FRICTIONAL LOSSES IN COMPLETELY DEVELOPED …...MOODY DIAGRAM Several features of the Moody diagram: 1) It is a log-log graph, 2) Correlates 3 dimensionless parameters (Reynolds Number

FRICTIONALLOSSES

IN COMPLETELY DEVELOPED

PRESSURIZED PIPEFLOWING FULL

FRICTIONAL LOSSES IN COMPLETELY DEVELOPED PRESSURIZED PIPE FLOWING FULL

Perhaps the most calculated quantity in pipe flowis the head loss due friction. The whole ofpressurized pipe systems design and analysismainly depends on this head loss determination.Knowing that, the wall shear stress in a developedflow causes the pressure change along the pipeand hence creates energy head losses within thesystem through friction.

Total Frictional Head Loss ‘hLoss’ occurring withinthe pipe flow system has two components:

I- The MAJOR Head Loss ‘hMajor’

II- The MINOR Head Loss ‘hMinor’

FRICTIONAL LOSSES IN COMPLETELY DEVELOPED PRESSURIZED PIPE FLOWING FULL

MAJOR LOSSES

I- The MAJOR Head Loss ‘hMajor’:It is the head loss occurring in the direction of theflow along the pipes’ length. The loss is mainlyoccurring due to the internal friction that takes placebetween the fluids’ molecules viscosity effect withthe inner circumferential surface of the pipe causingthe shearing stress that develops velocity retardationto the flow.

MAJOR LOSSES

The Darcy-Weisbach Equation is used to determine the major loss value.

g2

vh

pΔ2av

MajorossLD

Lf

γ where ‘f ’ is the friction factor coefficient (dimensionless) .

The friction factor ‘f ’, depends on the various quantities:

* fluid characteristics (ρ, μ)

* flow characteristics (vav

)

* conduit characteristics (D, ks)

)k,D,v,,(f savμρ f

MAJOR LOSSES

(In some books λ is used as a symbol instead of f )

The dimensional analysis suggests and the experimentsproved that the friction factor ‘f ’ in general, is a functionof both roughness coefficient (‘ks/D’ or ‘D/ks’ and theReynolds number (Re =

μ

ρ Dvav)

Experimental data that relates the friction factor to theReynolds number have been obtained for fully developedpipe flow over a wide range of wall roughness.

MAJOR LOSSES

a) LAMINAR Flow[f Re( < 2100)] (Linear relationship) {(ks/D) has no effect}

Re

64f

b) TURBULENT Flow

i- Smooth pipe { ks ≈ 0} [f Re]

41

Re

3164.0f formula Blasius -

8.0f.Relog0.2f

110 Nikuradse and Prandtl -

5146.1Re10log80.1f

1 Jain -

(EXPLICIT EQN.)

(EXPLICIT EQN.)

(IMPLICIT EQN.)

MAJOR LOSSES

ii) Transition zone from Smooth to Fully Rough pipe {f [(D/ks), Re]}

- Colebrook – White formula (Semi-empirical)

9.0

s10

Re

25.21

D

klog 0.214.1

f

1

- Akalank - Jain formula (Semi-empirical)

D

k27.0

f.Re

51.2log 0.2

f

1 s10

iii) Fully Rough pipe [f (ks/D)]

- Karman – Prandtl equation (sometimes referred as ‘QUADRATIC LAW’

136.1k

Dlog 0.2

f

1

s

10

iii) Fully Rough pipe [f (ks/D)]

MAJOR LOSSES

In order to apply one of the widely used transition zoneequations Colebrook – White or Akalank-Jain the wallroughness of the pipe and Reynolds number parametersshould be used togather.

For Smooth pipe flow, this equation simplifies dueks=0.

For fully turbulent flow, this equation simplifies sinceRe→∞.

MAJOR LOSSES

These available equations are usually implicit andlogarithmic. For the pipes of different characteristics,(Smooth, Rough and Smooth to Rough transition cases),the determination of friction factor ‘f’ was tried to beunified by Lewis F. Moody (1880-1953) where hedeveloped a simplified, general resistance diagram,which bears his name and is valid for the uniform flowin conduits and is presented in Figure 2.1.

