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 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT

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 V 1 2 / 2 + p 1 /  + gz 1 =  V 2 2 /2 + p 2 /  + gz 2 + h lT - PowerPoint PPT Presentation


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 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 ,

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slide2
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.

slide3
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.

slide4
V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hm;hlm = KV2/2

inlets, sudden enlargements & contractions; gradual contractions and exits

slide5
Minor losses due to inlets:

hlm= p/ = K(V2/2); V2 = mean velocity in pipe

If K=1, p =  V2/2

slide7
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!

slide8
vena contracta

separation

K = 0.78

unconfined mixing

as flow decelerates

r

K = 0.04

r/D > 0.15

D

slide9
V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits

slide10
Minor losses due to sudden area change:

hlm= p/ = K(V2/2); V2 = faster mean velocity pipe

  • hlm head losses are primarily due to separation
  • Energy is dissipated deceleration after separation leading to violent mixing in the separated zones
slide11
NOTE SOME BOOKS (Munson at al.):

hlm = K V2/(2g) our Hlm!!!

slide12
AR < 1

AR < 1

hlm = ½ KV2fastest

V1

V2

slide13
AR < 1

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

slide14
V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits

slide15
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

hydrogen bubbles

hydrogen bubbles

Only diffuser can help by reducing V.

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

slide16
V2 ~ 0

Which exit has smallest Kexpansion?

slide17
K =1.0

K =1.0

K =1.0

K =1.0

MYO

slide18
V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits

slide19
GRADUAL CONTRACTION

AR < 1

Where average velocity is fastest

slide21
V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = p/ = (Cpi – Cp) 1/2V12

gentle expansions ~ diffusers

slide22
ugly

DIFFUSERS

20 cm/sec

3 cm/sec

bad

good

slide23
assume

fully 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?

slide24
P1, V1

P3, V3

P1, V1

P2, V2

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?

slide25
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.

slide26
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

slide27
Cp = (p2 – p1) / (1/2 V12 )

Cpideal = 1 – 1/AR2

AR = A2/A1

> 1

p1 + ½ V12 = p2 + ½ V22 (BE - ideal)

p2/ – p1/ = ½ V12 - ½ V22

A1V1 = A2V2 (Continuity)

V2 = V1 (A1/A2)

p2/ – p1/ = ½ V12 - ½ [V1(A1/A2)]2

p2/ – p1/ = ½ V12 - ½ V12(1/AR)2

(p2 – p1)/( ½  V12) = 1 – 1/AR2

Cpi= 1 – 1/AR2

slide28
Relating Cp to Cpi and hlm

p1 / + ½ V12 = p2/ + ½ V22 + hlm (z1 = z2 = 0)

hlm = V12/2 - V22/2 – (p2 – p1)/

hlm = V12/2 {1 + V22/V12 – (p2 – p1)/( 1/2V12)}

A1V1 = A2V2

Cp = (p2 – p1)/( 1/2V12) (Cp is positive & < Cpi)

hlm = V12/2 {1 - A12/A22 – Cp}

Cpi = 1 – 1/AR2

hlm = V12/2 {Cpi – Cp} Q.E.D.

(see Ex. 8.10)

slide30
N/R1 = 0.45/(.15/2) = 6

*

AR ~ 2.7

Cp 0.62

Pressure drop fixed, want to max Cp

to get max V2; minimum hlm

slide31
If flow too fast or angle too big may get flow separation.

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

slide32
V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm; hlm = f(Le/D)V2/2

valves and fittings

slide35
Head loss of a bend is greater

than if pipe was straight (again due to separation).

slide37
A

Neglecting friction, is flow faster at A or B or same?

slide38
= 0

V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT

VA2/ 2 + patm/ + d = VB2/ 2 + patm/ + d

A

If flow at B did not equal flow at A then could connect and make perpetual motion machine.

A

slide39
C

d

slide40
C

d

0

V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT

= 0

= 0

VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d

slide41
?

neglect friction

V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT

slide42
C

d

Nozzle

0

V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT

= 0

= 0

VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d

slide44
Pipe Flow Examples ~

Solving for pressure drop in horizontal pipe

slide45
V1avg2/2 + p1/ + gz1 – (V2avg2/2 + 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

slide46
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 - V2avg2/2 - p2/ - gz2

= f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2]

p1/ - p2/ = f [L/D][V2/2] = hlm

slide47
p2- p1 = ?; Know L, D, Q, e,  , , z2, z1

p1/ - p2/ = f [L/D][V2/2] = hl

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

slide48
Pipe Flow Examples ~

Solving for pressure drop in non-horizontal pipe

slide49
p2- p1 = ?; Know L, D, Q, e,  , , z2, z1

Oil with  = 900 kg/m3 and = 0.00001 m2/s flows

at 0.2 m3/s through 500m of 200 mm-diameter cast

iron pipe. Determine pressure drop if pipe slopes

down at 10o in flow direction.

V1avg2/2 + p1/ + gz1 - V2avg2/2 - 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

slide50
p2- p1 = ?; Know L, D, Q, e,  , , z2, z1

p1/ + gz1 - p2/ - gz1= f [L/D][V2/2]

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

slide51
If know everything but pressure drop or L

then can use Moody Chart without iterations

slide52
Pipe Flow Examples ~

Solving for V in horizontal pipe

slide53
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 - V2avg2/2 - p2/ - gz2

= f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2]

p1/ - p2/= f [L/D][V2/2]

slide54
p1/ - p2/ = f [L/D][V2/2]

V = (0.7245/f)

e/D = 0.0008

Guess fully

rough regime

f ~ 0.19

f ~ 0.19

slide55
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

slide56
Pipe Flow Examples ~

Solving for D in horizontal pipe

slide57
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.

HlT = hl/g = [length] = 4.5 ft

V1avg2/2g + p1/g + z1 - V2avg2/2g - 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

slide58
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

slide59
(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 ~ 252,000

From (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

slide61
Assume D >> d;

turbulent flow;

Atm press. at top & bottom

f = 0.01

Find Ve as a function of g & d

V12/2 + p1/ + gz1 – (V22/2 + p2/ + gz2)

= hlT = hl + hlm

= f [L/D][V2/2] +  f [Le/D][V2/2] + K[V2/2]

slide62
V12/2 + p1/ + gz1 – (V22/2 + 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])(V22/2)

V02/2 = 0; V22/2 = 1V22/2

gz0–gz2–V22/2={f [L/D]+K1+K2)}(V22/2)

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

slide63
transitional

turbulent

laminar

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