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TL2101 Mekanika Fluida I

- Benno Rahardyan

Pertemuan

(No Transcript)

Pipes are Everywhere!

Owner City of Hammond, INProject Water Main

RelocationPipe Size 54"

Pipes are Everywhere!Drainage Pipes

Pipes

Pipes are Everywhere!Water Mains

Types of Engineering Problems

- How big does the pipe have to be to carry a flow

of x m3/s? - What will the pressure in the water distribution

system be when a fire hydrant is open?

FLUID DYNAMICSTHE BERNOULLI EQUATION

The laws of Statics that we have learned cannot

solve Dynamic Problems. There is no way to solve

for the flow rate, or Q. Therefore, we need a new

dynamic approach to Fluid Mechanics.

The Bernoulli Equation

By assuming that fluid motion is governed only by

pressure and gravity forces, applying Newtons

second law, F ma, leads us to the Bernoulli

Equation. P/g V2/2g z constant along a

streamline (Ppressure g

specific weight Vvelocity ggravity

zelevation) A streamline is the path of one

particle of water. Therefore, at any two points

along a streamline, the Bernoulli equation can be

applied and, using a set of engineering

assumptions, unknown flows and pressures can

easily be solved for.

Free Jets

The velocity of a jet of water is clearly related

to the depth of water above the hole. The

greater the depth, the higher the velocity.

Similar behavior can be seen as water flows at a

very high velocity from the reservoir behind the

Glen Canyon Dam in Colorado

Closed Conduit Flow

- Energy equation
- EGL and HGL
- Head loss
- major losses
- minor losses
- Non circular conduits

The Energy Line and the Hydraulic Grade Line

Looking at the Bernoulli equation again P/?

V2/2g z constant on a streamline

This constant is called the total

head (energy), H Because energy is assumed to be

conserved, at any point along the streamline, the

total head is always constant Each term in the

Bernoulli equation is a type of head. P/?

Pressure Head V2/2g Velocity Head Z elevation

head These three heads, summed together, will

always equal H Next we will look at this

graphically

Conservation of Energy

- Kinetic, potential, and thermal energy

head supplied by a pump

hp

head given to a turbine

ht

mechanical energy converted to thermal

hL

downstream

Cross section 2 is ____________ from cross

section 1!

irreversible

Point to point or control volume?

Why a? _____________________________________

Energy Equation Assumptions

hydrostatic

- Pressure is _________ in both cross sections
- pressure changes are due to elevation only
- section is drawn perpendicular to the streamlines

(otherwise the _______ energy term is incorrect) - Constant ________at the cross section
- _______ flow

kinetic

density

Steady

EGL (or TEL) and HGL

pressure head (w.r.t. reference pressure)

velocity head

downward

- The energy grade line must always slope

___________ (in direction of flow) unless energy

is added (pump) - The decrease in total energy represents the head

loss or energy dissipation per unit weight - EGL and HGL are coincident and lie at the free

surface for water at rest (reservoir) - If the HGL falls below the point in the system

for which it is plotted, the local pressures are

_____ ____ __________ ______

lower than reference pressure

Energy equation

Energy Grade Line

velocity head

Hydraulic G L

static head

pressure head

Why is static head important?

elevation

z

pump

z 0

datum

The Energy Line and the Hydraulic Grade Line

Lets first understand this drawing

Measures the Total Head

1 Static Pressure Tap Measures the sum of the

elevation head and the pressure Head. 2 Pilot

Tube Measures the Total Head EL Energy

Line Total Head along a system HGL Hydraulic

Grade line Sum of the elevation and the pressure

heads along a system

Measures the Static Pressure

1

2

1

2

EL

V2/2g

HGL

Q

P/?

Z

The Energy Line and the Hydraulic Grade Line

Understanding the graphical approach of Energy

Line and the Hydraulic Grade line is key to

understanding what forces are supplying the

energy that water holds.

Point 1 Majority of energy stored in the water

is in the Pressure Head Point 2 Majority of

energy stored in the water is in the elevation

head If the tube was symmetrical, then the

velocity would be constant, and the HGL would be

level

EL

V2/2g

V2/2g

HGL

P/?

