Fluid Mechanics and Applications Inter - Bayamon Lecture 4 Fluid Mechanics and Applications MECN 3110 Inter American University of Puerto Rico Professor:

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LectureLecture

44Fluid Mechanics and Applications

MECN 3110

Inter American University of Puerto RicoProfessor: Dr. Omar E. Meza Castillo

Chapter 6Fluid

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Viscous Flow in Ducts

Chapter 6

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To describe the appearance of laminar flow and turbulent flow.

State the relationship used to compute the Reynolds number.

Identify the limiting values of the Reynolds number by which you can predict whether flow is laminar or turbulent.

Compute the Reynolds number for the flow of fluids in round pipes and tubes.

State Darcy’s equation for computing the energy loss due to friction for either laminar and turbulent flow.

Define the friction factor as used in Darcy’s equation Determine the friction factor using Moody’s diagram for

specific values of Reynolds number and the relative roughness of the pipe.

Major and Minor losses in Pipe Systems.

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Course Objectives

Chapter 6Fluid

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on Introduction

This chapter is completely devoted to an important practical fluid engineering problem: flow in ducts with various velocities, various fluids, and various duct shapes. Piping Systems are encountered in almost very engineering design and thus have been studied extensively.

The basic piping problem is this: Given the pipe geometry and its added components (such as fitting, valves, bends, and diffusers) plus the desired flow rate and fluid properties, what pressure drop is needed to drive the flow? Of course, it may be stated in alternative form: Given the pressure drop available from a pump, what flow rate will ensue? The correlations discussed in this chapter are adequate to solve most such piping problems

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on Reynolds Number Regimes

As the water flows from a faucet at a very low velocity, the flow appears to be smooth and steady. The stream has a fairly uniform diameter and there is little or no evidence of mixing of the various parts of the stream. This is called laminar flow.

High-viscosity, low-Reynolds-number, laminar flow

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When the faucet is nearly fully open, the water has a rather high velocity. The elements of fluid appear to be mixing chaotically within the stream. This is a general description of turbulent flow.

Low-viscosity. High-Reynolds-number, turbulent flow

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The changeover is called transition to turbulent. Transition depends on many effects, such as wall roughness or fluctuations in the inlet stream, but the primary parameter is the Reynolds number.

Studies present the following approximate ranges that commonly occur:1. 0 < Re < 1: highly viscous laminar “creeping” motion2. 1 <Re<100: laminar, strong Reynolds number

dependence3. 100 <Re <103: laminar, boundary layer theory useful4. 103 <Re <104: transition to turbulence5. 104<Re<106: turbulent, moderate Reynolds number

dependence6. 106<Re<∞: turbulent, slight Reynolds number

dependence

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In 1883 Osborne Reynolds, British engineering professor was the first to demonstrate that laminar or turbulent flow can be predicted if the magnitude of a dimensionless number, now called the Reynolds number is known.

The following equation shows the basic definition of the Reynolds number, Re:

The value of 2300 is for transition in pipes. Other geometries, such as plates, airfoils, cylinders, and spheres, have completely different transition Reynolds numbers.

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VD

Re

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on Critical Reynolds Number

For practical applications in pipe flow we find that if the Reynolds number for the flow is less than 2000, the flow will be laminar.

Re < 2000 : Laminar flow

If the Reynolds number is greater than 4000, the flow can be assumed to be turbulent.

Re>4000 : Turbulent flow

In the range of Reynolds numbers between 2000 and 2000, it is impossible to predict which type of flow exists; therefore this range is called the critical region.

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Chapter 6Fluid

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Statement: Determine whether the flow is laminar or turbulent if glycerin at 25oC flows in a pipe with a 150-mm inside diameter. The average velocity of low is 3.6 m/s.

Solution:

Because Re=708, which is less than 2000, the flow is laminar

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708

10609

150631258

106091258

15063

1

3

13

s.Pax.

m.sm.mkgRe

s.Pax.,mkg

,m.D,sm.V

VDRe

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on Problem

Statement: Determine whether the flow is laminar or turbulent if water at 70oC flows in a 1-in Type K copper tube with a flow rate of 285 L/min.

Solution: For a 1-in Type K copper tube, D=0.02527m and A=5.017 x 10-4 m2. Then we have

Because Reynolds number is greater than 4000, the flow is turbulent.

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57

27

3

24

1082510114

025270479

10114

47960000

1

100175

285

x.x.

..VDRe

smx.

sm.minL

smx

mx.

minL

A

QV

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on Head Loss – The Friction Factor

In the general energy equation

Julius Weisbach in 1850 established that hf is proportional to (L/D), and G.H.L Hagen shown that for turbulent flow, hf is proportional to V2. The proposed correlation, still as effective today as in 1850, is

This expression is called Darcy’s Equation. The dimensionless parameter f is called the Darcy Friction factor.

