1

Chap.6 Flow in pipes The transport of a fluid (liquid or gas) in a closed conduit commonly called a pipe if it is of

round cross section is extremely important in our daily operations. The purpose of this chapter is

to understand the basic processes involved in such flow.

Consider a case where fluid runs from a tank into a pipe whose entrance section is fully rounded.

At the entrance, the velocity distribution is roughly uniform while the pressure head is lower by

V2/2g . As shown in below Figure ,the section from the entrance to just where the boundary layer

develops to the tube centre is called the inlet or entrance region, whose length is called the inlet

or entrance length.

For steady flow at a known flow rate, these regions exhibit the following:

Laminar flow: A local velocity constant with time, but which varies spatially due to

viscous shear and geometry.

Turbulent flow: A local velocity which has a constant mean value but also has a statistically

random fluctuating component due to turbulence in the flow. Typical plots of velocity time

histories for laminar flow, turbulent flow, and the region of transition between the two are shown

below .

Principal parameter used to specify the type of flow regime is the Reynolds number :

V - characteristic flow velocity

D - characteristic flow dimension

2

µ - dynamic viscosity

υ - kinematic viscosity

We can now define the critical or transition Reynolds number Recr

Recr is the Reynolds number below which the flow is laminar, above which the flow is turbulent

While transition can occur over a range of Re, we will use the following for internal pipe or duct

flow:

Typical criteria for the length of the entrance region are given as follows:

Le = length of the entrance region .The wall shear is constant, and the pressure drop linearly

with x, for either laminar or turbulent flow. All these details are shown in the below Figures.

Laminar flow:

computation by Boussinesq

L = O.06Red computation by Asao, Iwanami and

Mori

Turbulent flow:

L = 0.693Re1I4d computation by Latzko

Developing pressure changes in the entrance of a duct flow and the entrance region

Velocity distribution of Laminar Flow in pipe:

In the case of axial symmetry, when cylindrical coordinates are used , the momentum equation

become as following :

3

....(1)

....(2)

For the case of a parallel flow like this, the Navier-Stokes equation is extremely simple as

follows:

1. As the velocity is only u since v = 0, it is sufficient to use only equation1

2. As this flow is steady, u does not change with time, so ∂u/∂t = 0.

3. As there is no body force, so ρX = 0.

4. As this flow is uniform, u does not change with position, so ∂ul∂x = 0 and ∂2u/∂x

2=0

5. Since v= 0, the equation 2 simply expresses the hydrostatic pressure variation and has no

influence in the x direction. So, equation 1 becomes :

Integrating

According to the boundary conditions, since the velocity at r = 0 must be finite c1 = 0 and c2

is determined when u = 0 at r = ro:

Laminar flow in a circular pipe

From this equation, it is clear that the velocity distribution forms a parabolic with umax at r = 0:

The volumetric flow rate passing pipe Q becomes :

⇒��

��= �

�

�

�

��(��

��)

lnr

4

From this equation, the mean velocity v is :

Putting the pressure drop in length L as ∆p, the following equation is obtained :

( Hagen-Poiseuille formula )

Velocity distribution between parallel plates:

Let us study the flow of a viscous fluid between two parallel plates as shown in below Figure ,

where the flow has just passed the inlet length. The momentum equations in x and y directions as

in the following :

Under the same conditions as in the previous section , the upper equation (1) becomes :

Consider the balance of forces acting on the respective faces of an assumed small volume

dx dy (of unit width) in a fluid.

-------------( 1 )

-------------( 2 )

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Since there is no change of momentum between the two faces, the following equation is

obtained:

therefore

By integrating the above equation twice about y, the following equation is obtained:

---------(3)

Using u = 0 as the boundary condition at y = 0 and h, c1 and c2 are found as follows:

It is clear that the velocity distribution now forms a parabola. At y = h/2 , du/dy = 0 , so u

becomes umax :

The volumetric flow rate Q becomes :

---------(4)

From this equation, the mean velocity v is :

The shearing stress z due to viscosity becomes :

Putting L as the length of plate in the flow direction and ∆p as the pressure difference, and

integrating in the x direction, the following relation is obtained:

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Substituting this equation into eqn (4) gives :

As shown in the below Figure , in the case where the upper plate moves in the x direction at

constant speed U or -U, from the boundary conditions of u = 0 at y = 0 and u = U at y = h, c1 and

c2 in eqn (3) can be determined. Thus :

and

Couette-

Poiseuille

flow

Velocity distribution of turbulent Flow For two-dimensional flow, the velocity is expressed as follows:

where u and v are the timewise mean velocities and u

' and v

' are the fluctuating velocities.

