PIPE FLOW

Laminar Flow in Pipes

�Fluid is incompressible and Newtonian.

�Flow is steady, fully developed, parallel and,

symmetric with respect to pipe axis.

�Pipe is straight pipe and has a constant diameter.

Laminar Flow in Pipes

• Momentum Equation

z1 z2

x

dxθθθθ

τr

Datum

p+(dp/dx)dx

p

r0

+

γ=

γ

τ+=

γ

γ+−=−

τ=

γ+−

π=

=πτ−γ−−

=πτ−θγ−

+−

zp

h since r

2

dx

)zp(d

dx

dh

r

2

dx

)zp(d

)r Aby sidesboth (Divide

0rdx2Adx

dzdxdxA

dx

dp

0rdx2sinAdxAdxdx

dpppA

2

z

1

z

2

x

dx

θθθθ

τ

rp+(dp/dx)dx

pr0

(Constant velocity thus a=0)

Boundary Conditions

when r = 0 , τ = 0

r = ro , τ = τw o

w

r

2

r

2

dx

dh

γ

τ=

γ

τ=−

r

2

dx

)zp(d

dx

dh

γ

τ+=

γ

γ+−=−

ττττ = ττττw (1- r / r0)

τw.....wall shear stress

equations are equally applicable to both laminar

and turbulent flow in pipes

Laminar Flow

(2) 2

r

dx

)zp(d

(1) dr

du

dy

du

γ+−=τ

µ−=µ+=τ

µ

γ++=

2

r

dx

)zp(d

dr

du

τy

r

τw

CL

Boundary Conditions

µ

γ++=

2

r

dx

)zp(d

dr

du

C4

r

dx

)zp(d)r(u

2

+µ

γ+=

u = u(r) may be solved by integration

r = 0 , u = umax

( )

−

µ

γ+−=

2

o

2

o

r

r1

4

r

dx

zpdu Parabolic profile

Poiseuille Flow

r = ro ; u = 0

• Velocity:

• Average velocity:

• Maximum velocity:

• Wall shear stress:

• Shear stress:

• Flow rate:

• Head loss:

( )µ

γ+−=

4

r

dx

zpdu

2o

max

ow

r

V4µ=τ

ow

r

r

dr

duτ=µ−=τ

( )dx

zpd

8

rAVQ

4o γ+

µ

π−==

dx

)zp(d

dx

dh

L

hf

γ

γ+−=−=

( )µ

γ+−==∫==

8

r

dx

zpd

2

u

A

dau

A

QV

2omax

( )

−

µ

γ+−=

−=

2

o

2o

2

omax

r

r1

4

r

dx

zpd

r

r1uu

Turbulent Flow : Re ≥ 4000

R = D /2

V i s c o u s s u b l a y e r

V e l o c it y

p ro fi le , u = u (y )

y

x

δ s

Smooth wall

εδs

Rough wall

εε δs

(a) Smooth wall (b) Transitional flow (c) Rough wall

Velocity very small near wall

Thus flow must be laminar!!

This region is called

Viscous sub-layer

(a) smooth wall and (b) a rough wall.

Shear velocity

Average velocity in laminar flow

Viscous sublayer thickness

Smooth wall

Rough wall

Laminar and Turbulent

slope

slope

Laminar Turbulent

τwτw

τw,turb > τw,lam

Comparison of laminar and turbulent flow

Laminar

• Can solve exactly Flow is steady

• Velocity profile is parabolic

• Pipe roughness not important

Turbulent• Cannot solve exactly (too complex)

• Flow is unsteady , but it is steady in the mean

• Mean velocity profile is fuller (shape more like a top-hat profile, with very sharp slope at the wall)

• Vavg 85% of Umax (depends on Re a bit)

• Pipe roughness is very important

• No analytical solution, but there are some good semi-empirical expressions that approximate the velocity profile shape. Instantaneous

profiles

Darcy Weisbach Equation

• Consider a steady fully developed flow in a prismatic pipe

(A = constant along centerline)

