Viscous Flow in Pipes CHAPTER 8
Often required to size the i es and and dete determ rmin inee the the pumps, etc. Thus, we need to determine all sorts of friction losses. Sometimes we must use a judicious combination of experimental data an d JUDGEMENT, JUDGEMENT, along with theoretical considerations and dimensional analysis to obtain the desired results. Modified quote from TEXT!!!!
Pipe Flow
Flowing Full
Open Channel Flow Osborne Reynolds (1842-1912) a British scientist and engineer was the first to distinguish between these two classifications of flow ( LAMINAR LAMINAR and TURBULENT TURBULENT ). ).
Type of Flow
Laminar
Transitional Turbulent
Reynolds Number DIMENSIONLESS 1.
Laminar
2.
Transitional
3.
Turbulent
Re
VD VD
ρ= fluid density , μ = dynamic viscosity (with uni units of FL -2 T such as lb·s lb·s/f /ftt2 or N·s/m2), V is the averagevelocity veloci ty in i n the pipe, and and D is the INSIDE
pipe pipe diam diameter.
1
Reynolds Number Re
VD
Reynolds Number
Re
VD
Re
, 2 with units of lb·s/ft or N·s/m2. =
-2
,
Re
ν = kinematic viscosity = μ / ρ, with dimensions of
L 2 T, with units of
VD 3
FT / L2
MLL(T 2 ) L2
ft2·s or m2·s
3
ML4T 2
( L3T ) ML(T ) ML4T 2
( ML / T 2 )T / L2
DIMENSIONL ESS
MAKE SURE TO CHECK DEFINITION OF Re.
Entrance Region to pipe flow
Reynolds Number Re
VD
Reynolds number is less than 2100. 2100 The flow in a round pipe is turbulent if the Reynolds number is greater than approximately 4000. 4000 Figure 8.5 Entrance region, developing flow, and fully developed fl ow in a pipe system.
Length of the entrance region
Length of the entrance region
Laminar FLOW
l e D
0.06 Re
Turbulent FLOW
l e
D
4.4(Re)
1 6
2
Length of the entrance region
Laminar FLOW
e
D
Turbulent FLOW
e
D
0.06Re
Length of the entrance region For many practical engineering problems, 104
4.4(Re) Turbulent flow
For very low Reynolds numbers (Re = 10, the entrance length can be quite short (l e =0.6 D). For large Reynolds numbers (Re = 4000, it may take a length of several pipe diameters before the end of entrance region is reached (l e =17.5 D)
Fully Developed Flow
Flow that begins at the point that the velocity no longer changes with x, but only with r.
Flow will remain this way until there is a change in the pipe.
Pressure and Shear Stress
Pressure
Fully developed steady flow in a constant diameter pipe may be driven by: 1. 2.
GRAVITY or PRESSURE PRESSURE
FORCES
Pressure gradient Caused by VISCOUS effects
Horizontal pipe flow
Fully developed steady flow in a constant diameter pipe may be driven by gravity and/or pressure . , effect except for the pressure variation across the pipe, γ D, that is usually negligible. It is the pressure difference,
∆ p = p1 – – p p2 along
WHY negative sign?
∂ p/ ∂ x
= -∆ p/ l
the pipe which forces the fluid through the pipe.
3
Nature of Pipe Flow
Fully Developed Laminar Flow THREE methods to derive the equations describing fully developed LAMINAR flow.
Dependent on Type of Flow
Laminar
1.
Turbulent
2. 3.
Velocity Profile from F=ma
Pressure
F=ma
Shear stress, , acts on the surface of the cylinder. Viscous force is a function of the radius of the cylinder. = (r)
Newton’s law; F = ma applied directly to a fluid element From the Navier-Stokes equations of motion, Dimensional analysis methods.
If gravitational effects are neglected the pressure is constant across any vertical cross section (DOESN’T SIGNINFICANTLY CHANGE across the pipe diameter), but varies from one cross section to the next. Pressure decreases in the direction of flow.
and r (radial coordinate)
= Cr
2. At r =0 =0 (no shear stress0 3. At r =D/2 is a maximum, w , the wall shear stress 4. Therefore 2 w r
D
.
Balancing the forces in the x direction RESULTS in
p
2 r
EQ 8.3
4
Change in pressure p
2
EQ 8.3
r
p
How is shear stress related to velocity?
4 w
2 w r D
EQ 8.4
EQ 8.5
D
To further the analysis, we must prescribe how the shear stress τ, is related to the velocity
Velocity and Shear Stress
2
EQ 8.3 and
r
du d
du dr
du dr
Find C1 Where do weknow thevelocity? At thewall, the velocity u is zero (r = D/2 ).
p r dr 2
du
We obtain
Which we can integrate to obtain p 2 r C 1 u 4
Velocity and Shear Stress p 2 r C 1
u
at r D / 2, u 0 thus C 1
proportional to the velocity gradient. For pipe flow: velocity decreases from the pipe centerline to the pipe wall.