2

Q

P/?

Z

1

Z

Bernoulli Equation Assumption

- _________ (viscosity cant be a significant

parameter!) - Along a __________
- ______ flow
- Constant ________
- No pumps, turbines, or head loss

Frictionless

streamline

Steady

density

Why no a? ____________

point velocity

Does direction matter? ____

no

Useful when head loss is small

Pipe Flow Review

- We have the control volume energy equation for

pipe flow. - We need to be able to predict the relationship

between head loss and flow. - How do we get this relationship? __________

_______.

dimensional analysis

Example Pipe Flow Problem

cs1

D20 cm L500 m

Find the discharge, Q.

100 m

valve

cs2

Describe the process in terms of energy!

Flow Profile for Delaware Aqueduct

Rondout Reservoir (EL. 256 m)

70.5 km

West Branch Reservoir (EL. 153.4 m)

Sea Level

(Designed for 39 m3/s)

Need a relationship between flow rate and head

loss

Ratio of Forces

- Create ratios of the various forces
- The magnitude of the ratio will tell us which

forces are most important and which forces could

be ignored - Which force shall we use to create the ratios?

Inertia as our Reference Force

- Fma
- Fluids problems (except for statics) include a

velocity (V), a dimension of flow (l), and a

density (r) - Substitute V, l, r for the dimensions MLT
- Substitute for the dimensions of specific force

Dimensionless Parameters

- Reynolds Number
- Froude Number
- Weber Number
- Mach Number
- Pressure/Drag Coefficients
- (dependent parameters that we measure

experimentally)

Problem solving approach

- Identify relevant forces and any other relevant

parameters - If inertia is a relevant force, than the non

dimensional Re, Fr, W, M, Cp numbers can be used - If inertia isnt relevant than create new non

dimensional force numbers using the relevant

forces - Create additional non dimensional terms based on

geometry, velocity, or density if there are

repeating parameters - If the problem uses different repeating variables

then substitute (for example wd instead of V) - Write the functional relationship

Friction Factor Major losses

- Laminar flow
- Hagen-Poiseuille
- Turbulent (Smooth, Transition, Rough)
- Colebrook Formula
- Moody diagram
- Swamee-Jain

Laminar Flow Friction Factor

Hagen-Poiseuille

Darcy-Weisbach

Pipe Flow Dimensional Analysis

- What are the important forces?______,

______,________. Therefore ________number and

_______________ . - What are the important geometric parameters?

_________________________ - Create dimensionless geometric groups______,

______ - Write the functional relationship

viscous

pressure

Inertial

Pressure coefficient

Reynolds

diameter, length, roughness height

l/D

e/D

Other repeating parameters?

Dimensional Analysis

- How will the results of dimensional analysis

guide our experiments to determine the

relationships that govern pipe flow? - If we hold the other two dimensionless parameters

constant and increase the length to diameter

ratio, how will Cp change?

Cp proportional to l

f is friction factor

Laminar Flow Friction Factor

Hagen-Poiseuille

Darcy-Weisbach

-1

Slope of ___ on log-log plot

Viscous Flow in Pipes

Viscous Flow Dimensional Analysis

- Two important parameters!
- R - Laminar or Turbulent
- e/D - Rough or Smooth

Where

and

Laminar and Turbulent Flows

- Reynolds apparatus

inertia

damping

Transition at R of 2000

Boundary layer growth Transition length

What does the water near the pipeline wall

experience? _________________________ Why does

the water in the center of the pipeline speed up?

_________________________

Drag or shear

Conservation of mass

Pipe Entrance

Non-Uniform Flow

Need equation for entrance length here

Images - Laminar/Turbulent Flows

Laser - induced florescence image of an

incompressible turbulent boundary layer

Laminar flow (Blood Flow)

Simulation of turbulent flow coming out of a

tailpipe

Laminar flow

Turbulent flow

http//www.engineering.uiowa.edu/cfd/gallery/lim-

turb.html

Laminar, Incompressible, Steady, Uniform Flow

- Between Parallel Plates
- Through circular tubes
- Hagen-Poiseuille Equation
- Approach
- Because it is laminar flow the shear forces can

be quantified - Velocity profiles can be determined from a force

balance

Laminar Flow through Circular Tubes

- Different geometry, same equation development

(see Streeter, et al. p 268) - Apply equation of motion to cylindrical sleeve

(use cylindrical coordinates)

