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CHEM-E7160Fluid flow in process units

Lecture 2

Summary, lecture 1

• Energy balance contains all terms related toenergy, and are thus valid in all cases

• For fluid flow problems, it is convenient to splitenergy balance in two parts; mechanical energyand thermal energy

• Most fluid flow problems can be solved byusing mechanical energy balance only

• Bernoulli equation is a special case of these

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Summary, lecture 1• In practical cases, the Bernoulli equation must

be extended to contain shaft work by pumps andfriction in piping systems

• Pressure drop originates from straight pipecontributions and local losses. These can becalculated with appropriate correlations ortables.

• Pump curve describes pressure increase andpiping curve pressure drop as a function of flowrate. At the operating point these are equal

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4

W

zza

( ) ( ) ( ) f2aa

2bbababpp hvv21zzgppW r+a-ar+-r+-=rh

Increasespressure

Elevatesfluid

May increasevelocity

Compensatesfrictional losses

Pump

)zz(h2

v2

v)pp(g1

gW

abf

2a

a

2b

babpp -+

úúû

ù

êêë

é+÷

÷ø

öççè

æa-a+

r-

=h

Review of lecture 1

Mechanical energy balance, units (m)

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Pressure increase by the pump and pressure drop in thepiping

DpPump curve

åD=r+D+D+D=D phpppp fkinpotp

Volumetric flow m3/h

Piping curve

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Pressure dropIs often divided in two parts:

2v

DLh

2

if ÷øö

çèæ z+

Dxr=r å

Loss in the straight pipe Local loss coefficients

Dependency in turbulent flow Dp µ v2

Coefficients x and z can be estimated based on flowconditions and piping system structure. The pressure dropterm remains similar, but these parameters vary.

Outline of lecture 2

• Review of fluid flow basics:– shear stress– shear rate– viscosity– laminar flow (examples)

• Momentum balances- Hagen-Poiseuille equation

Learning outcomes for this lecture

• Know some fluid dynamic related terms: shearstress, shear rate, viscosity

• Understand momentum balance concept• Is capable of solving simple fluid flow problems

with momentum balances

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A classical introduction to viscousflow: fluid between two plates

Fluid is initially at rest between two parallel plates

y

x

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Flow between platesThe upper plate remains stagnant, but the lower ispulled to the right with force FA velocity profile is starting to appear between theplates. Fluid adjacent to the lower plate moves with theplate velocity, and drags fuid above it to the right. Fluidadjacent to the upper plate remains stagnant

F

y

x

Make a schetch of velocity profiles shortly after start and after long time

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Shear stress, shear rate and viscosityVelocity gradient (shear rate) dv/dy isproportional to the force F and inverselyproportional to the plate area A. Finalvelocity profile is linear (why?) dy

dvAF xµ

y

x F

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Ratio between force F and plate area A iscalled shear stress (t).What is its dimension?http://presemo.aalto.fi/fluidflow2What is the difference between shearstress and pressure?

t=AF

y

xF

Shear stress, shear rate and viscosity

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Shear stress

dydv

m-=t Velocity profilegradient, shearrate

Shear stress,force dividedby area

Proportionality coefficient

Linear momentum is transported tolower velocity direction (compare toheat and mass transfer!)

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Newtonian fluid

For Newtonian fluids, shear stress is directlyproportional to velocity gradient (shear rate). Inaddition, it is zero if velocity gradient is zero.

Viscosity does not depend shear stress, shear rate, ortime for Newtonian fluids. It can still depend (andoften does) on temperature, composition etc...

dydvx

yx m-=tlinear momentum in x –direction(velocity and momentum are vectorquantities)will be transported iny –direction

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(Dynamic) viscosity

dydv

AF

m=

h

Dimension in SI units (mass, time, length)?

m or

What is kinematic viscosity and its SI unit?

http://presemo.aalto.fi/fluidflow2

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dydv

AF

m=

22

2

2 mskg

ms

mkg

mN:

AF

=×

=

Dimension of dynamicviscosity in SI -units?

Force F ( ) N Area A ( ) m2

Velocity v ( ) m/s Distance y ( ) m

s1

ms/m:

dydv

=mskg

s1

mskg

:2=m

Derived unit

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Kinematic viscosity

Dynamic viscosity divided bydensity

Dimension m2/s, the same as fordiffusion coefficient and thermaldiffusivity (one concequence oftransport phenomena analogy)

rm

=n

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Flow in a circular pipe

We want to know thevelocity profile in acircular pipe.Furthermore, we areinterested in pressure dropas a function of geometry,flow rate, and physicalproperties

How this problem couldbe approached?

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Flow in a circular pipeWe need to formulate a linear momentum balance for adifferential control volume

Control volume is betweentwo cylindrical shaped areasshown in the picture. Theshape of this control volumeis obtained from”knowledge” that the flowdepends only on the distancefrom the wall, not angularposition

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Flow in a circular pipe

Inner diameter of thecontrol volume r (distancefrom the center)

Outer diameter r+dr

Length DL

rdr

L

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Linear momentum balance

• What is linear momentum?• What is its dimension?• What are the dimension of the terms in time

dependent linear momentum balances?• How many of them we need in a general case?

