pipe flow - unimasr · loss in moving water at outlet ... design of single pipe flow ......

Lecture 17

Pipe Flow

Pipe Flow and the Energy EquationFor pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through diameter changes and corners take head (energy) out of a system that theoretically conserves energy. Therefore, to correctly calculate the flow and pressures in pipe systems, the Bernoulli Equation must be modified.

P1/γ + V12/2g + z1 = P2/γ + V2

2/2g + z2 + Hmaj + Hmin

Hmaj

Energy line with no losses

Energy line with major losses

1 2

Major Losses

Major losses occur over the entire pipe, as the friction of the fluid over the pipe walls removes energy from the system.Each type of pipe has a friction factor, f, associated with it.

Hmaj = f (L/D)(V2/2g)

Hmaj

Energy line with no losses

Energy line with major losses

1 2

Minor LossesUnlike major losses, minor losses do not occur over the length of the pipe, but only at points of momentum loss. Since Minor losses occur at unique points along a pipe, to find the total minor loss throughout a pipe, sum all of the minor losses along the pipe. Each type of bend, or narrowing has a loss coefficient, KL to go with it.

Minor Losses

Major and Minor LossesMajor Losses: Hmaj = f (L/D)(V2/2g)

f = friction factor L = pipe length D = pipe diameterV = Velocity g = gravity

Minor Losses: Hmin = KL(V2/2g)

KL = sum of loss coefficients V = Velocity g = gravity

When solving problems, the loss terms are added to the system at the second analysis point

P1/γ + V12/2g + z1 = P2/γ + V2

2/2g + z2 + Hmaj + Hmin

Minor Head Loss

Minor loss at entrance (h1) Minor loss at exit (h2)

Loss in submerged discharge Loss in moving water at outlet

Minor loss at contraction or expansion (h3) Loss at sudden contraction Loss at gradual contraction

hL1 = losses due to entrance

Minor (secondary) losses in pipes (cont.)

pipe

reservoir

reservoir

pipe

hL 2 = losses due to exit

gvkhl 2

22

2 =

v2= velocity of exit

gvkhl 2

2

1 =

5.04.0 →≅kv1= velocity of entrance 0.0≈k

entrance losses = g

v2

5.02

1- Pipe with reservoir

0.1 ≅k

Secondary losses (Minor or transient losses)

2- Sudden enlargement

v2

v1

H. G. LT. E. Lv

2g1

2

v2

2gpiezometric head line

2

( )

gvk

gv

DD

gv

AA

gv

vv

gvvhL

221

21

21

2

21

21

22

2

1

21

2

2

1

21

2

1

2

221

=

−=

−=

−=

−=

For Gradually enlargement 0.0≈k

Secondary losses (Minor or transient losses) (cont.)

3. Sudden contraction

v2v1 A1

2AcA

gvk

gv

Dd

gv

AAh c

L 22)1(

21

22

22

2

222

2

2

=−=

−=

0.1 2

1 <

=

DDk φ

Minor Loss

Minor loss at fittings Loss in valve Loss in flange

Minor loss in bends and elbows

3. Losses due to bends or elbows


gvkhL 2

2

=

For bend( )RDk ,, θφ=

bendDv

R

v elbowFor elbow

( )θφ =k

4. Losses due to valves


gvkhL 2

2

=

plug valve butter fly valve gate valve

Loss

Coefficients

Bernoulli's Equation with Major and Minor Loss

LBB

AAA h

gV

DLfzP

gVzP

gV

++++=++222

222

γγB

54321 hhhhhhL ++++=

gVkh ii 2

2

=

Minor Loss Term

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?γoil= 8.82 kN/m3

f = .035

If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the 15 cm diameter smooth pipe, what is the elevation of the oil surface in the upper reservoir?

Include major losses along the pipe, and the minor losses associated with the entrance, the two bends, and the outlet.

Kout=1

r/D = 0

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?

If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the 15 cm diameter smooth pipe, what is the elevation of the oil surface in the upper reservoir?

Include major losses along the pipe, and the minor losses associated with the entrance, the two bends, and the outlet.

Kout=1

r/D = 0

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?

Kout=1

r/D = 0

Apply Bernoulli’s equation between points 1 and 2:Assumptions:

P1 = P2 = Atmospheric = 0 V1 = V2 = 0 (large tank)

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

Hmaj = (f L V2)/(D 2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)

Hmaj= 5.85m

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?

Kout=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin

Hmin= 2KbendV2/2g + KentV2/2g + KoutV2/2g

From Loss Coefficient table:

Kbend = 0.19 Kent = 0.5 Kout = 1

Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8)

Hmin = 0.24 m

Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1 = ?

Kout=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m

Z1 = 136.1 m

Numerical Example: Minor LossConvert the piping system to an equivalent length of 6” pipe.