Laminar Flow through Circular Tubes Equations

a is radius of the tube

Max velocity when r 0

Velocity distribution is paraboloid of revolution

therefore _____________ _____________

average velocity (V) is 1/2 umax

Vpa2

Q VA

Laminar Flow through Circular Tubes Diagram

Laminar flow

Shear at the wall

True for Laminar or Turbulent flow

Laminar flowContinue

- Momentum is
- Massvelocity (mv)
- Momentum per unit volume is
- ?vz
- Rate of flow of momentum is
- ?vzdQ
- dQvz2prdr
- but
- vz constant at a fixed value of r

Laminar flow

Laminar flowContinue

Hagen-Poiseuille

The Hagen-Poiseuille Equation

cv pipe flow

Constant cross section

h or z

Laminar pipe flow equations

(No Transcript)

- Prof. Dr. Ir. Bambang Triatmodjo, CES-UGM
- Hidraulika I, Beta Ofset Yogyakarta, 1993
- Hidraulika II, Beta Ofset Yogyakarta, 1993
- Soal-Penyelesaian Hidraulika I, 1994
- Soal-Penyelesaian Hidraulika II, 1995

- Air mengalir melalui pipa berdiameter 150 mm dan

kecepatan 5,5 m/det.Kekentalan kinematik air

adalah 1,3 x 10-4 m2/det. Selidiki tipe aliran

- Minyak di pompa melalui pipa sepanjang 4000 m dan

diameter 30 cm dari titik A ke titik B. Titik B

terbuka ke udara luar. Elevasi titik B adalah 50

di atas titik A. Debit 40 l/det. Debit aliran 40

l/det. Rapat relatif S0,9 dan kekentalan

kinematik 2,1 x 10-4 m2/det. Hitung tekanan di

titik A.

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- Minyak dipompa melalui pipa berdiameter 25 cm dan

panjang 10 km dengan debit aliran 0,02 m3/dtk.

Pipa terletak miring dengan kemiringan 1200.

Rapat minyak S0,9 dan keketnalan kinematik

v2,1x 10-4 m2/det. Apabila tekanan pada ujung

atas adalah p10 kPA ditanyakan tekanan di ujung

bawah.

(No Transcript)

Turbulent Pipe and Channel Flow Overview

- Velocity distributions
- Energy Losses
- Steady Incompressible Flow through Simple Pipes
- Steady Uniform Flow in Open Channels

Turbulence

- A characteristic of the flow.
- How can we characterize turbulence?
- intensity of the velocity fluctuations
- size of the fluctuations (length scale)

mean velocity

instantaneous velocity

velocity fluctuation

t

Turbulent flow

- When fluid flow at higher flowrates, the

streamlines are not steady and straight and the

flow is not laminar. Generally, the flow field

will vary in both space and time with

fluctuations that comprise "turbulence - For this case almost all terms in the

Navier-Stokes equations are important and there

is no simple solution - ?P ?P (D, ?, ?, L, U,)

Turbulent flow

All previous parameters involved three

fundamental dimensions, Mass, length, and time

From these parameters, three dimensionless

groups can be build

Turbulence Size of the Fluctuations or Eddies

- Eddies must be smaller than the physical

dimension of the flow - Generally the largest eddies are of similar size

to the smallest dimension of the flow - Examples of turbulence length scales
- rivers ________________
- pipes _________________
- lakes ____________________
- Actually a spectrum of eddy sizes

depth (R 500)

diameter (R 2000)

depth to thermocline

Turbulence Flow Instability

- In turbulent flow (high Reynolds number) the

force leading to stability (_________) is small

relative to the force leading to instability

(_______). - Any disturbance in the flow results in large

scale motions superimposed on the mean flow. - Some of the kinetic energy of the flow is

transferred to these large scale motions

(eddies). - Large scale instabilities gradually lose kinetic

energy to smaller scale motions. - The kinetic energy of the smallest eddies is

dissipated by viscous resistance and turned into

heat. (___________)

