-Poiseuille Law: relates flow velocity to pressure drop in circular pipes due to VISCOUS effects for laminar flows

−∆=

2

22

z

r1

L 4v

R

PR

µ

max

2

z 0.5v L 8

v z,

PR =∆=µ

L 8

v.v.Q4

z2

z µππ PR

RA∆===

L 23

v2

z µPD∆= 4

4 LQ 128

L 128Q

DP

PD

πµ

µπ =∆⇒

∆=

2L

Pr∆=τL0 PPP −=∆

∫= dAA

vz v

L

PRw 2

∆=τ

18

Laminar Flow in a Pipe - text

White Fig 6.7, sect 6.3 (also see figure 6.15 for annulus flow)

Potter (PWR) Fig 7.4, sect 7.3.1 (also see example 7.2 for annulus flow)

19

⇒∆=

L 128Q

4

µπ PD

⇒ )000,10)(09.0(128

)000,15()3.0( 4π

20

Cross-sectional area of annulus

= π (r+∆r)2 - π r2

= π (r2 + 2 r ∆r + (∆r)2) - π r2 = 2 π r ∆r + π (∆r)2

(∆r)2 can be neglected as it is very small compared to 2 r ∆r

Therefore, cross-sectional area of annulus

dA= 2 ππππ r ∆r

L 8 v

4L 2)

44

2(

L 2)

42(

L 2 v

r1

L 2

2r1

L 4 v v

2

222

2

42

2

2

0

22

2

0

2

µ

µµµ

µππ

µ

PR

RPRRP

R

RRP

drr

R

Pdr

R

r

R

PRdA

A

vRr

r

Rr

r

z

∆=⇒

∆=−∆=−∆=⇒

−∆=

−∆=⇒= ∫∫∫

=

=

=

=

Laminar Flow in a Pipe: Q and Vav

L 8

v.v.Q4

z2

z µππ PR

RA∆===

−=

2

2

z

r1 v2v

Rmax

2

z 0.5v L 8

v z,

PR =∆=µ

−=

2

2

z

r1 vv

RCL

R

r

∆r

∫= dAA

vz v Sub in vz, dA, and A=ππππR2

21

L 8 v

4L 2)

44

2(

L 2)

42(

L 2 v

r1

L 2

2r1

L 4 v v

2

222

2

42

2

2

0

22

2

0

2

µ

µµµ

µππ

µ

PR

RPRRP

R

RRP

drr

R

Pdr

R

r

R

PRdA

A

vRr

r

Rr

r

z

∆=⇒

∆=−∆=−∆=⇒

−∆=

−∆=⇒= ∫∫∫

=

=

=

=

Laminar Flow in a Pipe: Q and Vav

L 8

v.v.Q4

z2

z µππ PR

RA∆===

−=

2

2

z

r1 v2v

Rmax

2

z 0.5v L 8

v z,

PR =∆=µ

−=

2

2

z

r1 vv

RCL

R

r

∆r

∫= dAA

vz v

22

Laminar Flow in a Pipe

2

v

L 8v

2

zz, CLPR =∆=

µ

4

4 LQ 128

L 128Q

DP

PD

πµ

µπ =∆⇒

∆=

L0 PPP −=∆

Anywhere (r=r)

Wall (r = R) Centreline(r=0)

Shear Stress

Velocity (flow dir)

2L

Pr∆=τL

PRw 2

∆=τ 00 =τ

−∆=

2

22

z

r1

L 4v

R

PR

µ 0vz = L 4

2

, µPR

v CLz

∆=

Assume Newtonian fluid, incompressible, fully developed flow

−=

2

2

zz

r1 v2v

R

P0 PL

Rr

τ =τw = max

z=0 z=L

z τ = 0

23

Turbulent Flow in a Pipe

The observed velocity profile for turbulent flow is flatter than for laminar flow.

Due to the complex nature of the flow, no expression has been derived for

the variation in velocity across the radius.

However, at high Reynolds’ numbers in smooth pipes, it has been found that

velocity is proportional to 1/7 power of the distance from the wall.

Eq 9. Prandtl 1/7 Power Law

As for laminar flow, the flowrate can be calculated by considering an annulus

and integrating over the pipe radius, which gives:

The average velocity is closer to the centre-line

velocity than for laminar flow

So, Eq 10.

7/1

CL

z

R

r1

v

v

−=

7/1

z

R

r1 1.22

v

v

−=

cLvv 82.0=

24

Fluid velocities in PipesTypical velocity profiles for laminar and turbulent flow in a pipe of 50 mm diameter and with a bulk velocity of 1.5m/s:

-25

-20

-15

-10

-5

0

5

10

15

20

25

0 0.5 1 1.5 2 2.5 3

Rad

ius

(mm

)

velocity (m/s) Laminar Flow

Turbulent Flow

What is the Re if water is the fluid ?Which velocity profile is the correct one to use if the fluid is water ?

−=

2

2

z

r1 v2v

R

7/1

z

R

r1 1.22

v

v

−=

25

Fluid velocities in PipesIf the driving force behind a fluid is gravity. The required flowrate will dictate choice of pipe diameter. If the fluid is to be pumped, economics (annual operating cost) will dictate choice of pipe diameter.

As a guideline, the optimum velocities that should be maintained in a pipe line, in terms of fluid density, are given below.

As an even more general rule (easy to remember):

For liquids such as water, velocity should be approx 1 m/s.

For gases such as air, velocity should be approx 10 m/s.

Fluid density (kg/m3) Velocity (m/s)

1600 2.4

800 3.0

160 4.9

16 9.4

0.16 18

0.016 34

26

Frictional losses during flow in pipesLaminar Flow -remember dimensional analysis

When a fluid flows in a pipe there is a loss of energy due to friction, which occurs as a

result of viscous shear caused by the walls. When energy is being lost due to friction,

there is a drop in pressure that for laminar flow is given by the Hagen-Poiseuille equation.