R

W

z1 z2

CV

p2A2p1A1

Ff

1

2

Wsinθθθθ

θ

V2 x

L

V1

Wcosθθθθsθθθθ

Relation between wall shear stress and head loss

)zz(AsinAL

sinALsinW

21 −γ=θγ

θγ=θPLF wf τ=

0PL)zz(AApAp w212211 =τ−−γ+−

Constant VA AV AV Q 2221 ====

)VV(QFsinWApAp 1122f2211 β−βρ=−θ+−

A

LPpz

pz w2

21

1γ

τ=

γ−−

γ+

12 VV =

)A/(1 γ12 AAA ==

0

Relation between wall shear stress and head loss

4

D

P

AR h ==

A

LPpz

pz w2

21

1γ

τ=

γ−−

γ+

H

ww

R

L

A

LP

γ

τ=

γ

τ

H

w22

11

R

Lpz

pz

γ

τ=

γ−−

γ+

2D4

Aπ

= DP π=

Hydraulic Radius

Relation between wall

shear stress and head loss

H

w22

11

R

Lpz

pz

γ

τ=

γ−−

γ+ f

22

11 h

pz

pz =

γ−−

γ+

D

L4h w

fγ

τ=

Energy equation

between section 1 and 2

D

L4

R

L2

R

Lh ww

H

wf

γ

τ=

γ

τ=

γ

τ=

Darcy – Weisbach Friction Factor

Laminar Flow: Re ≤ 2000

u(r)

umax

r

g2

V

D

L

R

64

D

VL32

2V

2V

D

VL32

r

VL8h

2

e222

o

avf =

γ

µ=

γ

µ=

γ

µ=

g2D

LVfh

2

f =

R

64f

e

=

Darcy Weisbach Equation

D

L4h w

fγ

τ=

ow

r

V4µ=τ

Friction factor for Turbulent Flows

For hydraulically smooth pipe,

f=f(Re) only

For frictionally transition zone:

f=f(Re, ε/D)

For fully rough pipe:

f=f(ε/D) only.

Formula for friction factors in Turbulent flows

−=

fR

51.2log2

f

1

e

ε+−=

D7.3fR

51.2log2

f

1

e

ε=

D,Rfuncf e

Smooth Pipe and Hydraulically Smooth Flow

Colebrook - White – Transitional Flow

ε−=

D7.3log2

f

1

ε=

Dfuncf

Rough Pipe-

Hydraulically Rough Flow

( )eRfuncf =

Swamee – Jain Formula (Explicit)

2

9.0eR

74.5

D7.3ln

325.1f

+

ε

=

8e

26 10R5000and10D/10 <<<ε< −−for the range of

Moody’s Diagram

Moody diagram. (From L.F. Moody, Trans. ASME, Vol.66,1944.) (Note: If e/D = 0.01

and Re = 104, the dot locates ƒ = 0.043.)

Determination of Friction Loss (hf):

• Darcy - Weisbach Equation

• Hazen-Williams Equation

22

25

2

f KQg2

Q16

D

Lf

g2

V

D

Lfh =

π==

85.185.1

87.485.1

85.1

165.185.1f KQQD

L

C

6.10V

D

L

C

8.6h ===

85.1

165.185.1f VD

L

C

8.6h =

g2

V

D

Lfh

2

f =

Roughness Coefficients

Material Hazen-Williams

C

Manning’sCoefficient

n

Darcy-WeisbachRoughness

Height

εεεε (mm)

Asbestos cementBrassBrick Cast-iron, newConcrete:

Steel formsWooden formsCentrifugally spun

CopperCorrugated metalGalvanized ironGlassLeadPlasticSteel:

Coal-tar enamelNew unlinedRiverted

Wood stave

140135100130140120135135---

120140135150148145110120

0.0110.0110.0150.0120.0110.0150.0130.0110.0220.0160.0110.0110.0090.0100.0110.0190.012

0.00150.0015

0.60.260.180.6

0.360.0015

450.15

0.00150.00150.00150.00480.0450.9

0.18

,

Total Head Loss

hf – Friction (Viscous, Major) loss

hm – Local (Minor) loss

h= hf + hm

COMPUTATION OF FLOW IN SINGLE PIPES

VariableType of the problem

Type I Type II Type III

Fluid DensityViscosity

GG

GG

GG

Pipe DiameterLengthRoughness

GGG

GGG

D or GG

G or D

Flow Flowrate, or Average velocity G D G

Pressure Pressure Drop, or Head loss D G G

G- Given D- to be determined

1) Determination of Head Loss (Type I)

• Given : Q (or V), L, D, νννν, εεεεFind : hf

1)

2)

3) εεεε/D

4) f(Re,εεεε/D) is determined (from Moody Chart or Eqs.)