The shearing stress τ of a turbulent flow is :

τ1 = laminar flow shear stress

τt= turbulent shearing where numerous rotating eddies mix with each other. stress

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Now, let us examine the turbulent shearing stress only. the upward eddy motion of fluid particles

in a layer of lower velocity which passes in unit time in the y direction through dA parallel

to the x axis is ρv' dA. Since this fluid is at relative velocity u' , the momentum is pv' dA u

'. By

the movement of this fluid, the upper fluid increases its momentum per unit area by ρ u'

v' in the positive direction of x per unit time. Therefore, a shearing stress develops on face

dA. It is found that the shearing stress due to the turbulent flow is proportional to ρ u' v

' . they

are called Reynolds stresses or turbulent stresses.

A fundamental difference between laminar and turbulent flow is that the shear stress for

turbulent flow is a function of the density of the fluid, For laminar flow, the shear stress is

independent of the density, leaving the viscosity, as the only important fluid property. Below

Figure shows the shearing stress in turbulent flow between parallel flat plates. Expressing the

Reynolds stress as follows as in the case of laminar flow

� = �� ��

��

produces the following as the shearing stress in

turbulent flow:

This vt is called the turbulent kinematic viscosity. Vt is not the value of a physical property

dependent on the temperature or such, but a quantity fluctuating according to the flow condition.

Prandtl assumed the following equation in which, for rotating small parcels of fluid of turbulent

flow (eddies) traveling average length, the eddies assimilate the character of other eddies by

collisions with them:

Prandtl called this lm the mixing length.

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and the wall shear stress can be expressed as

Where y is the distance from the wall (note that y= R- r for a circular pipe).

and friction velocity is:

the velocity profile in the viscous sublayer can be expressed in dimensionless form as

this equation is known as The Law Of The Wall, and it is found to satisfactorily

correlate with experimental data for smooth surfaces for

Therefore, the thickness of the viscous sublayer is roughly

Turbulent flow along a wall can be considered to consist

of four regions, characterized by the distance from the

wall. The very thin layer next to the wall where viscous

effects are dominant is the viscous sublayer. The velocity

profile in this layer is very nearly linear, and the flow is

streamlined. Next to the viscous sublayer is the buffer

layer, in which turbulent effects are becoming

significant, but the flow is still dominated by viscous

effects.

Above the buffer layer is the overlap layer, in which the

turbulent effects are much more significant, but still not

dominant. Above that is the outer (or turbulent) layer in

the remaining part of the flow in which turbulent effects

dominate over molecular diffusion (viscous) effects.

The thickness of the viscous sublayer is very small

(typically, much less than 1 percent of the pipe diameter).

The wall dampens any eddy motion, and thus the flow in

this layer is essentially laminar and the shear stress

consists of laminar shear stress which is proportional to

the fluid viscosity. The velocity profile in this layer to be

very nearly linear. Then the velocity gradient in the

viscous sublayer remains

nearly constant at du/dy= u/y

.....1

.......2

.......3

.......4

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where uδ is the flow velocity at the edge of the viscous sublayer. The quantity ν/u* has

dimensions of length and is called the viscous length; it is used to nondimensionalize the

distance y from the surface. In boundary layer analysis, it is convenient to work with

nondimensionalized distance and nondimensionalized velocity defined as

Then the law of the wall becomes simply:

Dimensional analysis indicates and the experiments confirm that the velocity in the overlap layer

is proportional to the logarithm of distance, and the velocity profile can be expressed as:

where k and B are constants whose values are determined experimentally to be about 0.40 and

5.0, Equation 7 is known as The Logarithmic Law. Substituting the values of the constants, the

velocity profile is determined to be :

A good approximation for the outer turbulent layer of pipe flow can be obtained by evaluating

the constant B in Eq. 7 from the requirement that maximum velocity in a pipe occurs at the

centerline where r= 0. Solving for B from Eq. 7 by setting y= R- r= R and u= umax, with k= 0.4

gives:

.......5

.....6

............7

.............8

Eq.3

Eq.8

It turns out that the logarithmic law in Eq. 8

satisfactorily represents experimental data for the

entire flow region except for the regions very

close to the wall and near the pipe center, as

shown in Fig., and thus it is viewed as a universal

velocity profile for turbulent flow in pipes or over

surfaces.

