Reynold Num.pdf

1586_book.fm Page 1 Friday, May 7, 2004 3:56 PM

39 Boundary Layers Edwin R. Braun University of North Carolina, Charlotte

Pao-lien Wang University of North Carolina, Charlotte

39.1 39.2 39.3 39.4 39.5

Theoretical Boundary Layers Reynolds Similarity in Test Data Friction in Pipes Noncircular Channel Example Solutions

39.1 Theoretical Boundary Layers In a simple model of a solid, material deformation is proportional to the strain. In a simple model of a fluid, the deformation is proportional to the rate of strain or the change in velocity over a small distance. The mathematical term describing this phenomenon is the last term in the following boundary layer equation: È du du du ˘ dp d Ê du ˆ rÍ + u + v ˙ = - + Ám ˜ dx dy ˚ dx dy Ë dy ¯ Î dt

(39.1)

where u is the velocity in the x direction as a function of x, y, and t. The values r and m are the density and dynamic viscosity for the fluid, respectively. This equation is good for all situations with no pressure (P) change present in the direction normal to the wall. The left-hand side of Equation (39.1) represents time kinetic energy in flow. The pressure term is a potential energy term. The rate of strain term that represents this energy dissipates through viscous losses. When the dissipation term is significant compared to the others, a boundary layer must be considered as part of the flow analysis. For a straight-channel, steady (not time-dependent) flow, Equation (39.1) becomes m

d 2u dp = dy 2 dx

(39.2)

which has the solution u=-

1 dp 2 (D - y 2 ) 2m dx

(39.3)

where d is the distance toward the wall measured from the centerline and the velocity u is zero at the walls. This velocity equation is parabola. Note that when y = D, Equation (39.3) becomes

© 2005 by CRC Press LLC

1586_book.fm Page 2 Friday, May 7, 2004 3:56 PM

u cL = -

D 2 dp 2m dx

(39.4)

Thus, if the pressure loss over a distance X is measured along with the centerline velocity (ucL), the viscosity can be determined. Similarly, if the velocity is known at the centerline, the pressure loss per unit length can be calculated.

39.2 Reynolds Similarity in Test Data As a boundary layer develops, it starts in a smooth, or laminar, state. Downstream, it transforms into a turbulent state, where the flow is irregular and contains eddies. Various physical conditions, such as wall or surface roughness or upstream turbulence, will affect the speed of this transition. In smooth-walled pipes, laminar flow occurs for Reynolds numbers (Re) of less than 2000, with fully developed turbulence for Re greater than 4000. The Reynolds number is a dimensionless number developed from dynamic similarity principles that represents the ratio of the magnitudes of the inertia forces to the friction forces in the fluid. Re =

inertia force friction force

where inertia force = rVc2 L2c and friction force = mVc L2c . Then, Re =

rVc L c Vc L c = m n

(39.5)

where Vc and Lc are characteristic or representative velocities and lengths, respectively. For a pipe or similar narrow channel, Lc is the internal diameter (ID) of the pipe and Vc is the average or bulk velocity obtained by dividing the mass flow rate (M) by the cross-sectional area and density of the fluid: Vc =

M rA

(39.6)

Using the Reynolds number as a similarity parameter, test data can be correlated into generalized charts for frictional losses. For the flat plate (Figure 39.1) case, Vc is taken as the free stream velocity outside the boundary layer, and Lc is the length measured along the wall standing from the leading edge.

39.3 Friction in Pipes The energy equation for steady flow between any two points in a pipe can be written as V22 - V12 P2 - P1 + + Z 2 - Z1 - h f = 0 2g rg

(39.7)

where hf is a head loss due to friction. This equation neglects other minor losses (such as elbows, valves, exit and entrance losses, and bends). It is useful to define the head loss in terms of a friction factor ( f ) such that this nondimensional friction factor ( f ), known as the Darcy friction factor, can be determined experimentally as a function of the dimensionless Reynolds numbers and a relative roughness parameter e/D, as shown in Figure 39.2. Rough factors, e, are given in Table 39.1. © 2005 by CRC Press LLC

1586_book.fm Page 3 Friday, May 7, 2004 3:56 PM

39-3

Boundary Layers

Vc

Undisturbed flow

Vc

y

Boundary layer Lc (a)

Vc

Vc Vc

u

Flat plate

Laminar boundary layer

Turbulent boundary layer

Transition region (b)

FIGURE 39.1 (a) Boundary layer along a smooth plane. (b) Laminar and turbulent boundary layers along a smooth, flat plate. (Vertical scales greatly enlarged.) 0.1

0.08

Laminar Critical Transition Zone Zone Flow

Complete Turbulence, Rough Pipes 0.05

0.07

0.04

Lam

0.06

0.03

inar

∆P L ( ρV 2 D 2 ) Friction Factor, f =

0.04

64 = R e Flow f

0.05

0.02 0.015 0.01 0.008

Recr

0.006

0.03 0.004 0.025

0.002

0.02

0.015

0.01 0.009 0.008

Material K, ft Riveted Steel 0.003–0.03 Concrete 0.001–0.01 Wood Stave 0.0006–0.003 Cast Iron 0.00085 Galvanized Iron 0.0005 Asphalted Cast Iron 0.0004 Commercial Steel or Wrought Iron 0.00015 Drawn Tubing 0.000005 103