MAJOR LOSSES

In this chart, the curves show the variation of ‘f’ withrespect to ‘Re’ and ‘D/ks’ (three-parameter graph).

It should be emphasised that, Moody chart is onlyapplies to the average velocity and friction occurring forfully developed flow conditions. Consequently, if thefriction factor ‘f ’ is known, one can find the head loss‘hL’ and then the pressure drop ‘Δp’.

MAJOR LOSSES

Figure 2.1: Moody Diagram Note that : (D/ks=100) and Re=104, f = 0.043

Undefin

ed (critical) zo

ne

0.043

MOODY DIAGRAMSeveral features of the Moody diagram:

1) It is a log-log graph,

2) Correlates 3 dimensionless parameters (ReynoldsNumber ‘Re’, relative roughness ‘D/ks’ and the frictionfactor ‘f ’,

3) The graph composed of 4 distinct regions:Laminar [Re < ~ 2300],Undefined (Critical) [~ 2300 < Re < ~4000],Transition andFully turbulent (Complete turbulent) [sufficiently large

Re above which friction factor is constant] where theviscous effects are not significant.

4) For the larger relative roughness D/ks values it isobserved that, as Re number decreases, the frictionfactor ‘f ’ increases in the transition zone and eventuallybecomes the same curve as that of a smooth pipe.

5) Even for very smooth surfaces, the friction factor is notzero and can easily be determined from the appropriatediagram.

MOODY DIAGRAM

3

6s

Re

10

D

k2000010055.0f

Instead of using Moody Chart widely used suggested equations are:

*** Simplified Moody Chart (explicit) equation ( ‘f’ within 5%): valid for 4x103 < Re < 1x107 and ks/D < 0.01)

*** Barr (1975) (explicit) equation (valid for ‘f’ within 1% for Re > 1x105):

89.0

s

Re

1286.5

D7.3

klog2

f

1

2

9.0s

Re

74.5

D7.3

kln

325.1f

*** Swamee-Jain (explicit) equation:

PROBLEM TYPES IN PRESSURIZED PIPES

Pressurized, Steady, Uniform, fully developed fluid flowingturbulent in a single pipe system can be calculated in threedifferent ways due to the known parameters.

There are basically three categories of problems involved with uniform, steady, pressurized, fully developed

turbulent flow in a single pipe:

Category Known Unknown

1 Q, D, ks ,ν hL

2 hL, D, ks, ν Q

3 Q, ks, ν, hL D

Asked Type

Type 1:

Determination of the head loss hL is asked,Given: type (ks) and size (D) of the pipe and the flow rate

(Q=vav.A).

Solution Procedure:

Head loss is solved straight forward by obtaining ‘f’ value from Moody Diagram (or any relevant equation) by the help of the given data values through Re and (ks/D) parameters.

Problem Type 1

Question 2.1 Water at 23 oC is transported for 500 m in a 4 cm-diameter wrought iron (commercialsteel) horizontal pipe with a flow rate of 3 lt/sec. Calculate the pressure drop over the500-m length of this pipe, using Moody diagram. or by relevant equation.

Solution

Category 1 type problem

The average flow velocity in the pipe can be obtained by using the discharge relation,

A.vQ av sec/m387.202.0

103

A

Qv

2

3

av

π

To find the Reynolds number υ @ 23 o

C is needed.

Using the properties of water water table and applying double interpolation

oC υ m

2/s

20 1.004 x 10-6

23 ? 30− 23

30−20=

8.009𝑥10−7 − 𝜐?

8.009𝑥10−7 −1.004𝑥10−6 υ = 9.431 x 10

-7 m

2/s

30 8.009 x 10-7

The Reynolds number is: 5

7av 1001.1

10431.9

04.0387.2DvRe

ν

Obtaining ks from the Moody diagram ks= 0.046 mm so 00115.0mm40

mm046.0

D

ks

The friction factor is read from the Moody diagram to be 023.0f

Checking by the equation 023.0f OK.

The head loss is than calculated as

g2

v

D

Lfh

pΔ2av

L γ

m49.8381.92

387.2

04.0

500023.0h

2

L

o

C ρ kg/m3

20 998.2

23 ? 30− 23

30−20=

995.7 − 𝜌?

