v 1 2 / 2 + p 1 / + gz 1 = v 2 2 /2 + p 2 / + gz 2 + h lt
DESCRIPTION
V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT. H E A D L O S S. h lT = h l + h m. Convenient to break up energy losses, h lT , in fully developed pipe flow to major loses, h l , due to frictional effects along the pipe and minor losses, h lm , - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/1.jpg)
V12/ 2 + p1/ + gz1= V2
2/2 + p2/ + gz2 + hlT
hlT = hl + hm
HEADLOSS
![Page 2: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/2.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,…
Minor losses not necessarily < Major loss , hl, due to pipe friction.
![Page 3: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/3.jpg)
Minor losses traditionally calculated as: hlm = KV2/2
(K for inlets, exits, enlargements and contractions) where K is the loss coefficient or
hlm = (Cpi – Cp)V2/2 (Cpi & Cp for diffusers)
where Cp is the pressure recovery coefficient or hlm = f(Le/D)V2/2
(Le for valves, fittings, pipe bends)
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.
![Page 4: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/4.jpg)
hlT = hl + hm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
![Page 5: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/5.jpg)
Minor losses due to inlets: hlm= p/ = K(V2/2); V2 = mean velocity in pipe
If K=1, p = V2/2
![Page 6: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/6.jpg)
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hm
![Page 7: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/7.jpg)
Head is lost because of viscous dissipation when flow is slowed down (2-3) and in violent mixing in the separated zones
For a sharp entrance ½ of the velocity head is lost at the entrance!
![Page 8: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/8.jpg)
vena contracta
unconfined mixingas flow decelerates
separationK = 0.78
K = 0.04 r/D > 0.15
r
D
![Page 9: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/9.jpg)
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V12/ 2 + p1/ + gz1= V2
2/2 + p2/ + gz2 + hlT
![Page 10: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/10.jpg)
• hlm head losses are primarily due to separation
• Energy is dissipated deceleration after separation leading to violent mixing in the separated zones
Minor losses due to sudden area change: hlm= p/ = K(V2/2); V2 = faster mean velocity pipe
![Page 11: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/11.jpg)
NOTE SOME BOOKS (Munson at al.): hlm = K V2/(2g) our Hlm!!!
![Page 12: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/12.jpg)
AR < 1
V2 V1
AR < 1 hlm = ½ KV2
fastest
![Page 13: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/13.jpg)
Why is Kcontraction and Kexpansion = 0 at AR =1?
AR < 1
![Page 14: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/14.jpg)
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
![Page 15: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/15.jpg)
Entire K.E. of exiting fluid is dissipated through viscous effects, V of exiting fluid eventually = 0
so K = 1, regardless of the exit geometry.
hlm = KV2/2
Only diffuser can help by reducing V.
Water, velocity = 14 cm/s, width of opening = 30 mm, Re = 4300
hydrogen bubbles
hydrogen bubbles
![Page 16: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/16.jpg)
Which exit has smallest Kexpansion?
V2 ~ 0
![Page 17: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/17.jpg)
MYO
K =1.0
K =1.0K =1.0
K =1.0
![Page 18: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/18.jpg)
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
![Page 19: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/19.jpg)
GRADUAL CONTRACTION
Where average velocity is fastest
AR < 1
![Page 20: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/20.jpg)
breath
![Page 21: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/21.jpg)
V1avg2/ 2 + p1/ + z1=V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hlm; hlm = p/ = (Cpi – Cp) 1/2V12
gentle expansions ~ diffusers
![Page 22: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/22.jpg)
DIFFUSERS
good bad
3 cm/sec 20 cm/sec
ugly
![Page 23: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/23.jpg)
…..
assumefully developed
P1 P2
?><=
Fully developed laminar flow, is: P1 greater, less or equal to P2?
What if fully developed turbulent flow?What if developing flow?
![Page 24: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/24.jpg)
P1, V1 P2, V2P1, V1 P3, V3
Is P2 greater, less than or equal to P1?Is P2 greater, less than or equal to P1?Is P likely to be greater, less than or equal to P?
![Page 25: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/25.jpg)
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
[ hlm = p/ = (Cpi – Cp) 1/2V12]
The greatest that Cp can be is Cpi, the case of zero friction.
