8-4: shear stress disribution in fully developed pipe flow

8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW

rx = (r/2)(dp/dx) + c1/r FROM FORCE BALANCE

c1 = 0 or else rx becomes infinite

rx = (r/2)(dp/dx) wall = (R/2)(dp/dx)

HAVE NOT USED LAMINAR FLOW RELATION: rx = (du/dr)

rx = (r/2)(dp/dx) wall = (R/2)(dp/dx)

TRUE FOR BOTH LAMINAR AND TURBULENT FLOW!!!

Even though rx = (du/dr) + u’v’ in turbulent flow!!!

u’v’DO NOT KNOW AS FUNCTION OF MEAN VELOCITY!!!!!!!!

u’v’ modeled as duavg/dy and lm2(duavg/dy)2

but and lm vary from flow to flow and from place to place within a flow

u’v’ = 0if duavg/dy = 0

-u’v’ = +if duavg/dy 0

Davies

MYO

Viscous sublayer very thin: for 3”id pipe and uavg = 10 ft/sec,

about 0.002” thick

Viscous Sublayer

rx = (du/dr) + u’v’rx / = (du/dr) + u’v’

[rx / ]1/2 has units of velocity[wall / ]1/2 = u*

(friction of shear stress velocity) u*~u’ near the wall

Re = uavgD/

Re near wall = u*y/

u’v’ = 0 at the wall (no eddies) and at centerline (no mean velocity gradient) and approximately constant around

r/R=0.1

u’v’/(u*)2

u’v’/(u*)2

Data from Laufer 1954

y+= yu*/

y/a

u* = [wall / ]1/2

By a careful use of dimensional analysis in 1930 Prandtl deduced that near the wall:

u = f(, wall, , y)u+ = u/u* = f(y+= yu*/)

LAW OF THE WALL – inner layer

Near wall variables:

y+ = yu*/ y = R-r

u+ = u/u*

u+/y+ = (u/u*)(/yu*)

= u /(yu*u*) = (u /y)(wall/)-1

= (u/y)(wall)-1 wall = du/drwall = u/y

u+/y+ ~ 1 0y*5-7 viscosity dominatesViscous Sublayer

Rr

In 1937 C. B. Millikan showed thatin the overlap layer the velocity must vary

logarithmically with y:

u/u* = (1/)ln(yu*/) + BExperiments show that:

0.41 and B0.5u/u* = 2.5 ln(yu*/) + 5.0

LOGARITHMIC OVERLAP LAYER

For pipe at center:u/u* = 2.5 ln(yu*/) + 5.0 (a)Becomes: Uc/l/u* = 2.5 ln(Ru*/) + 5.0 (b)

(b)-(a):(Uc/l-u)/u* = 2.5ln[(yu*/)/ (Ru*/)] (Uc/l-u)/u* = 2.5ln[y/R] Eq.(8-21)

y = 0 no good!

Aside ~Again using dimensional reasoning

in 1933 Karman deduced that far from the wall:

u = f(, wall, )independent of viscosity,

is boundary layer thickness

(u=U for y > ) (U-u)/u* = g(y/)

VELOCITY DEFECT LAW

u/u* = 2.5 ln(yu*/) + 5.0Serendipitously, experiments show that the overlap layer extends throughout most of the velocity profile particularly for decreasing pressure gradients. The inner layer that is governed by the law of the wall typically is less than 2% of the velocity profile.

PIPE VELOCITY PROFILES

0

10

20

30

40

50

0 200 400 600 800 1000

PIPE VELOCITY PROFILES

0

10

20

30

40

50

1 10 100 1000

u+

= u/u*

u+

= u/u*

u+=y+

u+=y+

U*=2.5lny+ + 5

U*=2.5lny+ + 5

Pipe r=0

Empirically for smooth pipes it is found that the mean velocity profile can be expresses to a good approx. as: u(r)/Uc/l = (y/R)n = ([R-r]/R)1/n = (1-r/R)1/n

n = -1.7 + 1.8log(ReUcenterline)From experiment

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1

r/R

u/U

Laminar Flowu/Uc/l = 1-(r/R)2

n=6-10

Power law profile good near wall?

u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n

Power law profile deviates close to wall

u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n

u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n

Power law profile good at r = 0?

