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
τ x
r0 p+(dp/dx)dx
r
p
θ dx
z1 z2
Datum
dp
pA − p + dx A − γAdx sin θ − τ2πrdx = 0 (Constant velocity thus a=0)
dx
dp dz τ x
− dxA − γdx A − τ2πrdx = 0 0 r
rp
dx dx p+(dp/dx)dx
2 d
(Divide both sides by A = πr ) θ x
d ( p + γz ) 2 τ z z
− = 1 2
dx r
dh d ( p + γz ) 2τ p
− =− =+ since h = + z
dx γdx γr γ
Boundary Conditions
dh d( p + γz) 2τ
− =− =+
dx γdx γr
when r = 0 , τ = 0 dh 2τ 2τ w
− = =
r = ro , τ = τw dx γr γro
τ = τw (1- r / r0)
τw.....wall shear stress
equations are equally applicable to both laminar
and turbulent flow in pipes
Laminar Flow
τw
y
τ
r CL
du du
τ = +µ = −µ (1)
dy dr
d(p + γz ) r
τ=− (2)
dx 2
du d(p + γz ) r
= +
dr dx 2µ
Boundary Conditions
du d(p + γz ) r
=+
dr dx 2µ
u = u(r) may be solved by integration
d(p + γz) r 2
u (r ) = +C
dx 4µ
r = ro ; u=0 r=0 , u = umax
d (p + γz ) r 2 r
2
u=− o
1 − Parabolic profile
dx 4µ ro Poiseuille Flow
r 2 d(p + γz ) ro2 r
2
• Velocity: u = umax 1 − = − 1 −
r
o dx 4 µ ro
Q ∫ uda umax d(p + γz ) ro2
• Average velocity: V= = = =−
A A 2 dx 8µ
• Maximum velocity: d(p + γz ) ro2
umax =−
dx 4µ
4µV
• Wall shear stress: τw =
ro
• Shear stress: du r
τ = −µ = τw
dr ro
• Flow rate: πro4 d(p + γz )
Q=V A=−
8µ dx
• Head loss: hf dh d(p + γz )
=− =−
L dx γdx
Turbulent Flow : Re ≥ 4000
Velocity very small near wall
V e l o c it y
Thus flow must be laminar!!
R = D /2 p r o fi le , u = u(y )
y
This region is called
δs
Viscous sub-layer
x
V i s c o u s s ub l ay e r
δs ε ε δs
ε
Smooth wall Rough wall
(a) Smooth wall (b) Transitional flow (c) Rough wall
(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
Laminar Turbulent
slope
slope
τw τw
τw,turb > τw,lam
Comparison of laminar and turbulent flow
Laminar Turbulent
• Can solve exactly Flow is steady • Cannot solve exactly (too complex)
• Velocity profile is parabolic • Flow is unsteady , but it is steady in
• Pipe roughness not important 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
Instantaneous velocity profile shape.
profiles
Darcy Weisbach Equation
• Consider a steady fully developed flow in a prismatic pipe
(A = constant along centerline)
L
1
CV 2
V1 Wsinθ
θ
p1A1 p2A2 V2
x R
Ff
Wcosθ
θsθ
θ θ
W
z1 z2
Relation between wall shear stress and head loss
Q = V1A 2 = V2 A 2 = VA = Constant
p1A1 − p2 A 2 + W sin θ − Ff = ρ0Q(β2 V2 − β1V1)
W sin θ = γAL sin θ
Ff = τ w PL V2 = V1
γAL sin θ = γA(z1 − z 2 )
p1A1 − p 2 A 2 + γA (z1 − z 2 ) − τ w PL = 0
A = A 2 = A1 1 /(Aγ )
p1 p 2 τ w LP
z1 + − z 2 − =
γ γ γA
Relation between wall shear stress and head loss
p p τ LP π A D
A = D2 P = πD Rh = =
z1 + 1 − z 2 − 2 = w P 4
γ γ γA 4
τ w LP τ w L Hydraulic Radius
=
γA γR H
p1 p2 τw L
z1 + − z 2 − =
γ γ γR H
Relation between wall Energy equation
shear stress and head loss between section 1 and 2
p1 p τ L p1 p
z1 + − z2 − 2 = w z1 + − z 2 − 2 = hf
γ γ γR H γ γ
τ w L 2 τ w L 4τ w L
hf = = =
γR H γR γD
4τ w L
hf =
γD
Darcy – Weisbach Friction Factor
Laminar Flow: Re ≤ 2000
u(r)
4τ w L r Darcy Weisbach Equation
hf =
γD LV
2
hf = f
4µV D2g
τw =
ro umax
2
hf =
8LµVav
=
32LµV
=
2V 32LµV
=
64 L V 64
γro
2
γD2 2V γD2 Re D 2g f=
Re
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
Smooth Pipe and 2.51
1 f = func (Re )
Hydraulically
= −2 log
Smooth Flow f Re f
Colebrook - White
1 2.51 ε ε
– Transitional
Flow
= −2 log + f = func R e ,
D
f Re f 3.7D
Rough Pipe-
Hydraulically
Rough Flow 1 ε ε
= −2 log f = func
D
f 3.7D
Swamee – Jain Formula (Explicit)
1.325
f= 2
ε 5 . 74
ln + 0.9
3.7D R e
for the range of 10 −6 < ε / D < 10 −2 and 5000 < Re < 108
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
L V2 L 16 Q2
