Laminar Flow
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Transcript of Laminar Flow
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Reynolds apparatus
==
VDR
Viscous forceViscous force
inertiainertia
Consider a free body of fluid of the form of an elementary parallelopiped
y=0
y=c
x
yz
p
Velocity
Profile
dx
dp
dy
d=
u
y
=
dxdp
dy
ud
2
2
= Substituting
pp x
x
+
yy
+
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The flow fluid is:
a. Newtonian
b. Isothermal
c. Incompressible (dose not depend on the pressure)
d. Steady and uniform flow
e. Laminar flow (the velocity has only one single component)
Steady Laminar Flow through Circular pipes
Laminar Flow through Inclined pipes
Laminar Flow through Annulus
Laminar Flow between parallel plates both
plates at rest
Laminar Flow between parallel flat plates One
plate moving and the other at rest COUETTE
FLOW
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2 2
1 1 2 21 2
1 2
Bernoulli's equation, 2 2
frictional head loss
f
f
p V p Vz z h
w g w g
p p ph
w w
+ + = + + +
= = =
P1
P2
R: radius, D: diameter
L: pipe length
w: wall shear stress
Pressure gradient must exist in the direction of flow to overcome the resistance to
flow
ph f
2[ ( )]( ) (2 )p p dp r r dx + =2
r
dx
dp=
dv
dr = ( )22
4
1rR
dx
dpv
=
r
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( )224
1rR
dx
dpv
=
=2
max 1R
rvv2
max4
1R
dx
dpv
=
Integrating vdAdQ =
44
1288D
dx
dpR
dx
dpQ
=
=
22
mean32
1
8
1D
dx
dpR
dx
dp
A
QV
=
==
maxmean2
1vV =
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22
mean32
1
8
1D
dx
dpR
dx
dp
A
QV
=
==
2
32
D
V
dx
dp =
4221
12832
D
QL
D
VLpp
==
2
21 32
wD
VL
w
pph f
=
=
From DARCY-WEISBACH EquationgD
fLVh f
2
2
=eR
f64
=
For a horizontal pipe the flowrate is:
directly proportional to the pressure drop,
inversely proportional to the viscosity,
inversely proportional to the pipe length, and
proportional to the pipe diameter to the fourth power.
With all other parameters fixed, an increase in diameter by a factor of 2 will increase the
flowrate by a factor of 24 = 16 the flowrate is very strongly dependent on pipe size.
1 20
2 4
p pdp R D
dx L
= =
( ) 2 20
4 8 8
fwh D wfV fV
L g
= = =
8*
0 fVV ==
VF == work doing of ratePower
(-dp/dx) = average force per unit volume of fluid
Total Force = (-dp/dx) x AL
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)( 21 ppQALVdx
dp=
( )L
pp
dx
dpAVQ 21 and , discharge since
=
=
The flowrate, Q, of corn syrup through the horizontal pipeshown in Fig. is to be monitored by measuring the pressuredifference between sections (1) and (2). It is proposed that Q=K.p where the calibration constant, K, is a function oftemperature,T, because of the variation of the syrups viscosityand density with temperature. These variations are given inTable. Plot K(T) versus T for 60FT160 F
Determine the wall shear stress and the pressure drop, for Q = 0.5ft3/s and T = 100 F and
For the above conditions, determine the net pressure force, and thenet shear force, on the fluid within the pipe between the sections (1)and (2).
