7 steady flow in pipes - vocational training counciltycnw01.vtc.edu.hk/cbe2023/7_pipe.pdf · fluid...

Fluid Mechanics Chapter 7 – Steady Flow in Pipes

P.7-1

77 SSTTEEAADDYY FFLLOOWW IINN PPIIPPEESS 77..11 RReeyynnoollddss NNuummbbeerr

Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment:

– Laminar flow – Turbulent flow

Reynolds’ apparatus

dye

water

outlet

dye filament

Laminar flow

at low velocity

dye filament remained intact throughout the length of the tube

fluid particles move in a straight line

considered as moving in layers

Dye injection

Dye filament


P.7-2

Turbulent flow

at high velocity

dye diffused over the whole cross-section fluid particles do not move in a straight line velocity in average sense

The transition of flow is due to change of velocity, size of pipe and properties of fluid.

Reynolds explained the phenomena by considering the forces acting

on the fluid particle.

When the motion of a fluid particle in a stream is disturbed, its inertia will tend to carry it on in the new direction, but the viscous forces due to the surrounding fluid will tend to make it conform to the motion of the rest of the stream.

The criterion that determines whether flow will be viscous or turbulent

is therefore the ratio of the inertial force to the viscous force acting on the particle. Hence, for a particular flow,

i.e. force viscousforce inertial = constant (7.1)

By using dimensional analysis, Reynolds derived a criterion to

distinguish between laminar and turbulent flow.

Reynolds Number, Re = forceviscousforce inertial =

µρvd (7.2)

where ρ = density of the fluid, kg/m3

v = velocity of the flow, m/s d = diameter of the pipe, m µ = dynamic viscosity, Ns/m2

Dye injection

Dye filament


P.7-3

Since ν = ρµ (7.3)

where ν = kinematic viscosity, m2/s

Hence Reynolds Number, Re = νvd (7.4)

Reynolds Number is a dimensionless number

In general, the flow is – laminar when Re - “small” – transitional when Re - “intermediate” – turbulent when Re - “large”

The flow in pipe can be treated as – laminar when Re < 2000 – transitional when 2000 < Re < 4000 – turbulent when Re > 4000


P.7-4

Worked example: What is the critical velocity of a water flow through a circular pipe of diameter 2 cm so that the flow is laminar? (dynamic viscosity of water is 0.89*10-3 Ns/m2, density of water = 1000 kg/m3) Answer As the Reynolds number is

Re = µρvd

= 310*89.002.0*v*1000

− < 2000

i.e. v = 02.0*100010*089*2000 3−

= 8.9 cm/s Hence the critical velocity is 8.9 cm/s.


P.7-5

77..22 LLaammiinnaarr FFllooww iinn PPiippeess 77..22..11 HHaaggeenn -- PPooiisseeuuiillllee EEqquuaattiioonn It was discovered independently by:

– G.H.L. Hagen - a German engineer in 1893 – J.L.M. Poiseuille - a French physician in 1840

It states that the head loss experienced by the water when it flows

through a pipe is

– directly proportional to the rate of flow, (Q), and – inversely proportional to the fourth power of the diameter of the

pipe (d4).

i.e. hf = k 4dQ where k is a constant (7.5)

For a laminar flow, the shear stress on the cylindrical surface is given by

dxdp

2r =τ

For Newtonian fluid,

drdv µ=τ

Equating the two equations,

dxdp

2r

drdv

=µ

When integrating the above equation with respect to r with the

boundary condition v = 0 when r = R (i.e. no slip condition), the result is

)rR(dxdp

41- v 22 −µ

= (7.6)

From (7.10), we can see that the velocity distribution is in parabolic

form with maximum velocity at r = 0.


P.7-6

umax

The maximum velocity at centre (r = 0)

vmax = 2Rdxdp

41µ

−

by putting dpdx

= -∆pL

vmax = R pL

2

4∆µ

(7.7)

The corresponding discharge, Q is Q = 20 πrvdrR

∫

= 2 14

2 20 π

µr dp

dxR r drR [ ( )]− −∫

= - πµR dp

dx

4

8

by putting dpdx

= - ∆pL

& R = d/2

Q = πµ

d pL

4

128∆ (7.8)

- known as Hagen-Poiseuille equation

The mean velocity, v = A area, tionalsecx

Q rate, flow−

= d pL

2

32∆µ

(7.9)

When compare with vmax, maxv21v = (7.10)

From Hagen-Poiseuille law,


P.7-7

hf = k * 4dQ

By substituting the expression of Q in (7.8) and consider the change of pressure head as head loss, i.e.

