Fluids 16

CE 319 F Daene McKinney

Elementary Mechanics of Fluids Flow in

Pipes

Reynolds Experiment •

Reynolds Number

•

Laminar flow: Fluid moves moves in smooth smooth streamlines

•

Turbulent flow: Violent mixing, fluid velocity at a point varies randomly with time

•

Transition to turbulence in a 2 in. pipe p ipe is at =2 ft/s, so most pipe flows are turbulent V =2

Laminar

Turbulent

  2000 Laminar flow   VD  Re  2000  4000 Transition flow    Turbulent f low   4000

h f   V  h f   V 2

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  F s  0   pA  ( p  dp

0  0 

ds

 D

4

 0  

[

•

ds

sA    As d  ds

 (

 p  

s ) A  W  sin     0 (  D)s dz ds

  0 (  D )s

  z )]

 D  dh

4 ds 4 L 0

h1  h2  h f  

•

dp

  D

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

Darcy-Weisbach Equation  0 

  

V 

 

 D

e

ML-1T-2

ML

LT-1

ML-1T-1

L

L

-3

 0  F (  , V ,  , D, e) h f  

 4  F ( 1 ,  2 ) Repeating variables :   ,V , D  1  Re;  2   0   V 

2

 F (Re, 2

e  D e

 D

 0    V  F (Re,

;  3 

) e

 D

)



 0   V 

2

4 L   D

4 L   D

 0

2   V  F (Re,

e  D

)

2

 L V   e  8F (Re, )    D 2 g   D  h f    f 

 L V 2  D 2 g

Darcy-Weisbach Eq.

 f   8F (Re,

e  D

)

Friction factor

Laminar Flow in Pipes •

Laminar flow -- Newton’s law of viscosity is valid:      dV  dy dV  dr 

dy

 

dV   V  

dV 



r   dh

2 ds

dV  dr 

r   dh

2   ds r   dh

2   ds 2

r    dh

4   ds

dr 

 C 

C   

   r   2  1     V    4   ds   r 0     

2

r 0   dh

4   ds

2

r 0   dh

   r   2  V   V max 1       r 0     

•

Velocity distribution in a pipe (laminar flow) is parabolic with maximum at center.

Discharge in Laminar Flow V   

  dh

4   ds

( r 02  r 2 )

  dh 2 Q   VdA  0r 0  ( r 0  r 2 )( 2 rdr ) 4   ds



r  2 2 0

2

  dh ( r   r 0 )

4   ds

Q



V  

4

 r 0 dh

8  ds 4   D dh

128  ds

Q  A

V   

2   D dh

32  ds

2

0

Head Loss in Laminar Flow V    dh ds

2   D dh

32  ds

 V 

32    D

dh  V  h2  h1  V 

32  2   D

32  2   D

32  LV  2   D



2

h1  h2  h f 

h f  

h f  

32  LV    D

32  LV    V  2 / 2   D

ds

( s2  s1 )

2

2

2   V   / 2

 

 L )( )  V  2  / 2   V  D  D

 64( 

64  L ( )  V  2  / 2 Re  D

h f    f 

 L   V   D

2

2

 f  

64 Re

 Nikuradse’s Experiments •

In general, friction factor  f   F (Re,

 – 

•

 D

)

Function of  Re and roughness  f  

Laminar region  f  

 – 

•

e

64

k 

Re1 / 4

Rough

Blausius

Re

Independent of roughness

Turbulent region  – 

Smooth pipe curve •

 – 

 f  

64 Re

Rough pipe zone •

 f  

All curves coincide @ ~Re=2300 All rough pipe curves flatten out and become independent of Re