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For steady, uniform flow, the momentum balance in s for the fluid
cylinder yields
with
and
we solve for to get:
( ) srwW = 2
( )2
rwzp
ds
d+=
22
r
ds
dhw
rz
w
p
ds
dw =
+=( )[ ]zwph +=
PiezometricPiezometric headhead
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Velocity for laminar flow in pipes
Using the result for , we substitute
Integration yields
dr
dv =
2
r
ds
dhw
dr
dv=
( )224
1rR
ds
dhwv
=
Glycerin at 20C flows upward in a vertical 75-
mm-diameter pipe with a centerline velocity of
1.0 m/s. Determine the head loss and pressure
drop in a 10-m length of the pipe. [ = 1260
kg/m3; = 1.50 N.s/m2]
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dx
dr
r
R1
R2
( )[2 ( ) ]dr r dr dxr
+ +
(2 )r dx (2 )p r dr ( )(2 )
pp dx r dr
x
+
(2 ) 2 (2 ) 2 ( ) 0p
p r dr p dx rdr r dx dr r dr dxx r
+ + + + =
dr
2 (2 ) (2 ) (2 ) 0pdx rdr dr dx dr r dx dr dr dxx r r
+ + =
Dividing throughout by the volume of the element Dividing throughout by the volume of the element (2(2r) r) drdr dxdxDividing throughout by the volume of the element Dividing throughout by the volume of the element (2(2r) r) drdr dxdx
0p
x r r
+ + = ( )1 0p r
x r r
+ =
Integrating Integrating w.r.tw.r.t. . rrIntegrating Integrating w.r.tw.r.t. . rr
2
12
r pr C
x
+ =
2
12
r p vr C
x r
=
v
r
=
2
1 2log4
e
r pv C r C
x
= +
Dividing by Dividing by rr and then Integrating and then Integrating w.r.tw.r.t. . rrDividing by Dividing by rr and then Integrating and then Integrating w.r.tw.r.t. . rr
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( )( )
( )2 2
2 12 2
2 2
2 1
1log
4 loge
e
R Rpv R r R r
x R R
=
Boundary Conditions:
r = R1 v = 0
r = R2 v = 0
Location of the point for maximum velocity:Location of the point for maximum velocity:Location of the point for maximum velocity:Location of the point for maximum velocity:
( )( )
2 2
2 1
2 1
1 10 2
4 loge
R Rv pr
r x r R R
= = +
( )( )
1/ 22 2
2 1
2 12 loge
R Rr
R R
=
The discharge Q passing through any cross-section of the annulus
2
1
2
R
R
Q r vdr= ( )
( )
22 2
2 14 4
2 1
2 18 log e
R RpQ R R
x R R
=
Mean velocity:Mean velocity:Mean velocity:Mean velocity:
( )( )
( )
2 2
2 12 2
2 12 22 12 1
1
8 log e
R RQ pV R R
x R RR R
= = +
If R1 = 0 laminar flow through circular pipe
If annulus is inclined replace by( / )p x [ ( ) / ]p wz x +
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B
Y
X
Z
dy
y
dx
( ) p dy dz( )
pp dx dy dz
x
+
( ) dx dz
( ) dy dx dzy
+
( ) ( ) 0p
pdydz p dx dydz dy dxdz dxdzx y
+ + + =
p
y x
=
2
2
p v
x y
=
v
y
=
Integrating Integrating w.r.tw.r.t. . yyIntegrating Integrating w.r.tw.r.t. . yy
2
1 2
1
2
p yv C y C
x = + +
C2 = 0 for v = 0 at y = 0
for v = 0 at y = B1 2
B pC
x =
( )212
pv By y
x =
VelocityVelocity isis parabolicparabolic
distributiondistribution curvecurve withwith
itsits vertexvertex atat midmid--wayway
betweenbetween platesplates
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Bdy
y
dx
( ) p dy dz ( ) p dp dy dz+( ) dx dz
( ) d dx dz +
vmax
2
max8
B pv
x =
Discharge Discharge qq per unit width:per unit width:Discharge Discharge qq per unit width:per unit width:
dq vdy= ( )20 0
1
2
B Bp
q vdy By y dyx
= = 3
12
B pq
x =
Mean velocity of flow Mean velocity of flow V:V:Mean velocity of flow Mean velocity of flow V:V:
2
12
q B pV
B x = =
max
2
3V v=
ReRe--arrangingarrangingReRe--arrangingarranging
2
12p V
x B
= Bdy
y
dx
( ) p dy dz ( ) p dp dy dz+( ) dx dz
( ) d dx dz +
x1
x2
L
( )2 2
1 1
1 2 2
12p x
p x
Vp p p x
B
= =
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( )1 2 2 1212 V
p p x xB
=
1 2 2
12 VLp p
B
= 1 2
2
12f
p p VLh
w wB
= =
Shear stress distributionShear stress distribution::Shear stress distributionShear stress distribution::
v
y
=
( )21
2
pBy y
y x
=
2
p By
x
=
Bdy
y
dx
( ) p dy dz ( ) p dp dy dz+( ) dx dz
( ) d dx dz +
02
p B
x
=
2
p By
x
=
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Two parallel flat plates kept 75 mm apart have
laminar flow of glycerine between them with a
maximum velocity of 1 m/s. Calculate the
discharge per meter width, the shear stress at
the plates, the difference in pressure between
two points 25 m apart, the velocity gradients at
the plates and velocity at 15 mm from the plate.
Take viscosity of glycerine as 8.35 poise.