viscosity

inertia

head loss

Velocity Distributions

- Turbulence causes transfer of momentum from

center of pipe to fluid closer to the pipe wall. - Mixing of fluid (transfer of momentum) causes the

central region of the pipe to have relatively

_______velocity (compared to laminar flow) - Close to the pipe wall eddies are smaller (size

proportional to distance to the boundary)

constant

Turbulent Flow Velocity Profile

Turbulent shear is from momentum transfer

h eddy viscosity

Length scale and velocity of large eddies

Dimensional analysis

y

Turbulent Flow Velocity Profile

increases

Size of the eddies __________ as we move further

from the wall.

k 0.4 (from experiments)

Log Law for Turbulent, Established Flow, Velocity

Profiles

Integration and empirical results

Turbulent

Laminar

Shear velocity

y

x

Pipe Flow The Problem

- We have the control volume energy equation for

pipe flow - We need to be able to predict the head loss term.
- We will use the results we obtained using

dimensional analysis

Friction Factor Major losses

- Laminar flow
- Hagen-Poiseuille
- Turbulent (Smooth, Transition, Rough)
- Colebrook Formula
- Moody diagram
- Swamee-Jain

Turbulent Pipe Flow Head Loss

Proportional

- ___________ to the length of the pipe
- Proportional to the _______ of the velocity

(almost) - ________ with surface roughness
- Is a function of density and viscosity
- Is __________ of pressure

square

Increases

independent

Smooth, Transition, Rough Turbulent Flow

- Hydraulically smooth pipe law (von Karman, 1930)
- Rough pipe law (von Karman, 1930)
- Transition function for both smooth and rough

pipe laws (Colebrook)

(used to draw the Moody diagram)

Pipe Flow Energy Losses

Horizontal pipe

Dimensional Analysis

Darcy-Weisbach equation

Turbulent Pipe Flow Head Loss

Proportional

- ___________ to the length of the pipe
- ___________ to the square of the velocity

(almost) - ________ with the diameter (almost)
- ________ with surface roughness
- Is a function of density and viscosity
- Is __________ of pressure

Proportional

Inversely

Increase

independent

Surface Roughness

Additional dimensionless group ?/D need to be

characterize

Thus more than one curve on friction

factor-Reynolds number plot

Fanning diagram or Moody diagram Depending on the

laminar region. If, at the lowest Reynolds

numbers, the laminar portion corresponds to f

16/Re Fanning Chart or f 64/Re Moody chart

Friction Factor for Smooth, Transition, and Rough

Turbulent flow

Smooth, Transition, Rough Turbulent Flow

- Hydraulically smooth pipe law (von Karman, 1930)
- Rough pipe law (von Karman, 1930)
- Transition function for both smooth and rough

pipe laws (Colebrook)

(used to draw the Moody diagram)

Moody Diagram

0.10

0.08

0.05

0.04

0.06

0.03

0.05

0.02

0.015

0.04

0.01

0.008

friction factor

0.006

0.03

0.004

laminar

0.002

0.02

0.001

0.0008

0.0004

0.0002

0.0001

0.00005

0.01

smooth

1E03

1E04

1E05

1E06

1E07

1E08

R

Fanning Diagram

f 16/Re

Swamee-Jain

- 1976
- limitations
- ?/D lt 2 x 10-2
- Re gt3 x 103
- less than 3 deviation from results obtained with

Moody diagram - easy to program for computer or calculator use

no f

L

hf

Each equation has two terms. Why?

Colebrook Solution for Q

Colebrook Solution for Q

Swamee D?

Pipe roughness

pipe material

pipe roughness

?