Discuss first with a friend and then together

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Linear momentum balanceLinear momentum is mass times velocityLet’s write a time dependent balance for linearmomentum

( ) mavmdtdvm

dtdmv

dtmvd

+=+= &

mass flow times velocity mass timesacceleration is force

Dimension of each term much thus be the same as for force,i.e. N or kgm/s2

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Momentum balance

Momentumflow in

Momentumflow out

Shear stress

Pressure at theinlet p0

Pressure at theoutlet, pL

• Assumptions:- Pipe diemeter assumed relatively small (hydrostatic pressure

neglected)- Steady state- Incompressible flow

• Source terms: Flow in and out, pressure affecting at bothends, and shear stress from the core side and outside

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Force due to shear stress.Note that area is not constant!

AF t=( ) rrzLr2in tDp=( ) rrrzLr2out D+tDp=

What is the ”plate” that is pulled with a force here?

Now try to formulate yourself the linear momentum balance!Momentum flow in and out, force due to pressure (in and out)…

F pA=

Force due to pressuresat inlet and outlet

Area of a ”ring” wherepressure affects

vmF &=Momentum flowwith the fluid

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Momentum flow withthe fluid

vmF &=( )( ) 0zzz vvrr2in =rDp=

( )( ) Lzzz vvrr2out =rDp=

Incompressible flow, constant diameter ® r constant ®momentum flow in = momentum flow out

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Force due to pressureat the inlet and at the

outlet

pAF =( ) 0prr2in Dp=( ) Lprr2out Dp=

What would be different in vertical pipe?

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Balance

( )( ) Lrrrz

0rrz

rpr2Lr2

rpr2Lr20

Dp-tDp-

Dp+tDp=

D+

Momentum transfer due to viscous forces causepressure drop

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Balance( ) ( ) ( )L0rrzrrrz pprr2Lr2Lr2 -Dp=tDp-tDp D+

( ) ( )r

Lpp

rrr

0rlim L0rrzrrrz

÷øö

çèæ

D-

=÷÷ø

öççè

æ

D

t-t

®DD+

( ) rLppr

drd L0

rz ÷øö

çèæ

D-

=t

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Shear stress (a closure model for the balance)

( ) rLppr

drd L0

rz ÷øö

çèæ

D-

=tdrdv

rz m-=t

Second order differential equation (d2v/dr2)® two boundary conditions needed

Which would they be?

rLpp

drdvr

drd L0 ÷

øö

çèæ

D-

=÷øö

çèæ m-

One solution method

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rLpp

drdvr

drd L0 ÷

øö

çèæ

D-

=÷øö

çèæ m-

Assume a velocity profile, e.g. second order polynomialv = a + b × r + c × r2. This choise is based on ”educated guess”

Insert that into the differential equation and use also boundaryconditions t(r=0) = 0 → dv/dr(r=0) = 0, and v(r=R) = 0

Use these three equations to solve for three unknownparameters a, b and c

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Laminar flow

Velocity profile is parabolic in well developed laminar pipe flow

This is one of the few examples, where hand calculation can be used toanalytically derive pipe friction coefficient for mechanical energy balances.In general, those are either measured experimentally or calculated withcomputational fluid dynamics (discussed later on this course)

-1

-0.5

0

0.5

1

0 0.2 0.4 0.6 0.8 1

( ) ÷÷ø

öççè

æ÷øö

çèæ-÷÷

ø

öççè

æDm-

=-÷÷ø

öççè

æDm-

-=2

2L022L0

Rr1R

L4ppRr

L4ppv

Calculate total volumetric flow and average velocityfrom local velocities

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2vR

L8pp

RVv max2L0

2 =m-

=p

=&

ò p=R

0

rdr2vV& 2RVvp

=&

( ) ( ) ÷÷ø

öççè

æ÷øö

çèæ-÷÷

ø

öççè

æDm-

=-÷÷ø

öççè

æDm-

-=2

2L022L0

Rr1R

L4ppRr

L4pprv

Solve for pressure drop as a function of pipediameter

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2vR

L8pp

RVv max2L0

2 =Dm-

=p

=&

vD

L32vR

L8p 22

Dm=

Dm=D This is the well-known

Hagen-Poiseuille law

Solve Hagen-Poiseuille law for Darcy friction factor

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2v

DLph

2

fD

rx=D=r

vD64rm

=x

vD

L32p 2

Dm=D

This was presentedalready in previous lecturewith Moody diagramRe

64=

Example of connection between momentumbalances and mechanical energy balances

Summary• Shear stress appears when force is applied to fluid

adjacent to surface. Pressure is force perpendicularto a surface.

• Proportionality factor between shear stress andshear rate (velocity gradient) is called viscosity

• Fluid flow problems can be solved with momentumbalances. Analytical solutions possible only forsome simple cases, such as laminar pipe flow(Hagen-Poiseuille equation)

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