995.7 −998.2 ρ = 997.45 kg/m

3

30 995.7

The pressure drop is LhpΔ γ

2L N/m 81694.8 49.8381.9 x 45.997 hpΔ γ or 695.81pΔ kPa

0.045 0.001125

= 816965.9 =816.97

D/ks=888.89

0.045

Type 2:Determination of discharge amount (Q=vav A) within the pipe,Given: head loss hloss , pipe type (ks) and diameter (D) with temperature

or fluid viscosity.

Only by using trial and error approach the discharge (Q) can be calculated. Since discharge (Q) is unknown, the average flow velocity (vav) also not known hence, Re number can not be determined. For this reason, an initial approximate ‘f ’ value has to be determined and until the correct value is obtained, the trial and error procedure should be continued.

Solution Procedure:1- by assuming the flow to be fully turbulent!, with the help of relative

roughness parameter (D/ks) the approximate ‘f’ value was determined.2- Substituting this f value is into the head loss equation , the average flow

velocity (vav) can be calculated.

3- Since the average flow velocity determined , Re can be calculated andwith the help of (D/ks) a new f value is detected.

4- Until ‘fprevious’ ≈’flatest’ satisfies, the latest f value is continiously reused.

Category 2:

Determination of the flow rate (Q) is solved by trial and error through succesive iterations. Initial

approximation of ‘f’ is required. Since ‘f’ depends on Re and Re is unknown since average velocity (or

discharge) not known.

The solution procedure is:

1. assume the flow is fully turbulent, as an initial approximation and only using (ks/D) value obtain a

value for ‘f’,

2. substitute this value to the loss equation and obtain a value for average velocity (vav);

g2

v

D

Lfh

2av

Major ossL

3. Since velocity average is computed now obtain a new value for ‘f’ using Re and ks/D.

4. Repeat this until convergence (‘fprevious’ ≈ ‘flatest’).

Question 2.2Question 2.2

A pressure drop of 700 kPa is measured over a 300-m length of horizontal, 10-cm diameter

galvanized iron pipe that transports oil of sg = 0.9 and ν = 10-5

m2/s. Calculate the flow rate

using Moody diagram. Take γw=9810 N/m3.

Solution


The relative roughness is 0015.0mm 100

mm 15.0

D

ks

Since no other direct information is given an assumption and then checking approach should be

adopted. Hence assume that directly the friction factor ‘f ’; 022.0f


g2

v

D

Lh

pΔ2av

L fγ

m 28.799.09810

700000pΔh

oilL

γ note that

water

oilsgγ

γ

The velocity can be calculated from the head loss equation of Darcy Weisbach

g2

v

D

Lfh

2av

L L f

h D g 2v L

av sec/m 855.4300022.0

28.791.081.92v

5.0

av

using the equation or the Moody Diagram.

Question 2.2





Solution



mm 15.0

D

ks




g2

v

D

Lh

pΔ2av

L fγ

m 28.799.09810

700000pΔh

oilL

γ note that

water

oilsgγ

γ


g2

v

D

Lfh

2av

L L f

h D g 2v L

av sec/m 855.4300022.0

28.791.081.92v

5.0

av

Question 2.2

Question 2.2





Solution



mm 15.0

D

ks




g2

v

D

Lh

pΔ2av

L fγ

m 28.799.09810

700000pΔh

oilL

γ note that

water

oilsgγ

γ


g2

v

D

Lfh

2av

L L f

h D g 2v L

av sec/m 855.4300022.0

28.791.081.92v

5.0

av

adopted. Hence, the value of ‘f’ is initially estimated by assuming fully turbulent flow; f = 0.022

then it will be checked by trial and error procedure.

Checking the assumption by finding Reynolds number:

4

5av 10855.4

10

1.0855.4DvRe

ν

using this Reynolds number and ks/D =0.0015, the Moody diagram gives the friction factor as

026.0f ≠ 0.022 (the initially assumed one)

The new value of f is now will be used to calculate velocity of flow and Reynolds number as

sec/m 47.4300026.0

28.791.081.92v

5.0

av

This provides a Reynolds number of

4

51047.4

10

1.047.4Re

From the Moody diagram check f. f = 0.026 (same as the last assumption).

