v 1 2 / 2 + p 1 / + gz 1 = v 2 2 /2 + p 2 / + gz 2 + h lt h lt = h l + h m headlossheadloss

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 ~