![Page 26: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/26.jpg)
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
[ hlm = p/ = (Cpi – Cp) 1/2V12]
Cp will get from empirical data charts. It is not difficult to show that the ideal
(frictionless) pressure recovery coefficient is:Cpi = 1 – 1/AR2, where AR = area ratio
![Page 27: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/27.jpg)
Cpideal = 1 – 1/AR2
p1 + ½ V12 = p2 + ½ V2
2 (BE - ideal)p2/ – p1/ = ½ V1
2 - ½ V22
A1V1 = A2V2 (Continuity)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
Cpi = 1 – 1/AR2
Cp = (p2 – p1) / (1/2 V12 )
AR = A2/A1
> 1
![Page 28: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/28.jpg)
Relating Cp to Cpi and hlm
p1 / + ½ V12 = p2/ + ½ V2
2 + hlm (z1 = z2 = 0)
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) (Cp is positive & < Cpi)
hlm = V12/2 {1 - A1
2/A22 – Cp}
Cpi = 1 – 1/AR2
hlm = V12/2 {Cpi – Cp} Q.E.D. (see Ex. 8.10)
![Page 29: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/29.jpg)
hlm = (Cpi – Cp)V2/2; Cpi = 1 – 1/AR2
Cp
![Page 30: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/30.jpg)
N/R1 = 0.45/(.15/2) = 6
Cp 0.62
Pressure drop fixed, want to max Cp to get max V2; minimum hlm
*AR ~ 2.7
![Page 31: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/31.jpg)
If flow too fast or angle too big may get flow separation.Cp for Re > 7.5 x 104, “essentially” independent of Re
![Page 32: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/32.jpg)
V1avg2/ 2 + p1/ + z1=V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hlm; hlm = f(Le/D)V2/2
valves and fittings
![Page 33: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/33.jpg)
![Page 34: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/34.jpg)
hlm = f(Le/D)V2/2
![Page 35: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/35.jpg)
Head loss of a bend is greater than if pipe was straight (again due to separation).
![Page 36: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/36.jpg)
Nozzle Problem
![Page 37: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/37.jpg)
Neglecting friction, is flow faster at A or B or same?
A
![Page 38: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/38.jpg)
If flow at B did not equal flow at A then could connect and make perpetual motion machine.
A
A
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
VA2/ 2 + patm/ + d = VB
2/ 2 + patm/ + d
= 0
![Page 39: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/39.jpg)
C
d
![Page 40: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/40.jpg)
VT2/ 2 + patm/ + d = VC
2/ 2 + patm/ + d
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
= 0 = 0
C
d
0
![Page 41: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/41.jpg)
?
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
neglect friction
![Page 42: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/42.jpg)
VT2/ 2 + patm/ + d = VC
2/ 2 + patm/ + d
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
= 0 = 0
C
d
0
Nozzle
![Page 43: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/43.jpg)
breath
![Page 44: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/44.jpg)
Pipe Flow Examples ~
Solving for pressure drop in horizontal pipe
![Page 45: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/45.jpg)
V1avg2/2 + p1/ + gz1 – (V2avg
2/2 + p2/ + gz2) = hlT = hl + hlm
= f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
Laminar flow ~ f = 64/ReD
Turbulent flow ~ 1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5)
(f = 0.316/ReD0.25 for Re < 105)
p2- p1 = ?; Know hlT , L, D, Q, e, , , z2, z1
![Page 46: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/46.jpg)
p2- p1 = ?; Know L, D, Q, e, , , z2, z1
Compute the pressure drop in 200 ft of horizontal 6-in-diameter asphalted cast-iron pipe carrying water with a mean velocity of 6 ft/s.*
V1avg2/2 + p1/ + gz1 - V2avg
2/2 - p2/ - gz2 = f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
p1/ - p2/ = f [L/D][V2/2] = hlm
![Page 47: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/47.jpg)
p1/ - p2/ = f [L/D][V2/2] = hl
p2- p1 = ?; Know L, D, Q, e, , , z2, z1
f(Re, e/D); ReD = 270,000 & e/D = 0.0008
1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5); f = 0.0197
f ~ 0.02
p2 – p1 = hl = f(Re, e/D) [L/D][V2/2] = 280 lbf/ft2
![Page 48: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/48.jpg)
Pipe Flow Examples ~
Solving for pressure drop in non-horizontal pipe
![Page 49: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/49.jpg)
p2- p1 = ?; Know L, D, Q, e, , , z2, z1
Oil with = 900 kg/m3 and = 0.00001 m2/s flowsat 0.2 m3/s through 500m of 200 mm-diameter castiron pipe. Determine pressure drop if pipe slopes down at 10o in flow direction.