2(103) 3 4 5 6 7 8 9104

0.001 0.0008 0.0006 0.0004

Sm

Relative Roughness, k D

0.09

oo

th

Pi

0.0002

pe

s 0.0001 0.000,05

2(104) 3 4 5 6 78 9105

2(105) 3 4 5 6 7 8 9106

Reynolds Number, ReD =

2(106) 3 4 5 6 7 8 9107

VDρ VD V(4Rh ) µ = ν = ν

0.000,01 2(107) 3 4 5 6 7 8 9108 0.000,005 0.000,001

FIGURE 39.2 Friction factors for commercial pipe. (Source: Moody, L. F. 1944. Friction factors for pipe flow. Trans. ASME. 66:672. With permission.)

© 2005 by CRC Press LLC

1586_book.fm Page 4 Friday, May 7, 2004 3:56 PM

39-4

The Engineering Handbook, Second Edition

TABLE 39.1 Roughness Surface

e, ft

e, m

Glass, plastic Drawn tubing Commercial steel, wrought iron or aluminum sheet Galvanized iron Cast iron Concrete pipe Riveted steel pipe Wood

Smooth 5 ◊ 10-6 1.5 ◊ 10-4

Smooth 1.5 ◊ 10-6 4.6 ◊ 10-5

5 ◊ 10-4 8.5 ◊ 10-4 4 ◊ 10-3 .01 .001

1.2 ◊ 10-4 2.4 ◊ 10-4 1.2 ◊ 10-3 .003 .0003

There are two equations that describe the data shown in Figure 39.2. The first is the laminar line. For laminar flow in pipes with Re less than 2000, it can be shown through analysis that

f =

64 Re

(39.8)

The second is the Colebrook equation [Colebrook, 1938]: È e /D 2.51 ˘ = -2 log 10 Í + ˙ f ÍÎ 3.7 Re f ˙˚

1

(39.9)

which describes the turbulent region. Note that, as the roughness e approaches zero, we obtain the smooth pipeline and the equation becomes È Re f ˘ = 2 log 10 Í ˙ f ÍÎ 2.51 ˙˚

1

(39.10)

For fully developed turbulence, the Re approaches zero and the Colebrook equation simplifies to È 3.7 ˘ = 2 log 10 Í ˙ f Î e /D ˚

1

(39.11)

For turbulent flows in closed conduits with noncircular cross-sections, a modified form of Darcy’s equation may be used to evaluate the friction loss: Ê L ˆ Ê v2 ˆ hf = f Á ˜ Á ˜ Ë D ¯ Ë 2g ¯ where D is the diameter of the circular conduit.

39.4 Noncircular Channel In the case of noncircular cross-sections, a new term, R, is introduced to replace diameter D. R is defined as hydraulic radius, which is the ratio of the cross-sectional area to the wetted perimeter (WP) of the noncircular flow section. © 2005 by CRC Press LLC

1586_book.fm Page 5 Friday, May 7, 2004 3:56 PM

39-5

Boundary Layers

R=

A WP

For a circular pipe of diameter D the hydraulic radius R is pD 2 / 4 D A = = WP pD 4

R=

or D = 4R. Substitution of 4R for D in Darcy’s equation yields Ê L ˆ Ê v2 ˆ hf = Á ˜ Á ˜ Ë 4R ¯ Ë 2g ¯ The Reynolds number can be modified as Re =

v(4R)r m

or Re =

v(4R) n

39.5 Example Solutions Example 39.1 Refer to Figure 39.3. Water at 50∞C is flowing at a rate of 0.07 m3/sec. The pipeline is steel and has an inside diameter of 0.19 m. The length of the pipeline is 900 m. Assume the kinematic viscosity (n) is 5.48 ◊ 10-7 m2/sec. Find the power input to the pump if its efficiency is 82%; neglect minor losses. Given information is as follows: Q = 0.07 m3 ; DZ = 15 m;

L = 900 m;

T = 50∞C

n = 5.48 ◊10 -7 m2 / s;

D = 0.2 m

g at 50∞C = 9.69 kN / m3 Find power input to pump. 2

15 m

1 0

60

Pump 300 m

FIGURE 39.3 Pipeline in Example 39.1.

