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

H

E

A

D

L

O

S

S

hlT = hl + hm


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.


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

hlT = hl + hm;hlm = KV2/2

inlets, sudden enlargements & contractions; gradual contractions and exits


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= V2avg2/2 + 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 slowed down (2-3) and in violent mixing in the separated zones

separation

K = 0.78

unconfined mixing

as flow decelerates

r

K = 0.04

r/D > 0.15

D


slowed down (2-3) and in violent mixing in the separated zones V12/ 2 + p1/ + gz1= V22/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits


Minor losses due to sudden area change: slowed down (2-3) and in violent mixing in the separated zones

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


NOTE SOME BOOKS (Munson at al.): slowed down (2-3) and in violent mixing in the separated zones

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


AR < 1 slowed down (2-3) and in violent mixing in the separated zones

AR < 1

hlm = ½ KV2fastest

V1

V2


AR < 1 slowed down (2-3) and in violent mixing in the separated zones

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


slowed down (2-3) and in violent mixing in the separated zones V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits


Entire K.E. of exiting fluid is slowed down (2-3) and in violent mixing in the separated zones

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


V slowed down (2-3) and in violent mixing in the separated zones 2 ~ 0

Which exit has smallest Kexpansion?


K =1.0 slowed down (2-3) and in violent mixing in the separated zones

K =1.0

K =1.0

K =1.0

MYO


slowed down (2-3) and in violent mixing in the separated zones V1avg2/ 2 + p1/ + gz1= V2avg2/2 + p2/ + gz2 + hlT

hlT = hl + hlm;hlm = KV2/2

inlets,sudden enlargements & contractions; gradual contractions and exits


GRADUAL CONTRACTION slowed down (2-3) and in violent mixing in the separated zones

AR < 1

Where average velocity is fastest


breath slowed down (2-3) and in violent mixing in the separated zones


slowed down (2-3) and in violent mixing in the separated zones V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT

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

gentle expansions ~ diffusers


ugly slowed down (2-3) and in violent mixing in the separated zones

DIFFUSERS

20 cm/sec

3 cm/sec

bad

good


assume slowed down (2-3) and in violent mixing in the separated zones

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?


P slowed down (2-3) and in violent mixing in the separated zones 1, 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?


DIFFUSERS slowed down (2-3) and in violent mixing in the separated zones

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 slowed down (2-3) and in violent mixing in the separated zones

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


C slowed down (2-3) and in violent mixing in the separated zones p = (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


Relating C slowed down (2-3) and in violent mixing in the separated zones p 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)


h slowed down (2-3) and in violent mixing in the separated zones lm = (Cpi – Cp)V2/2; Cpi= 1 – 1/AR2

Cp


N/R slowed down (2-3) and in violent mixing in the separated zones 1 = 0.45/(.15/2) = 6

*

AR ~ 2.7

Cp 0.62

Pressure drop fixed, want to max Cp

to get max V2; minimum hlm


If flow too fast or angle too big may get flow separation. slowed down (2-3) and in violent mixing in the separated zones

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


slowed down (2-3) and in violent mixing in the separated zones V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT

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

valves and fittings


h slowed down (2-3) and in violent mixing in the separated zones lm = f(Le/D)V2/2


Head loss of a bend is greater slowed down (2-3) and in violent mixing in the separated zones

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


Nozzle Problem slowed down (2-3) and in violent mixing in the separated zones


A slowed down (2-3) and in violent mixing in the separated zones

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


= 0 slowed down (2-3) and in violent mixing in the separated zones

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


C slowed down (2-3) and in violent mixing in the separated zones

d


C slowed down (2-3) and in violent mixing in the separated zones

d

0

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

= 0

= 0

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


? slowed down (2-3) and in violent mixing in the separated zones

neglect friction

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


C slowed down (2-3) and in violent mixing in the separated zones

d

Nozzle

0

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

= 0

= 0

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


breath slowed down (2-3) and in violent mixing in the separated zones


Pipe Flow Examples ~ slowed down (2-3) and in violent mixing in the separated zones

Solving for pressure drop in horizontal pipe


slowed down (2-3) and in violent mixing in the separated zones 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


p slowed down (2-3) and in violent mixing in the separated zones 2- 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


p slowed down (2-3) and in violent mixing in the separated zones 2- 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


Pipe Flow Examples ~ slowed down (2-3) and in violent mixing in the separated zones

Solving for pressure drop in non-horizontal pipe


p slowed down (2-3) and in violent mixing in the separated zones 2- 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


p slowed down (2-3) and in violent mixing in the separated zones 2- 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


If know everything but pressure drop or L slowed down (2-3) and in violent mixing in the separated zones

then can use Moody Chart without iterations


Pipe Flow Examples ~ slowed down (2-3) and in violent mixing in the separated zones

Solving for V in horizontal pipe


Q = ?; Know L, D, Q, slowed down (2-3) and in violent mixing in the separated zones 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]


p slowed down (2-3) and in violent mixing in the separated zones 1/ - p2/ = f [L/D][V2/2]

V = (0.7245/f)

e/D = 0.0008

Guess fully

rough regime

f ~ 0.19

f ~ 0.19


f slowed down (2-3) and in violent mixing in the separated zones 1 = 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 ~ slowed down (2-3) and in violent mixing in the separated zones

Solving for D in horizontal pipe


Q = ?; Know L, D, Q, slowed down (2-3) and in violent mixing in the separated zones 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


Re slowed down (2-3) and in violent mixing in the separated zones D = 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) Re slowed down (2-3) and in violent mixing in the separated zones D = 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


Solve for V in vertical pipe with minor losses, h slowed down (2-3) and in violent mixing in the separated zones lm


Assume D >> d; slowed down (2-3) and in violent mixing in the separated zones

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]


slowed down (2-3) and in violent mixing in the separated zones 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


transitional slowed down (2-3) and in violent mixing in the separated zones

turbulent

laminar


The END ~ slowed down (2-3) and in violent mixing in the separated zones


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