Thus the flow rate is /sm 035.005.047.4AvQ 32av π

Type 3:Determination of Pipe diameter (D),Given: head loss hloss , pipe type (ks) and flow discharge (Q) with temperature

or fluid viscosity.In order to determine Re, (D/ks), f , vav ,and headloss hL the diameter of the pipe (D) should be known definitely. For this reason, an initial approximate ‘f ’ value has to be determined and until the correct value of pipe diameter (D) is obtained, the trial and error procedure continues.

Solution Procedure:1- Initially a value of f is assummed. Since the range is: 0.008 < f < 0.1 ; so f=0.03

is an appropriate value to start with! 2- Substituting this f value is into the head loss equation,

the pipe diameter (D) obtained. 3- Using this diameter value (D), Re and (D/ks) were determined and through

Moody’s Diagram f is obtained.4- This more correctly obtained f value is compared with the previous f value;

‘fprevious’ ≈’flatest’ 5- If these f values are not equal, the latest determined f value is used to

determine a more correct pipe diameter (D). Trial and error procedure continues until the f values be equal.

Category 2:

Determination of the flow rate (Q) is solved by trial and error through succesive iterations. Initial

approximation of ‘f’ is required. Since ‘f’ depends on Re and Re is unknown since average velocity (or

discharge) not known.

The solution procedure is:

1. assume the flow is fully turbulent, as an initial approximation and only using (ks/D) value obtain a

value for ‘f’,

2. substitute this value to the loss equation and obtain a value for average velocity (vav);

g2

v

D

Lfh

2av

Major ossL

3. Since velocity average is computed now obtain a new value for ‘f’ using Re and ks/D.

4. Repeat this until convergence (‘fprevious’ ≈ ‘flatest’).

Question 2.3

Drawn tubing pipe of what diameter should be selected to transport 0.002 m3/sec at 20

oC of water over a 400-

m length so that the head loss does not exceed 30-m? Use Moody diagram to solve the question.

Solution


In this problem D is unknown. Thus, a trial and error solution is required. The average velocity is related to D

by

22avD

00255.0

4

D

002.0

A

Qv

π

The friction factor and D are related as follows

g2

v

D

Lh

2av

L f 81.92

)/00255.0(40030

22

D

Df

But Reynolds number also related with D:

D

2540

10x004.1D

D00255.0DvRe

62av

ν

So all the relevant equations are correlated with D.

f65 1042.4D

Note that the pipe is horizontally placed.

Question 2.3

Drawn tubing pipe of what diameter should be selected to transport 0.002 m3/sec at 20

oC of water over a 400-

m length so that the head loss does not exceed 30-m? Use Moody diagram to solve the question.

Solution


In this problem D is unknown. Thus, a trial and error solution is required. The average velocity is related to D

by

22avD

00255.0

4

D

002.0

A

Qv

π

The friction factor and D are related as follows

g2

v

D

Lh

2av

L f 81.92

)/00255.0(40030

22

D

Df

But Reynolds number also related with D:

D

2540

10x004.1D

D00255.0DvRe

62av

ν

So all the relevant equations are correlated with D.

f65 1042.4D

This problem a trial end error procedure, since the diameter (D) is unknown.Catagory 3 type problem.

For this reason as an initial trial f=0.03 value is fairly good. Basedon this f value the pipe diameter D=0.042 m is obtained.Using Re and (ks/D) parameters through Moody’s Diagram a morecorrect f value f=0.02 is calculated.But this value should be compared (checked) with previously usedf value. ‘fprevious= 0.03 ≠ flatest’=0.02.Hence trial and error procedure should continue!For the next trial, the f value found latest should be use (beingprobably the more closer value) f=0.02 .Based on this f value the pipe diameter D=0.039 m is obtained.With Re and (ks/D) parameters through Moody’s Diagram a moreclose to correct f value f=0.02 is determined.This value is also compared with the latest assummed value‘fprevious= 0.02 = flatest’=0.02 . Since they are equal the trial and errorprocedure stops. Hence D=0.039 m is the correct value.Note that the standard diameter value to be purchased is 4 cm!