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fturbinepump hhzg

VPhz

g

VP

2

222

1

211

22

shapeduct,D,Refcnfwhere

g

V

D

Lfh Df 2

2

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on Friction Loss in Laminar Flow

Because laminar flow is so regular and orderly, we can derive a relation between the energy loss and the measurable parameters of the flow system.

This relationship is known as the Hagen-Pouseuille equation:

The Hagen-Pouseuille equation is valid only for laminar flow (Re<2000).

If the two previous relationships for hf are set equal to each other, we can solve for the value of the friction factor:

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2

32

gD

LVh f

2

2 32

2 gD

LV

g

V

D

Lf

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on Friction Loss in Laminar Flow

In summary, the energy loss due to friction in laminar flow can be calculated either from Hagen-Pouseuille equation or Darcy’s equation. The pipe friction factor decrease inversely with Reynolds number.

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Ref

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Ref

DVReif,

DVf

64

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Chapter 6Fluid

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Statement: Determine the energy loss if glycerin at 25oC flows 30 m through a 150-mm-diamter pipe with an average velocity of 4.0 m/s.

Solution: First, we must determine whether the flow is laminar or turbulent by evaluating the Reynolds number:

Because Re=768, which is less than 2000, the flow is laminar

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786

10609

150041258

106091258

15004

1

3

13

s.Pax.

m.sm.mkgRe

s.Pax.,mkg

,m.D,sm.V

VDRe

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Using Darcy’s Equation

This means that 13.2 NM of energy is lost by each newton of the glicerin as it flow along the 30 m of pipe.

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NNmorm..

.

..h

.Re

f

g

V

D

Lfh

f

f

2138192

04

150

300810

0810786

6464

2

2

2

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on Friction Loss in Turbulent Flow

For turbulent flow of fluids in circular pipes it is most convenient to use Darcy’s Equation to calculate the energy loss due to friction.

Turbulent flow is rather chaotic and is constantly varying. For these reasons we must rely on experimental data to

determine the value of f. The following figure illustrate pipe wall roughness

(exaggerated) as the height of the peaks of the surface irregularities.

Because the roughness is somewhat irregular, averaging techniques are used to measure the overall roughness value

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on Friction Loss in Turbulent Flow

For commercially available pipe and tubing, the design value of the average wall roughness has been determined as shown in the following table

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Chapter 6Fluid

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on Relative Roughness of Pipe Material

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on Moody Diagram

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It is the graphical representation of the function f(ReD, ε/D)

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on Moody Diagram

Several important observations can be made from these curves:1. For a given Reynolds number flow, as the relative

roughness is increased, the friction factor f decreases.2. For a given relative roughness, the friction factor f

decreases with increasing Reynolds number until the zone of complete turbulent is reached.

3. Within the zone of complete turbulence, the Reynolds number has no effect on the friction factor.

4. As the relative roughness increases, the value of Reynolds number at which the zone of complete turbulence begins alto increases.

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Chapter 6Fluid

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Check your ability to read the Moody Diagram correctly by verifying the following values for friction factors for the given values of Reynolds number and relative roughness:

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Chapter 6Fluid

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Statement: Determine the friction factor f if water at 70oC is flowing at 9.14 m/s in an uncoated ductile iron pipe having an inside diameter of 25 mm.

Solution: The Reynolds number must first be evaluated to determine whether the flow is laminar or turbulent:

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527

27

106510114

0250149

10114

0250149

x.smx.

m.sm.Re

smx.

,m.D,sm.V

VDRe

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Thus, the flow is turbulent. Now the relative roughness must be evaluated. From previous table we find ε=2.4x10-4 m. Then , the relative roughness is

The final steps in the procedure are as follows:1. Locate the Reynolds number on the abscissa of the

Moody Diagram.2. Project vertically until the curve for ε/D =0.00961538

is reached.3. Project horizontally to the left, and read f=0.038

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0096153800250

41042.

m.

mx.

D

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Statement: In chemical processing plant, benzene at 50oC (SG=0.86) must be delivered to point B with a pressure of 550 kPa. A pump is located at point A 21m below point B, and the two points are connected by 240 m of plastic pipe having an inside diameter of 50 mm. If the volume flow rate is 110 L/min, calculate the required pressure at the outlet of the pump.

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Solution: Using the energy equation we get the following relation:

Mass Balance

Energy Balance

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fBBB

AAA hz

g

VPz

g

VP

22

22

fBBB

AAA hz

g

VPz

g

VP

22

22

BABA

BBAA

VVthen,AAAs

AVAV

fABBA hzzPP

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The evaluation of the Reynolds number is the first step. The type of flow, laminar or turbulent, must be determined.

