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
K = 0.78
separation
unconfined mixing
as flow decelerates
r
K = 0.04
r/D > 0.15
D
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:
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.):
hlm = K V2/(2g) our Hlm!!!
AR < 1
V2
AR < 1
hlm = ½ KV2fastest
V1
AR < 1
Why is Kcontraction and Kexpansion = 0 at AR =1?
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
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
V2 ~ 0
Which exit has smallest Kexpansion?
K =1.0
K =1.0
K =1.0
K =1.0
MYO
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
AR < 1
Where average velocity is fastest
breath
V1avg2/ 2 + p1/ + z1=V2avg2/2 + p2/ + gz2 + hlT
hlT = hl + hlm; hlm = p/ = (Cpi – Cp) 1/2V12
gentle expansions ~ diffusers
ugly
DIFFUSERS
3 cm/sec
good
20 cm/sec
bad
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?
P1, V1
P2, V2
P1, 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
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
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)
hlm = (Cpi – Cp)V2/2; Cpi = 1 – 1/AR2
Cp
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
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=V2avg2/2 + 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
A
Neglecting friction, is flow faster at A or B or same?
V12/
2 + p1/ +
z1=V22/2
=0
+ 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
d
C
d
V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT
=0
0
=0
VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d
?
neglect friction
V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT
C
d
Nozzle
V12/ 2 + p1/ + z1=V22/2 + p2/ + gz2 + hlT
=0
0
=0
VT2/ 2 + patm/ + d = VC2/ 2 + patm/ + d
breath
Pipe Flow Examples ~
Solving for pressure drop in horizontal pipe
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
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
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
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 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
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
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 - V2avg2/2 - 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)
f ~ 0.19
e/D = 0.0008
Guess fully
rough regime
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.
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
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 ~ 252,000
From (3) get e/D = 0.0004/0.542 ~ 7.38 x 10-4
fnew ~ 0.0196 from plot;
ReDnew ~274,000;
fnewest ~ 0.0198 from plot
Dnew ~ 0.498;
e/D ~8.03 x 10-4
Dnewest~ 0.499
Solve for V in vertical pipe with minor losses, hlm
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]
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
laminar
transitional
turbulent
The END ~