8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE...
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Transcript of 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE...
![Page 1: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/1.jpg)
![Page 2: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/2.jpg)
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!!!
![Page 3: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/3.jpg)
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
![Page 4: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/4.jpg)
u’v’ = 0if duavg/dy = 0
-u’v’ = +if duavg/dy 0
Davies
![Page 5: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/5.jpg)
MYO
![Page 6: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/6.jpg)
Viscous sublayer very thin: for 3”id pipe and uavg = 10 ft/sec,about 0.002” thick
Viscous Sublayer
![Page 7: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/7.jpg)
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/
![Page 8: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/8.jpg)
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
![Page 9: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/9.jpg)
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
![Page 10: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/10.jpg)
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
![Page 11: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/11.jpg)
![Page 12: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/12.jpg)
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
![Page 13: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/13.jpg)
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!
![Page 14: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/14.jpg)
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
![Page 15: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/15.jpg)
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.
![Page 16: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/16.jpg)
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
![Page 17: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/17.jpg)
![Page 18: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/18.jpg)
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
![Page 19: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/19.jpg)
![Page 20: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/20.jpg)
Power law profile good near wall?
u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n
![Page 21: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/21.jpg)
Power law profile deviates close to wall
u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n
![Page 22: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/22.jpg)
u/Uc/l = (y/R)1/n = ([R-r]/R)1/n = (1-r/R)1/n
Power law profile good at r = 0?
![Page 23: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/23.jpg)
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
![Page 24: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/24.jpg)
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
![Page 25: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/25.jpg)
= 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
![Page 26: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/26.jpg)
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)}
![Page 27: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/27.jpg)
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)}
![Page 28: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/28.jpg)
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
![Page 29: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/29.jpg)
![Page 30: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/30.jpg)
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
=
+
![Page 31: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/31.jpg)
CONSERVATION of ENERGYChapter 4.8
Total energy Specific internal energy
![Page 32: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/32.jpg)
EQ. 4.56:
Conservation of Energy
also steady, 1-D, incompressible
only pressure work
Total energyInternal energy
![Page 33: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/33.jpg)
On surface: dWshear/dt = Fshear V = 0 since V = 0
![Page 34: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/34.jpg)
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
![Page 35: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/35.jpg)
1
![Page 36: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/36.jpg)
0 0 0 0
If no heat (Q) transfer then:
= 0
Mechanical energy
![Page 37: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/37.jpg)
![Page 38: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/38.jpg)
8-6: ENERGY CONSIDERATIONS IN PIPE FLOW
ENERGYEQUATION
dm/dt(Eq. 4.56)
![Page 39: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/39.jpg)
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.
![Page 40: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/40.jpg)
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)
![Page 41: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/41.jpg)
Energy Equation
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(Q/dt)/(dm/dt)= Q/dm
Dividing by mass flow rate, dm/dt, gives:
![Page 43: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/43.jpg)
p/ + V2/2 + gz represents the mechanical energy per unit mass at
a cross section
(u2 –u1) - Q/dm = hlTrepresents 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
![Page 44: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/44.jpg)
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.
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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
![Page 46: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/46.jpg)
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?
![Page 47: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/47.jpg)
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)
![Page 48: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/48.jpg)
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
![Page 49: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/49.jpg)
![Page 50: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/50.jpg)
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
![Page 51: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/51.jpg)
TURBULENT FLOW: Calculation (EMPIRICAL) of Head Loss
Dimension Analysis
p/ = hl (major head loss) Eq. (8.32)
![Page 52: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/52.jpg)
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)
![Page 53: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/53.jpg)
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
![Page 54: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/54.jpg)
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
![Page 55: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/55.jpg)
Fully rough zone where have flow separation over roughness elements and p ~ uavg
2
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~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
![Page 57: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/57.jpg)
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%
![Page 58: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/58.jpg)
Curves are from average values good to +/- 10%
![Page 59: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/59.jpg)
Original Data of Nikuradze
Stromungsgesetze in Rauhen Rohren, V.D.I. Forsch. H, 1933, Nikuradze
![Page 60: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/60.jpg)
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?
![Page 61: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/61.jpg)
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
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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
![Page 63: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/63.jpg)
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
![Page 64: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/64.jpg)
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
![Page 65: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/65.jpg)
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
![Page 66: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/66.jpg)
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 –
![Page 67: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/67.jpg)
![Page 68: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/68.jpg)
Minor Losses, hlm: fittings, bends, inlets, exits, changes in area
Minor losses not necessarily < Major loss , hl, due to pipe friction.
![Page 69: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/69.jpg)
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.
![Page 70: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/70.jpg)
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)
![Page 71: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/71.jpg)
Minor losses due to inlets and exits: hlm= p/ = K(uavg
2/2)
If K=1, p= uavg2/2
![Page 72: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/72.jpg)
vena contracta
unconfined mixingas flow decelerates
separation
![Page 73: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/73.jpg)
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!
![Page 74: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/74.jpg)
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
![Page 75: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/75.jpg)
Which exit has smallest Kexpansion?
V2 ~ 0
![Page 76: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/76.jpg)
MYO
![Page 77: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/77.jpg)
AR < 1
Why is Kcontraction and Kexpansion = 0 at AR =1?
![Page 78: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/78.jpg)
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!
![Page 79: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/79.jpg)
![Page 80: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/80.jpg)
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 !!!
![Page 81: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/81.jpg)
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 !!!
![Page 82: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/82.jpg)
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
![Page 83: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/83.jpg)
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)
![Page 84: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/84.jpg)
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 -
![Page 85: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/85.jpg)
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![Page 87: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/87.jpg)
V1 = 20 cm/secA1 = 40 cm
DIFFUSERS
good bad
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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
![Page 89: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/89.jpg)
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 )
![Page 90: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/90.jpg)
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.
![Page 91: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/91.jpg)
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
![Page 92: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/92.jpg)
Head loss of a bend is greater than if pipe was straight (again due to separation).
![Page 93: 8-4: SHEAR STRESS DISRIBUTION IN FULLY DEVELOPED PIPE FLOW rx = (r/2)(dp/dx) + c 1 /r FROM FORCE BALANCE c 1 = 0 or else rx becomes infinite rx.](https://reader036.fdocuments.in/reader036/viewer/2022062401/5a4d1b687f8b9ab0599b1dd7/html5/thumbnails/93.jpg)