pertemuan 13 - 14 closed conduit flow 1
DESCRIPTION
Pertemuan 13 - 14 CLOSED CONDUIT FLOW 1. Reynolds Experiment. Reynolds Number Laminar flow: Fluid moves in smooth streamlines Turbulent flow: Violent mixing, fluid velocity at a point varies randomly with time - PowerPoint PPT PresentationTRANSCRIPT
Pertemuan 13 - 14CLOSED CONDUIT FLOW 1
Bina Nusantara
Reynolds Experiment• Reynolds Number
• Laminar flow: Fluid moves in smooth streamlines• Turbulent flow: Violent mixing, fluid velocity at a point
varies randomly with time• Transition to turbulence in a 2 in. pipe is at V=2 ft/s,
so most pipe flows are turbulent
2lowfTurbulent4000
flowTransition40002000
flowLaminar2000
Re
Vh
VhVD
f
f
Laminar Turbulent
Bina Nusantara
Shear Stress in Pipes• Steady, uniform flow in a pipe: momentum flux is zero
and pressure distribution across pipe is hydrostatic, equilibrium exists between pressure, gravity and shear forces
D
Lhhh
ds
dhD
zp
ds
dD
sDds
dzsAsA
ds
dp
sDWAsds
dpppAF
f
s
021
0
0
0
0
44
)]([4
)(0
)(sin)(0
• Since h is constant across the cross-section of the pipe (hydrostatic), and –dh/ds>0, then the shear stress will be zero at the center (r = 0) and increase linearly to a maximum at the wall.
• Head loss is due to the shear stress.
• Applicable to either laminar or turbulent flow• Now we need a relationship for the shear stress in
terms of the Re and pipe roughness
Bina Nusantara
Darcy-Weisbach Equation
)(Re,
)(Re,
;Re;
,,:variablesRepeating
),(
),,,,(
20
20
20
321
214
0
D
eFV
D
eF
V
VD
e
DV
F
eDVF
V D e
ML-1T-2 ML-3 LT-1 ML-1T-1 L L
)(Re,82
)(Re,82
)(Re,4
4
2
2
2
0
D
eFf
g
V
D
Lfh
D
eF
g
V
D
L
D
eFV
D
L
D
Lh
f
f
Darcy-Weisbach Eq. Friction factor
Bina Nusantara
Laminar Flow in Pipes• Laminar flow -- Newton’s law of viscosity is valid:
2
0
20
20
2
14
44
2
2
2
r
r
ds
dhrV
ds
dhrCC
ds
dhrV
drds
dhrdV
ds
dhr
dr
dV
dr
dV
dy
dV
ds
dhr
dy
dV
• Velocity distribution in a pipe (laminar flow) is parabolic with maximum at center.
2
0max 1
r
rVV
Bina Nusantara
Discharge in Laminar Flow
ds
dhD
ds
dhrQ
rr
ds
dh
rdrrrds
dhVdAQ
rrds
dhV
r
r
128
8
2
)(
4
)2()(4
)(4
4
40
0
220
2
022
0
220
0
0
ds
dhDV
A
QV
32
2
Bina Nusantara
Head Loss in Laminar Flow
2
21
12212
2
2
2
32
)(32
32
32
32
D
VLh
hhh
ssD
Vhh
dsD
Vdh
DV
ds
dh
ds
dhDV
f
f
Re
64
2
2/)(Re
64
2/))((64
2/
2/32
32
2
2
2
2
2
2
2
fV
D
Lfh
VD
L
VD
L
DV
V
V
D
VL
D
VLh
f
f
Bina Nusantara
Nikuradse’s Experiments
• In general, friction factor
– Function of Re and roughness
• Laminar region
– Independent of roughness
• Turbulent region– Smooth pipe curve
• All curves coincide @ ~Re=2300
– Rough pipe zone• All rough pipe curves flatten
out and become independent of Re
Re
64f
Blausius
Re 4/1k
f Rough
Smooth
Laminar Transition Turbulent
Blausius OK for smooth pipe
)(Re,D
eFf
Re
64f
2
9.010Re
74.5
7.3log
25.0
D
ef
Bina Nusantara
Moody Diagram
Bina Nusantara
Pipe Entrance• Developing flow
– Includes boundary layer and core, – viscous effects grow inward from the wall
• Fully developed flow– Shape of velocity profile is same at all points