u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n

du/dy = Uc/l (1/n)R1/ny(1/n)-1 at y = 0 blows up

Power Law profile no good at r = 0

TO PROVE: V/Uc/l = {2n2/(2n+1)(n+1)}Eq. (8.24) Fox et al. Strategy: find V = Q/A

u/Uc/l = (y/R)1/n note: u is a f(y), V is not a f(y)

u /Uc/l = (y/R)n = ([R-r]/R)1/n = (1-r/R)1/n

Q = 2ru(r)dr = 2rUc/l(y/R)1/ndr from 0-R

y=R-r; r=0, y=R -dy=dr; r=R, y=0

Q=2rUc/l(y/R)1/ndr = 2(R-y)Uc/l(y/R)1/n(-dy) from R-0

= 2Uc/l{R(y/R)1/n(-dy) + (-y)(y/R)1/n(-dy)}from R-0

= 2Uc/l{R(y/R)1/n(dy) + (-y)(y/R)1/n(dy)} from 0-R

= 2Uc/l{R1-1/ny1/n+1/(1+1/n) - R-1/ny1+1/n+1/(1+1/n +1)}|0R

TO PROVE: V/Uc/l = {2n2/(2n+1)(n+1)}

Q=2rUc/l(y/R)1/ndr = 2(R-y)Uc/l(y/R)1/n(-dy) from R-0

Q: = 2Uc/l{R1-1/ny1/n+1/(1+1/n) - R-1/ny1+1/n+1/(1+1/n +1)}|0R

= 2Uc/l{R2/(1+1/n) – R2/(2+1/n)}

= 2Uc/l{nR2/(n+1) – nR2/(2n+1)}

= 2Uc/l{(2n+1)nR2/(2n+1)(n+1)–(n+1)nR2/(n+1)(2n+1)}

TO PROVE: V/Uc/l = {2n2/(2n+1)(n+1)}

Q:

= 2Uc/l{(2n+1)nR2/(2n+1)(n+1)–(n+1)nR2/(n+1)(2n+1)}

= 2Uc/l{nnR2/(2n+1)(n+1)}

V = Q/R2= 2Uc/l{n2/(2n+1)(n+1)}

V/Uc/l = uavg/uc/l = {2n2/(2n+1)(n+1)} Q.E.D.

TO PROVE: V/Uc/l = {2n2/(2n+1)(n+1)}

V/Uc/l = uavg/uc/l = 2n2/(n+1)(2n+1) = 2n2/(2n2+n+2n+1) = 2/(2+3/n +1/n2)As n , uavg /uc/l 1

For laminar flow uavg /uc/l = 1/2

With increasing n (Re)velocity gradient at wall

becomes steeper, wall increases

CONSERVATION of ENERGYChapter 4.8

rate of changeof total energy

of system

net rate ofenergy addition

by heat transfer to fluid

net rate ofenergy addition

by work done on fluid

=

+

CONSERVATION of ENERGYChapter 4.8

Total energy Specific internal energy

EQ. 4.56:

Conservation of Energy

also steady, 1-D, incompressible

only pressure work

Total energyInternal energy

On surface: dWshear/dt = Fshear V = 0 since V = 0

Wnormal = Fnormal ds work done on area element

d Wnormal / dt = Fnormal V = p dA V rate of work done on area element

- d Wnormal / dt = - p dA V = - p(v)dA V

+ for going in + for going out

1

1

0 0 0 0

If no heat (Q) transfer then:

= 0

Mechanical energy

8-6: ENERGY CONSIDERATIONS IN PIPE FLOW

ENERGYEQUATION

dm/dt(Eq. 4.56)

Velocity not constant at sections 1 and 2. Need to introduce acorrection factor, , that allows use of the average velocity, V, to compute kinetic energy at a cross section.

LAMINAR FLOW: the kinetic energy coefficient, , = 2 (HW 8.66)

FOR TURBULENT FLOW: the kinetic energy coefficient, (Re), 1 (HW 8.67)

More specifically, = (Uc/l/V)3(2n2)/[(3+n)(3+2n)]n = 6, = 1.08; n=10, =1.03

V = f(y)

V f(y)

Energy Equation

(Q/dt)/(dm/dt)= Q/dm

Dividing by mass flow rate, dm/dt, gives:

p/ + V2/2 + gz represents the mechanical energy per unit mass at

a cross section

(u2 –u1) - Q/dm = hlT

represents the irreversible conversion of mechanical energy per unit mass to thermal energy , u2-u1, per unit mass and the loss of

energy per unit mass, -Q/dm, via heat transfer

Energy lossper unit mass

of flowing fluid

“one of the most important and useful equationsin fluid dynamic”

What are units of hlT ? What are units of HlT?