Note from the figure that the logarithmic-law

velocity profile is quite accurate for y+ > 30, but

neither velocity profile is accurate in the buffer

layer, i.e., the region

5 < y+< 30.

.........9

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Numerous other empirical velocity profiles exist for turbulent pipe flow. Among those, the

simplest and the best known is the power-law velocity profile expressed as:

The value of n increases with increasing Reynolds number. The value n= 7 when

Re = 1* 105 generally approximates many flows in practice.

Major Losses (Losses By pipe Friction): Let us study the flow in the region where the velocity distribution is fully developed Thus,

the pressure drop, for steady, incompressible turbulent flow in a horizontal round pipe of

diameter D can be written in functional form as

It can be written in dimensionless form as:

by experiments, the only way that this can be true is if ι/ D the dependence

is factored out as

the quantity

is termed the friction factor, f. Thus, for a horizontal pipe

and

This equation is called The Darcy-Weisbach Equation', and the coefficient f is called the

friction coefficient of the pipe.

where for turbulent flow and for laminar flow

Laminar flow

In this case the equations and

f =

............10

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No effect of wall roughness is seen. The reason is probably that the flow turbulence caused by

the wall face coarseness is limited to a region near the wall face because the velocity and

therefore inertia are small, while viscous effects are large in such a laminar region.

Turbulent flow f generally varies according to Reynolds number and the pipe wall roughness.

Smooth circular pipe

The roughness is inside the viscous sublayer if the height ε of wall face ruggedness is

In the case of a smooth pipe, the following equations have been developed:

Rough circular pipe

If

whenever Re > 900(ε/d) , it turns out that

A good approximate equation for the turbulent region of the Moody chart is given

by Haaland’s equation:

For a new commercial pipe , f can be easily obtained from Moody diagram shown in

Fig.a using ε/d in Fig.b .

The Moody chart valid for all steady, fully developed, incompressible pipe flows.

f

f

f

f

Re√f

f

12

Fig,b

Fig,a Mody diagram

√f Re√f

f

13

Example 1 ( Laminar flow):

Water, ρ=998 kg/m3 , ν = 1.005 ×10

-6 m

2 /s

flows through a 0.6 cm tube diameter, 30 m

long, at a flow rate of 0.34 L/min. If the pipe

discharges to the atmosphere, determine the

supply pressure if the tube is inclined 10o

above the horizontal in the flow direction.

Example 2

An oil with ρ = 900 kg/m3 and ν = 0.0002 m

2 /s flows upward through an inclined pipe .

Assuming steady laminar flow (a) verify that the flow is up, (b) compute hf between 1 and 2 ,

and compute (c) V , (d) Q, and (e) Re. Is the flow really laminar?

HGL1 > HGL2 hence the flow is from 1 to 2 as assumed.

30*sin(10)

10o

30 m

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Example 3: (turbulent flow)

Oil , ρ = 900 kg/m3 , ν = 1 ×10

-5 m

2 /s ,

flows at 0.2 m3 /s through a 500 m length

of 200 mm diameter , cast iron pipe

ε=0.0013. If the pipe slopes downward

10o in the flow direction , compute hf ,

total head loss, pressure drop, and power

required to overcome these losses.

Note that for this problem, there is a negative gravity head loss ( i.e. a head increase ) and a

positive frictional head loss resulting in the net head loss of 29.8 m

V=2.7 m/s , Q=0.0076 m3/s and Re=810 the flow is laminar

500 m

d=200 mm

10o

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Minor losses in pipes

In a pipe line, in addition to frictional loss, head loss is produced through additional

turbulence arising when fluid flows through such components as 1-change of area 2-

change of direction 3- branching 4- junction 5- bend and 6- valve. The loss head for such cases

is generally expressed by the following equation:

υ is the mean flow velocity on a section

losses in a suddenly expanding pipe For a suddenly expanding pipe as shown in below Figure, assume that the pipe is horizontal,

disregard the frictional loss of the pipe, let h, be the expansion loss, and set up an equation of

energy between sections 1 and 2 as :

Apply the equation of momentum setting the control

volume as shown in the Figure . Thus :

Since Q = A1 v1 = A2 v2 , from the above equation,

Substituting into eqn( 1 ) :

This hs is called The Borda-Carnot Head Loss or simply the expansion loss.

