V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT
64
V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT h lT = h l + h m H E A D L O S S
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 Presentation
Transcript of V 1 2 / 2 + p 1 / + gz 1 = V 2 2 /2 + p 2 / + gz 2 + h lT
V12/ 2 + p1/ + gz1= V2
2/2 + p2/ + gz2 + hlT
hlT = hl + hm
HEADLOSS
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.
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.
hlT = hl + hm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
Minor losses due to inlets: hlm= p/ = K(V2/2); V2 = mean velocity in pipe
If K=1, p = V2/2
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hm
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!
vena contracta
unconfined mixingas flow decelerates
separationK = 0.78
K = 0.04 r/D > 0.15
r
D
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V12/ 2 + p1/ + gz1= V2
2/2 + p2/ + gz2 + hlT
• 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
NOTE SOME BOOKS (Munson at al.): hlm = K V2/(2g) our Hlm!!!
AR < 1
V2 V1
AR < 1 hlm = ½ KV2
fastest
Why is Kcontraction and Kexpansion = 0 at AR =1?
AR < 1
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
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
Which exit has smallest Kexpansion?
V2 ~ 0
MYO
K =1.0
K =1.0K =1.0
K =1.0
hlT = hl + hlm; hlm = KV2/2
inlets, sudden enlargements & contractions; gradual contractions and exits
V1avg2/ 2 + p1/ + gz1= V2avg
2/2 + p2/ + gz2 + hlT
GRADUAL CONTRACTION
Where average velocity is fastest
AR < 1
breath
V1avg2/ 2 + p1/ + z1=V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hlm; hlm = p/ = (Cpi – Cp) 1/2V12
gentle expansions ~ diffusers
DIFFUSERS
good bad
3 cm/sec 20 cm/sec
ugly
…..
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?
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?
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.
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
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
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)
hlm = (Cpi – Cp)V2/2; Cpi = 1 – 1/AR2
Cp
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
If flow too fast or angle too big may get flow separation.Cp for Re > 7.5 x 104, “essentially” independent of Re
V1avg2/ 2 + p1/ + z1=V2avg
2/2 + p2/ + gz2 + hlT
hlT = hl + hlm; hlm = f(Le/D)V2/2
valves and fittings
hlm = f(Le/D)V2/2
Head loss of a bend is greater than if pipe was straight (again due to separation).
Nozzle Problem
Neglecting friction, is flow faster at A or B or same?
A
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
C
d
VT2/ 2 + patm/ + d = VC
2/ 2 + patm/ + d
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
= 0 = 0
C
d
0
?
V12/ 2 + p1/ + z1=V2
2/2 + p2/ + gz2 + hlT
neglect friction
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
breath
Pipe Flow Examples ~
Solving for pressure drop in horizontal pipe
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
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
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
Pipe Flow Examples ~
Solving for pressure drop in non-horizontal pipe
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
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
If know everything but pressure drop or L then can use Moody Chart without iterations
Pipe Flow Examples ~
Solving for V in horizontal pipe
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]
p1/ - p2/ = f [L/D][V2/2] V = (0.7245/f)
e/D = 0.0008Guess fully
rough regime f ~ 0.19
f ~ 0.19
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
Pipe Flow Examples ~
Solving for D in horizontal pipe
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
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
(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
Solve for V in vertical pipe with minor losses, hlm
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]
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
laminar transitional turbulent
The END ~