For a 50-mm pipe, D=.050 m and A=1.963 x 10-3 m2. Then, we have

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4

4

3

334

23

33

333

105491024

86005009320

86010008601024

9320109631

10831

1083160000

1110

x.s.Pax.

mkgm.sm.VDRe

mkgmkg.,s.Pax.

:tableFrom

sm.mx.

smx.

A

QV

smx.minL

smminLQ

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For turbulent flow, Darcy’s equation should be used:

With the Reynolds number and the relative roughness we obtain the friction factor from the Moody’s Diagram f=0.018

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00000600050

1003

1003

2

7

7

2

.m.

mx.

D

.mx.:tableFrom

g

V

D

Lfh f

m..

.x

.x.

g

V

D

Lfh f 833

8192

9320

0500

2400180

2

22

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You should have the pressure as follows:

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kPaP

kPakPaP

m.mPa

kPasm.mkgkPaP

hzzPP

A

A

A

fABBA

759

209550

833211000

819860550

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on Fundamental Equation of Fluid Mechanics

In order to apply previous equation to a piping system, we must extend the Bernoulli equation to account for losses which result from pipe fittings, valves, and direct losses (friction) within the pipes themselves. The extended Bernoulli equation may be written as:

Additionally, at various points along the piping system we may need to add energy to provide an adequate flow. This is generally achieved through the use of some sort of prime mover, such as a pump, fan, or compressor.

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losseshz

g

VPz

g

VP2

222

1

211

22

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on Fundamental Equation of Fluid Mechanics

For a system containing a pump or pumps, we must include an additional term to account for the energy supplied to the flowing stream. This yields the following form of the energy equation:

Finally, if somewhere in the piping system a component extracts energy from the fluid stream, such as a turbine, the energy equation takes the form:

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losses

pumps hzg

VP

gm

Wz

g

VP2

222

1

211

22

losses

turbinespumps hgm

Wz

g

VP

gm

Wz

g

VP

2

222

1

211

22

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on Losses in Piping System

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Friction Factor: The total head loss hf in a piping system are typically categorized as major and minor losses. Major losses are associated with the pipe-wall

skin friction over the length of the pipe, and Minor losses in piping systems are generally

characterized as any losses which are due to pipe inlets and outlets, fittings and bends, valves, expansions and contractions, filters and screens, etc.

Minor losses are not necessarily smaller than major losses.

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on Major Losses

Major losses of head in a piping system are the direct result of fluid friction in pipes and ducting. The resulting head losses are usually computed through the use of friction factors. Friction factors for ducts have been compiled for both laminar and turbulent flows. Two widely adopted definitions of the friction factor are the Darcy and Fanning friction factors.

The head loss due to flow of a fluid at an average velocity V through a length L of pipe with a diameter D is

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cWDf g

V

D

Lfh

2

2

(Darcy-Weisbach) or

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on Major Losses

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Where:fD-W: Darcy-Weisbach friction factorfF: Fanning friction factorL: Length of considered pipeD: Pipe diameterV2/2gc: Velocity head

cFf g

V

D

Lfh

24

2

(Fanning)

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on Roughness Height (e or ε) for Certain Common Pipes

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ε

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on Friction Factor f

The Moody diagram is sufficient to determine the friction factor and, hence, the head loss for given flow conditions. If we should desire to generate a computer-based solution, the translation of the Moody diagram into tabular form to use in interpolation is awkward. To say the least. What is needed is a simple algebraic expression in the form f(ReD, ε/D). Historically, the implicit expression of Colebrook has been accepted as the most accurate in the turbulent zone.

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2512

73

1

fRe

.

D.log

f D

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Benedict suggests the expression proposed by Swamee and Jain, i.e.,

While for ε/D>10-4 Haaland recommends

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2

90

74573

250

.DRe

.D.

log

.f

2111

7396

30860

.

D D.Re.

log

.f

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For situations where ε/D is very small, as in natural-gas pipelines, Haaland proposes

Where n ~ 3 The use of the Swamee-Jain or Haaland provide

an explicit formula of the friction factor in turbulent flow, and is thus the preferred technique.

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2111

2

7377

30860

n.n

D D.Re.

log

n.f

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For laminar flow (Re<2000) the usual Darcy-Weisbach friction factor representation is

For turbulent flow in smooth pipes (ε/D=0) with 4000<Re<105 is

For turbulent flow (Re>4000) the friction factor can be founded from the Moody diagram

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DRe

.f

064

41

3160/

DRe

.f

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Churchill devised a single expression that represents the friction factor for laminar, transition and turbulent flow regimes. This expression, which is explicit for the friction factor given the Reynolds number and relative roughness, is

where

and

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121

51

1218

8

/

.D BARe

f

16

90 2707

14572

D.Reln.A .