Assume a value for f and get the other relations given above. Then read the Moody diagram; repeat until getting

acceptable values (assumed and determined values are approximately equal)

Trial 1: Let f = 0.03 and the related values are detailed in the following table. Note: the second guess is the

value for f found from the calculations of the first guess.

TRIAL f D (m) Re ks/D f (from Moody)

1

0.03

(INITIALLY ASSUMED)

0.042

6.05x104

0.000036

0.02

2

(RESULT OF PREVIOUS f VALUE

OBTAINED FROM MOODY)

0.02

0.039

6.51x104

0.000039

0.02

The value of f = 0.02 is acceptable since for the last trial assumed and calculated are same values. The diameter

is 39 mm. Since this diameter would undoubtedly not be a standard one a diameter of 4D cm would be the

pipe size selected.

This pipe would have a head loss less than the limit of 30-m as needed in the problem statement.

(EQUAL TO EACH OTHER)OK!

Question 2.4 :

For the given pipeline system, determine the average discharge ‘Qav’passing within the pipe. Ignore minor losses.

Pipe: • commercial steel• 135 m long• diameter ϕ = 28 cm

water @ 25 °C

average discharge Q = ?

138.65 m

103.20 m

A

B

Answer: Q=627.56 lt/sec

CATAGORY 2 TYPE PROBLEM

Question 2.5 (T)

For the given pipeline system, determine the average discharge ‘Qav’passing within the pipe. Ignore the minor losses.

A

Pipe is pvc of diameter ϕ = 20 cm; Length 390 m.

17.00 m

0.50 m


water @ 20 °C

Answer: Q= 115 lt/sec

CATAGORY 2 TYPE PROBLEM

Question 2.6 (T)

For the given pipeline system, determine the average discharge ‘Qav’passing within the pipe. Ignore the minor losses.

CATAGORY 2 TYPE PROBLEMQ=0.1419 m3/s ≡ 141.9 lt/s

Pipe is pvc of diameter ϕ = 20 cm; Length 390 m.

0.50 m


water @ 20 °C

AB

Pipe: • Galvanized iron• 1.45 km long• average discharge Qav = 23.5 lt/s

water @ 17 °C

diameter φ = ?

234.65 m

188.70 m

Question 2.7:For the given pipeline system, determine the nominal diameter ‘φ ’ of the pipe. Ignore minor losses.

Answer: D= 0.126 mCATAGORY 3 TYPE PROBLEM

MINOR LOSSES

2.2 Minor Losses in Pipe FlowThey also called local or form or secondarylosses. These losses are due to fitting typesalong the pipe. Each one of these fittingscauses change in the magnitude and/or thedirection of the velocity and hence results locallosses at that part. These are:• sudden enlargements, • sudden contractions, • elbows and bends, • valves, • reservoir inlets, • reservoir outlets…

MINOR LOSSESA minor loss is expressed in terms of a loss coefficient kL

with respect to the velocity head:

The values of kL are constant for that specific type and havebeen determined experimentally for the various fittings andgeometry changes.

Depending on the type of fitting, kL value is eitherexpressed with respect to entrance average velocity (vav ent)or with respect exit average velocity (vav ex).

Detailed in Figure 2.2

kL

MINOR LOSSES• For sudden enlargement (expansion),

(exit from a pipe to a tank, has different connection types like bell mouthed or submerged etc...)

This head loss can be expressed as with respect to pipe velocity just at the entrance since it is larger:

• For sudden contractions, (entry to a pipe from a tank, has different connection types: like bell mouthed or reenterant (projecting) etc...)

This head loss can be expressed with respect to pipe velocity at the exit since it is larger:

Detailed in Figure 2.2

kL

kL

MINOR LOSSES• Elbows (different types like 90°, 45°, U...),

• Tees (branching and connection) fittings,

• Valves (different types butterfly, gate, spherical) and their opening cases (full, ½, ¼ ...),

• Contraction (pipes with decreasing diameters connected one after another)...

• Expansion (pipes with increasing diameters connected one after another)...