For a pipe of constant diameter, D, and for incompressible flow, pressure varies with pipe

length :

2zvL 23

DP

µ=∆

µρ Dv=Re

×

=∆

D

L

DP .

v

322

2

v

z

2z

ρµρ

=∆2

2v

D

LfP

ρ

Dimensionless #

=∆2

v

Re

64 2zρ

D

LP

Re

64laminar =f

27

Frictional losses during flow in pipes

=∆2Re

64 2v

D

LP

ρ

D ∆Pl / (ρ V2 ) = fn [µ/(ρ V D)]

For Laminar Flow :

Recall from dimensionless analysis :

Generalise to any pipe flow regime:(i.e. including turbulence)

MFEE, Heads:

Darcy-Weisback equation

=∆2

2v

D

LfP

ρ

f= fn [ReD, roughness]

=g

v

D

Lfh f 2

2

ffrictionL hhg

Ph ==∆=

ρ

)0;0;;(

...0..

)( 2

2

2121

21

2

2

1

2

====

=+−

−+=

++−

++

qT

L

qTL

hhzzvv

hg

PP

hhhzg

v

g

Pz

g

v

g

P

ρ

αρ

αρ

Pipe flow head lossDarcy equation

28


White, Fig 6.12

=g

v

D

Lfh f 2

2

29

Alternative expression for f

=∆2

2v

D

LfP

ρ

φρτφ 8

2=⇒= f

vw

f = Moody Friction factor

Alternative:

Fanning friction factor (f’) is often

used in American texts.

4

1

2

2

=⇒

vfw

ρτ

2

8

vf w

ρτ=⇒

L

Rv

D

Lf

L

PR

222

2

=∆⇒

ρ

4/2' ff == φ

30


=∆2

2v

D

LfP

ρf= fn [ReD, roughness]

section pipe

2

2

=∆=g

v

D

Lf

g

Ph f ρ

Head loss is pipe = > sub into MFEE to account for friction losses in a pipe section

1 2

Evaluate losses for sections of pipes. V is constant for constant diameter (Continuity holds) but pressure decreases due to friction / viscous losses

21 PPP −=∆

31

Example

===∆

21.0

1000064.0

2v

mD

mLP

ρ

Calculate v from Re = ρvD/µ…v =µ.Re/ρD

To get answer for ∆P

1 2

10 km straight horizontal pipeline with 10 cm diameter carrying water at an Re = 100 . (a) What is the pressure drop if we used Bernoulli equation assuming inviscid flow ? (b) What is the pressure drop if we do not assume inviscid flow ?

021 =−=∆ PPP(a)

(b)

=0.01 m/s (check)

=6400 Pa (check)

32

Example No. 13Glycerin (density = 1260 kgm-3, viscosity = 0.9 Pa.s) is pumped through a

straight pipe of diameter 100 mm at 0.02 m3s-1. The pipe is 45 m long and rises

through a height of 12 m. Gauge pressure at the inlet to the pipe is 590 kPa.

Calculate the pressure at the end of the pipe.

33

Frictional losses during flow in pipesBack to MFEE/Extended Bernoulli

- Recall:

- a pump adds energy and a turbine extracts energy from flow.

- IF hturbine (hT or hs) or hpump (hq) included, they must lie b/n points 1 & 2

- There are many ways to present this equation (e.g. terms of energy per unit mass)

-Often hf is expressed per unit length of pipe.

-E.g. hf = 8.50 m/100 m length @ 0.04 L/s for 9.3 mm ID steel pipe

-(what value of friction factor does this correspond to ?)

(Available Head)IN + hq = (Available Head)OUT + hf + hT

(Available Head)IN + hq = (Available Head)OUT + f.(L/D)(v2/2g) + hT

Tq hg

v

D

Lfz

g

v

g

Phz

g

v

g

P ++

++=+

++

2

2

2

2

2

222

1

211 α

ρα

ρ

34

Frictional losses during flow in pipes2 methods to obtain f as a function of roughness and Re

1. The Moody Diagram (White Fig 7.13, PWR fig 6.13)

-see White table 6.1 for values of roughness

2. Implicit and explicit equations

- For Re > 4000, The Colbrook equation (White eq. 6.48, PWR eq. 7.6.28) requires an iterative solution for f:

This reduces to other developed equations for the 2 extremes:

+−=

f

D

f DRe

51.2

7.3

/log2

110

ε

( ) 8.0Relog21

10 −−= ff

d

−=7.3

/log2

110

D

f

ε

Smooth pipe flow (ε = 0)

Completely turbulent zone (Re = infinity)

35


White – Moody chart

36


Potter – fig 7.13 Moody diagram

37


Pipe Friction Chart

Data for friction factor is presented on a friction chart which shows plots of f against Re for

different values of pipe relative roughness. Everything is dimensionless.

Relative roughness is a ratio of the absolute roughness to the pipe diameter.

ie.

Values for ε can be found in standard texts. Some commonly used are:

Note:

In the turbulent region, f becomes independent of Re (see dotted line). This occurs earlier for

rougher pipes.

ddiameter pipe

roughness absolute roughness relative

ε==

e (m)

drawn tubing 1.5 ×10-6

commercial steel and wrought iron 4.5 ×10-5

cast iron 2.6 ×10-4

concrete 3.3 ×10-4

corroded pipes ~ 3.3 ×10-3

38


White – Table 6.1 – Recommended roughness values for commercial ducts

39

Minor LossesFrictional losses in pipe fittings

In addition to pipe walls causing energy losses, fittings within the piping system

also cause energy losses. Pipe fittings include bends, valves, expansions,

contractions, T-pieces etc. Darcy’s formula can be written as:

where,

and is the velocity head, as given in Bernoulli’s equation.

Different values of K obtained for different fittings, bends, etc.. These are

tabulated or put in terms of equivalent length of pipe Le, where

=

g 2

vK

2

fh

d

LfK

.=

g 2

v2

f

dKLe

.=

40

White Fig 6.17,

GateGlobe

angle

swing

disk

41

White Table 6.5

42

Minor Frictional losses in Pipe FittingsFrictional losses can be represented as a number of velocity heads. For pipe

fittings, experiments have been carried out to give the frictional loss as a number

of velocity heads. See table listing velocity heads.