along pipe
flowTurbulent 4.4Re
flowLaminar Re06.01/6D
Le
eL
Entrance length LeFully developed
flow region
Region of linear pressure drop
Entrance pressure drop
Pressure
x
Bina Nusantara
Entrance Loss in a Pipe• In addition to frictional losses, there are minor
losses due to – Entrances or exits– Expansions or contractions– Bends, elbows, tees, and other fittings– Valves
• Losses generally determined by experiment and then corellated with pipe flow characteristics
• Loss coefficients are generally given as the ratio of head loss to velocity head
• K – loss coefficent– K ~ 0.1 for well-rounded inlet (high Re)– K ~ 1.0 abrupt pipe outlet– K ~ 0.5 abrupt pipe inlet
Abrupt inlet, K ~ 0.5
g
VKh
g
V
hK L
L
2or
2
2
2
Bina Nusantara
Elbow Loss in a Pipe• A piping system may have many minor losses which
are all correlated to V2/2g • Sum them up to a total system loss for pipes of the
same diameter
• Where,
m
mm
mfL KD
Lf
g
Vhhh
2
2
lossheadTotalLh
lossheadFrictionalfh
mhm fittingfor lossheadMinor
mKm fittingfor tcoefficienlossheadMinor
Bina Nusantara
EGL & HGL for Losses in a Pipe• Entrances, bends, and other flow transitions cause the
EGL to drop an amount equal to the head loss produced by the transition.
• EGL is steeper at entrance than it is downstream of there where the slope is equal the frictional head loss in the pipe.
• The HGL also drops sharply downstream of an entrance
Bina Nusantara
Ex(10.2)Given: Liquid in pipe has = 8 kN/m3. Acceleration = 0. D =
1 cm, = 3x10-3 N-m/s2. Find: Is fluid stationary, moving up, or moving down? What
is the mean velocity?Solution: Energy eq. from z = 0 to z = 10 m
smV*x*
).*(.V
μL
γDhV
γD
μLVh
mh
h
h
zp
g
Vhz
p
g
V
-
L
L
L
L
L
L
/04.11010332
0108000251
32
32
upward) (moving25.1
108
90
108000
000,110
8000
000,200
22
3
2
2
2
22
22
211
21
1
1
2
Bina Nusantara
Ex (10.4)• Given: Oil (S = 0.97, m = 10-2 lbf-s/ft2) in 2 in pipe, Q =
0.25 cfs.• Find: Pressure drop per 100 ft of horizontal pipe.• Solution:
ftpsi/100791122
46111002103232
(laminar)36010
)12/2(*46.11*94/1*97.0Re
/46.114/)12/2(
25.0
22
2
2
.)/(
.**-*
D
μLVΔp
VD
sftA
QV
Bina Nusantara
Ex. (10.8)Given: Kerosene (S=0.94, =0.048 N-s/m2). Horizontal 5-cm
pipe. Q=2x10-3 m3/s.Find: Pressure drop per 10 m of pipe.Solution:
cfs
sftV
VV
VVg
VγD
μLV
g
α
g
Vα
γD
μLV
γD
μLVh
zγ
p
g
Vαhz
γ
p
g
Vα
L
L
32
5
22
2
522
222
2
22
22
2
22
22
211
21
1
10*23.14/(0.25/12)**1.6A*VQ
(laminar)129310*4
)12/25.0(*6.1*94.1*8.0Re
/60.1
01.1645.8
05.0)32/1(*4.62*8.0
10*10*4*32
2
2
05.032
2
002
325.000
32
22
Bina Nusantara
Ex. (10.34)
2/62.0,/300,12 mNsmN
Given: Glycerin@ 20oC flows commercial steel pipe. Find: hSolution:
mγD
μLVhh
VDVD
hzγ
pz
γ
ph
zγ
phz
γ
p
zγ
p
g
Vαhz
γ
p
g
Vα
L
L
L
L
42.2)02.0(*300,12
)6.0)(1)(62.0(3232
(laminar)5.2310*1.5
02.0*6.0Re
)(
22
22
4
22
11
22
11
22
22
211
21
1
Bina Nusantara
Ex. (10.43)Given: FigureFind: Estimate the elevation required in the upper reservoir to
produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?