The bigger avarage velocity of that point should be used.Detailed in Figure 2.2...

kL

Figure 2.2: Minor Loss coefficients kL for different fittings

MINOR LOSSESFor gradual enlargements such as conical connections, the head loss is expressed as:

where the values of K' as a function of cone angle are given in the Figure 2.3

Figure 2.3: Loss coefficient for conical connections

MINOR LOSSESQuestion 2.8If the flow rate through a 10-cm diameter wrought iron pipe is 0.04 m3/sec, find the difference in elevation H of the two reservoirs. Since the connection details are given consider them as well!

55 cm

1.35 m

25 cm

45 cm

Flow direction

(threaded)

threaded

MINOR LOSSESSolution:The energy (Bernoulli) equation written for a control volume between two reservoir surfaces:

where pressures at point 1 and point 2 are atmospheric so gage pressure = zero.

Similarly due very large reservoir cross-sectional areas their velocities are very low, hence velocity heads approaches to zero.

minor LLmajor2

22av2

1

21av1 hhz

g2

vPz

g2

vP

γγ

Patm=Pgage=0 Patm=Pgage=0

large reservoir surface area≈0

MINOR LOSSESThus letting z1 - z2 = H then,

Since the pipe diameter is constant everywhere, then the average velocity is constant, Reynolds number and relative roughness are

m/sec

0.0450.00045

threaded90°

0.10

k k k k

MINOR LOSSESUsing the Moody diagram 0173.0f , the total length of the pipe is L=50 m,

using the loss coefficient from the given tables, for an entrance (K=0.5),

a globe valve (K=5.7), screwed 10-cm diameter standard elbow (K=0.64),

and for an exit (K=1); there results

The minor losses are even greater than to the frictionallosses as expected, since there are five minor losselements in 100 mm diameters of pipe ~50 m long.

0.8 + 10 + 2(1.5) + 1

H=11.88 + 19.54 = 31.42 m

10 1.5

0.0178

50.550.0178

L=50.55 m

(k=0.5)

(k=10) (k=1.5)(k=1)

MINOR LOSSES

Any piping system where:L/D > 1000

the major loss componentmainly controls; hence theminor losses component can beignored!

FLOW WITHIN NON-CIRCULAR PIPES (CONDUITS)

a

aa

a

bφ

a

b

a

b

NON-CIRCULAR PIPES2.3 Losses in Non-circular conduitsDue to some engineering requirements, some conduits canbe manufactured with non-circular cross-sections.Since the generated friction equations are all based on fluidsflowing within a circular pipe that has a circular cross-section, so the velocity or the discharge flowing within theconduit forming circular cross-sections, but for those flowsthat are not forming circular cross-section, need to beapproximated so as to adopt the friction equation. Thisapproximation can be made by using the hydraulic radius‘RH’ approach which is defined as:

A: the cross-sectional area (m2)P: the wetted perimeter (the total contact length of the

fluid with the conduits inner wall) (m).

NON-CIRCULAR PIPES

For a circular pipe of radius R:

• Radius ‘R’ is twice of hydraulic radius ‘RH’.

• Diameter ‘D’, ‘φ’ is 4 times of the hydraulic radius ‘RH’.

D= 4‘RH’; (2*2RH).

NON-CIRCULAR PIPES

• Relative roughness:

• Reynolds Number:

Note that : Q = vav * (Area of the original (given) cross-section)Do not use 4RH concept for discharges!

NON-CIRCULAR PIPES

‘f’ value is:• ± 40 % uncertain (Laminar)• ± 15 % uncertain (Turbulent)

NON-CIRCULAR PIPES

AB

Conduit: • 7850 m long• Rectangular cross-section 15 cm X 20 cm • Smoothened concrete

zA= 100 m

zB= 70 m

For the given system, determine the average discharge ‘Qav’ of water@ 20o C. Ignore minor losses.