(mm)

glass, drawn brass, copper

0.0015

commercial steel or wrought iron

0.045

asphalted cast iron

0.12

galvanized iron

0.15

cast iron

0.26

concrete

0.18-0.6

rivet steel

0.9-9.0

corrugated metal

45

0.12

PVC

Solution Techniques

- find head loss given (D, type of pipe, Q)

- find flow rate given (head, D, L, type of pipe)

- find pipe size given (head, type of pipe,L, Q)

Exponential Friction Formulas

- Commonly used in commercial and industrial

settings - Only applicable over _____ __ ____ collected
- Hazen-Williams exponential friction formula

range of data

C Hazen-Williams coefficient

Head lossHazen-Williams Coefficient

- C Condition
- 150 PVC
- 140 Extremely smooth, straight pipes asbestos

cement - 130 Very smooth pipes concrete new cast iron
- 120 Wood stave new welded steel
- 110 Vitrified clay new riveted steel
- 100 Cast iron after years of use
- 95 Riveted steel after years of use
- 60-80 Old pipes in bad condition

Hazen-Williams vs Darcy-Weisbach

- Both equations are empirical
- Darcy-Weisbach is dimensionally correct, and

________. - Hazen-Williams can be considered valid only over

the range of gathered data. - Hazen-Williams cant be extended to other fluids

without further experimentation.

preferred

Non-Circular ConduitsHydraulic Radius Concept

- A is cross sectional area
- P is wetted perimeter
- Rh is the Hydraulic Radius (Area/Perimeter)
- Dont confuse with radius!

For a pipe

We can use Moody diagram or Swamee-Jain with D

4Rh!

Pipe Flow Summary (1)

- Shear increases _________ with distance from the

center of the pipe (for both laminar and

turbulent flow) - Laminar flow losses and velocity distributions

can be derived based on momentum and energy

conservation - Turbulent flow losses and velocity distributions

require ___________ results

linearly

experimental

Pipe Flow Summary (2)

- Energy equation left us with the elusive head

loss term - Dimensional analysis gave us the form of the head

loss term (pressure coefficient) - Experiments gave us the relationship between the

pressure coefficient and the geometric parameters

and the Reynolds number (results summarized on

Moody diagram)

Questions

- Can the Darcy-Weisbach equation and Moody Diagram

be used for fluids other than water? _____

Yes

- What about the Hazen-Williams equation? ___

No

- Does a perfectly smooth pipe have head loss? _____

Yes

- Is it possible to decrease the head loss in a

pipe by installing a smooth liner? ______

Yes

Darcy Weisbach

Major and Minor Losses

Major Losses Hmaj f x (L/D)(V2/2g) f

friction factor L pipe length D pipe

diameter V 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 point P1/? V12/2g z1

P2/? V22/2g z2 Hmaj Hmin

- Hitung kehilangan tenaga karena gesekan di dalam

pipa sepanjang 1500 m dan diameter 20 cm, apabila

air mengalir dengan kecepatan 2 m/det. Koefisien

gesekan f0,02 - Penyelesaian
- Panjang pipa L 1500 m
- Diameter pipa D 20 cm 0,2 m
- Kecepatan aliran V 2 m/dtk
- Koefisien gesekan f 0,02

- Air melalui pipa sepanjang 1000 m dan diameternya

150 mm dengan debit 50 l/det. Hitung kehilangan

tenaga karenagesekan apabila koefisien gesekan f

0,02 - Penyelesaian
- Panjang pipa L 1000 m
- Diameter pipa D 0,15 m
- Debit aliran Q 50 liter/detik
- Koefisien gesekan f 0,02

- Hitung kehilangan tenaga karena gesekan di dalam

pipa sepanjang 1500 m dan diameter 20 cm, apabila

air mengalir dengan kecepatan 2 m/det. Koefisien

gesekan f0,02 - Penyelesaian
- Panjang pipa L 1500 m
- Diameter pipa D 20 cm 0,2 m
- Kecepatan aliran V 2 m/dtk
- Koefisien gesekan f 0,02

- Air melalui pipa sepanjang 1000 m dan diameternya

150 mm dengan debit 50 l/det. Hitung kehilangan

tenaga karenagesekan apabila koefisien gesekan f

0,02 - Penyelesaian
- Panjang pipa L 1000 m
- Diameter pipa D 0,15 m
- Debit aliran Q 50 liter/detik
- Koefisien gesekan f 0,02

Example

Solve for the Pressure Head, Velocity Head, and

Elevation Head at each point, and then plot the

Energy Line and the Hydraulic Grade Line

Assumptions and Hints P1 and P4 0 --- V3 V4

same diameter tube We must work backwards to

solve this problem

1

?H2O 62.4 lbs/ft3

R .5

4

R .25

2

3

4

1

Point 1 Pressure Head Only atmospheric ? P1/?