Can also express the energy loss in a pipe fitting as an equivalent length of pipe

to be substituted into Darcy’s equation. Table below shows how much to

increase the pipe length by, in terms of d (the pipe diameter).

Fitting equivalent length of pipe

45° elbow 15d

90° rounded elbow 30-40d

90° square elbow 60d

entry into one leg of a T-piece 60d

entry into two legs of a T-piece 90d

gate valve, fully open 70d

gate valve, ½ open 200d

gate valve, ¼ open 800d

Le

=

g 2

vk

2

fhd

Lf e=k

43

Minor Frictional losses in Pipe FittingsSudden Enlargement:

When a change in diameter is gradual, little energy is lost due to turbulence.

When a change in diameter is great and sudden, the flow patterns are very

turbulent, resulting in a lot of energy being lost to turbulence.

Equating forces and using Bernoulli’s

equation, it has been shown that:

As A2 → ∞, A1/A2 → 0 and

22

2

11

A

A-1

g 2

vfh

=

g 2

v 21

fh =

Therefore, when fluid goes from a pipe into a vessel, the energy loss is equal to

one velocity head and is known as the EXIT LOSS.

u1 A1u2 A2

44

Frictional losses in Pipe FittingsSudden Contraction:

For flow from a large vessel into a pipe, a vena contracta is formed and most of

the energy is lost between between the vena contracta and the wall.

In this case,

where Cc = Avena contracta/A2

The value of is related to the ratio of d2/d1 as shown:

22

1 C

1

g 2

v

c

2fh

−=

2

1 C

1

c

−

u1 A1 u2 A2

vena contracta

d2/d1 0 0.2 0.4 0.6 0.8 1.0

(1/Cc-1)2 0.5 0.45 0.38 0.28 0.1 0.0 Therefore, when fluid goes from a vessel into a pipe, d2/d1 → 0, and the energy

loss is equal to ½ velocity head and is known as the ENTRY LOSS.

45

Potter (PWR)Table 7.2 Nominal loss coefficient K (turbulent flow)

46

Minor Frictional losses

Pipe systems contain more than straight pipe sections: valves, bends, tees, entry/exit points

These introduce Minor Losses

=

g

vKh LL 2

2

KL = function (geometry, Re,D)

Minor losses are also expressed in terms of equivalent pipe length

≡

=

g

v

g

vKL 2D

Lf

2h

2eq

2

L f

DKL L

eq =

47

Extended Bernoulli Equation

With Viscous Pipe Losses and minor Losses

qhg

v

g

v

D

Lfz

g

v

g

Pz

g

v

g

P −

+

=

++−

++ ∑ 2

K222

2

lossesminorall

L

2

2

22

22

1

21

11 α

ρα

ρ

qhD

Lf

g

vz

g

u

g

Pz

g

u

g

P −

+

=

++−

++ ∑

sectionlossesminorall

L

2

2

22

22

1

21

11 K

222α

ρα

ρ

Determine for each section of flow that has different average v

48

Pipe flow sections

1section lossesminorall

L

21 K

2

+

∑D

Lf

g

v

2section lossesminorall

L

22 K

2

+

∑D

Lf

g

v

d

Lf e.

( )1section

21

2

+

∑ eLL

D

f

g

v ( )2section

21

2

+

∑ eLL

D

f

g

v

v1A1=v2A2

49

Pipe flow sections

Q A=QB= Q1=Q2=Q3

Q A=QB= Q1 + Q2 + Q3

∑ ∑=

→

+

=

3

1 lossesminor

2

, 2i i

iiiBAf K

D

Lf

g

vh

∑ ∑=

→

+

=

3

1 lossesminor

2

, 2iei

i

iiBAf LL

D

f

g

vh

50



qhg

v

g

v

D

Lfz

g

v

g

Pz

g

v

g

P −

+

=

++−

++ ∑ 2

K222

2

lossesminorall

L

2

2

22

22

1

21

11 α

ρα

ρ

qeii

ii hLLD

f

g

vz

g

v

g

Pz

g

v

g

P −

+

=

++−

++ ∑ ∑

= section i lossesminor

2

2

22

22

1

21

11

222α

ρα

ρ

Determine for each section of flow between 1 and 2 that has different average v ….

51



Type I: Q is given. Calculate DP (or hL, or hS, etc...)

v is known everywhere. Simple problem since friction factors are uniquely related to v via Moody diagram (or Colebrook equation)

Type II: DP (or hL) is given. Calculate Q.

Requires an iterative procedure, since f is a function of Q (f = function(ReD) Rapid convergence however.

Type III: DP and Q are given. Calculate pipe diameter D.

Both v and f are functions of D.

qhg

v

g

v

D

Lfz

g

v

g

Pz

g

v

g

P −

+

=

++−

++ ∑ 2

K222

2

lossesminorall

L

2

2

22

22

1

21

11 α

ρα

ρ

52

Example No. 14

Water (density = 1000 kg/m3, viscosity = 1 ×10-3 Pa.s) is to be pumped from a

storage vessel, which is at atmospheric pressure, into a header tank, also at

atmospheric pressure. The pipe has a diameter of 150 mm, a total length of 150

m and a roughness of 0.03 mm. In the pipework there are two globe valves, both

fully open, four 90° standard radius bends and an in-line filter with a frictional loss

equivalent to 15 velocity heads. The entry to the header tank is 40 m above the

water level in the storage vessel. The required flowrate is 120 tonnes/hr. What

power needs to be supplied to the pump assuming an overall efficiency of 65%?

53

Frictional losses during flow in pipesExtension to non-circular channels; Hydraulic diameter (Section 6.6)

Perimeter (wetted) P πD

Area (wetted) A πD2/4

Hydraulic diameter DH is such that perimeter to surface area ratio are equal between non-circular and circular cross-sections.

Note: Use velocity based on actual cross-sectional area.