Blausius OK for smooth pipe

0.25

 5.74     e log10  3.7  0.9     D Re   

2

Laminar

Transition

Turbulent

Smooth

Moody Diagram

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 Pressure

 0.06 Re Laminar flow  1/6 Turbulent flow  D 4.4Re

 Le

Fully developed flow region

Entrance length Le

Entrance pressure drop

Region of linear pressure drop

 Le

 x

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  

h L V 

2

or

h L  K 

V 

2

2g

2g

•

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

Elbow Loss in a Pipe •

A piping system may have many minor losses which are all correlated to V 2 /2g

•

Sum them up to a total system loss for pipes of the same diameter h L  h f    hm  m

•

V 2 

  L   K m   f   2 g   D m 

Where, h L  Total head loss

h f   Frictionalhead loss hm  Minor head loss for fitting m K m  Minor head loss coefficien t for fitting m

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

Ex(10.2) Liquid in pipe has  = 8 kN/m3. Acceleration = 0.  D = 1 cm,  = 3x10-3 N-m/s2.

Given: Find:

Is fluid stationary, moving up, or moving down? What is the mean velocity?

Solution:

 1

V 12

2g

Energy eq. from z = 0 to z = 10 m



200,000 8000 90

 p1  

  z1  h L   2

 h L 

110,000 8000

V 22

2g



 p2  

2

  z 2

 10

h L 

 10 8 h L  1.25 m (moving upward) h L 

32 μLV 

V   h L

γD

2

γD

2

32 μL

V   1.25

2 8000*( 0.01 )

32*3 x10-3*10 V   1.04 m / s

1

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:

V  

Q  A

Re   Δp 

0.25



2

 11.46  ft  / s

 ( 2 / 12)  / 4

  VD  



32 μLV   D

2

0.97 * 1 / 94 * 11.46 * ( 2 / 12) 2



10 32*10-2*100*11.46 ( 2 / 12 )

2

 360 (laminar)

 91.7 psi/100 ft

Ex. (10.8) Given: Kerosene (S=0.94, =0.048 N-s/m2). Horizontal 5cm pipe. Q=2x10-3 m3 /s. Find: Pressure drop per 10 m of pipe. Solution: 2

α1

V 1

2g

h L 



 p1 γ

  z1  h L  α2

γD

2 2g

 p2 γ

  z 2

2

32 μLV  γD

2g

2g



32 μLV 

0  0  0.5  α2

2

V 2

V 22  2 V 2 

32 μL γD

2

2

 α2

V 22

2g

00

V   0.5  0

 32 * 4 * 10 5 * 10

0.8 * 62.4 * (1 / 32) 2

V   0.5  0

V 22  8.45V   16.1  0 V   1.60  f t  / s

Re 

0.8 * 1.94 * 1.6 * (0.25 / 12)  4 * 10 5

 1293(laminar)

 Q  V * A  1.6 *   * (0.25/12)2  / 4  1.23 * 10 3 cf s

Ex. (10.34) Given: Glycerin@ 20oC flows commercial steel pipe. Find: h Solution:    12,300 N  / m,    0.62 Ns / m2 2

2

V   p V   p α1 1  1   z1  h L  α2 2  2   z2 2g 2g γ γ  p1

γ

  z1  h L 

h 

 p1

Re 

  VD

γ

 p2

  z1  (

  z2

 p2

VD

γ

 



  z2 )  h L 0.6 * 0.02

 23.5 (laminar) 4 5.1 * 10 32 μLV  32(0.62)(1)(0.6) h  h L    2.42 m 2 12,300 * (0.02) 2 γD  



γ

Ex. (10.43) Given: Figure Find: 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:

2 2 V   p V   p α1 1  1   z1   h L  α2 2  2   z2 2g 2g γ γ

2 2 V   p V   p α1 1  1   z1   h L  αb b  b   zb 2g 2g γ γ

0  0   z1   h L  0  0   z 2

2 V   p 0  0   z1   h L  1 * b  b   zb 2g γ

2

 L   V     h L   K e  2 K b  K  E    f    D  2 g  

K e  0.5; K b  0.4 (assumed); K  E   1.0;  f  V  

Q  A



10 2   / 4 * 1

 L  D

 0.025 *

 12.73  f t  / s

 z1  100  0.5  2 * 0.4  1.0  10.75

12.732 2 * 32.2

 133 f t 

430 1

 pb

 10.75

γ

  z1   zb 

2

2

 L   V      K e  K b   f   2 g    D  2 g

V b

2

300  12.73    133  110.7  1.0  0.5  0.4  0.025  1   2 * 32.2    1.35  f t   pb  62.4 * ( 1.53)  0.59  psig Re 

VD  



12.73 * 1  9 * 105 5 1.14 * 10

Ex. (10.68) 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.

   1.22 x105  ft 2 / s; k s  1.5 x104  ft 

Solution:

Assume  f = 0.015 h f    f 

 L V 

1

 D  D  8.06  f t 

1.5 x10 4 8.06

 0.00002

VD   Q

(  / 4) D 2

 D

 



Q

(  / 4) D 

300 (  / 4)(8.06)1.22 x10

1000 (Q /(  / 4) D 2 ) 2

33,984 5

Re 

Re  2g



Get better estimate of  f 

2

 D

k s  D



 D 2 g

1  0.015 *

Relative roughness:

 f=0.010

1

22,656

 D 5  D  7.43  ft  89 in.

Use a 90 in pipe

5

 3.9 x106

Ex. (10.81) 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: h f    f 

Assume  f = 0.020 1 / 5

   fL Q 2     D   8  h f   2 g     

1 / 5

  0.02 140 (0.025) 2     8 2  14.45   9.81     Relative roughness:

2

 L V 

 D 2 g

  f 

2 2

k s  D



0.15 100

 0.100 m

 0.0015

 L (Q /(  / 4) D )  D

2g

Friction factor:

 f   0.022

1 / 5

   fL Q 2     D   8  h f   2 g     

1 / 5

 0.022   D  0.100   0.020 

2

2

V   p V   p α1 1  1   z1   h L  α2 2  2   z2 2g 2g γ γ

300,000

 h f  

9810 h f   14.45 m

60,000 9810

 10

Use 12 cm pipe

 0.102 m

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: 2

α1

V 1

2g



 p1

γ

  z1   h L  α2

2

V 2

2g



 p2

γ

  z2 2

 L V  0  0  12  (1  K e  K v  4 K b   f  ) 00  D 2 g

D = 10-cm and assume f = 0.025 24 g  (1  0.5  10  4 * 0.9  0.025 2

V  

1000 0.1

2

 p A

)V 2

24 g

265.1 V   0.942 m / s

2

V   p V   p α A  A   A   z A   h L  α2 2  2   z2 2g γ 2g γ

γ  p A

γ

 15  (2 K b   f 

 L V 2 )  D 2 g

 ( 2 * 0.9  0.025

500 (0.942) 2 )  15  9.6 m 0.1 2g

 p A  9810 * ( 9.26)  90.8 kP a

Q  VA  0.942(  / 4)(0.10) 2  0.0074 m 3 / s

Re 

VD  



0.942 * 0.1  7 x104 6 1.31 x10

So f = 0.025

Near cavitation pressure, not good!

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: Horsepower Solution: 2

α1

V 1



2g

 p1

γ

  z1  h p  α2

0  0  30  h p  V   2 V 2

2g

Q  A



2

V 2

2g



 p2

γ

2 V 2 0  60  (1  0.5  4 K b

2g

2 (  / 4)(1 / 2) 2

 10.18  f t  / s

VD  



10.18 * (1 / 2) 1.22 x10

5

  f 

So f = 0.0135  L  D

)

h p  60  30  1.611(1  0.5  4 * 0.19  0.0135

 107.6  f t   p 

 1.611  f t 

Re 

  z2   h L

 4.17 x105

Q h p

550

 24.4 hp

1700 (1 / 2)

)