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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9
A classical introduction to viscousflow: fluid between two plates
Fluid is initially at rest between two parallel plates
y
x
10
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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