D

1653037

DRe

,B

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In our discussion so far we have been concerned only with circular pipes, but for a variety of reason conduit cross sections often deviate from circular. The appropriate characteristic length to use in evaluating the Reynolds number for noncircular cross-sectional areas is the hydraulic diameter. The hydraulic diameter is defined as

To use the hydraulic diameter concept, the Reynolds number is defined as

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perimeterwetted

areationalseccrossDH

4

HHD

VDRe

51

6

542

10497..T

x

μ (m2/s)T (oC)

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on Example 2

Find the head loss due to friction in galvanized-iron pipe 30 cm diameter and 50 m long through which water is flowing at a velocity of 3 m/s assume that water flowing at 20oC.

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ε

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on Minor Losses

Minor losses are due to the change of the velocity of the flowing fluid in magnitude or direction. They are most often calculated using the concept of a loss coefficient or equivalent friction length method. In the loss coefficient method, a constant or variable factor K is defined as:

The associated head loss is related to the loss coefficient through

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22 212 V

P

gV

hK f

g

VKh f 2

2

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on Minor Losses

The Minor Losses occurs at: Valves Tees Bends Reducers Valves And other appurtenances

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on Minor Losses

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on Minor Losses: Typical Constant K-Factors

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on Minor Losses

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on Head Loss Due to a Sudden Expansion (Enlargement)

Or:

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g

VKh LL 2

21

2

2

11

A

AKL

g

VVhL 2

221

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on Head Loss Due to a Sudden Contraction

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g

VKh LL 2

22

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on Head Loss Due to Gradual Enlargement (Conical diffuser)

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g

VVKh LL 2

221

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on Head Loss Due to Gradual Contraction (Reducer or nozzle)

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g

VVKh LL 2

212

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on Head Loss at the Entrance of a Pipe (Flow leaving a Tank)

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g

VKh LL 2

2

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on Another Typical Values for various amount of Rounding of the Lip

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on Head Loss at the Exit of a Pipe (flow entering a tank)

The entire kinetic energy of the exiting fluid (velocity V1) is dissipated through viscous effects as the stream of the fluid mixes with the fluid in the tank and eventually comes to rest (V2)

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g

VhL 2

2

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on Head Loss Due to Bends in Pipes

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g

VKh bb 2

2

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on Head Loss Due to Mitre Bends

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on Head Loss Due to Piping Fittings (Valves, Elbows, Bends, and Tees)

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g

VKh vv 2

2

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on Head Loss Due to Piping Fittings (Valves, Elbows, Bends, and Tees)

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on The loss coefficient for elbows, bends, and tees

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on General Equation

The basis for any analysis or design in the energy equation written between any two points and incorporating multiple pumps, turbines, and majors and minor losses. The general representation that we shall use is:

The relation between head loss and pressure drop is given by

And the relation between power and pressure drop is given by

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g

gh

g

gh

g

gW

g

gWz

g

VPz

g

VP cL

)or(minl

fc

K

)majors(k

fc

J

)pumps(j

sc

I

)turbines(i

s lkji

1111

2

222

1

211

22

fhp

ff hmQhpQpower

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on Piping Network: HVAC Piping System

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on Piping Network

Most engineering systems are comprised of more than one section of pipe. In fact in most systems a complex network of piping is required to circulate the working fluid of a particular thermal system. These networks consist of series, parallel, and series-parallel configurations.

Pipe flow problems fall into three categories. In Category I problems the solution variable is the head loss or pressure drop Δp. The problem is specified such that the volumetric flow Q, the length of pipe L, the size or diameter D, are all known along with other parameters such as the pipe roughness and fluid properties. These types of problems yield a direct solution for the unknown variable Δp.

67

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on Piping Network

In a Category II problem, the head loss (h or Δp) is specified and the volumetric flow Q is sought. Finally in a Category III problem, both the head loss and volumetric flow are specified, but the size or diameter of the pipe D is sought. Category I and Category II problems are considered analysis problems since the system is specified and only the flow is calculated. Whereas Category III problems are considered design problems, as the operating characteristics are known, but the size of the pipe is to be determined. Both Category II and Category III problems require an iterative approach in solution.

68

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on Piping Network

Depending upon the nature of the flow (and solution process), it may be required to recompute other parameters such as the relative roughness at each iterative pass, since the ε/D ratio will change as the pipe diameter changes. However, with most modern computational software, we may solve “iterative” problems rather efficiently and need not resort to classic methods such as Gaussian elimination.