Accuracy of head loss calculation using DH is ~ 15%

P

ADH

4=

54

Non-circular ductsSo far, all calculations have been related to flow in circular pipes. Other possible flow areas also need to be considered. For example, flow through a rectangular duct (used for air-conditioning), and flow through an annulus. As long as the shape is not too complex, all previous formulae can still be used whereby the equivalent length of diameter is replaced by the ‘hydraulic mean diameter’ or sometimes known as the ‘equivalent diameter’.

perimeter wetted

flow of area sectional-cross 4 d diameter, mean hydraulic m

×=

For a circular pipe: of course!

For a rectangular duct:

For an annulus:

( )d

d

d/2 4 dm == π

π 2

a

b b) 2(a

b a 4 dm +

=

di

do

( )( )io

io

ioio

io

2i

2o

io

2i

2o

m d - d d d

d - dd d

d d

d - d

d d

4

d -

4

d4

d =+

+=+

=π+π

ππ

=

55

Example No. 15Air (density = 1.2 kg/m3, viscosity = 1.7 ×10-5 Pa.s) flows through a rectangular

duct with internal dimensions of 1 m by 0.5 m. The duct is 25 m long. Assuming

a relative roughness of 0.0001, calculate the pressure loss due to friction in the

duct for a flowrate of 8 m3/s.

56

Main issues:• List the major pipe flow regimes• Velocity profiles• What is the Moody Chart for?• What leads to minor losses in pipes?• How do we deal with non-circular ducts? You should now try a range of typical pipe

flow problems


57

Supplementary slides

Viscous Pipe Flow

58

Kinetic Energy Correction FactorIn the Bernoulli equation, kinetic energy per unit mass (J/kg) is given by v2/2.

Considering the velocity profiles previously derived, it is apparent that the sum

of kinetic energy over the velocity profile may not be the same as simply using

the mean, or bulk, velocity in the expression v2/2.

Consider first the kinetic energy (KE) when using the bulk velocity:

KE (J/s) = mass flowrate x v2/2

= ρ π R2 v . v2/2

KE (J/s) = (ρ π R2 v3)/2 Eq. 11

Now, consider the case when using velocity profiles….


59

Kinetic Energy Correction FactorLaminar Flow

Consider again the annulus at radius r of thickness dr where the velocity of the annulus is v(r).

Change in KE of fluid flowing through the annulus (J/s) = mass flowrate x v2/2

= ρ 2 π r dr v x v2/2

= ρ π r dr v3

From Eq 8. we know the relationship between ur and u:

Substituting this gives:

Change in KE =

Integrating with respect to r from r=0 to r=R gives:

KE (J/s) = ρ π R2 v3

Comparing this to Eq 11 shows that the kinetic energy for laminar flow is actually double that based on the bulk velocity.

−=

2

2

R

r1 2

v

v

drR

r1 v 2r

3

2

2

−πρ


60

Kinetic Energy Correction FactorTurbulent Flow

Again,

Change in KE of fluid flowing through the annulus (J/s) = ρ π r dr ur3

From Eq 10. we know the relationship between ur and u:

Carrying out the same analysis as for laminar flow gives:

KE (J/s) = 0.523 ρ π R2 u3

Comparing this to Eq 11 shows that the kinetic energy for laminar flow is 1.046 times that based on the bulk velocity.

7/1

=R

r - R 1.22

u

ur

To compensate for the higher kinetic energy, a correction factor (α) is introduced

to Bernoulli’s equation. The term for kinetic energy becomes αu2/2, where:

α = 2 for laminar flow

α = 1 for turbulent flow


61

Frictional losses during flow in pipesTurbulent Flow

Due to the complex nature of turbulent flow, a semi-empirical approach is taken.

Consider a short length of pipe.

P0 PLd

τ

L

As for laminar flow, force balance gives:

While for laminar flow, , for turbulent flow we don’t have a

solution for τ. Instead, we re-arrange the equation as follows: dr

dvrµτ −=

=

∆2

2f

v

v

d

4

L

P

ρρw

τ

=

2

v

d

8 2

2 vρ

ρτ

w

d

4

r

2

L

Pf ww

ττ==

∆


62


The quantity is a dimensionless group known as the friction factor ϕ and

has been determined experimentally for a range of Re and degrees of wall

roughness.

=>

And, Eq 13. Darcy’s Equation

Pipe Friction Chart

Data for ϕ is presented on a friction chart which shows plots of ϕ against Re for

different values of pipe relative roughness. Everything is dimensionless.

Relative roughness is a ratio of the absolute roughness to the pipe diameter.

ie.

2u ρτ

w

=

∆

2

v

d

8 2 L

P ρφ

=

=∆

2..

2

v8P

22 v

d

Lf

d

L ρφ ρ

d

e

diameter pipe

roughness absolute roughness relative ==


63

Laminar Flow in a Pipe: MFEE

P0 PL

Rr

τ =τw = max

z=0 z=L

z τ = 0

qsf hhhzg

v

g

Pz

g

v

g

P −+=

++−

++ )(

2

22

2

1

2

αρ

αρ

No shaft work or heat transfer effects, assume fully developed flowα1 = α2 ; Q1=Q2 = constant = v1A1=v2A2 =>v1=v2

=> hf =(P1-P2)/ρg =∆P/ρg


64

FLUID MECHANICSFLUID MECHANICSLECTURE 10

Fluid Flow: Non-Newtonian pipe flow

http://www.egr.msu.edu/~steffe/freebook/STEFFE.pdf

65

Fluid Mechanics: OutlineAIM: To be able to predict the behaviour (processing implications) of fluids in simple flows

under conditions relevant to many situations encountered by process engineers

Module 1 : Properties of Fluids and Fluid Statics

Lecture 1. Fluid Properties

Lecture 2. Dimensional Analysis

Lecture 3. Non-Newtonian Fluids

Lecture 4. Fluid Statics: Fluids at Rest

Module 2: Fluid Dynamics: Fluids in Motion

PART A – Mass (continuity), Momentum and Energy Conservation

Lecture 5 Regimes, Continuity, Momentum (Bernoulli)

Lecture 6 Bernoulli, Flow measurement

Lecture 7 Mechanical Form of Energy Equation (MFEE)

PART B – Fluid Flow in Pipes

Lecture 8-9 Viscous Pipe Flow

Lecture 10 Non-Newtonian Pipe Flow

Lecture 11 Pumping

Lecture 12 Compressibility

Examples of Pipe flow of non-Newtonian Fluids

http://www.youtube.com/watch?v=GX4_3cV_3Mw

This one isnt pipe flow, but an interesting shear thinning phenomena called the Kaye effect:

http://dsc.discovery.com/videos/time-warp-non-newtonian-fluid.html

http://www.youtube.com/watch?v=dm29GAgPyn8&feature=related

http://www.youtube.com/watch?v=TIqMCHS7uHc&feature=related

This is another quick sand video....