∆p = ρghf

then πρµ

=g

L128k

Therefore, k depends on the properties of fluid, ρ and µ and pipe length, L.

Hagen-Poiseuille equation can be expressed in terms of head loss, hf

and average velocity,

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛⎟

⎠

⎞⎜⎝

⎛ρµ

=g2

v*dL*

dv*64h

2

f

Since Re = µρvd

Therefore head loss in Hagen-Poiseuille equation

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

g2v*

dL*

Re64h

2

f (7.11)

If hf can be expressed as ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

g2v*

dL*fh

2

f (7.12)

where f - friction factor

Then the friction factor in laminar flow is

f = Re64 (7.13)

Noted that in UK publication, hf is often written as

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛=

g2v*

dL*f4h

2

f

and hence f = Re16


P.7-8

Worked examples: 1. For a laminar flow, d = 250 mm, P1 - P2 = 20 kN/m2 L = 200 m, µ = 10-3 Ns/m2

Determine the flow rate, Q in m3/s. Answer From Hagen-Poiseuille equation,

Q = L128

)PP(d 214

µ−π

= 200*10*128

)10*20(*)25.0(*3

34

−

π m3/s

= 9.59 m3/s


P.7-9

2. Oil of viscosity 0.101 Ns/m2 and specific gravity of 0.85 flows through 3000 m of 300 mm-dia. pipe at the rate of 0.0444 m3/s. What is the lost in head in the pipe?

Answer

Since Q = L128

)PP(d 214

µ−π

Now, Q = 0.0444 m3/s, µ = 0.101 Ns/m2, L = 3000 m, d = 0.3 m

then 3000*101.0*128

)PP()3.0(0444.0 214 −π

=

Hence P1 - P2 = 4)3.0(*3000*101.0*128*0444.0

π

= 67.67 kN/m2 = lost in pressure

Lost in head = gPP 21

ρ−

= 81.9*850

67.67 m of oil

= 8.12 m of oil


P.7-10

77..33 TTuurrbbuulleenntt FFllooww iinn PPiippeess 77..33..11 DDaarrccyy -- WWeeiissbbaacchh FFoorrmmuullaa

Darcy, Weisbach and others found that a formula for pipe friction loss could be expressed as

hf = f Ld

vg

⎛⎝⎜

⎞⎠⎟

2

2 (7.14)

where f is friction factor

The above equation can be applied in both laminar (refer 7.2.2) and turbulent flows and is known as Darcy - Weisbach formula.

It is found that friction factor depends on

– density of the fluid, ρ – velocity of the flow, v – diameter of the pipe, d – viscosity of the fluid, µ – wall roughness, ε ε

wall roughness

i.e. f = f(ρ, v, d, µ, ε)

By dimensional analysis,

f = f ρµ

εvdd

,⎛⎝⎜

⎞⎠⎟

= f ( Reynolds number, relative roughness)

where Reynolds number = µρvd and

relative roughness = εd

(7.15)


P.7-11

Once the Reynolds number and relative roughness have been determined, the corresponding value of the friction factor can be obtained from a graphical relationship known as the Moody diagram.

Typical values of surface roughness

New pipe surface Roughness, ε (m) Glass, brass, copper and lead smooth Wrought iron, steel 0.46*10-4

Cast iron 2.6*10-4 Concrete 3*10-4 to 30*10-4

77..33..22 MMooooddyy DDiiaaggrraamm

Moody diagram has been used extensively in solving pipe flow problems.

Two equations are related to the Moody diagram

– for laminar flow, the friction factor is

f = Re64

This is the straight portion of the diagram when Re < 2000. – for a turbulent flow, friction factor is

1f

= ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

fd

Re51.2

7.3log*0.2 ε

(7.16)

This is known as Colebrook-White formula.