Energy lossper unit weight

of flowing fluid

Fox et al.

Energy lossper unit mass

of flowing fluid

“one of the most important and useful equationsin fluid dynamic”

Units of hlT are L2/t2, If divide by g, get units of length for HlT Unfortunately both hlT and HLT are referred

to as total total head loss.

Energy lossper unit weight

of flowing fluid

Energy lossper unit mass

of flowing fluid

“one of the most important and useful equationsin fluid dynamic”

What provides energy input to match energy loss?

Energy lossper unit mass

of flowing fluid

“one of the most important and useful equationsin fluid dynamic”

Pumps, fans and blowersprovide ppump/ = hpump

(pump supplies pressure to overcome head loss, not K.E.)(power supplied by pump = Q ppump [m3t-1Fm-2 = Wt-1)

Convenient to break up energy losses, hlT, in fully developed pipe flow to major loses, hl, due to frictional effects along the pipe and minor losses, hlm, associated with entrances, fittings, changes in area,…

For fully developed flow of constant pipe area:

=

0 if pipe horizontal

LAMINAR FLOW: Calculation (THEORETICAL) of Head Loss

Eq. 8.13c

(p1 –p2)/ = p/ = hl (major head loss) Eq. (8.32) p = hl

units of energy per unit mass.

hl = p/ = (64/Re)(L/D)(uavg2/2);

(p/L)D / (1/2 uavg2) = fD = 64/Re

TURBULENT FLOW: Calculation (EMPIRICAL) of Head Loss

Dimension Analysis

p/ = hl (major head loss) Eq. (8.32)

has to be determined by experiment.As might be expected, experiments find that the nondimensional major head loss, hl/(V2), is directly proportional to L/D.

Rememberfor laminar flow

TURBULENT FLOW: Calculation (EMPIRICAL) of Head Loss

hl [L2/t2] = (L/D)f(1/2)uavg2 Hl [L] = (L/D)f(1/2)(uavg

2/g)

f must be determined experimentally

f = 2 (Re,e/D)

TURBULENT PIPE FLOW*: hl [L2/t2] = (L/D)f(1/2)V2

f = hl /{(L/D)(1/2)V2} p/ = hl (major head loss) Eq. (8.32)

fF = (p/L)D/{(1/2) V2}

LAMINAR PIPE FLOW*: hl [L2/t2] = (64/Re)(L/D)(1/2)V2 EQ. 8.33

p/ = (64/Re)(L/D)(1/2)V2 (p/L)D/{(1/2) V2} = fF = 64/Re

f(Re,e/D) determine experimentally

f (Re) derived from theory; independent of roughness (if e/D 0.1, then likely “roughness” important even in laminar flow)

V = uavg

In laminar flow and not “extremely” rough flow moves overroughness elements, in turbulent flow if roughness elements“stick out” of viscous sublayer get separation and p ~ uavg

2

Fully rough zone where have flow separation over roughness elements and p ~ uavg

2

~1914

f = = (p/L)D/{(1/2) V2}

Note: y-axis is not log so laminar and turbulent relationships not straight lines

Similarity of Motion in Relation to the Surface Friction of Fluids Stanton & Pannell –Phil. Trans. Royal Soc., (A) 1914

fF = (p/L)D/{(1/2) V2} Darcy friction factor

ReD = UD/

For new pipes, corrosionmay cause e/D for old pipesto be 5 to 10 times greater.

Curves are from average values good to +/- 10%

Curves are from average values good to +/- 10%

Original Data of Nikuradze

Stromungsgesetze in Rauhen Rohren, V.D.I. Forsch. H, 1933, Nikuradze

Question?Looking at graph – imagine that pipe diameter and

kinematic viscosity and density is fixed.Is there any region where an increase in uavg

results in an increase in pressure drop?

Question?Looking at graph – imagine that pie diameter and

kinematic viscosity and density is fixed.Is there any region where an increase in uavg

results in an increase in pressure drop?

