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).