69

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on Series Piping Systems

70

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The series flow arrangement is the simplest to analyze. In a series arrangement of pipes, the volumetric flow at any point in the system remains constant assuming the fluid is incompressible. Thus, for an arrangement of N pipes, the volumetric flow is given by

Or

The head loss in the system is the sum of the individual losses in each section of pipe. That is

71

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The series flow arrangement is the simplest to analyze. In a series arrangement of pipes, the volumetric flow at any point in the system remains constant assuming the fluid is incompressible. Thus, for an arrangement of N pipes, the volumetric flow is given by

Or

The head loss in the system is the sum of the individual losses in each section of pipe. That is

72

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on Example 1-1 (Book)

Apply the energy equation to the situation sketched in the following figure.

74

1

Water La, Da, Va

Lb, Db, Vb

Lc, Dc, Vc

2

H

Figure 1-1

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The energy equation will be applied from the free surface at position 1 of the upper reservoir to the free surface at position 2 of the lower reservoir. The results, with each loss term identified, are

From the figure, we obtain

75

)frictionpipe(

b

b

bb

)elbow(

bb

frictionpipe

a

a

aa

)entrance(

aa g

V

D

Lf

g

VK

g

V

D

Lf

g

VKz

g

VPz

g

VP

222222

2222

2

222

1

211

)exit(

ce

)valve(

cv

)frictionpipe(

c

c

cc

)elbow(

cc g

VK

g

VK

g

V

D

Lf

g

VK

2222

2222

Hzz

VV

PP

21

21

21

0

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And from the continuity equation for incompressible steady flow, we can find that

Which, for circular pipes, becomes

Substitution of the preceding into the energy equation yields, after applying some algebra,

Until additional information is specified, we can not proceed any further than this

76

ccbbaa AVAVAV

222ccbbaa DVDVDV

ev

v

ccc

c

a

b

bbb

b

a

a

aaa

a KKD

LfK

D

D

D

LfK

D

D

D

LfK

g

VH

4

4

4

42

2

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77

ft75Hand,ft80L,ft20L,ft100L,DDD cbacba

evbcbaa

2

KKK2LLLD

fK

g2

Vft75

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78

0.1K:exitsand,f55K:valve,f30K:elbows,78.0K:entrance eTvTba

78.1f115

D

f200

g2

Vft75 T

2

Nominal size ½” ¾” 1” 1 1/4 ” 1 1/2” 2” 2 ½, 3”

Friction factor fT 0.027 0.025 0.023 0.022 0.021 0.019 0.018

Nominal size 4” 5” 6” 8-10” 12-16” 18-24”

Friction factor fT 0.017 0.016 0.015 0.014 0.013 0.012

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79

D(ft)

ε/D V(ft/s)

ReD f fT (200f/D+115fT+1.78) H(ft)

0.1 0.00150 25.46 1.819X105 0.0228 0.023 50.036 504.20

0.2 0.00075 6.37 9.095X104 0.0213 0.018 25.126 15.82

0.15 0.00100 11.32 1.213X105 0.0216 0.020 32.894 65.48

0.14 0.00107 12.99 1.299X105 0.0218 0.020 35.152 92.91

0.145 0.00103 12.10 1.254X105 0.0217 0.020 33.959 77.42

0.147 0.00102 11.78 1.237X105 0.0216 0.020 33.534 72.37

0.146 0.00103 11.95 1.246X105 0.0217 0.020 33.770 74.90

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on Generalized Series Piping System Software

80

Generalized Series Piping System Schematic

Figure 1-2

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81

'1T

'1

1

11

21

b

2bb

a

2aa KfC

D

Lf

g2

Vz

g2

VPz

g2

VP1

g

gWKfC

D

Lf

g2

V cs

'2T

'2

2

22

22

2

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82

iTii

ii4

i

2J

1i2ab

abcs KfC

D

Lf

gD

Q8zz

PP

g

gW

i

222

211 D

4VD

4VQ

2211 AVAVQ

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The example 1-2 in generic form appears as

For a simple-pipe system, the symbolic form of the generic energy equation becomes

83

KCfD

Lf

gD

Q8zz

PP

g

gW T4

2

2ababc

s

78.1f115

D

f200

g2

V750 T

2

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on Parallel Piping Systems

84

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Flow in parallel piping elements is also easy to analyze. In a parallel arrangement the total head loss or pressure drop across the system is constant. That is

On the other hand, the volumetric flow through the system is the sum total of the individual flow in each pipe. That is

85

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Parallel systems are often used to reduce the pumping power required for process-control system.

The following figure shows a generalized parallel system.

86

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The increase in head across the pump, Ws, is such that the pressures at a and b are equal. The change in pressures (or heads) are equal for each leg of the parallel arrangement; hence Ws is also the change in head across each leg required for Pa = Pb. If QT is the total flow rate, the conservation of mass yields

The energy equation for line i, under the assumptions of Pa = Pb and Va = Vb, can be expressed

87

321T QQQQ

iTi

i

ii4

i

2i

2abc

s KfCD

Lf

gD

Q8zz

g

gW

i

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Analysis or design calculations for parallel piping require the use of the conservation mass and an energy equation, for each line. For example for a system composed of two parallel lines the requirement expressions are:

Together with the usual friction-factor and Reynolds-number definitions.