Instead of non-dimensionalizing p by ½ uavg

2; use D3 /( 2L) Laminar flow

Turbulent flow

transition

From Tritton

Laminar flow: fF = 64/ReD

Turbulent flow: fF depends of ReD and roughness, e/D

fF = -2.0log([e/D]/3.7 + 2.51/(RefF0.5)]

If first guess is: fo = 0.25[log([e/D]/3.7 + 5.74/Re0.9]-2

should be within 1% after 1 iteration

For turbulent flow in a smooth pipe and ReD < 105,can use Blasius correlation: f = 0.316/ReD

0.25 which can be rewritten as wall = 0.0332 V2 (/[RV])1/4)

For laminar flow (Re < 2300), wall = 8(V/R)sometimes written as uavg (= V) is not a function of y

For turbulent flow and Re < 105; can use Blasius correlation: fF = 0.316/Re0.25

Which can be rewritten as:

wall =0.0332 V2 (/[RV])1/4

fF = (p/L)D/{(1/2) V2}wall = (R/2)(dp/dx)

wall/(1/8 V2) = fF= 0.316/Re0.25

fF = (p/L)D/{(1/2) V2}= (p/L)2R2/2{ ½ V2}fF = 4wall /{(1/2) V2} = wall /{(1/8) V2} = 4 fD

PROOF

For turbulent flow and Re < 105; can use Blasius correlation: f = 0.316/Re0.25

Which can be rewritten as:

wall =0.0332 V2 (/[RV])1/4

PROOF

wall/(1/8 V2) = 0.316 1/4 / (V1/4 D1/4) wall = (0.0395 V2) [ 1/4 / (V1/4 (2R)1/4)wall = (0.0332 V2) [ / (VR)]1/4 QED

wall = (R/2)(dp/dx) = (R/2)(p/L) = (R/2)(hl/L) wall = (R/2) (/L)hl = (R/2) (/L) f(L/D)(V2/2)

f(ReD, e/D)

wall = (R/2) (/L) f(L/2R)(V2/2) = -(1/8)f(V2)

wall = -(1/8)f(V2) = -(1/8)(V2) 0.316/(VD/)0.25

wall = -(1/8)(V2) 0.316/(VD/)0.25

wall = -(0.316/[(8)(2.25)])(V2)[0.25/(VR)]0.25

wall = -(0.0332)(V2)[0.25/(VR)]0.25 (Eq. 8.29)

Want expression for wall for turbulent flow and Re < 105 –

Minor Losses, hlm: fittings, bends, inlets, exits, changes in area

Minor losses not necessarily < Major loss , hl, due to pipe friction.

Minor losses traditionally calculated as: hlm = Kuavg2/2

where K is the loss coefficient or hlm = f(Le/D)uavg2/2

where Le is the equivalent length of pipe. Both K and Le must be experimentally determined* and will depend on geometry and Re, uavgD/. At high flow rates weak dependence on Re.

*Theory good for sudden expansion because can ignore viscous effects at boundaries.

These additional head losses are primarily due to separation,

Energy is dissipated by violent mixing in the separated zones. (pg 341 Fox…)

Energy is dissipated by deceleration after separation. (pg 66 Visualized Flow)

Minor losses due to inlets and exits: hlm= p/ = K(uavg

2/2)

If K=1, p= uavg2/2

vena contracta

unconfined mixingas flow decelerates

separation

Some KE at (2) is lost because of viscous dissipation when flow is slowed down (2-3)

For a sharp entrance ½ of the velocity head is lost at the entrance!

Entire K.E. of exiting fluid is dissipated through viscous effects, K = 1, regardless of

the exit geometry.

Only diffuser can help by reducing V.

Water, velocity = 14 cm/s, width of opening = 30 mm, Re = 4300

hydrogen bubbles

hydrogen bubbles

Which exit has smallest Kexpansion?

V2 ~ 0

MYO

AR < 1

Why is Kcontraction and Kexpansion = 0 at AR =1?

Minor losses dueto enlargements and contractions:

hlm= p/ = K(uavg

2/2)

Note: minor loss coefficients based on the larger uavg = V(e.g. uavg2 = V2 for inlet and uavg1 = V1 for outlet)• AR = 1, is just pipe flow so hlm = 0• AR = 0, is square edge inlet so K = ½ • AR = 0, is exit into reservoir so K = 1

Kexpan can be predicted!