Flow in pipes : At the outlet of the pipe as shown in the right

Figure, since v2 = 0, the above equation becomes

hs =k

hs = k

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Flow contraction

The loss is similar to that in the case of sudden expansion . Like eqn ( 1 ) , it is expressed by:

Here Cc = Ac / A2 is a contraction coefficient.

Inlet of pipe line The loss of head in the case where fluid enters from a large vessel is expressed by the following

equation:

k is the inlet loss factor and v is the mean flow velocity in the pipe. The value of k will be the

value as shown in below Figure.

hs = k

k= k= k=

k= k= k=

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Divergent pipe or diffuser The head loss for a divergent pipe as shown in below Figure. is expressed in the same manner as

for a suddenly widening pipe:

Appling Bernolli equation :

Putting p2th for the case where there is no loss,

The pressure recovery efficiency η for a diffuser :

Substituting this equation in equation ( 1 ) :

The value of k varies according to θ .

Loss whenever the flow direction changes Bend In a bend, in addition to the head loss due to pipe friction, a loss due to the change in flow

direction is also produced. The total head loss hb is expressed by the following equation:

Here, k is the loss factor due to the bend effect. In a bend, secondary flow is produced as shown

in the figure owing to the introduction of the centrifugal force, and the loss increases. If guide

blades are fixed in the bend section, the head loss can be very small. Below table shows values

of k for the bends.

hs= k

----------- ( 1 )

1 - k 1 - k

hb=( f + k )

Table , loss factor k for bends (smooth wall Re=225000, coarse wall face Re=146000 )

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Elbow The section where the pipe curves sharply is called an elbow. The head loss hb is given in the

same form as above equation of the bend . Since the flow separates from the wall in the curving

part, the loss is larger than in the case of a bend. Below table shows values of k for elbows.

Pipe branch and pipe iunction

Pipe branch As shown in below Figure , a pipe dividing into separate pipes is called a pipe branch. Putting

hs1 as the head loss produced when the flow runs from pipe 1 to pipe 3 , and hs2 as the head

loss produced when the flow runs from pipe 1 to pipe 2 , these are respectively expressed as

follows:

the loss factors k1 , k2 vary according to

the branch angle θ

Pipe junction Two pipe branches converging into one are called a pipe

junction. Putting hs2 as the head loss when the flow runs from pipe 1 to pipe 3, and hs2 as the

head loss when the flow runs from pipe 2 to pipe 3 ,

these are expressed as follows:

Valve and cock Head loss on valves is brought about by changes in their section areas, and is expressed by

this equation provided that v indicates the mean flow velocity at the point not affected by the

valve .

k Table , Loss factor k for elbows

k

hs1= k1 hs2= k2

hs1= k1 hs2= k2

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Gate valve

The values of k for the various valves such as relief valve , needle valve ,pool valve , disc valve

ball valve..etc are also depend on the ratio of the valve area to pipe area .

Total Losses Along A Pipe Line

or

These equations would be appropriate for a single pipe size ( with average velocity V ) . For

multiple pipe/duct sizes, this term must be repeated for each pipe size.

Example

Water, ρ=1000 kg/m3 and ν = 1.02 ×10

-6m

2/s , is pumped between two reservoirs at 0.0508

m3/s through 122 m of 5.08 cm diameter pipe and several minor losses, as shown in the table

below. The roughness ratio is ε/d = 0.001. Compute the pump power required.

hs =k

k

Global valve

k

cock

k

ht = hf + ∑hs

ht

ht

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Write the steady-flow energy equation between sections 1 and 2, the two reservoir surfaces:

where hp is the head increase across the pump.

A=π×0.05082 /4

= 139000.95

the flow is turbulent and Haaland’s equation can be used to determine the friction

factor:

f= 0.0214

But since p1 = p2 = 0 and V1 =V2 = 0, solve the above energy equation for the pump head :

Loss element Ki

Sharp entrance 0.5

Open globe valve 6.9

bend, R/D = 2 0.15

Threaded, 90Þ, reg.,

elbow

0.95

Gate valve, 1/2 closed 2.7

Submerged exit 1

Z1=6 m

Z2=36 m

122 m of pipe , d=5.08 c

m

hs

V=2.81 m/s

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Z2 = 36 m , Z1 = 6 m , L = 122 m

hP = 55.78 m The power required to be delivered to the fluid is give by :

= 3119 W

If the pump has an efficiency of 80 %, the power requirements would be specified

Pin= Pf / η = 3119 /0.8

Pin= 3898.75 W

Multiple-Pipe Systems

Series Pipe System: The indicated pipe system has a

steady flow rate Q through three

pipes with diameters D1, D2, & D3.