88

21T QQQ

1T1

1

114

1

21

2abc

s KfCD

Lf

gD

Q8zz

g

gW

1

2T2

2

224

2

22

2abc

s KfCD

Lf

gD

Q8zz

g

gW

2

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Two types of parallel-system analysis problems are evident in the last equations: (1) given Ws, find QT, Q1 and Q2; and (2) given QT, find Ws, Q1

and Q2. Type 1 problems are straightforward, as they can

be solved by category II problem methodology. Type 2 problems are more complex, since the

total flow rate must be apportioned to each of the parallel lines in such manner that the pressure drops across each line are equal. We can list a simple sequence to systematically accomplish this apportionment:

89

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1. Assume a discharge Q’i through pipe i of the parallel system. Solve for the head loss h’fi

(or pressure drop

Δp’i) through pipe i; this is a category I pipe-flow solution.

2. Using h’fi= h’fj

(i≠j), solve for the Q’i (i≠j). This is a category

II pipe=flow solution.

3. Redistribute the total flow rate QT by the simple ratio process

where K is the total number of pipes

5. Check the hfk (k=1, K) for the equality using the Qk

(k=1,K) obtained from the step 3. Repeat steps 1 through 4 until convergence is obtained.

90

)K,1l,K,1k(QQ

QQ T

l

'l

'k

k

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91

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on Example 1-5(Book)

Consider the parallel flow network of the following figure with the following specifications:

92

(1)

(2)

● BA ● s/ft00003.0

slugs/ft 22

3

L1=3000 ft, PA=80 psia, L2=3000 ft

D1=1 ft, ZA=100 ft, D2=8 in

ε1 =0.001 ft, ZB=80 ft, ε2 =0.0001 ft

ε1/D1=0.001, ε2/D2=0.00015

QA=5.3 ft3/s

Find Q1, Q2 and PB

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Solution: This is a type 2 problem Step1: Assume Q’1 = 3 ft3/s. Apply the energy

equation along line 1 from A to B to obtain

Assuming that VA=VB. Finding h’f1with Q’1 specified is

a category I problem, and we have

93

g

ghzzKfC

D

Lf

gD

Q8zz

g

gW c

fab1T11

114

1

21

2abc

s 11

s/ft82.3A

QV

1

'1'

1

5

1

1'

1D 10x273.1

DVRe

1

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So f’1=0.0022, and

Step2: The loss for pipe 2 must then become

Finding Q’2 for h’f2=14.97 ft-lbf/lbm is category II

problem. We shall use V’2 as the iteration variable. The results are given in the following table

94

lbm

lbfft97.14

D

Lf

gD

Q8'h

1

1'14

1

21

2f1

lbm

lbfft97.14'h'h

12 ff

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Step3: From step2, Q’2=V’2A2=1.411ft3/s and ΣQ’l=1.141+3=4.141ft3/s. We find the corrected values by using

95

sftQ /84.33.5141.4

3 31

V’2 (ft/s) ReD f’2 h’f 2(ft-lbf/lbm)

1 22,233 0.0265 1.859

4 88,932 0.0192 21.47

3.27 72,702 0.0200 14.94

sftQ /46.13.5141.4

141.1 32

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Step4: Using Q1=3.84ft3/s and Q2=1.46ft3/s, we compute V1, V2, ReD1

, ReD2, f1 and f2, to find

The heat losses in the pipes then agree to 0.54 percent, which is sufficient.

With the flow rates in each line and the head loss for the parallel segments knows, Ws is computed from the first equation in step1.

96

lbm

lbfft85.23h

1f

lbm

lbfft98.23h 2f

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Thus, a pump with an increase in head of 3.91 ft-lbf/lbm would be required for the system to pass 5.3 ft3/s while maintaining PA=PB.

97

lbm

lbfft91.3W

ft91.3ft91.23ft20sft

174.32

slbflbmft

174.32

lbm

lbfft91.23ft100ft80

g

ghzz

g

gW

s

2

2c

fabc

s 1

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on Generalized Parallel Piping System Software

98

21T QQQ

1T1

1

114

1

21

2abc

s KfCD

Lf

gD

Q8zz

g

gW

1

2T2

2

224

2

22

2abc

s KfCD

Lf

gD

Q8zz

g

gW

2

21 QQ3.5

1

114

1

21

2c

s D

Lf

gD

Q820

g

gW

2

224

2

22

2c

s D

Lf

gD

Q820

g

gW

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on Series-Parallel Network

99

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In a series-parallel pipe network as shown in previous figure, we must apply rules which are analogous to the analysis of an electric circuit. In previous figure only shows the pipe network in 2-Dimensions. In reality, a pipe network is most often three dimensional. Thus, the elevations of each nodal point need to be considered when writing the extended Bernoulli equation. The following rules apply in any network of pipes:

100

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Application of these rules leads to a complex set of equations which must be solved numerically. These are easily dealt with in most mathematical/numerical analysis programs. However, a method of hand calculation know as the Hardy-Cross Method may also be applied. This method is the basis for most computer software developed for analyzing piping systems.