5) hf is computed

)4

DV Q(or

4 2

2

π

π==

D

QV

0hcesinhh and hHH mf21 ==+= λλ

HHh 21f −=

g2

VpzH

2

+γ

+= 2

52f

2

f QDg

fL8hthenA/QVcesin

g2

V

D

Lfh

π===

νπ=

ν D

Q4VD = Re

Example 1 (Type-I problem):

• A galvanized iron pipe with a roughness height of 5x10-6 m with a diameter of 0.05 m and a length of 100 m carries a discharge of 0.003 m3/s. Calculate the head loss.

• Cross section area A = 3.14159 * (0.025) * (0.025) = 0.001963 m2

• Velocity V = 0.003 / 0.001963 = 1.528 m/s

• Reynolds Number = 1.528 * 0.05 * 106 = 76,394

• Iron Pipe =====� Relative roughness = 5x10-6 / 0.05 = 0.0001

• f = 0.0195 from the Chart

νπ=

ν D

Q4VD = Re

Moody diagram. (From L.F. Moody, Trans. ASME, Vol.66,1944.) (Note: If e/D = 0.01

and Re = 104, the dot locates ƒ = 0.043.)

Example 1

π - 3.141592

EXAMPLE g m/s29.81

for water ρ kg/m31000

for water µ kg/m.s 0.001

for water ν m2 /s 0.000001

A galvanized iron pipe with a roughness height of 0.000005 m

with a diameter of 0.05 m

and a length of 100 m

carries a discharge of 0.003 m3 /s

its x-section is 0.001963 m2

and water flows with a velocity of 1.528 m/s

Flow Reynolds number is 76394

and the relative roughness is 0.0001

using S-J formula the friction factor is obtained as 0.01941

The head loss in the pipe is found to be 4.620 m

2) Determination of average velocity (Type II)

Given: hf, L, D, υυυυ, εεεε

Find : V

Since f depends on V through Re, and V is unknown a priori, iteration is needed

g2

V

D

Lfhh , hHH

2

f21 ==+= λλ

( ) fL

gD2h

fL

gD2HHV f21 =−=

(hm ≈ 0)

Re=VD/ν

ε/D

V,D,ν

Moody Chart or Eqs.

νπ=

ν D

Q4VD = Re

Solution procedure (Type II):

1. Calculate relative roughness

2. Select friction factor,

(assume completely rough turbulent

flow); f(i) = f(0)

3. Calculate velocity;

4. Calculate Reynolds number;

5. Determine f by using data from

Step1 and 4; f(i+1) (use Moody

Chart, or Equation)

6. Check if f(i+1) = f(i);?

7. no, go to step 3 with f(i+1)

8. yes, continue

9. Calculate Q or V

D

ε

fL

g2DhV f=

4

DVQ

2π=

f(i) = f(0)

ν=

VDRe

Given: hf, L, D, υυυυ, εεεε Find : V

Iteration Table

f (i) V Re f (i+1)*

f(0) Assumed calculated calculated f (1)-determined

f (1) calculated calculated f (2) -determined

f (2) calculated calculated f (3) -determined

.

. iteration is

.

.

. stopped

.

.

. when

.

.

. f(i)=f(i+1)

.

f (i) f (i+1)

* obtained from Moody Chart, or determined using equations.

fL

g2DhV f=

ν=

VDRe

• Given: hf, L, D, υυυυ, εεεε.• Find : V

Example 2 (Type-II problem):

• Example 2.2 (Type-II problem): A galvanized iron pipe with a roughness height of 5x10-6 m with a diameter of 0.05 m and a length of 100 m has experienced head loss of 10 m. Calculate the flow rate.