Two important rules apply to this

problem.

1. The flow rate is the same through each pipe section.

2. The total frictional head loss is the sum of the head losses through the various sections.

Example: Given a pipe system as shown in the previous figure. The total pressure drop is

Pa – Pb = 150 kPa and the elevation change is Zb – Za = 5 m. Given the following data ,

determine the flow rate of water through the section.

The fluid is water, ρ = 1000 kg/m

3 and ν= 1.02 ×10

-6 m

2/s. Calculate the flow rate Q in m

3/h

through the system.

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……….(1)

Begin by estimating f1 , f2 , and f3 from the Moody-chart fully rough regime

Substitute in Eq. (1) to find :

V1=0.58 m/s , V2=

1.03 m/s , V3= 2.32 m/s

Hence, from the Moody chart, e/d with Re

Substitute in Eq. (1) :

Parallel Pipe System:

Example : Assume that the same three pipes in above Example are now in parallel . The

total pressure drop is PA – PB = 150 kPa and the elevation change is ZB – ZA = 5 m. Given

the following data . Compute the total flow rate Q in m3/h, neglecting minor losses.

The fluid is water, ρ = 1000 kg/m

3 and ν= 1.02 ×10

-6 m

2/s.

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Guess fully rough flow in pipe 1: f1 = 0.0262, V1= 3.49 m/s; hence Re1= 273,000.

From the Moody chart Re with ε/d

f1 =0.0267; recomputed V1 =3.46 m/s , Q1 = 62.5 m3/h.

Next guess for pipe 2: f2 =0.0234 , V2 = 2.61 m/s ; then Re2 =153,000,

From the Moody chart Re with ε/d

f2 = 0.0246 , V2 = 2.55 m/s , Q2 = 25.9 m3 /h .

Finally guess for pipe 3: f3 = 0.0304, V3=2.56 m/s ; then Re3 = 100,000

From the Moody chart Re with ε/d

f3 =0.0313 , V3 = 2.52 m/s, Q3 = 11.4 m3/h.

This is satisfactory convergence. The total flow rate is

These three pipes carry 10 times more flow in parallel than they do in series.

Branched pipes Consider the third example of a three-reservoir pipe junction as shown in the figure . If all flows

are considered positive toward the junction, then

………….(1)

which obviously implies that one or two of the flows must be away from the junction. The

pressure must change through each pipe so as to give the same static pressure pJ at the junction.

In other words, let the HGL at the junction have the elevation

where pJ is in gage pressure for simplicity. Then the head loss through each , assuming

P1 = P2 = P3 = 0 (gage) at each reservoir surface, must be such that

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We guess the position hJ and solve the above Equations for V1 , V2 , and V3 and hence Q1 , Q2,

and Q3 , iterating until the flow rates balance at the junction according to Eq.(1). If we guess hJ

too high, the sum Q1 + Q2 + Q3 will be negative and the remedy is to reduce hJ , and vice versa.

Example : Take the same three pipes as in the previous example , and assume that they connect three

reservoirs at these surface elevations

Find the resulting flow rates in each pipe, neglecting minor losses.

As a first guess, take hJ equal to the middle reservoir height , Z3 = hJ = 40 m. This saves one

calculation (Q3 = 0) and enables us to get the lay of the land :

Since the sum of the flow rates toward the junction is negative, we guessed hJ too high. Reduce

hJ to 30 m and repeat :

This is positive Q, and so we can linearly interpolate to get an accurate guess: hJ = 34.3 m.

Make one final list :

Hence we calculate that the flow rate is 52.4 m3/h toward reservoir 1, balanced by 47.1 m3/h

away from reservoir 2 and 6.0 m3/h away from reservoir 3. One further iteration with this

problem would give hJ = 34.53 m, resulting in Q1= 52.8, Q2= 47.0, and Q3 =5.8 m3/h, so that

Q = 0 to three-place accuracy.