101

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on Hardy-Cross Method

The Hardy-Cross formulation is an iterative method for obtaining the steady-state solution for any generalized series-parallel flow network. Its great advantage is systematicness.

The Hardy-Cross method can be systematically applied to any fluid flow network, and if the guidelines are followed, a converged solution will always be obtained.

The basis for any Hardy-Cross analysis technique is the same as for series-parallel flow network analysis (1) conservation of mass at a node and (2) uniqueness of the pressure at a given point in the loop.

102

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A modified version of the head-loss of the Darcy-Weishbach expression called Hazen-Williams expression

Where K and n are determined either by experiment or by curve fits using the Moody diagram.

C is called the Hazen-Williams coefficient

nf KQh

852.18704.4852.1

1f Q

DC

Lkh

852.1nand,DC

LkK

8704.4852.11

103

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Table 1-5 Hazen-Williams Coefficients

Types of pipe C

Extremely smooth and straight pipes 140

New, smooth cast iron pipes 130

Average cast iron, new riveted steel pipes 110

Vitrified sewer pipes 110

Cast iron pipes, some years in service 100

Cast iron, in bad condition 80

104

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Table 1-6 values of k1 for Different Units

Units of Q k1

CFS (ft3/s) 4.727

MGD (million gals/day) 10.63

CMS (m3/s) 10.466

105

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106

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on Example 1-8

Use the Hardy-Cross method to obtain the flow rates in each lines of the network as follow. (C=130)

3 cfs

1 cfs

2 cfs

4000’

6”

3000’

6”

2000’

12”1000

’8”

3000’

8”

2000’

8”

2000’

8”

107

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on Example 1-8

Solution: Step 1:We begin by dividing the network into loops

and numbering all pipes and nodes in each loop.

The system has 6 nodes (s=6) and 7 lines (r=7)

3 cfs

1 cfs

2 cfs

A

F

B

E

D

C

3

6

2

7

4

5

1

Loop 1

Loop 2

108

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on Example 1-8

Steps 2 and 3:Determine the zeroth estimate for the flow rate and obtain the flow rates for all lines in such manner that conservation of mass is satisfied at each node.

To start the process, we shall assume that Q5=1.0cfs. Then visiting each node in turn, we obtain:

Node B:Q5=1.0 cfs

Q6=1.0 cfs

Node C:

Q6=1.0 cfs

Q7=-1.0 cfs

2 cfs

Node D:

Q1=?

Q2=?

Q7=-1.0 cfs

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on Example 1-8

Here we make the second assumption, Q1=-0.8cfs.

Node F:Q5=1.0 cfs

Q3=-1.2 cfs

Node E:

Q3=-1.2 cfsQ2=-0.2 cfs

1 cfs

Node D:

Q1=-0.8cfs

Q2=-0.2cfs

Q7=-1.0 cfs

Q4=-1.2 cfs

Node A:

Q1=0.8cfs

Q4=-1.2cfs

Q5=1.0 cfs

3 cfs

110

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on Example 1-8

Steps 4 and 5: Now determine a correction factor ΔQi for each loop

Where J is the number of line in the loop and Kj equals the constant for jth line.

After obtaining ΔQi for each loop, obtain algebraically a new value for the flow rate in each line; that is

J

1jJ

1j

1n0jj

1n0j

0jj

i

QKn

QQKQ

i0j

1j QQQ

111

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on Example 1-8

Converged solution

3 cfs

1 cfs

2 cfs

Q1=1.8596 cfs

Q3=-0.2431 cfs

Q6=0.8973 cfs

Q4=-0.2431 cfs

Q5=0.8973 cfs

Q2=0.7569 cfs

Q7=-1.1027 cfs

112

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on Generalized Hardy-Cross Analysis

The Hardy-Cross analysis has been restricted to flow networks in which the pipe wall friction represented the only loss (i.e., minor loss has been neglected).

If the line lengths are short enough so that minor losses are important, the equivalent-length approach can be used to include the losses due to fittings.

The equivalent length is additional length of pipe needed to give the same head loss (or pressure drop) as a fitting at a giving flow rate. Hence the equivalent length Leq is obtained by equating the loss-coefficient expression to conventional head-loss expression, i.e.,

c

2eq

c

2

g2

V

D

Lf

g2

VK

f

DKLeq

113

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The generalized Hardy-Cross analysis can be used for piping network in which the lines can contain devices that result either in additional pressure drop (a heat exchanger or turbine, for example) or in a pressure increase ( a pump, for example).