© 2005 by CRC Press LLC

m

1586_book.fm Page 6 Friday, May 7, 2004 3:56 PM

39-6

The Engineering Handbook, Second Edition

Solution First, determine the Reynolds number:

Re =

vD ; n

Re =

4Q 4(0.07) = pDn p(0.2)(5.48 ◊10 -7 )

v=

Q A

= 8.13 ◊105 Second, determine e/D ratio and friction factor f. Roughness (e) for steel pipe = 4.6 ◊ 10-5 m. e 4.6 ◊10 -5 m = = 0.000242 D 0.19 m From Moody’s diagram with values of NR and e/D, f = 0.0151. Next, determine head loss due to friction: Ê L ˆ Ê v2 ˆ h f = 0.0151Á ˜ Á ˜ , Ë D ¯ Ë 2g ¯ h f = 0.0151

v=

Q A

8LQ 2 p 2 gD5

= 0.0151

8(900)(0.07)2 p 2 (9.81)(0.02)5

= 0.0151

35.28 = 17.2 m 0.031

Finally, determine power input into pump: kN ˆ Ê m3 ˆ Ê PA = h A gQ = 17.2Á 9.69 3 ˜ Á 0.07 ˜ Ë s ¯ m ¯Ë = 11.67 ep =

kN ◊ m = 11.67 kW s

PA P1

e p = Pump efficiency PA = Power delivered to fluid PI = Power input into pump PI =

© 2005 by CRC Press LLC

PA 11.67 kW = = 14.23 kW ep 0.82

1586_book.fm Page 7 Friday, May 7, 2004 3:56 PM

39-7

Boundary Layers

50 mm 25 mm Dia.

50 mm

FIGURE 39.4 Duct in Example 39.2. TABLE 39.2 Dynamic Viscosity of Liquids (m) (mPa ◊ sec) Liquid Water Mercury Methanol Isobutyl acetate Toluene Styrene Acetic acid Ethanol Ethylene glycol

-25∞∞C

0∞∞C

25∞∞C

50∞∞C

75∞∞C

1.793

0.890 1.526 0.544 0.676 0.560 0.695 1.056 1.074 16.1

0.547 1.402

0.378 1.312

0.493 0.424 0.507 0.786 0.694 6.554

0.370 0.333 0.390 0.599 0.476 3.340

1.258

0.793

1.165

0.778 1.050

3.262

1.786

100∞∞C

0.286 0.270 0.310 0.464 1.975

Example 39.2 Air with a specific weight of 12.5 N/m3 and dynamic viscosity of 2.0 ◊ 10-5 N ◊ sec/m2 flows through the shaded portion of the duct shown in Figure 39.4 at the rate of 0.04 m3/sec. (See Table 39.2 or Figure 39.5 for dynamic viscosities of some common liquids.) Calculate the Reynolds number of the flow, given that g = 12.5 N/m3, m = 2.0 ◊ 10-5 N ◊ sec/m2, Q = 0.04 m3/sec, and L = 30 m. Solution r = g /g =

12.5 N / m3 = 1.27 N ◊ s 2 /m4 or kg / m3 9.81 m / s 2

p A(shaded) = (0.05 m)2 - (0.025 m)2 4 = 0.0025 m2 - 0.00049 m2 = 0.002 m2 Wet parameter (WP)= 4(0.05 m) + p(0.025 m) = 0.2 m + 0.0785 m = 0.279 m Hydraulic radius (R) = =

A WP 0.002 m2 0.279 m

= 0.00717 m

© 2005 by CRC Press LLC

1586_book.fm Page 8 Friday, May 7, 2004 3:56 PM

39-8

The Engineering Handbook, Second Edition

0.5 0.4 0.3

il ro sto Ca

yc Gl

0.2

n eri

0.1

E SA

SA

E

10

0.06

30

oil

0.04 0.03

oi

l

0.02

Absolute viscosity µ, (N.s)/m2

0.01 6

Crud

e oil

An

4 3

ilin

Ke ro

(SG

e

sin

2

Car

bon

1 × 10−3

tetr ach

e

4 3

lco

e

line (

2

hol

Benz ene

Gaso

)

Mercury

Eth yl a

lorid

6

0.86

Wa te

SG 0

r

.68)

1 × 10−4 6 4 3

Air Helium

2

ide

Carbon diox

1 × 10−5

Hydrogen

5 −20

0

20

40 60 Temperature, °C

80

100

120

FIGURE 39.5 Absolute viscosity of common fluids at 1 atm. (Source: White, F. 1986. Fluid Mechanics, 2nd ed. McGraw-Hill, New York. With permission.)

v=

Q 0.04 m3 / s = = 20 m / s A 0.002 m2

Reynolds number (Re)=

NR =

4(0.00717 m)(20 m / s)(1.27 N ◊ s / m4 ) 2.0 ◊10 -5 N ◊ s / m2

= 3.64 ◊10 4

© 2005 by CRC Press LLC

4Rvr m

1586_book.fm Page 9 Friday, May 7, 2004 3:56 PM

Boundary Layers

39-9

References Colebrook, C. F. 1938. Turbulent flow in pipes with particular reference to the transition points between smooth and rough laws. ICF Journal. 2:133–156. Moody, L. F. 1944. Friction factors for pipe flow. Trans. ASME. 66:672. White, F. 1986. Fluid Mechanics, 2nd ed. McGraw-Hill, New York.

© 2005 by CRC Press LLC