Slurry pumping

Product filling

Rheology

Inelastic

(De ~ 0)

� = ��

viscosity µ = constant

Viscoelastic

(De > 0 )

� = ��

= ��

viscosity η varies with shear rate

� = ��,� ��

= ��,� ��

viscosity η varies with shear rate and time

� = ��,� �, � �

= ��,� �, � �

viscosity η varies with shear rate, time and thermal history

� = ��… � + “Yield stress”

Any of these can also have yield stress

For Non-Newtonian Fluids, the relationship between stress and strain

is not linear.

Stressτ

(Pa)

Shear rate (s-1)

Newtonian fluid

Shear-thinning fluid

Shear-thickening fluid Yield stress

fluid

PWR 1.5

Rheology: Non-Newtonian Fluids

…alternative names you will see….

Rheology: Non-Newtonian Fluids

Stressτ

(Pa)

Shear rate (s-1)

Newtonian fluid

Pseudoplastic

DilatantIdeal Plastic,Bingham fluid

PWR 1.5

Lecture 3 – Rheology, Recap

Time independent Non-Newtonian fluidsi.e. shear stress is nonlinear function of shear rate, E.g.:

Most non-Newtonian fluids are shear thinning and can be described according to a power law:

K = consistency coefficient (Pasn), n = power law index

Many fluids also possess an apparent yield stress (σy, a minimum stress

must be applied before flow occurs, e.g. mineral slurries, mayonnaise, paint…, The Herschel-Bulkley equation is often used (reduces to Bingham if n=1):

nKγσ &= 1/)()( −==⇒ nKγγγσγη &&&&

ny Kγσσ &+=

)(γσ &f=

γγσγη&

&&

)()( =

Stress vs. Shear rate (Linear axis)

Shear rate (s-1)

0 100 200 300 400 500 600

She

ar S

tres

s (P

a)

0

100

200

300

400

Concentrated Orange JuiceStirred YogurtHellmans MayoHoney, NewtonianMilk, NewtonianHerschel-Buckley ModelPower Law Model

Viscosity vs. Shear rate (Log axis)

Shear rate (s-1)

10 100

Vis

cosi

ty (

Pas

) 0.01

0.1

1

10

Concentrated Orange Juice Stirred YogurtHellmans MayoHoneyMilk

Lecture 3 – Rheology, Recap

Rheology flow curves of some common fluid foods, obeying Newtonian, Power law or Herschel-Buckley models

γγση&

&)(=

Relevance of shear rate

Steffe

Incr

easi

ng s

hear

rat

e

Why do we want to characterise rheology ?

Putting it into context

Show flow curve in connection to processing conditions for

Foods and Personal care products

Slurry flow

Coatings

Time Independent Non-Newtonian Fluids in Pipe Flow

What do we need to consider for time independent fluidsShear stress is a non-linear function of shear rate

Note, can also put it in terms of viscosity:

We need a Generalised Approach to obtain velocity profile

)(γσ &nf=

Textbook: see Blackboard link to Free Rheology book by Steffe, Chapter 2 – Tube Viscometry

)()( γ

γγση &&

& nf==

82

Laminar Flow in a Pipe of NNF

P0 PL

Rr

σ=σ w = max

z=0 z=L

z σ = 0

From last lecture, using either a force or mtm balance, the stress in pipe flow can be obtained as function of the radius (r)

Note: this does not assume any particular fluid constitutive model or flow regime => applies to laminar and turbulent flow as well as to time-independent inelastic non-Newtonian fluids.

2L

Pr∆=σ

2L

PR∆=wσ

r=0

r=R

r

z

83

Laminar Flow in a Pipe of NNF

P0 PL

Rr

σ=σ w = max

z=0 z=L

z σ = 0

2L

Pr∆=σ2L

PR∆=⇒ wσ

r=0

r=R

r

z

rRw σσ ==∆

⇒2L

P

84

NEWTONIAN: Substitute fluid constitutive model,

No slip Boundary condition, i.e. r=R, vz = 0

Average pipe velocity

Volumetric flow rate:

Rearrange equation for average velocity

This gives the pressure drop (energy loss) along the pipe section due to viscous forces (not accounted for in Bernoulli) . We use this in the MFEE:

dr

dvzµσ −=

−=

2

z

r1 2v

Rvz

Laminar Pipe Flow of Newtonian fluid: recap

∫= dAA

vz v

L 8v

2

z µPR∆=

L 8

Q4

µπ PR∆=zv.Q A=

2

2

=∆ v

D

LfP

ρ

g

Ph f ρ

∆=Re

64laminar =f

85

Time independent non-Newtonian Fluid ;

Substitute fluid constitutive model (Power Law),

No slip Boundary condition, i.e. r=R, vz = 0



Rearrange equation for average velocity to get pressure drop in terms of velocity, pipe dimensions, etc.. Then obtain frictional losses to use in MFEE (hf);

Laminar Pipe Flow of Power Law fluidn

zn

dr

dvKK

−=⇒= σγσ &

∫= dAA

vz v

zv.Q A=

( ) ( ) nn

n Rnn

11

132LKP v

+

+∆=⇒

( ) ( )( )nn

Rr

nn

zz vv1

1113

+

−= ++

( ) ( ) nn

n RQ nn

131

132LKP

+

+∆=⇒ π

;2

2

g

v

D

Lfh f =⇒

2

2

=∆ v

D

LfP

ρ

( )n

nn

n

nn

PL K

Dv13

41

2

8Re +−

−

= ρPLf Re/64 =

86

( )( )nn

Rr

nn

z

z

v

v 1

1113

+

−= ++Power Law Fluid:

r/R

-1.0 -0.5 0.0 0.5 1.0

v/v av

erag

e

0.0

0.5

1.0

1.5

2.0

2.5

n=2

n=1

n=0.5

n=0.25

n=0.1

This demonstrates how ‘shear thinning’ (n < 1) and ‘shear thickening’ (n > 1) significantly affect the velocity profile in pipe flow