P.7-13


P.7-14

Worked examples: 1. Determine the head loss for flow of 140 L/s of oil, ν = 0.00001 m2/s,

through 400 m of a 200 mm diameter cast iron pipe. Answer Given Q = 140 L/s = 0.14 m3/s d = 200 mm = 0.2 m ν = 0.00001 m2/s = 10-5 m2/s L = 400 m ε = 0.26 mm (cast iron)

v = Qdπ 2 4

= 0140 2 42.

* .π m/s = 4.456 m/s

Re = νvd = 0 2 4 456

10 5. * .

− = 8.912*104

εd

= 2.010*6.2 4−

= 0.0013

From the Moody diagram, f(89120, 0.0013) = f = 0.0238

i.e. hf = 0.0238*2.0

400 *81.9*2

456.4 2

m of oil

= 48.17 m of oil


P.7-15

2. Solving the previous example by using Colebrook-White formula. Answer

As 1f

= − +⎛⎝⎜

⎞⎠⎟

2 03 7

2 51. log.

.Re

ε df

where εd

= 0.0013; Re = 89120

i.e. 1f

= − +⎛⎝⎜

⎞⎠⎟

2 0 0 00133 7

2 5189120

. log ..

.f

or 1f

+ 2log(1 + 0 0801.f

) – 6.908 = 0

Since this is a non-linear equation, it has to be solved by trial & error

or iterations. f = 0.023365

Hence hf = 0.023365* 4000 2.

* 4 4562 9 81

2.* .

m of oil

= 47.29 m of oil


P.7-16

77..44 MMiinnoorr LLoosssseess

In section 7.3, the head loss in long, straight sections of pipe can be calculated by use of the friction factor obtained from Moody diagram or the Colebrook – White equation. This is called friction loss or major loss.

Most pipe systems consist of considerably more than straight pipes.

These pipe fittings add to the overall head loss of the system. These losses are called minor losses.

In some cases, the minor losses may be greater than the friction loss.

Since the flow pattern in fittings and valves is quite complex, the

theory is very week. The losses are commonly measured experimentally and correlated with the pipe flow patterns.

77..44..11 DDiiffffeerreenntt TTyyppeess ooff MMiinnoorr LLoosssseess

Minor losses are losses due to the inclusion of a pipe fittings in a pipeline.

Some examples are

– entrances or exit of a pipe – expansions or contractions of a pipe – bends, elbow and tees – valves of open or partially closed – gradual expansions or contractions

Minor losses is given by

hL = K* vg

2

2 (7.17)

where K is a constant


P.7-17

Component K a. Elbows Regular 90°, flanged 0.3 Regular 90°, threaded 1.5 Long radius 90°, flanged 0.2 Long radius 90°, threaded 0.7 Long radius 45°, flanged 0.2 Regular 45°, threaded 0.4 b. 180° return bends 180° return bends, flanged 0.2 180° return bends, threaded 1.5 c. Tees Line flow, flanged 0.2 Line flow, threaded 0.9 Branch flow, flanged 1.0 Branch flow, threaded 2.0 d. Valves Globe, fully open 10 Gate, fully open 0.15 Ball valve, fully open 0.05 e. Others Entrance loss 0.5 Exit loss 1.0

77..44..22 MMooddiiffiieedd BBeerrnnoouullllii’’ss EEqquuaattiioonn

The original Bernoulli’s equation should be extended to include the friction loss and minor losses.

i.e. total energy at 1 = total energy at 2 + energy loss on the way

2

222

1

211 z +

g2v + p =z +

g2v + p

γγ+Σ f L

dv

gi i

i

22

2+ΣK v

gi2

2

2

(7.18)


P.7-18

Worked examples: 1. Find the discharge through the pipe in the figure below. The minor loss

coefficient for entrance is 0.5. The pipe diameter is 15 mm and the pipe roughness produces a friction factor of 0.025.