hf = f =f 5 2 = KQ2
D 2g D π 2g
• Hazen-Williams Equation
6 .8 L 1.85 10.6 L
hf = 1.85 1.165 V = 1.85 4.87 Q1.85 = KQ1.85
C D C D
Roughness Coefficients
Material Hazen- Manning’s Darcy-Weisbach
Williams Coefficient Roughness
C n Height
ε (mm)
Asbestos cement 140 0.011 0.0015
Brass 135 0.011 0.0015
Brick 100 0.015 0.6
Cast-iron, new 130 0.012 0.26
Concrete: 140 0.011 0.18
Steel forms 120 0.015 0.6
Wooden forms 135 0.013 0.36
Centrifugally spun 135 0.011 0.0015
Copper --- 0.022 45
Corrugated metal 120 0.016 0.15
Galvanized iron 140 0.011 0.0015
Glass 135 0.011 0.0015
Lead 150 0.009 0.0015
Plastic 148 0.010 0.0048
Steel: 145 0.011 0.045
Coal-tar enamel 110 0.019 0.9
New unlined 120 0.012 0.18
Riverted
Wood stave
6. 8 L
hf = 1.85 1.165 V1.85 L V2
hf = f
, C D D 2g
Total Head Loss
h= hf + hm
hf – Friction (Viscous, Major) loss
hm – Local (Minor) loss
COMPUTATION OF FLOW IN SINGLE PIPES
G- Given D- to be determined
Type of the problem
Variable Type I Type II Type III
Fluid Density G G G
Viscosity G G G
Pipe Diameter G G D or G
Length G G G
Roughness G G G or D
Flow Flowrate, or Average velocity G D G
Pressure Pressure Drop, or Head loss D G G
1) Determination of Head Loss (Type I)
H1 = H 2 + h λ and h λ = h f sin ce h m = 0
hf = H1 − H2
p V2 L V2 8fL
H=z+ + hf = f sin ce V = Q / A then hf = 2 5 Q2
γ 2g D 2g gπ D
• Given : Q (or V), L, D, ν, ε
Find : hf
4Q VπD 2
1) V= (or Q = )
πD 2 4
VD 4 Q
2) R e = =
ν πD ν
3) ε/D
4) f(Re,εε/D) is determined (from Moody Chart or Eqs.)
5) hf is computed
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
VD 4 Q
• Reynolds Number Re = = = 1.528 * 0.05 * 10 6 = 76,394
ν πD ν
• Iron Pipe ===== Relative roughness = 5x10-6 / 0.05 = 0.0001
• f = 0.0195 from the Chart
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/s2 9.81
for water ρ kg/m3 1000
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)
L V2 (hm ≈ 0)
H1 = H2 + hλ , hλ = h f = f
D 2g
V= (H1 − H2 ) 2gD = hf
2gD
Given: hf, L, D, υ, ε
fL fL
Find : V
VD 4 Q
ε/D Re = =
ν πD ν
V,D,ν Since f depends on V
Moody Chart or Eqs. through Re, and V is
unknown a priori, iteration
Re=VD/ν is needed
Solution procedure (Type II):
Given: hf, L, D, υ, ε Find : V
1. Calculate relative roughness ε
2. Select friction factor, D
(assume completely rough turbulent f(i) = f(0)
flow); f(i) = f(0) hf D2g VD
V= Re =
3. Calculate velocity; fL ν
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
πD2
9. Calculate Q or V Q=V
4
Iteration Table
• Given: hf, L, D, υ, ε. hf D2g VD
V= Re =
• Find : V fL ν
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.
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/s2 9.81
for water ρ kg/m3 1000
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-06
4 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 m
g = 9.81 m/s2 hf D2g R = VD ε
L = 100m V= e = 5*10-6 / 0.05 = 0.0001
fL ν D
hf = 10m
ε = 5 * 10-6 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, ν, ε. Find : D
1. Assume f(i) = f(0) (arbitrarily 0.02)
1/ 5
8LQ 2
8LQ 2 1/ 5
2. Calculate pipe diameter D=5 f 2
= 2
f
h λπ g h λπ g
VD 4Q
3. Calculate Reynolds number Re = =
ν πDν
ε
4. Calculate relative roughness
D
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
Iteration Table
1/ 5
8LQ 8LQ 1/ 5
2 2
VD 4Q ε
D=5 f 2
= f
2
Re = =
h λπ g h λπ g ν πDν D
f (i) D Re ε/D f (i+1)*
f(0) calculated calculated calculated f (1)-determined
Assumed
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
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 100
under 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-09
4 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