Laminar Pipe Flow of Power Law Fluid

nKγσ &=

87

Power Law Fluid:

2D Graph 1

r/R

-1.0 -0.5 0.0 0.5 1.0

v/v a

vera

ge

0.0

0.5

1.0

1.5

2.0

2.5

r/R vs n=0.1 r/R vs n =0.25 r/R vs n=0.5 r/R vs n=1 r/R vs n=2

r/R

-1.0-0.5

0.00.5

1.0

v/vaverage

0.0

0.5

1.0

1.5

2.0

2.5

n=2

n=1

n=0.5

n=0.25

n=0.1

This demonstrates how ‘shear thinning’ (n < 1) and ‘shear thickening’ (n > 1) significantly affect the velocity profile in pipe flow

( )( )nn

Rr

nn

z

z

v

v 1

1113

+

−= ++


88

Other key equations for Power law fluid;

Shear stress and Shear rate at the wall, r = R

Determining power law relationship in Laminar Pipe Flow

nww

n KK γσγσ && =⇒=

2L

PR∆=wσ

( )Γ= + .413

nn

wγ&

=

=ΓD

v

R

Q 843π

This is the shear rate for a Newtonian fluid

Task: put vav and Q in terms of wall shear stress

∆P

Q

( ) ( ) ( ) ( ) ( )( ) ( ) ( )( )( )Dn

nw

nn

K

nw

nn

Kwnn

RK

RR

nn

nnn

v813

4

131

131

132LKPR

11

111

v

v

+

+

++∆

=⇒

=⇒

==

γγ

σ

&

&

89

Laminar Pipe Flow of NNF: General Approach

Γ

+=n

nw 4

13

γ&

Plot Log-log of σσσσw vs ΓΓΓΓ to determine relationship to get:

This gives actual wall shear rate:

Plot σσσσw vs wall shear rate to obtain rheological relationship between stress and shear rate.

Alternative: plot ∆P vs Q…

nKγσ &=

2L

PR∆=wσ

=

=ΓD

v

R

Q 843π

∆P

Q or vav

'LogKIntercept

nSlope

==

nww

nww

n

K

KK

Γ=

=⇒=

'σ

γσγσ &&

n

n

nKK

+=

13

4'

90

Example, Steffe p 143Determine the rheological properties of bread dough using the data from a ‘capillary viscometer’ (i.e Q and P from pipe flow using various L & Ds

Measured Q & ∆PLink to excel sheet

91

92


Measured Q & ∆PLink to excel sheet

2L

PR∆=wσ

93


Link to excel sheet

=Γ3

4

R

Q

π

94

95

n = 0.28K’=eIntercept

K’ = 27.6σw=27.6 Γw

0.28

K= 24.07

n

n

nKK

+=

13

4'

28.0

28.0

24

24

γσγσ

γσ

&

&

&

=⇒

=⇒

=

ww

nK

Generalised shear rate (Γ s-1)

0 200 400 600 800 1000

Wal

l She

ar S

tres

s ( σ

w)

60

80

100

120

140

160

180

200

220

y = 27.6 x0.28

Ln Γ

3.5 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5

Ln( σ

w)

4.2

4.4

4.6

4.8

5.0

5.2

5.4

Intercept = 3.32slope = 0.28

r ² 0.9764522388

96


Link to excel sheet

My answer: 28.024γσ &=

( )Γ= + .413

nn

wγ&

97

Reynolds Number: Power Law Fluid

For Power Law Fluid:

ηρ Dv=Re

( )n

nn

n

nn

PLn

w K

DvK 13

41

21

8Re +−

−− =⇒= ργη &

( )( )3

44

13R

Qn

nw π

γ +=&

Still valid, but now viscosity depends on shear rate

98

Time independent non-Newtonian Fluid with yield stress;

Substitute fluid constitutive model (Bingham),

Solve



Rearrange equation for average velocity to get pressure drop in terms of velocity, pipe dimensions, etc.. Then obtain frictional losses to use in MFEE (hf);

Laminar Pipe Flow of Bingham Fluid

∫= dAA

vz v

zv.Q A=

;2

2

g

v

D

Lfh f =⇒

γησσ &By +=

( )

−∆−−∆

=2

2

, 2

Pryyw

BBz LP

Lv σσσ

η

+−

∆=4

31

34

4

18 w

y

w

y

BL

PRQ

σσ

σσ

ηπ

2Re6

)Re6.(64

B

B Hef

+=B

B

Dv

ηρ=Re 2

2

B

y DHe

ηρσ

=

99

Reynolds Number, Power Law fluid: Critical Value

Critical value for transition from laminar to turbulent flow depends on power law index. Different approaches/experiments used. (note, other factors for NNF such as fluid elasticity can also affect this critical

transition)

Newtonian fluids

Steffe, p 108.

100

Example

A Newtonian oil (specific gravity = 0.9, µµµµ = 0.1 Pas) is being pumped into a mixer at v=1 m/s along a pipe of diameter of D=0.20 m. The process has been altered so that an oil-in-water emulsion is used in place of the oil. The scientist claims they have matched the viscosity so that flow conditions will be the same.

What is the wall shear rate ?What is the Re ?What flow regime is it in?

You observe differences in the pressure drop for the emulsion, and later discover that the scientist matched the viscosity using a rheometer at a shear rate of 1s-1. But when you measure the emulsion rheology using a rheometer, you find it follows a power-law with K=0.1 Pas0.5 and n = 0.5.

What flow regime is the emulsion in ?

Determine apparent and actual wall shear rate, the apparent viscosity at the wall shear rate and Re.