1

2

150m

15m

Answer

Applying Bernoulli’s equation between pt.1 and 2

2

222

1

211 z +

g2v + p =z +

g2v + p

γγ+ fL

dv

g2

2

2+ K v

g2

2

2

15 = (1 + K + fLd

)* vg2

2

2

= (1+0.5+0.025*150/0.015) * vg2

2

2

= 251.5 vg2

2

2

or v2 = 1.082 m/s

Hence Q = A2*v2 = 1.082*π*0.0152/4 m3/s

= 0.1912 L/s


P.7-19

2. Find the discharge through the pipe in the figure below for H = 20 m. The minor loss coefficients for entrance, elbows and globe valve are 0.5, 0.8 (each) and 10 respectively. The kinematic viscosity of water is 1.02*10-6 m2/s.

40m

30m 40m

20melbows

globevalve

dia. = 150mmε = 0.0003

12

Answer

Using 2

222

1

211 z +

g2v + p =z +

g2v + p

γγ+Σ f L

dv

gi i

i

22

2+ΣK v

gi2

2

2

Σf Ld

vg

i i

i

22

2 = f

dv

gLi

22

2Σ

= f v015 2 9 81

30 20 4022

. * .( )+ +

= 30.58 f v22

ΣK vgi2

2

2 = v

gKi

22

2Σ

= v22

2 9 8105 2 08 10

* .( . * . )+ +

= 0.617 v22

As P1-P2 = 0, v1 = 0, z1-z2 = 20 m,

i.e. 20 = vg2

2

2+Σ f L

dv

gi i

i

22

2+ΣK v

gi2

2

2

= (0.051+30.58f + 0.617)v22

= (0.668+30.58f) v22


P.7-20

or v2 = 5471 4578

..+ f

Since Re = 1.47*105 v2 ; ε/D = 0.002

f v2 (m/s) Re fcal. 0.030 3.551 5.2*105 0.0230 0.0230 3.818 5.6*105 0.0235 0.0235 3.797 5.6*105 0.0235 (ok)

∴ v2 = 3.797 m/s

Since Q = A2*v2 = π*0.152/4 * 3.797 m3/s = 0.067 m3/s


P.7-21

77..55 PPiippee SSyysstteemmss

In many pipe systems, there is more than one pipe involved.

The governing mechanisms for the flow in multiple pipe systems are the same as for the single pipe system discussed previously.

77..55..11 RReessiissttaannccee CCooeeffffiicciieennttss ffoorr PPiippeelliinneess iinn SSeerriieess aanndd PPaarraalllleell

In general, the equation of head loss can be expressed as hf = k*Q2 (7.19)

Pipes are in series if they are connected end to end so the fluid flow in a continuous line is a constant.

Q

Q

QQ

Q

h1

h2

h3

hn

By continuity of flow, Q is same for each pipe.

The total loss of the system is given as hf = h1 + h2 + h3 +…+ hn = k1*Q2 + k2*Q2 + k3*Q2 + … + kn*Q2 = (k1 + k2 + k3 + … + kn)*Q2 (7.20)

The effective resistance coefficient is keff = k1 + k2 + k3 + … + kn (7.21) i.e. the total head loss is the summation of the individual pipe.


P.7-22

For pipes connected in parallel, the fluid can flow from one to the other by a number of alternative routines.

Q1

Q2

Q3

Qn

Q Q

hf

The head loss for individual pipe is the same as the total head loss.

The total flow rate is the summation of the individual pipe.

Since hf = ki*Qi2

or Qi = hk

f

i

Total Q = Q1 + Q2 + Q3 + … + Qn (7.22)

= hk

f

1+ h

kf

2+ h

kf

3+…+ h

kf

n

= ( 1

1k+ 1

2k+ 1

3k+…+ 1

kn) hf

= hk

f

eff

Hence 1keff

= 1

1k+ 1

2k+ 1

3k+…+ 1

kn (7.23)


P.7-23

Worked examples: 1. Two reservoirs are connected by a pipeline which is 150 mm in

diameter for the first 6 m and 225 mm in diameter for the remaining 15 m. The water surface in the upper reservoir is 6 m above that in the lower. By neglecting any minor losses, calculate the rate of flow in m3/s. Friction coefficient f is 0.04 for both pipes.