Question 2.9:

This is a Type 2 problem!Q = 0.0218 m3/s ≡ 21.8 lt/s

For the given pipeline system, oil of sg=0.83, ν=10-5 m2/s @ 20 oC isbeing transferred to reservoir B with the discharge Q=107 lt/s.i- determine the nominal diameter ‘φ ’ of the pipe:

a) ignore minor losses.b) considering the total minor loss coefficient ΣK=179.0

ii- if a squared conduit cross-section with the same material is expectedto be designed what will be its side sizes? Consider the total losscoefficient Σk=179.

Question 2.10 T:

i- a) Φ =27.4 → 28 cmb) Φ =30.0 → 30 cm

ii- L=20.7 cm

A

m

zA= 58.25 m

Below given system is proposed to carry the discharge of 45.7 lt/s of water at 23 °C from reservoir A along point M to point

B that discharges freely to a lake through a single conduit which is made up of commercial steel. This water is in fact

flowing between the two semi-circular cross-sections (between the inner and the outer). The outer diameter ‘ϕouter’ is twice

the diameter of the inner ‘ϕinner’ one.

a) determine the outer diameter ‘ϕouter’ to satisfy this requirement by considering the effect of the minor losses as

well.

b) check the cavitation occurance at point M which is located 2500 m away from reservoir A, ignore the effect of

minor losses.

Note: Whole system is functioning under the absolute atmospheric pressure Patm =100.93 kPa (abs). Letter ‘z’ indicates the

topographic elevation of that point.

Question 2.11 T:

Given system is proposed to discharge of 85.7 lt/s of water at 23 °C from reservoir A to

reservoir B with a single commercial steel semi-circular shaped conduit. Water is expected to

flow between the two semi-circular cross-sections (the inner and the outer). The outer

diameter ‘ϕouter’ is twice the diameter of the inner ‘ϕinner’ one. Ignoring the minor losses.

Determine the outer diameter to satisfy this requirement. The system is functioning under the

absolute atmospheric pressure Patm=100.93 kPa (abs).

Previous Exam Question:

Question: (100 p)

A fluid of kinematic viscosity ‘ν = 5.02x10-7 m2/s’ is flowing within the areal gap of the concentric pipes

(cylinders) which are fixed to a cylindrical reservoir as detailed below. The length of this concentric galvanized iron

pipes are 45 m, having outer diameter of ϕouter = 10 cm and inner diameter of ϕinner = 6 cm. These pipes are connected

to a gradually varying contraction connection of a negligible length that has a circular opening of ϕ = 1.8 cm where

the flow is discharging out. Considering only the given minor losses, determine the discharge ‘Q’ flowing within

this system.

φouter =10 cm

φinner = 6 cm

Section A-A

Question 2.12 T:

GENERALREVIEW

Every component of Bernoulli’s equationfor any piping system,hydro-mechanically,get effected.

Any piping system hydro-mechanically get effected from every component of Bernoulli’s equation.

pressure head variation due upper and/or lower end of the system

velocity head variation due upper and/or lower end of the system

vav ≈ 0

vav > 0


elevation head variation due upper and/or lower end of the system




i- pipe length ‘L’ii- pipe diameter ‘D’iii- flow average velocity within the pipe ‘vav’iv- friction factor ‘f’


L=1000 mL=2000 m

i- pipe length ‘L’ as the length of the pipe increases, losses withinthe system increases



D =35 cm D=20 cm

ii- pipe diameter ‘D’as the diameter of the pipe varies definitely losses within the pipe changes


v =1.35 m/s v=2.0 m/s

iii- flow average velocity within the pipe ‘vav’as the discharge varies definitely frictionallosses changes


iv- friction factor ‘f’ Re : velocity, diameter, viscosity, temperature ks/D : pipe material

Any change of above the parameters implicitelyaffects ‘f’ as detailed in Moody’s chart.

ks: cast ironμ @ T 25°C

ks: smoothμ @ T 10°C

The piping system hydro-mechanically get effected from each component

II- The MINOR Head Loss ‘hMinor’

Depending on number and type of localfittings and the average flow velocity withinthat point, directly effects the energy losses

kL

Documents

FRICTIONAL LOSSES IN COMPLETELY DEVELOPED …...MOODY DIAGRAM Several features of the Moody diagram: 1) It is a log-log graph, 2) Correlates 3 dimensionless parameters (Reynolds Number