Hint: shear rate = 8v/D

101

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.1 1 10 100

Vis

cosi

ty (

Pas

)

Shear rate

oil

Emulsion

The feed line is being altered to inject an oil-in-water emulsion at the same velocity (V=1 m/s , D= 0.20 m). The emulsion was measured on a rheometer to be described by a power-law with K=0.1 Pas0.5 and n = 0.5. What flow regime do you reckon it is in ? Estimate the Re ?

102

MFEE (energy equation): Power Law fluid

MFEE can be applied to time independent non-Newtonian fluids. The friction term (hf ) can be calculated in a similar way

For Laminar flow of Newtonian fluids:

For Laminar flow of Power Law fluids:

sfq hhzg

v

g

Phz

g

v

g

P ++

++=+

++ 2

222

1

211

2

2α

ρα

ρ

Re/64 ;2

2

==⇒ fg

v

D

Lfh f

g

Ph f ρ

∆=

2zvL 23

DP

µ=∆

PLf fg

v

D

Lfh Re/64 ;

2

2

==⇒

( )( )Dv

nn

nww

w

K

R

L

84

13w

2P

+==

=∆

γγσ

σ

&

&

103

Note on nomenclature:

I use:

σ = shear stressσy = yield stressRe = Reynolds No.

Also used in various texts and/or tutorial problems:τ = shear stressτ0 = τy = yield stress

I also use µ = viscosity of Newtonian fluidη = viscosity of Non-Newtonian fluid (e.g. Shear rate dependent)

Steffe’ uses:

σ = shear stressσ0 = yield stressNRe = Reynolds No.

104

φρ

σ2

eunit volumper energy kinetic

pipein stressshear

2'

2===

vf w

MFEE – calculating friction factor

Also recall:

f’ = Fanning friction factor (Steffe uses this but uses the symbol f)

f = Moody friction factor (which we use)

φρσ

8'48

2=== f

vf w

105

Laminar Pipe Flow of Non-Newtonian Fluids

Yield stress fluids:A minimum stress is required before flow occurs.

106

Laminar Flow in a Pipe of NNF: Yield Stress Fluids

Can you think of examples of Yield stress fluids ?

These are fluids that behave as a solid (i.e. don’t flow) until a stress is applied to a sample that is greater than a critical value. Constitutive models used to describe such fluids:

Bingham model

Herschel-Buckley modeln

y Kγσσ &+=

γησσ &By +=

107


Pipe flow of Yield Stress Fluids :• the fluid may NOT flow. This will be the case if:

σw < σy (wall shear stress is below the yield stress)

g

Ph f ρ

∆=

gD

Lh y

f ρσ4

<

gD

Lh

gR

Lh

R

L

wf

wf

w

ρσ

ρσσ

4

2P

2

=⇒

=⇒∆=

hf in terms of wall shear stress:

108

Example

Crude oil is being pumped along a 1000m horizontal pipeline of inside diameter 0.5m. It has partially crystallised (i.e waxy crude oil) and has a yield stress of 500 Pa. Assuming negligible frictional and other losses, what is the minimum pressure drop required for flow to occur

R

L yw

σσ

2P

2L

PR =∆⇒∆=

Pa4000025.0

500)1000(2P ==∆⇒

Add a wick and we have a giant candle

R

L y

yw

σσσ

2P=∆⇒

=

109


Bingham model

Flowing regionσ > σ ySolid plugσ < σ y

P

20 ∆

=L

R yσ

Herschel-Buckley model

2L

PR∆=wσ

ny Kγσσ &+=

γησσ &By +=

2L

Pr∆=σ

110

General equation for Q for time-independent fluids:

In a Bingham Fluid, shear rate is defined as:

The function is discontinuous because there is no shearing flow is regions when the stress is less than the yield stress

(inner portion of tube)

(outer portion of tube)

Need to integrate for each region to get Q or average velocity, and since the shear rate = 0 when the stress is below the yield stress.

Laminar Flow in a Pipe of Bingham: General Approach

B

yfη

σσσγ

−== )(&

wy

y

for )(

0for 0)(

σσση

σσσ

σσσ

<≤−

=

<<=

B

yf

f

∫

−=

w

y

dR

Q

B

y

w

σ

σ

ση

σσσ

σπ2

3

1

∫= w

dfR

Q

w

σσσσ

σπ 0

233

)(1

+−

∆=4

31

34

4

18 w

y

w

y

B L

PRQ

σσ

σσ

ηπ

111


Pipe flow of Yield Stress Fluids (e.g. Bingham or H-B) :• When fluid does flow, there will be a ‘solid’ plug-like core in the centre of the tube (where the shear stress < yield stress)

Velocity Profile, Bingham:

Velocity of plug determined by setting r=R0=σy2L/∆P

2L

Pr∆=σStress as fn of radius:

Radius (R0 ) when σ > σy:

P

20 ∆

=L

R yσ

( )B

ywplugBinghamz

P

Lv

ησσ

∆−

=2

,

( )

−∆−−∆

=2

2

, 2

Pryyw

BBz LP

Lv σσσ

η

113


Bingham model

2L

Pr∆=σ

Flowing regionσ > σ ySolid plugσ < σ y

P

20 ∆

=L

R yσ

Herschel-Buckley model

2L

PR∆=wσ

ny Kγσσ &+=

γησσ &By +=

( )

−∆−−+∆

=+

+n

n

nyyw

n

HBz LKP

Lv

1

1

1

11

1, 2

Pr

)1(

2 σσσ ( )n

n

KP

Lv

n

ywHBplugz 1

1

)1(

21

1

,+∆

−=

+σσ

..reduces to Bingham if K = ηB and n=1

114

MFEE – Mechanical For of energy equation: NNF

Bingham Reynolds number

Hedstrom number

Friction factor for Bingham, Laminar flow:

BB

Dv

ηρ=Re

HeDDDDvDv

B

By

B

w

B

Dv

BBflowonset 8

1

2812

8128

81 )/()()()(

Re ======η

ησρηγρ

ηρ

ηρ

ηρ &

2

2

B

y DHe

ηρσ

=

2Re6

)Re6.(64

B

B Hef

+= g

v

D

Lfh f 2

2

=

σ at onset of flow

115

( ) 4.0'Re4'