EXAMPLE

e - 2.718282

π - 3.141592

g m/s29.81

for water ρ kg/m31000

for water µ kg/m.s 0.001

for water ν m2 /s 0.000001

A galvanized iron pipe with a roughness height of ε 0.000005 m

with a diameter of D 0.05 m

and a length of L 100 m

has experienced head loss of hf 10 m

its x-section is A 0.00196 m2

and the relative roughness is ε/D 0.0001

carrying out an iterative solution procedure i fi V Re fi+1 ∆f

0 0.02 2.215 110736.2 0.018105 -0.00189

1 0.01811 2.328 116387.2 0.017944 -0.00016

2 0.01794 2.338 116909.4 0.017929 -1.4E-05

3 0.01793 2.339 116956.2 0.017928 -1.3E-064 0.01793 2.339 116960.4 0.017928 -1.1E-07

one obtains the friction factor and velocity (m/s) 5 0.01793 2.339 116960.8 0.017928 -1E-08

hence the discharge is obtained as Q 0.00459 m3 /s

Example 2 (Type-II problem):

• Example 2.2 (Type-II problem): A galvanized iron pipe with a roughness height of 5x10-6 m with a diameter of 0.05 m and a length of 100 m has experienced head loss of 10 m. Calculate the flow rate.

D = 0.05 mg = 9.81 m/s2

L = 100mhf = 10m

= 5 * 10-6

fL

g2DhV f=

ν=

VDRe

εεεε

D

ε= 5*10-6 / 0.05 = 0.0001

f = 0.012 assuming fully rough flow

V2 = [ (10*0.05*2*9.81) / (0.012*100) ] = 8.175 V= 2.8592 m/s

Re = 2.8592 * 0.05 / 10-6 = 1.43 * 105

f = 0.0165 Since not the same with the previous continue iteration

Initial f V (m/s) Re (106) Final f Del f

0.012 2.8592 0.143 0.0165 0.0045

0.0165 2.4383 0.122 0.017 0.0005

0.017 2.4022 0.120 0.017 0

Moody diagram. (From L.F. Moody, Trans. ASME, Vol.66,1944.) (Note: If e/D = 0.01

and Re = 104, the dot locates ƒ = 0.043.)

........................................................................................................

3) Determination of Diameter (Type III)

• Given: hf, L, Q, νννν, εεεε. Find : D

1. Assume f(i) = f(0) (arbitrarily 0.02)

2. Calculate pipe diameter

3. Calculate Reynolds number

4. Calculate relative roughness

5. Determine friction factor, f(i+1) use Moody Chart or Equations

6. Check if f(I+1) = f(i) ; ?

7. if no, go to step 2 with f(i+1)

8. if yes, stop. Diameter Calculated at Step 2 is the result.

9. Select the next larger commercially available pipe diameter size

5/1

5/1

2

2

52

2

fgh

LQ8

gh

LQ8fD

π=

π=

λλ

νπ=

ν=

D

Q4VDRe

D

ε

Iteration Table

f (i) D Re εεεε/D f (i+1)*

f(0)

Assumed

calculated calculated calculated f (1)-determined

f (1) calculated calculated calculated f (2)-determined

f (2) calculated calculated calculated f (3) -determined

..

.

.

. iteration is

.

.

. stopped

.

.

. when

.

.

. f(i)=f(i+1)

.

f (i) calculated calculated calculated f (i+1)- determined

5/1

5/1

2

2

52

2

fgh

LQ8

gh

LQ8fD

π=

π=

λλνπ

=ν

=D

Q4VDRe

D

ε

Example 2.3 (Type-III problem):

A galvanized iron pipe with a roughness height of 0.00005 m and a length of 100 m under the head loss of 10 m delivers discharge of 0.003 m3/s. Calculate pipe diameter.

EXAMPLE

e - 2.71828

π - 3.14159

g m/s2 9.81

ρ kg/m3 1000

µ kg/m.s 0.001

ν m2 /s 1E-06

A galvanized iron pipe with a roughness height of ε m 0.00005

with a length of L m 100under the head loss of hf m 10

delivers discharge of Q m3 /s 0.003

carrying out an iterative solution procedure i fi D A V Re ε/D fi+1 ∆f

0 0.02 0.0431 0.00146 2.05624 88624.3 0.00116 0.02312 0.00312

1 0.02312 0.0444 0.00155 1.94025 86088.5 0.00113 0.02308 -4E-05

2 0.02308 0.0444 0.00155 1.94169 86120.4 0.00113 0.02308 5.2E-07

3 0.02308 0.0444 0.00155 1.94168 86120 0.00113 0.02308 -6E-094 0.02308 0.0444 0.00155 1.94168 86120.1 0.00113 0.02308 7.7E-11

one obtains the friction factor and velocity (m/s) 5 0.02308 0.0444 0.00155 1.94168 86120.1 0.00113 0.02308 -9E-13

hence the discharge is obtained as D 0.0444 m