V1avg2/2 + p1/ + gz1 - V2avg
2/2 - p2/ - gz2 = f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
p1/ + gz1 - p2/ - gz2= f [L/D][V2/2] = hlm
![Page 50: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/50.jpg)
p1/ + gz1 - p2/ - gz1= f [L/D][V2/2]
p2- p1 = ?; Know L, D, Q, e, , , z2, z1
f(Re, e/D); ReD = 128,000 & e/D = 0.0013
1/f0.5 = -2.0 log{(e/D)/3.7 + 2.51/(ReD f0.5) f = 0.0227
f ~ 0.023
p2 – p1 = hl - g500(sin 10o) = 265,000 kg/m-s2
![Page 51: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/51.jpg)
If know everything but pressure drop or L then can use Moody Chart without iterations
![Page 52: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/52.jpg)
Pipe Flow Examples ~
Solving for V in horizontal pipe
![Page 53: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/53.jpg)
Q = ?; Know L, D, Q, e, , , z2, z1, p1, p2
Compute the average velocity in 200 ft of horizontal 6-in-diameter asphalted cast-iron pipe carrying water with a pressure drop of 280 lbf/ft2.
V1avg2/2 + p1/ + gz1 - V2avg
2/2 - p2/ - gz2 = f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
p1/ - p2/ = f [L/D][V2/2]
![Page 54: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/54.jpg)
p1/ - p2/ = f [L/D][V2/2] V = (0.7245/f)
e/D = 0.0008Guess fully
rough regime f ~ 0.19
f ~ 0.19
![Page 55: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/55.jpg)
f1 = 0.19; V1 = (0.7245/f1)1/2 = 6.18 ft/s ReD1 = 280,700 f2 = 0.198; V2 = (0.7245/f1)1/2 = 6.05 ft/s
![Page 56: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/56.jpg)
Pipe Flow Examples ~
Solving for D in horizontal pipe
![Page 57: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/57.jpg)
Q = ?; Know L, D, Q, e, , , z2, z1, p1, p2
Compute the diameter of a 200 m of horizontal pipe, e = 0.0004 mm, carrying 1.18 ft3/s, = 0.000011ft2/s and the head loss is 4.5 ft.
V1avg2/2g + p1/g + z1 - V2avg
2/2g - p2/g - z2 = f [L/D][V2/2g] + f [Le/D][V2/2g] + K[V2/2g]
p1/ - p2/ = f [L/D][V2/2]
f = function of ReD and e/D
HlT = hl/g = [length] = 4.5 ft
![Page 58: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/58.jpg)
ReD = VD/ = 4Q/(D)
or ReD = 136,600/D
f [L/D][V2/2] = hl
f = hl [D/L]2/[(4Q/D2)2] f ={2/8}{hlD5/(LQ2)} = 0.642D
or D = 1.093 f 1/5
e/D = 0.0004/D
![Page 59: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/59.jpg)
(1) ReD = 136,600/D
(2) D = 1.093 f 1/5
(3) e/D = 0.0004/D
Guess f ~ 0.03; then from (2) get D ~ 1.093(0.03)1/5 ~ 0.542 ft
From (1) get ReD ~ 136,600/0.542 ~ 252,000From (3) get e/D = 0.0004/0.542 ~ 7.38 x 10-4
fnew ~ 0.0196 from plot; Dnew ~ 0.498; ReDnew ~274,000; e/D ~8.03 x 10-4
fnewest ~ 0.0198 from plot Dnewest~ 0.499
![Page 60: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/60.jpg)
Solve for V in vertical pipe with minor losses, hlm
![Page 61: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/61.jpg)
Assume D >> d; turbulent flow;Atm press. at top & bottomf = 0.01Find Ve as a function of g & d
V12/2 + p1/ + gz1 – (V2
2/2 + p2/ + gz2) = hlT = hl + hlm
= f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
![Page 62: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/62.jpg)
V12/2 + p1/ + gz1 – (V2
2/2 + p2/ + gz2) = hlT = hl + hlm
= f [L/D][V2/2] + f [Le/D][V2/2] + K[V2/2]
V02/2 + patm/ + gz0
- V22/2 - patm/ - gz2
= f [L/D](V22/2)+(K1+K2+f[L/d])(V2
2/2)
V02/2 = 0; V2
2/2 = 1V22/2
gz0–gz2–V22/2={f [L/D]+K1+K2)}(V2
2/2)
V22 = 2g(z0-z2)/[1 + K1 + K2 + f(L/d)]
V22 =2g140d/(1+0.5+1.0+1.0+001x100)1/2
V2 = (80gd)1/2
![Page 63: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/63.jpg)
laminar transitional turbulent
![Page 64: V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT](https://reader035.vdocuments.us/reader035/viewer/2022062723/56813bd7550346895da5008a/html5/thumbnails/64.jpg)
The END ~