0 Velocity Head In a large tank, V1 0 ?

V12/2g 0 Elevation Head Z1 4

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

Point 4 Apply the Bernoulli equation between 1

and 4 0 0 4 0

V42/2(32.2) 1 V4 13.9 ft/s Pressure Head

Only atmospheric ? P4/? 0 Velocity Head

V42/2g 3 Elevation Head Z4 1

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

Point 3 Apply the Bernoulli equation between 3

and 4 (V3V4) P3/62.4

3 1 0 3 1 P3 0 Pressure Head P3/?

0 Velocity Head V32/2g 3 Elevation Head Z3

1

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

Point 2 Apply the Bernoulli equation between 2

and 3 P2/62.4 V22/2(32.2)

1 0 3 1 Apply the Continuity

Equation (?.52)V2 (?.252)x13.9 ? V2 3.475

ft/s P2/62.4 3.4752/2(32.2) 1 4 ? P2

175.5 lbs/ft2

Pressure Head P2/? 2.81 Velocity Head

V22/2g .19 Elevation Head Z2 1

1

?H2O 62.4 lbs/ft3

4

R .5

R .25

2

3

4

1

Plotting the EL and HGL

Energy Line Sum of the Pressure, Velocity and

Elevation heads Hydraulic Grade Line Sum of the

Pressure and Velocity heads

V2/2g.19

EL

P/? 2.81

V2/2g3

V2/2g3

Z4

HGL

Z1

Z1

Z1

Pipe Flow and the Energy Equation

For 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/? V12/2g z1 P2/? V22/2g z2

Hmaj Hmin Major losses Hmaj 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 as a friction

factor, f, associated with it.

Energy line with no losses

Hmaj

Energy line with major losses

1

2

Pipe Flow and the Energy Equation

Minor Losses Hmin Momentum losses in Pipe

diameter changes and in pipe bends are called

minor losses. Unlike 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

Minor Losses

- We previously obtained losses through an

expansion using conservation of energy, momentum,

and mass - Most minor losses can not be obtained

analytically, so they must be measured - Minor losses are often expressed as a loss

coefficient, K, times the velocity head.

High R

Head Loss Minor Losses

- Head loss due to outlet, inlet, bends, elbows,

valves, pipe size changes - Flow expansions have high losses
- Kinetic energy decreases across expansion
- Kinetic energy ? ________ and _________ energy
- Examples ________________________________

__________________________________________ - Losses can be minimized by gradual transitions

potential

thermal

Hydraulic jump

Vehicle drag

Minor losses!

Vena contracta

Minor Losses

- Most minor losses can not be obtained

analytically, so they must be measured - Minor losses are often expressed as a loss

coefficient, K, times the velocity head.

High Re

Head Loss due to Gradual Expansion (Diffusor)

Sudden Contraction

V2

V1

flow separation

- losses are reduced with a gradual contraction

Sudden Contraction

Entrance Losses

- Losses can be reduced by accelerating the flow

gradually and eliminating the

vena contracta

Head Loss in Bends

High pressure

- Head loss is a function of the ratio of the bend

radius to the pipe diameter (R/D) - Velocity distribution returns to normal several

pipe diameters downstream

Possible separation from wall

R

D

Low pressure

Kb varies from 0.6 - 0.9

Head Loss in Valves

- Function of valve type and valve position
- The complex flow path through valves can result

in high head loss (of course, one of the purposes

of a valve is to create head loss when it is not

fully open)

Solution Techniques

- Neglect minor losses
- Equivalent pipe lengths
- Iterative Techniques
- Simultaneous Equations
- Pipe Network Software

Iterative Techniques for D and Q (given total

head loss)

- Assume all head loss is major head loss.
- Calculate D or Q using Swamee-Jain equations
- Calculate minor losses
- Find new major losses by subtracting minor losses

from total head loss

Solution Technique Head Loss

- Can be solved directly

Solution TechniqueDischarge or Pipe Diameter

- Iterative technique
- Set up simultaneous equations in Excel

Use goal seek or Solver to find discharge that

makes the calculated head loss equal the given

head loss.

