Laminar Flow

download Laminar Flow

of 20

description

Laminar flow through pipes

Transcript of Laminar Flow

  • 9/6/2014

    1

  • 9/6/2014

    2

    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

    +

  • 9/6/2014

    3

    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

  • 9/6/2014

    4

    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

  • 9/6/2014

    5

    ( )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 =

  • 9/6/2014

    6

    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

  • 9/6/2014

    7

    )( 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).

  • 9/6/2014

    8

    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

  • 9/6/2014

    9

    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]

  • 9/6/2014

    10

    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

  • 9/6/2014

    11

    ( )( )

    ( )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 +

  • 9/6/2014

    12

    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

  • 9/6/2014

    13

    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

    = =

  • 9/6/2014

    14

    ( )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

    =

  • 9/6/2014

    15

    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))

  • 9/6/2014

    16

    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 =

  • 9/6/2014

    17

    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

  • 9/6/2014

    18

    ++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.

  • 9/6/2014

    19

    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.

  • 9/6/2014

    20

    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.