Velocity
Combining
p
Recall from Chapter 1 that in laminar flow of a Newtonian fluid,
p 16
D 2
Velocity 2 2 pD 2 2r 2r 1 V c 1 16 D D
u (r )
EQ 8.7
Where Vc is the centerline or maximum velocity and
pD 2 V c 16
5
Velocity
Velocity
2 2r 2 pD 2 2r 1 V c 1 u (r ) 16 D D
p
4 w
EQ 8.5
D
D r u (r ) w 1 4 R
2
Volume Flow Rate in a Horizontal Pipe (In terms of maximum velocity)
Flow is axisymmetric about the centerline. Velocity is constant on small elements of radiusr and thickness dr .
Q u dA
r R r 0
u (r ) 2 r dr 2 V c
Q
R
0
R 2 V c
By definition, the average velocity, V V Q / A Q /( R 2 ) with Q V Q
2
R 2 V c
2
R 2 V c
V c pD 2 1 2 32 R 2
D 4 p
EQ 8.8
EQ 8.9 Poiseuille’s LAW
128 Reynolds number is less than 2100 in a HORIZONTAL PIPE.
Directly proportional to the pressure drop
Inversely proportional to viscosity
Inversel
Proportional to pipe diameter to the 4th 4 power.
ro ortional to i e len th
D p
Volume Flow Rate in a Horizontal Pipe (In terms of average velocity)
r
1 r dr R
For Laminar Horizontal Pipe Flow the Flowrate Is:
Q
The velocity profile, plotted in Figure8.9 is parabolic in the radial coordinate, has a maximum velocity at the pipe centerline and a minimum velocity (zero) at the pipe wa .
EQ 8.7
EQ 8.9 Poiseuille’s LAW
128
For Laminar Flow RESTRICTED TO REYNOLDS NUMBERS LESS THAN APPROXIMATELY 2100 APPROXIMATELY 2100 IN IN A HORIZONTAL PIPE.
Re
VD
Remember: MEAN V=V c / 2
6
Non-horizontal laminar pipe flow
Balancing the forces
p sin
2 r
(From force balance)
EQ 8.10
emem er: = - u r an o ow ng t e same eveopment as horizontal flow:
V
Q
Non-horizontal laminar flow
Driving force PRESSURE drop in the flow direction
1.
Component of WEIGHT in the flow direction.
2.
If flow is downhill, gravity helps the flow and a smaller pressure smaller drop is required, sin θ < 0.
If flow is uphill, gravity works against the against flow and a larger pressure drop is required, sin θ > 0.
p sin D 2 32
p sin D 4
V Q / A Re
VD
(2 x105 m 3 / s) 0.0637m / s / 4(0.020m) 2
Given an oil with a viscosity of μ = 0.40 N·s/m2 and density ρ = 900 kg/m3 flows in a pipe diameter D = 0.020m. What pressure drop is required to producea flow rate of Q = 2x10-5 m3/s if the pipe is horizontal and the pipe length is 10 meters? (2 x105 m 3 / s) 0.0637m / s V Q / A / 4(0.020m) 2 0.020m (900kg / m 3 )(0.0637m / s )(0.020m) 0.40 N s / m 2 Re 2.87 Hence la min ar 128 Q EQ 8.9 Poiseuille’s LAW p 4
VD
D
128(0.40 N s / m 2 )(2 x105 m 3 / s)(10 m) p (0.02m) 4
p 20.37 kPa p 20.37 kPa 2.037kPa / m 10m
sin
p sin D 4
128
p sin D 4
128
or
128 Q sin D 4
Re 2.87 Hence la minar
Q
Same problem. How steepa hill(θ ) must thepipe be on if the oil is to flow through the pipe at the samerate but no pressure loss?
Q
(900kg / m 3 )(0.0637m / s)(0.020m) 0.40 N s / m 2
EQ 8.12
128
Same problem. How steep ahill(θ ) must the pipe be on if the oil is to flow through thepipe at the samerate but no pressure loss?
EQ 8.11
EQ 8.12
sin
128 Q g D 4
128(0.40 N s / m 2 )(2 x105 m 3 / s) (900kg / m 3 )(9.81m / s 2 )(0.02m)4
sin 0.23074 13.34
7
Check Reynol ds number with water, same pipe and same flow. Viscosity of μ = 1.12 E - 3 N·s/m2 and density ρ = 999 kg/m3 flows in a pipe diameter D = 0.020m. Flow rateof Q = 2x10-5 m3/s . V Q / A
Re
(2 x105 m 3 / s) 0.0637m / s / 4(0.020m) 2
0.020m
3
. . (1.12 x103) N s / m 2
Re 1136 What if the velocity were1 m/s (3.2 ft/s)? What is theReynolds number? VD (999kg / m 3)(1.0m / s)(0.020m) Re (1.12 x103) N s / m 2 Re 17800
8