Answer

The velocities v1 and v2 are related by the continuity equation. i.e. A1v1 = A2v2

v1 = v2AA

2

1

⎛⎝⎜

⎞⎠⎟ = v2

dd

2

1

2⎛⎝⎜

⎞⎠⎟

= 225150

2⎛⎝⎜

⎞⎠⎟

v2 = 2.25 v2

Friction in the 150 mm pipe

hf1 = g2

vdLf 2

1

1

11 = 0 04 6015 2

12. *

.v

g

= 1.6 vg12

2 = 1.6*2.252* v

g2

2

2 = 8.1 v

g2

2

2

Similarly, friction in the 225 mm pipe

hf2 = g2

vdLf 2

2

2

22 = 0 04 150 225 2

22. *

.v

g

= 2.67 vg

22

2

Hence, total head loss = hf1 + hf2 = 10.77 vg

22

2

Applying Bernoulli’s equation between the two top water surfaces, p1 = p2 = 0 (Patm) v1 = v2 = 0 (water surfaces) z1 = 6 m; z2 = 0

2

222

1

211 z +

g2v + p =z +

g2v + p

γγ+hL


P.7-24

or z1 – z2 = hL

6 = 10.77 vg

22

2

or v2 = 6 2 9 8110 77

* * ..

= 3.31 m/s

Hence Q = A2v2

= π * . * .0 2254

3 312

m3/s

= 0.132 m3/s


P.7-25

2. Two reservoirs have a difference of level of 6 m and are connected by two pipes laid in parallel. The first pipe is 600 mm diameter of 3000 m long and the second one is 300 mm diameter of 2000 m long. By neglecting all the minor losses, calculate the total discharge if f = 0.04 for both pipes.

Answer

For parallel pipes, hf = g2

vdLf 2

1

1

11 = g2

vdLf 2

2

2

22

Apply Bernoulli’s equation to the points on the free surfaces and from the result of the previous worked example, level difference = head loss

H = g2

vdLf 2

1

1

11 = g2

vdLf 2

2

2

22

6 = 0 04 30000 6 2 9 81

12. *

. * .v = 0 04 2000

0 3 2 9 812

2. *. * .

v

Therefore, v1 = 0.767 m/s v2 = 0.664 m/s

Hence Q1 = A1v1 = π * . * .0 64

0 7672

= 0.217 m3/s

Q2 = A2v2 = π * . * .0 34

0 6642

= 0.047 m3/s

Total discharge, Q = Q1 + Q2 = 0.217 + 0.047 m3/s = 0.264 m3/s


P.7-26

77..55..22 BBrraanncchheedd--ppiippee PPrroobblleemm

h1

h3

h2

J

reservoir 1

reservoir 3

reservoir 2pipe1, k1

pipe2, k2

pipe 3, k3

Q1Q2

Q3

Assume h1 > h2 > h3 and the 3 pipes intersect at junction J.

As h1 is the highest head, the flow in pipe 1 must be toward J.

As h3 is the lowest head, Q3 is flowing from J to the reservoir 3.

The flow Q2’s direction is unknown because it depends on the head at junction J.

If hJ be the head at junction J. There are two possible cases

(i) h1 > hJ > h2, or (ii) h2 > hJ > h3

For case (i), h1 > hJ > h2, Q2 is from J to reservoir 2.

Q1 - Q2 - Q3 = 0 h1 – hJ = k1*Q1

2

hJ - h2 = k2*Q22 (7.24)

hJ - h3 = k3*Q32

For case (ii), h2 > hJ > h3, Q2 is from reservoir 2 to J. Q1 + Q2 - Q3 = 0 h1 – hJ = k1*Q1

2

h2 – hJ = k2*Q22 (7.25)

hJ - h3 = k3*Q32


P.7-27

Both sets of equations have 4 unknowns Q1, Q2, Q3 and hJ. We have to determine which case controls the problem.

It is determined by assuming hJ = h2, i.e. no flow from J to reservoir 2.

Therefore Q1’ = h hk

1 2

1

− (7.26)

Q3’ = h hk

2 3

3

− (7.27)

If Q1’ > Q3’, Q2 is from J to reservoir 2 - case (i).

If Q1’ < Q3’, Q2 is from reservoir 2 to J - case (ii).