Example Minor and Major Losses

- Find the maximum dependable flow between the

reservoirs for a water temperature range of 4ºC

to 20ºC.

25 m elevation difference in reservoir water

levels

Water

Reentrant pipes at reservoirs

Standard elbows

2500 m of 8 PVC pipe

Sudden contraction

Gate valve wide open

1500 m of 6 PVC pipe

Directions

- Assume fully turbulent (rough pipe law)
- find f from Moody (or from von Karman)
- Find total head loss
- Solve for Q using symbols (must include minor

losses) (no iteration required) - Obtain values for minor losses from notes or text

Example (Continued)

- What are the Reynolds number in the two pipes?
- Where are we on the Moody Diagram?
- What value of K would the valve have to produce

to reduce the discharge by 50? - What is the effect of temperature?
- Why is the effect of temperature so small?

Example (Continued)

- Were the minor losses negligible?
- Accuracy of head loss calculations?
- What happens if the roughness increases by a

factor of 10? - If you needed to increase the flow by 30 what

could you do? - Suppose I changed 6 pipe, what is minimum

diameter needed?

Pipe Flow Summary (3)

- Dimensionally correct equations fit to the

empirical results can be incorporated into

computer or calculator solution techniques - Minor losses are obtained from the pressure

coefficient based on the fact that the pressure

coefficient is _______ at high Reynolds numbers - Solutions for discharge or pipe diameter often

require iterative or computer solutions

constant

Loss Coefficients

Use this table

to find loss coefficients

Head Loss due to Sudden ExpansionConservation

of Energy

z1 z2

What is p1 - p2?

Head Loss due to Sudden ExpansionConservation

of Momentum

A2

A1

x

1

2

Apply in direction of flow

Neglect surface shear

Pressure is applied over all of section

1. Momentum is transferred over area

corresponding to upstream pipe diameter. V1 is

velocity upstream.

Divide by (A2 g)

Head Loss due to Sudden Expansion

Mass

Energy

Momentum

Contraction

EGL

HGL

Expansion!!!

V1

V2

vena contracta

- losses are reduced with a gradual contraction

Questions

- In the rough pipe law region if the flow rate is

doubled (be as specific as possible) - What happens to the major head loss?
- What happens to the minor head loss?
- Why do contractions have energy loss?
- If you wanted to compare the importance of minor

vs. major losses for a specific pipeline, what

dimensionless terms could you compare?

Entrance Losses

reentrant

- Losses can be reduced by accelerating the flow

gradually and eliminating the vena contracta

Head Loss in Valves

- Function of valve type and valve position
- The complex flow path through valves often

results in high head loss - What is the maximum value that Kv can have? _____

?

How can K be greater than 1?

Questions

EGL

HGL

- What is the head loss when a pipe enters a

reservoir? - Draw the EGL and HGL

Example

cs1

100 m

valve

cs2

D40 cm L1000 m

D20 cm L500 m

Find the discharge, Q. What additional

information do you need? Apply energy

equation How could you get a quick estimate?

_________________ Or spreadsheet solution find

head loss as function of Q.

Use S-J on small pipe

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

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.

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

Apply Bernoullis equation between points 1 and

2Assumptions P1 P2 Atmospheric 0 V1

V2 0 (large tank) 0 0 Z1 0 0 130m

Hmaj Hmin Hmaj (fxLxV2)/(Dx2g)(.035 x 197m x

(1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj 5.85m

Pipe Flow Example

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

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

?oil 8.82 kN/m3 f .035

1

Z1 ?

2

Z2 130 m

60 m

Kout1

7 m

r/D 0

130 m

r/D 2

0 0 Z1 0 0 130m Hmaj Hmin 0 0

Z1 0 0 130m 5.85m 0.24m Z1 136.09

meters

Pipa ekivalen

- Digunakan untuk menyederhanakan sistem yang

ditinjau - Ciri khasnya adalah memiliki keserupaan hidrolis

dengan kondisi nyatanya ? Q, hf sama - Pipa ekivalen dapat dinyatakan melalui ekivalensi

l,D,f

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