2
1 2
1
2
dp yv C y C
dx = + +
Velocity distributionVelocity distribution::Velocity distributionVelocity distribution::
C2 = 0 for v = 0 at y = 0
for v = V at y = B
C2 = 0 for v = 0 at y = 0
for v = V at y = B12
V B dpC
B dx =
( )212
V dpv y By y
B dx =
VelocityVelocity dependsdepends onon
bothboth VV andand ((dpdp//dxdx))
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Bdy
y
dx
( ) p dy dz ( ) p dp dy dz+( ) dx dz
( ) d dx dz +
Moving
plateVV
Stationary plate
( )212
V dpv y By y
B dx =
Vv y
B= When dp/dx = 0
Simple (or plain) Simple (or plain) CouetteCouette flow or Simple shear flowflow or Simple shear flow
Shear stress distributionShear stress distribution::Shear stress distributionShear stress distribution::
dv
dy = ( )21
2
d V dpy By y
dy B dx
=
2
V dp By
B dx = +
At y = 010 2
V dp B
B dx = = +
At y = B/2V
B =
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At 0 =
At y = B
1
2
B Vy
B dp dx
= +
20 2
V dp B
B dx = =
B
Moving plateVV
Stationary plate
B/2
1
2
B Vy
B dp dx
= +
20 2
V dp B
B dx =
10 2
V dp B
B dx = +
( )212
V dpv y By y
B dx =
2
12
v y B dp y y
V B V dx B B = +
If
2
2
B dpP
V dx =
Dimensionless pressure gradient
1v y y y
PV B B B
= +
Non-dimensional velocity distribution
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++PP indicatesindicates pressurepressure
dropdrop inin thethe directiondirection ofof
flowflow
--PP indicatesindicates increaseincrease inin
pressurepressure inin thethe directiondirection
ofof flowflow
ItIt hashas beenbeen observedobserved thatthat
forfor PP --11,, therethere existsexists aa
backwardbackward flowflow forfor somesome
layerslayers ofof fluidfluid
ReasonReason:: adverseadverse
pressurepressure gradientgradient
ForFor aa givengiven PP,, thethe maximummaximum valuevalue ofof (v/V)(v/V) occursoccurs atat aa pointpoint wherewhere thethe shearshear stressstress
isis equalequal toto zerozero ii..ee.. atat
1
2
B Vy
B dp dx
= +
A viscous fluid (specific weight = 80 lb/ft3; viscosity = 0.03 lb.s/ft3 iscontained between two infinite, horizontal parallel plates as shown inFig. The fluid moves between the plates under the action of a pressuregradient, and the upper plate moves with a velocity U while the bottomplate is fixed. A U-tube manometer connected between two points alongthe bottom indicates a differential reading of 0.1 in. If the upper platemoves with a velocity of 0.02 ft/s, at what distance from the bottomplate does the maximum velocity in the gap between the two platesoccur? Assume laminar flow.
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A vertical shaft passes through a bearing and is lubricated withan oil having a viscosity of 0.2 N.s/m2 as shown in Fig. Assumethat the flow characteristics in the gap between the shaft andbearing are the same as those for laminar flow between infiniteparallel plates with zero pressure gradient in the direction offlow. Estimate the torque required to overcome viscousresistance when the shaft is turning at 80 rev/min.
Oil flows through the horizontal pipe shown in
Fig. under laminar conditions. All sections are
the same diameter except one. Which section of
the pipe (A, B, C, D, or E) is slightly smaller in
diameter than the others? Explain.
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A viscous fluid flows in a 0.10-m-diameter pipe
such that its velocity measured 0.012 m away
from the pipe wall is 0.8 m/s. If the flow is
laminar, determine the centerline velocity and
the flowrate.
An oil with a viscosity of = 0.40 Ns/m2 and density = 900 kg/m3
flows in a pipe of diameter D = 0.02 m
FIND (a) What pressure drop (p1-p2), is needed to produce a flowrateof Q= 2 10-5 m3/s if the pipe is horizontal with x1 = 0 and x2 = 10 m?
[p = 20.4 kPa]
(b) How steep a hill, , must the pipe be on if the oil is to flow throughthe pipe at the same rate as in part (a), but with p1 = p2
[ = 13.34]
(c) For the conditions of part (b), if p1 = 200 kPa, what is the pressureat section x3 = 5 m, where x is measured along the pipe?
[p1 = p2=p3 = 200 kPa]
For the horizontal pipe it is the work done by the pressure forces that overcomes For the horizontal pipe it is the work done by the pressure forces that overcomesthe viscous dissipation. For the zero-pressure-drop pipe on the hill, it is the changein potential energy of the fluid falling down the hill that is converted to theenergy lost by viscous dissipation. Note that if it is desired to increase the flowrateto Q = 10-4 m3/s with p1=p2, the value of is as sin = 1.15. Since the sine of anangle cannot be greater than 1, this flow would not be possible. The weight of thefluid would not be large enough to offset the viscous force generated for theflowrate desired. A larger diameter pipe would be needed.