Numerical Example: Minor LossSolution

This problem will be solved by using the Bernoulli equation, A to M, datum M, as follows:

(0+0+h) = 0+0+0+(0.025x150/1+8.0+2x0.5+0.7+1.0)(V12

2/(2g)+(0.02x100/0.5+0.7+6.0+2x0.5+3.0+1.0)(V6

2/(2g)

)22

(2222

26

2

212

1

26

2

22

212

1

112

22

21

12

1

gVk

gVk

gV

DLf

gV

DLfzP

gVzP

gV

ici

ic∑ ++++++=++γγ

0 0 h 0 0 0

Numerical Example: Minor Loss

SolutionThenh =14.5V12

2/(2g) + 15.7V62/(2g) =

=(14.5x(1/16)+15.7) [V62/(2g)]=16.6V6

2/(2g)For any available head h, the lost head is 16.6V6

2/(2g).The lost head in LE ft of 6” pipe is f(LE/D)[V6

2/(2g)]Equating the two values,16.6V6

2/(2g)=0.02(LE/0.5)[(V62/(2g)] LE=415ft

V12 A12 = V6A6 4V12 = V6

V122 = (1/16)V6

2

Design of Single Pipe Flow

Type I: Head loss problemGiven flow rate and pipe combinations, determine the total head loss in a system

Type II: Discharge problemGiven allowable total head loss and pipe combinations, determine the flow rate

Type III: Sizing problemGiven allowable total head loss and flow rate, determine the pipe diameter

Type I: Head loss problems

Given the pipe material, pipe geometry, and flow rate, find out how much friction loss we may encounter over the pipe.

Application: pumping station design.

Type I: Head loss Problem A 20-in-diameter galvanized iron pipe 2

miles long carries 4 cfs of water at 60oF. Find the friction loss using Moody diagram

Unit Conversion of Length

1 mile = 1760x3 = 5,280 ft 1 ft = 12 in = 30.42 cm 1 in = 2.54 cm 1 mile =1.6 Km

Type I: Head loss Problem

e = 0.0005 ft, so e/D = 0.0005 (12)/20 = 0.0003L= 2 mile=2*(5280 ft/mile)= 10,560 ft

fpsDQV 8333.1

)12/20()4(44

22 ===ππ

for water at 60oF: ν = 1.217x10-5 ft2/sec

55 1051.2

10217.18333.1)12/20(

×=×

== −υDVRe

Turbulent flow

Absolute Pipe Roughness

Type I: Head loss Problem

Based on e/D and Re, we find our f = 0.0172 in the Moody Diagram, then we may obtain:

ftg

VDLfhf 69.5)

)2.32(28333.1)(

12/2010560(0172.0

2

22

===

Type II: Discharge Problem

Given the pipe material and friction loss gradient, find out how much fluid we may send over the pipe.

Application: leak detection.


Water at 20oC flows in a 500-mm diameter welded steel pipe.

If the friction loss gradient is 0.006, determine the flow rate.


For welded steel pipe:e = 0.046 mm; e/D =0.046/500 = 0.000092

At 20oC, = 1.003x10-6 m2/sec hf /L = 0.006 is given

2/1

2

2

243.081.9)2(5.0

006.0

21

fV

fVg

VD

fLh

S f

=

=

==

υ

Dimensional Analysis of Pipe FlowMoody Chart

Laminar

Marks Reynolds Number independence

Dimensional Analysis of Pipe FlowMoody Chart

Absolute Pipe Roughness

Type II: Discharge Problem For e/D = 0.000092 fmin~0.0118 (Moody Diagram) Try f = 0.0118 then V = 0.243/(0.0118)0.5 = 2.23 m/sec

With e/D =0.000092 and Re=1.114x106; we may get f = 0.0131

Try one more time and show this results have converged

Q = AV = 0.416 m3/sec

66 10114.1

10003.1)23.2)(5.0(

×=×

== −υVDRe

Type III: Sizing Problem

Given the boundary condition of water head in the pipe system and the allowable head loss, determine the pipe diameter.

Application: designing the pipe system between the reservoir and the end user.


The allowable head loss is 5 m/km of pipe length.

Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 500 l/sec of water at 10oC.

Type III: Sizing ProblemSolution Bernoulli’s equation can be applied to pipe

section 1 km apart

For a uniform, horizontal pipe with no localized head losses

V1=V2 z1 = z2

LB

BB

AAA h

gV

DLfzP

gVzP

gV

++++=++222

222

γγB


The Bernoulli’s equation reduces to

52

2

22

22 8)4/(22 Dg

fLQDgQ

DLf

gV

DLfhf ππ

===

fgfLQD 13.4

58

2

25 ==

π

(per each Km)mhPPf

BA 5==−γγ


Therefore L = 1000m, and hf = 5m. At 10oC, v =1.31x10-6 m2/sec. Assuming welded-steel roughness to be in

the lower range of riveted steel, e = 0.9 mm. Apply try and error procedure:

Assume D = 0.8 m


sec/995.0)4.0(

sec/5.02

3

mm

mAQV ===

π

e/D = (0.9mm)/(800mm) = 0.0011

Entering these value to the Moody diagram,

we get f = 0.021

56 101.6

1031.1)8.0)(995.0(

×=×

== −υVDRe


A better estimate of D can be obtained by substituting the latter values into:

A new iteration provides V = 1.71 m/sec, Re = 8.0x105, e/D = 0.0015, f = 0.022, and D = 0.62m

fg

QLfD 13.45

82

25 ==

π

Which gives D = 0.61m

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pipe flow - unimasr · loss in moving water at outlet ... design of single pipe flow ......

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