P.7-28

Worked example: Three reservoirs are connected as the figure below. Determine the flow, Q1, Q2 and Q3 with k1 = 3.058, k2 = 8.860 and k3 = 0.403 s2/m5 and hf = ki*Qi

2.

J

reservoir 1

reservoir 3

reservoir 2Q1

Q2

Q3

200m

140m

180m

Answer Step 1

Pipe hi (m) ki (s2/m5) 1 200 3.058 2 180 8.860 3 140 0.403

Step 2 - calculate Q1’ and Q3’

Q1’ = h hk

1 2

1

− = 200 1803058−

. = 2.557 m3/s

Q3’ = h hk

2 3

3

− = 180 1400 403−

. = 9.963 m3/s

Q1’ < Q3’ ⇒ case (ii) i.e. h2 > h > h3


P.7-29

Step 3 - set up the equations For case (ii), we have 200 - h = 3.058Q1

2

or Q1 = 2003 058

− h.

m3/s

180 - h = 8.860Q2

2

or Q2 = 1808 860

− h.

m3/s

h - 140 = 0.403Q3

2

or Q3 = h −1400 403.

m3/s

Since Q1 + Q2 - Q3 = 0

therefore 2003 058

− h.

+ 1808 860

− h.

- h −1400 403.

= 0

Step 4 - solve for h (180 < h < 140) by iterations

h (m) Q1 + Q2 + Q3 (m3/s) Error 180 2.557 + 0 – 9.963 -7.406 160 3.617 + 1.502 – 7.045 -1.926 150 4.044 + 1.840 - 4.981 +0.903 153 3.920 + 1.746 - 5.680 -0.014 152.9 3.925 + 1.749 - 5.658 +0.016 152.95 3.922 + 1.747 - 5.669 +0.000

Therefore, the head h at the junction is 152.95 m and Q1 = 3.922 m3/s (towards J) Q2 = 1.747 m3/s (towards J) Q3 = 5.669 m3/s (towards reservoir C)


P.7-30

77..55..33 HHaarrddyy--CCrroossss MMeetthhoodd ((RReeffeerreennccee oonnllyy))

loop 1 loop 2 loop 3

loop 4 loop 5 loop 6

waterin

water out

The supply of water system to a city is a complicated network of pipelines. The commonly used technique is the Hardy-Cross method.

The two assumptions made by Hardy-Cross method are:

– The algebraic sum of head loss around each loop must be zero.

1

2 3

4

+ve

h14 + h43 - h23 - h21 = 0 (7.28)

– The net flow out of each junction must be equal to zero.

71

2 34

56

Q17

+ Q47 - Q76 - Q75 - Q72 - Q73 = 0 (7.29)


P.7-31

Method of Analysis

– Assume hi = ki*Qi

2

– The sign convention is clockwise positive for the discharge and head loss.

– Initially a flow rate, Q is assumed. A correction for discharge, ∆Q

is then evaluated and the new flow rate is Q + ∆Q.

– the head loss for each member can be approximated as

hi’ = ki*(Qi + ∆Q)2 = ki*[Qi

2 + 2Qi∆Q + (∆Q)2] ≈ ki*Qi

2 + 2 ki*Qi*∆Q

– Summation around the loop Σ hi’ = Σ[ki*Qi

2 + 2ki*Qi* ∆Q] = 0 (by assumption 1)

i.e. Σhi + 2∆Q*Σ(hi / Qi) = 0

– hence ∆Q = − 12

Σ

Σ

hh

Q

i

ii

( ) (7.30)

Σhi is the sum of head loss around a loop which can be +ve or -ve.

Σ(hi / Qi) is the sum of the ratio (head loss/flow) for each member of

the loop. The ratio is a magnitude and therefore is +ve only.

∆Q is the correction flow for a loop. Each pipe within the loop will have this correction. Any pipe belonging to 2 or more loops, the correction for that particular pipe will contribute from every loop.


P.7-32

Worked examples: 1. Determine the flow in each branch of the loop as shown below by

using (i) equivalent pipe method, and (ii) Hardy-Cross method.