Solution:
ftz
sftA
QV
D
LfKKK
g
V
D
LfKKKh
zhz
zγ
p
g
Vαhz
γ
p
g
Vα
Ebe
EbeL
L
L
1332.32*2
73.1275.100.14.0*25.0100
/73.121*4/
10
75.101
430*025.0;0.1;(assumed)4.0;5.0
22
0000
22
2
1
2
221
22
22
211
21
1
55
2
22
1
2
1
2
11
21
1
10*910*14.1
1*73.12Re
59.0)53.1(*4.62
35.1
2.32*2
73.12
1
300025.04.05.00.17.110133
22
2*100
22
VD
psigp
ft
g
V
D
LfKK
g
Vzz
γ
p
zγ
p
g
Vhz
zγ
p
g
Vαhz
γ
p
g
Vα
b
beb
bb
bbb
L
bbb
bL
Bina Nusantara
Ex. (10.68)
ftxksftx s425 105.1;/1022.1
ftDD
g
DQ
D
g
V
D
Lfh f
06.8
984,331
2
))4//((1000*015.01
2
5
22
2
Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1 ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the diameters are expressed in inches (i.e., 10 in, 12 in, etc.).
Find: Diameter.Solution:
Assume f = 0.015
Relative roughness: 00002.006.8
105.1 4
x
D
ks
Get better estimate of f
65
2
109.31022.1)06.8)(4/(
300Re
)4/()4/(
Re
xx
D
QD
D
Q
VD
f=0.010
.8943.7
656,221
5
inftDD
Use a 90 in pipe
Bina Nusantara
Ex. (10.81)
mh
h
zγ
p
g
Vαhz
γ
p
g
Vα
g
Q
h
fLD
g
DQ
D
Lf
g
V
D
Lfh
f
f
L
f
f
45.14
109810
000,60
9810
000,300
22
8
2
))4//((
2
22
22
211
21
1
5/1
2
2
222
Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m from the main? Assume galvanized-steel pipe is to be used and that the pressure required at the factory is 60 kPa gage at a point 10 m above the main connection.
Find: Size of pipe.Solution:
Assume f = 0.020
m
g
Q
h
fLD
f
100.081.9
)025.0(140
45.14
02.08
8
5/1
2
2
5/1
2
2
Relative roughness:
022.0
0015.0100
15.0
f
D
ks
Friction factor:
mD 102.0020.0
022.0100.0
5/1
Use 12 cm pipe
Bina Nusantara
Ex. (10.83)Given: The 10-cm galvanized-steel pipe is 1000 m long and
discharges water into the atmosphere. The pipeline has an open globe valve and 4 threaded elbows; h1=3 m and h2 = 15 m.
Find: What is the discharge, and what is the pressure at A, the midpoint of the line?
Solution:
002
)41(1200
222
22
22
211
21
1
g
V
D
LfKKK
zγ
p
g
Vαhz
γ
p
g
Vα
bve
L
D = 10-cm and assume f = 0.025
46
32
2
2
1071031.1
1.0*942.0Re
/0074.0)10.0)(4/(942.0
/942.01.265
24
)1.0
1000025.09.0*4105.01(24
xx
VD
smVAQ
smV
gV
Vg
So f = 0.025
kPap
mgγ
p
g
V
D
LfK
γ
p
zγ
p
g
Vαhz
γ
p
g
Vα
A
A
bA
LAAA
A
8.90)26.9(*9810
6.9152
)942.0()
1.0
500025.09.0*2(
2)2(15
22
2
2
22
22
2
2
Near cavitation pressure, not good!
Bina Nusantara
Ex. (10.95)
Given: If the deluge through the system shown is 2 cfs, what horsepower is the pump supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is smooth.
Find: HorsepowerSolution:
55
22
2
22
22
22
211
21
1
1017.41022.1
)2/1(*18.10Re
611.12
/18.10)2/1)(4/(
2
)45.01(2
6003000
22
xx
VD
ftg
V
sftA
QV
D
LfK
g
Vh
hzγ
p
g
Vαhz
γ
p
g
Vα
bp
Lp
hphQ
p
ft
h
p
p
4.24550
6.107
))2/1(
17000135.019.0*45.01(611.13060
So f = 0.0135