Conservation of Mass/t CV dVol + CS V•dA = 0 (4.12)Assumptions – steady, one-dimensional, incompressible

V1A1 = V2V2 = V3A3

From geometry & “intuition”: p1 = p2 = pa = pb = pc; V1 = V2; A2 = A3

Kexpan. can be predicted !!!

Conservation of Momentum:FSx + FBx = /t CV Vx dVol + CS Vx V•dA (4.18a)Additional assumptions – no body forces, ignore shear forces

p2A2 – p3A3 = - V22A2 + V3

2A3

p1A3 – p3A3 = -V2V1A2 + V32A3 ? See MYO

p1A3 – p3A3 = -V3V1A3 + V32A3

From geometry & intuition: p1 = p2 = pa = pb = pc; V1 = V2; A2 = A3

Kexpan. can be predicted !!!

From geometry & intuition: p1 = p2 = pa = pb = pc; V1 = V2; A2 = A3

p1A3 – p3A3 = - V3V1A3 + V32A3 = V3A3 (V3 – V1)

Kexpan. can be predicted !!!

Conservation of Energy:p1+ V1

2/2 = p3+ V32/2 + gHl

Eq.8.29

hl

p1+ V12/2 = p3+ V3

2/2 + gHl

Hl = (p1-p3)/(g) + (V12-V3

2)/(2g)

p1A3 – p3A3 = V3A3 (V3 – V1)p1 – p3 = V3(V3 – V1)

Hl = V3(V3 – V1)/(g) + (V12-V3

2)/(2g)Hl = 2V3(V3 – V1)/(2g) + (V1

2-V32)/(2g)

Hl = {2V32 – 2 V3V1 + V1

2- V32}/(2g)

Hl = {V32 – 2 V3V1 + V1

2}/(2g) Hl = {V1 – V3}2 /(2g)

Hl = hl/g = {V1 – V3}2 /(2g)

Hlm has units of length or energy per unit weightHlm = hlm/g

hlm has units of velocity squared or energy per unit mass(both Hlm and hlm are referred to as head loss)

hlm = K V2/2K =hlm2/V1

2 = Hlmg2/V12

Hl = {V1 – V3}2 /(2g)K = (1 - V3/V1)2

V1A1 = V2A2 = V3A3 ; V3/V1 = A1/A3

K = (1 - A1/A3 ) 2 QED

For present problem:

In general -

V1 = 20 cm/secA1 = 40 cm

DIFFUSERS

good bad

DIFFUSERS

Diffuser data usually presented as a pressure recovery coefficient, Cp,

Cp = (p2 – p1) / (1/2 V12 )

Cp indicates the fraction of inlet K.E. that appears as pressure rise

It is not difficult to show that the ideal (frictionless) pressure recovery coefficient is:

Cp = 1 – 1/AR2, where AR = area ratio

V1 V2

Cpideal = 1 – 1/AR2

p1 + ½ V12 = p2 + ½ V2

2 (ideal)p2/ – p1/ = ½ V1

2 - ½ V22

A1V1 = A2V2

V2 = V1 (A1/A2)p2/ – p1/ = ½ V1

2 - ½ [V1(A1/A2)]2

p2/ – p1/ = ½ V12 - ½ V1

2 (1/AR)2

(p2 – p1)/( ½ V12) = 1 – 1/AR2

Cp = 1 – 1/AR2

Cp = (p2 – p1) / (1/2 V12 )

Relating Cp to Cpi and hlm

p1 / + ½ V12 = p2/ + ½ V2

2 + hlm

hlm = V12/2 - V2

2/2 – (p2 – p1)/ hlm = V1

2/2 {1 + V22/V1

2 – (p2 – p1)/ ( 1/2 V12)}

A1V1 = A2V2

Cp = (p2 – p1)/ ( 1/2 V12)

hlm = V12/2 {1 + A1

2/A22 – Cp}

hlm = V12/2 {Cpi – Cp} Q.E.D.

Smaller the divergence, Cp closer to Cpi. If flow too fast or angle too big may get flow separation.

Cp for Re > 7.5 x 104, “essentially” independent of Re

Head loss of a bend is greater than if pipe was straight (again due to separation).