2m /s3 2m /s3

pipe 1, k

pipe 2, 2k

Q1

Q2

Answer (i) As a parallel system hf1 = hf2

k*Q1

2 = 2k*Q22

or Q1 = 2 Q2

As Q = 2m3/s, = Q1 + Q2 2 = 2 Q2 + Q2 (1+ 2 )Q2 = 2 Q2 = 0.8284 m3/s

Q1 = 2 Q2 = 2 *0.8284

= 1.1716 m3/s


P.7-33

(ii) Assume Q1 = 1.5 m3/s Q2 = 0.5 m3/s

2m /s3 2m /s3

1.5m3 /s

-0.5m3 /s

+ve

1st iteration

Pipe ki Qi ±hi abs(hi/Qi) 1 1 1.5 2.25 1.5 2 2 -0.5 -0.5 1.0 Σ = 1.75 2.5

∆Q = − 12

Σ

Σ

hh

Q

i

ii

( )

= - 1752 2 5

.* .

m3/s

= -0.35 m3/s 2nd iteration

Pipe ki Qi ±hi abs(hi/Qi) 1 1 1.15 1.3225 1.15 2 2 -0.85 -1.445 1.70 Σ = -0.1225 2.85

∆Q = - −012252 2 85

.* .

m3/s

= 0.0215 m3/s


P.7-34

3rd iteration

Pipe ki Qi ±hi abs(hi/Qi) 1 1 1.1715 1.3724 1.1715 2 2 -0.8285 -1.3728 1.655 Σ = -0.0004 2.8265

∆Q = - −0 00042 2 8265

.* .

m3/s

= 0.000071 m3/s

∴ Q1 = 1.1715 m3/s Q2 = 0.8285 m3/s (after 3 iterations)


P.7-35

2. Find the flow in the pipeline using Hardy Cross Method. K for vertical members are 3 and horizontals are 5.

A B C

D E F

0.5m /s3

0.1 0.2 0.2m /s3 m /s3 m /s3

300m200mm

300m200mm

300m200mm

500m200mm

500m200mm

500m200mm

500m200mm

Answer Assume the flow rates are

A B C

D E F

0.5m /s3

0.1 0.2 0.2m /s3 m /s3 m /s3

0.2 0.05

0.2 0.15

0.050.3 0.15

m /s3 m /s3

m /s3

m /s3m /s3

m /s3 m /s3

1st Iteration

Pipe ki Qi ±hi Abs(hi/Qi) AB 5 0.20 0.2000 1.00 BE 3 0.15 0.0675 0.45 DE 5 -0.20 -0.2000 1.00 AD 3 -0.30 -0.2700 0.90

-0.2025 3.35 BC 5 0.05 0.0125 0.25 CF 3 0.05 0.0075 0.15 EF 5 -0.15 -0.1125 0.75 BE 3 -0.15 -0.0675 0.45

-0.16 1.6

∆Q1 = − −0 20252 3 35

.* .

= 0.03 m3/s

∆Q2 = − −0162 16

.* .

= 0.05 m3/s


P.7-36

2nd Iteration


-0.04657 3.2 BC 5 0.100 0.05 0.5 CF 3 0.100 0.03 0.3 EF 5 -0.100 -0.05 0.5 BE 3 -0.130 -0.05087 0.390672

-0.02087 1.690672

∆Q1 = − −0 46572 3 2

.* .

= 0.0071 m3/s

∆Q2 = − −0 20872 16907

.* .

= 0.0062 m3/s

3rd Iteration


-0.00498 3.181479 BC 5 0.106 0.056364 0.530868 CF 3 0.106 0.033818 0.318521 EF 5 -0.094 -0.04402 0.469132 BE 3 -0.131 -0.05174 0.393979

-0.00557 1.7125

∆Q1 = − −0 4982 31815

.* .

= 0.00078 m3/s

∆Q2 = − −0 05572 17125

.* .

= 0.00163 m3/s


P.7-37

4th Iteration


-0.00128 3.176597 BC 5 0.108 0.058105 0.539005 CF 3 0.108 0.034863 0.323403 EF 5 -0.092 -0.0425 0.460995 BE 3 -0.130 -0.05108 0.391444

-0.00061 1.714847

∆Q1 = − −0 01282 31766

.* .

= 0.00020 m3/s

∆Q2 = − −0 00612 17148

.* .

= 0.00018 m3/s

5th Iteration


-0.00014 3.176062 BC 5 0.108 0.058298 0.539896 CF 3 0.108 0.034979 0.323938 EF 5 -0.092 -0.04234 0.460104 BE 3 -0.131 -0.05109 0.391515

-0.00016 1.715453

∆Q1 = − −0 000142 31761

.* .

= 2.2*10-5 m3/s

∆Q2 = − −0 000162 17155

.* .

= 4.6*10-5 m3/s

Therefore OK


P.7-38

Therefore, the flow rates after five iterations will be given by

A B C

D E F

0.5m /s3

0.1 0.2 0.2m /s3 m /s3 m /s3

0.238 0.108

0.162 0.0920.1080.262 0.131

m /s3 m /s3

m /s3

m /s3m /s3

m /s3 m /s3


P.7-39

Class Exercise 7.1: A steady push on the piston causes a flow rate of 0.4 cm3/s through the needle. The fluid has S.G. = 0.9 and µ = 0.002 Ns/m2. Determine the head loss at the needle and hence the force F required to maintain the flow. Neglect the head loss in the piston only. (F = 0.012 kN)

D=1cm

D=0.25mm

3cm1.5cm

QF


P.7-40

Class Exercise 7.2: Oil, with density = 900 kg/m3 and ν = 1*10-5 m2/s, flows at 0.2 m3/s through a 20-cm diameter pipe 500 m long cast-iron pipe. The roughness of iron is 0.26 mm. Determine the head loss in the pipe. (hf = 117 m)


P.7-41

Class Exercise 7.3: Water flows at a velocity of 1 m/s in a 150 mm new ductile iron pipe. Estimate the head loss over 500 m using Darcy-Weisbach equation. (ε = 0.26 mm and µ = 10-3 Ns/m2 ). (hf = 4.04 m)


P.7-42

Class Exercise 7.4: The distance between two sections A and B of a 300 mm diameter pipe is 300 m. The elevations of A and B are 90 m and 75 m and the pressures are 280 kPa and 350 kPa respectively. Find the direction of flow of water and calculate the head loss due to friction and the value of the friction factor for the pipe if the flow is 142 L/s. (A to B, h = 7.87 m, f = 0.0386)


P.7-43

Class Exercise 7.5: Three pipes A, B & C are interconnected and discharge water from a reservoir as shown below. With the provided pipe characteristics, determine the flow rate in each pipe. Neglect all the minor loss. Pipe Diameter,

mm Length,

m f

A 150 600 0.02 B 100 500 0.032C 200 1250 0.024

(QA = 0.0752 m3/s, QB = 0.0236 m3/s, Qc = 0.0988 m3/s)


P.7-44

Tutorial: Steady Pipe Flow 1. Glycerine of viscosity 0.9 Ns/m2 and density 1260 kg/m3 is pumped

along a horizontal pipe 6.5 m long of diameter d = 0.01 m at a flow rate of Q = 1.8 L/min. Determine the flow Reynolds number and verify whether the flow is laminar or turbulent. Calculate the pressure loss in the pipe due to frictional effects.

2. If oil (ν = 4*10-5 m2/s, S.G. = 0.9) flows from the upper to the lower

reservoir at a rate of 0.028 m3/s in the 15-cm smooth pipe, determine the elevation of the oil surface in the upper reservoir. (K for entrance, exit and bend are 0.5, 1 and 0.19 respectively)

60m

130m7m

130mPD

3. Two reservoirs having a difference of surface level of 24 m are

connected by two parallel pipes each 1600 m long and of diameters 450 mm and 300 mm. To repair a length of 120 m of the 450 mm diameter pipe midway between the reservoirs, the total flow is diverted over this length to the 300 mm pipe. Calculate the percentage reduction in discharge resulting from the diversion. Consider only the friction losses and take f = 0.04.

450mm

300mm1600m

24m

450mm

300mm

24m 300mm

450mm120m 300mm

7 steady flow in pipes - vocational training counciltycnw01.vtc.edu.hk/cbe2023/7_pipe.pdf · fluid...

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