1 −= fLogf

µρ Dv=Re

BB

Dv

ηρ=Reγησσ &By +=

nKγσ &=

γµσ &=

MFEE – Friction factor, turbulent flow

'4 ff =

( )2.1

)1(

75.0

4.0'Re

4

'

12

nfLog

nf

n

PL −

= −

3.2)'(Re53.4)1(53.4'

1 −+−= fLogcLogf

B

Newtonian

Power Law

Bingham fluid

w

ycσσ

=

Just like turbulence of Newtonian fluids, equations are available for time independent inelastic fluids, i.e obtain f from equations or charts if you know Re; for example:

Steffe

( )n

nn

n

nn

PL K

Dv13

41

2

8Re +−

−

= ρ

116

MFEE – Friction factor, turbulent flow

'4 ff = Steffe

117

Steffe, p 132

f’

'4 ff =( )n

nn

n

nn

PL K

Dv13

41

2

8Re +−

−

= ρ

118

Steffe, p 130

f’

'4 ff =

2

2

B

yHe

DN

ηρσ

=

BB

Dv

ηρ=Re

119

MFEE – Kinetic energy correction α for Laminar flow

Steffe

Just like Newtonian fluids, corrections for kinetic energy term in MFEE is required due to the non-uniform flow field. For example:

120

MFEE – Kinetic energy correction α for Laminar flow

121

Non-Newtonian Pipe Flow, summary

Provided with the tools to predict the velocity profile of fluids

Can now account for Viscous Pipe Losses and minor Losses, as well as time-independent non-Newtonian fluids in pipe flows in the MFEE / extended Bernoulli

qhg

v

g

v

D

Lfz

g

uPz

g

uP −

+

=

++−

++ ∑ 2

K222

2

lossesminorall

L

2

2

22

22

1

21

11 α

γα

γ

=

=ΓD

v

R

Q 843π

∆P

Q or vavnKγσ &=

2L

PR∆=wσFluid model: e.g.

Mea

sura

ble

Mea

sura

ble

122

Laminar Pipe Flow of Non-Newtonian Fluids

Step – by – step


123

NEWTONIAN: Substitute fluid constitutive model, i.e. Newtonians Law of viscosity:

⇒Parabolic velocity profile

Time independent non-Newtonian Fluid:

Power Law fluid:

From previous page

Integration gives:

No slip Boundary condition, i.e. r=R, v = 0…=>c =……

dr

dvzµσ −=

−∆=

22

z

r1

L 4v

R

PR

µ

n

zn

dr

dvKK

−=⇒= σγσ &

zn

nn

z dvKdrdr

dvK .

2L

Pr

2L

Pr 11

−=

∆⇒

−=∆=σ

( ) cvr

dvdrr z

n

zn

n n

n +−=+

⇒−=

∆ +∆

12LK

P1

1

2LKP

11

1

1


124

Integration gives:

Average Velocity

( ) ( )( )nn

nn

n rRv nn

z

111

12LKP

++

−= +∆

∫= dAA

vz v

Recall from last lecture for shell (annulus) of fluid:

dA= 2 π r dr

A = π R2

drrrRR

dAA

vnn

nn

z ∫∫++

−==121

2constant

v2

constant

R

nn

nn

nn rr

R0

12

12

12

12

1

+−

+

+++


125

( ) ( )( )

( ) ( )

( ) ( )

( )

+=

+−+

+−

++

+−=

+

+

++

++

∆

+∆

+∆

+

+

+∆

13

)13(

2)13(

13

2

)13(

)13(

12 v

11

11

11

1

12

1

2

1

2LKP

12LKP

12LKP

12

122

12LKP

n

nR

n

nnR

n

nR

n

Rn

RRR

nn

n

nn

n

nn

nn

n

nn

nn

n

nn

nn

nnRn

n

126

( )

= +34

13 4

R

Qn

nw π

γ&2L

PR∆=wσ

This puts the wall shear stress and shear rate as a direct function of the measurable pressure drop and flow rate.

For a Newtonian fluid, i.e. n=1, this reduces to

Therefore, to simplify equations for Power law fluid, this shear rate is used as a reference, i.e. apparent shear rate is Γ:

=3

4

R

QNewtonian π

γ&

( )Γ= + .413

nn

wγ&

=

=ΓD

v

R

Q 843π


nww

n KK γσγσ && =⇒=

127

( ) ( ) nn

n Rnn

11

132LKP v

+

+∆=⇒

Average Velocity for Power Law Fluid:

Volumetric Flow rate , Q:

Note, wall shear stress is directly proportional to the pressure drop:

and the wall shear rate can be determined by subbing into equation for Q:

( ) ( ) nn

n RRQ nn

11

132LKP2.A v

+

+∆==⇒ π

( ) ( ) nn

n RQ nn

131

132LKP

+

+∆= π

( )nww K γσ &=∆=

2L

PR

nw

w K

1

=⇒σγ& ( )

=⇒ +34

13 4

R

Qn

nw π

γ&

Power law model


128

General expression can be used for any fluid model :

(see derivation in Steffe)

Use Leibnitz’ rule to change variables to differentiate as function of σw and solve for

shear rate at the wall, gives:

∫= w

dfR

Q

w

ττσσ

σπ 0

233

)(1

)(ln

)(ln:

)( 41

43

xd

yd

dx

dy

y

xNB

d

df

wwww

=

Γ+Γ==σ

σσγ&

=

=ΓD

v

R

Q 843π

Γ

+=⇒Γ=

Γ

Γ+=

'4

1'3

)(ln

)(ln

'

1

)(ln

)(ln41

43

n

n

d

d

n

d

d

ww

ww

γσ

σγ

&

&

n’ can be obtained directly from ∆P and Q data !For Power Law, n’ = n

Laminar Pipe Flow of NNF: General Approach

Rabinowitsch-Mooney equation

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Fluid Mechanics - Lecture 7 - 9 _viscous + NNF Pipe Flowx