Device

Lj

Dj

Qj

114

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Consider a typical line, line j with length Lj, and some device in the line that causes either a head decrease or increase.

In general, the change in head hfD across the device

will depend on the flow rate

The functional dependence can be represented by the polynomial expression

)Q(gh jD j

M

1m

mjm0jD QBBh

j

115

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Where M represents the degree of polynomial. The Bjm’s can have positive, negative or zero values depending of the particular device being described.

The head loss through a fitting is typically described by

The coefficients in the polynomial expression then take the values

2j

2j

cj

c

2j

jD A

Q

g2

1K

g2

VKh

j

2jc

j2j1j0j Ag2

1KB,0B,0B,2M

116

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Since this represents a positive head loss for loop flow in the positive flow direction, Bj2>0. If an ideal backward curved blade centrifugal pump is used, the representation is

Note that hDj must be less than zero, because a pump

represents a negative head loss (i.e., an increase in head).

j1j0jD QBBhj

J

1j

M

1m

1m

jjm

1n

jj

J

1j

M

1m

1m

jjjm0jj

1n

jjj

i

QmBQnK

QQBB)Qsgn(QQK

Q

117

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Where sgn(Qj)=1 When Qj>0 and sgn(Qj)=-1 When Qj<0.

118

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119

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on Example 1-9

For the network of Example 1-8, investigate the effects of adding

a) A pump with hD =-(50-0.4Q) ft-lbf/lbm to line 6

b) A heat exchanger with hD =50Q2 ft-lbf/lbm to line 6

c) A very large pump with hD =-1000 ft-lbf/lbm to line 6

120

0.0B,4.0B,0.50B 626160

50B,0B,0B 626160

0B,0B,1000B 626160

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on Example 1-9

a) A pump with hD =-(50-0.4Q) ft-lbf/lbm to line 6

3 cfs

1 cfs

2 cfs

Q1=0.963 cfs

Q3=-0.208 cfs

Q6=1.828 cfs

Q4=-0.208 cfs

Q5=1.828 cfs

Q2=0.792 cfs

Q7=-0.172 cfs

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b) A heat exchanger with hD =50Q2 ft-lbf/lbm to line 6

3 cfs

1 cfs

2 cfs

Q1=0.963 cfs

Q3=-0.259 cfs

Q6=0.558 cfs

Q4=-0.259 cfs

Q5=0.558 cfs

Q2=0.741 cfs

Q7=1.442 cfs

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c) A very large pump with hD =-1000 ft-lbf/lbm to line 6

3 cfs

1 cfs

2 cfs

Q1=4.658 cfs

Q3=0.230 cfs

Q6=7.888 cfs

Q4=0.230 cfs

Q5=7.888 cfs

Q2=1.230 cfs

Q7=5.888 cfs

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If the fluid is not water, the procedure outlined in the previous slides can be used, but a more direct approach in to develop the major losses in a friction-factor form rather than a Hazen-William form

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Economic considerations are important in most energy systems, and estimating the cost of a project in an integral part of any bidding procedure.

www.RSMeans.com Typically values for the purchase and installation of

schedule 40 commercial steel piping are as follows:

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Nominal diameter (in.) Cost per foot ($)

1 11

2 20

3 30

4 40

5 55

6 70

8 90

10 130

12 160

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The values do not include the cost of purchase of land, trenching, backfilling or disposal of old (replaced) systems.

Pump cost are more varied, but an order-of-magnitude estimate is given by the following expression:

Where hp is the power output of the pump and the pump cost is in dollars.

The energy charge is based on the kW-h (kilowatt-hour. The cost of kW-h varies $0.02 for large industrial costumers to as such as $0.15.

Typical value $8-10 per kW

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5.0130/hp6000tcosPump

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The best way to select an appropriate type of pipeline is to compare the inside diameter of various pipe materials that might be part of the design of a project.

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DIPS: Ductile Iron PipePCCP: Prestressed Concrete Cylinder PipeHDPE: High Density Polyethylene

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For a 24-inch nominal diameter water transmission pipeline that is 30,000 feet long with water flowing at 6,000 gpm, the cost is $0.06/kW-h, 70 percent of pumping efficiency and the pump will operate 24 hours per day. Calculate the pumping Costs of all type of pipes.

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Due, Wednesday, March ??, 2011

Omar E. Meza Castillo Ph.D.

Homework5 Webpage

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Documents

Fluid Mechanics and Applications Inter - Bayamon Lecture 4 Fluid Mechanics and Applications MECN 3110 Inter American University of Puerto Rico Professor: