civil 253 mechanics of fluids and hydraulics
Download
Skip this Video
Download Presentation
MECHANICS OF FLUIDS AND HYDRAULICS

Loading in 2 Seconds...

play fullscreen
1 / 156

MECHANICS OF FLUIDS AND HYDRAULICS - PowerPoint PPT Presentation


  • 79 Views
  • Uploaded on

Fluid mechanics for civil engineering students

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'MECHANICS OF FLUIDS AND HYDRAULICS' - zdadach


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
what is fluid mechanics
WHAT IS FLUID MECHANICS?
  • FLUID MECHANICS AND HYDRAULICS DEAL WITH THE BEHAVIOR OF FLUIDS:

 AT REST ( FLUID STATIC)

 IN MOTION ( FLUID DYNAMIC)

what is fluid
WHAT IS FLUID?
  • FLUIDS ARE SUBSTANCES THAT ARE CAPABLE TO FLOW AND CONFORM TO THE SHAPE OF CONTAINING VESSELS
  • FLUIDS CAN BE GASES OR LIQUIDS
  • GASES ARE COMPRESSIBLE AND LIQUIDS ARE PRACTICALLY INCOMPRESSIBLE ( WHY?)
what is density of fluid
WHAT IS DENSITY OF FLUID?
  • AMOUNT OF MASS PER UNIT VOLUME OF SUBSTANCE:
units of density
UNITS OF DENSITY
  • THE UNITS OF DENSITY ARE KILOGRAMS PER CUBIC METER IN THE SI SYSTEM AND SLUGS PER CUBIC FOOT IN THE US SYSTEM
  • EXAMPLE: DENSITY OF WATER IS 1.94 SLUGS/FT3 ( US) OR 1000 KG/M3 ( SI) AT 40C
what is specific weight
WHAT IS SPECIFIC WEIGHT?
  • IS THE AMOUNT OF WEIGHT PER UNIT VOLUME OF A SUBSTANCE
units of specific weight
UNITS OF SPECIFIC WEIGHT
  • THE UNITS FOR SPECIFIC WEIGHT ARE NEWTONS PER CUBIC METER ( SI) OR POUNDS PER CUBIC FOOT( US)
  • EXAMPLE: SPECIFIC WEIGHT OF WATER AT 00C IS 62.4 LB/FT3 OR 9.81 kN/M3
what is specific gravity
WHAT IS SPECIFIC GRAVITY?
  • IT IS OFTEN CONVENIENT TO INDICATE THE SPECIFIC WEIGHT OR DENSITY OF A FLUID IN TERMS OF ITS RELATIONSHIP TO THE SPECIFIC WEIGHT ORDENSITY OF A COMMON FLUID
  • THE COMMON FLUID USED IS WATER AT 40C.
relationship
γ-ρ RELATIONSHIP
  • Often, the specific weight of a substance must be found when its density is known and vice versa:
example 1
EXAMPLE #1
  • CALCULATE THE WEIGHT OF A RESERVOIR OF OIL IF IT HAS A MASS OF 825 KG

(TAKE g= 9.81 m/s2)

example 2
EXAMPLE #2
  • IF THE RESERVOIR FROM EXAMPLE 1 HAS A VOLUME OF 0.917 M3, COMPUTE THE DENSITY, SPECIFIC WEIGHT AND SPECIFIC GRAVITY OF THE OIL
example 3
EXAMPLE #3
  • GLYCERINE AT 200C HAS A SPECIFIC GRAVITY OF 1.263 . COMPUTE ITS DENSITY AND SPECIFIC WEIGHT
example 4
EXAMPLE #4
  • A PINT OF WATER WEIGHTS 4.632 N , FIND ITS MASS
specific gravity in degre baume
SPECIFIC GRAVITY IN DEGRE BAUME
  • Fluids lighter than water:
  • Fluids heavier than water
specific gravity in degre api
SPECIFIC GRAVITY IN DEGRE API
  • LIQUIDS LIGHTER THAN WATER:
viscosity of fluids1
VISCOSITY OF FLUIDS
  • The ease with which a fluid pours is an indication of its viscosity
  • Oils pours more slowly than water because it has a higher viscosity
  • Cold oil pours more slowly than warm oil because the viscosity increases when temperature decreases
dynamic or absolute viscosity
DYNAMIC OR ABSOLUTE VISCOSITY
  • Figure 2-1 page 27 from book
  • A force F is used to move the upper plate
  • The upper plate moves at velocity v
  • The lower plate is stationary
  • If A is the area of the upper plate, we define the shear stress:

τ= F/A

dynamic or absolute viscosity1
DYNAMIC OR ABSOLUTE VISCOSITY
  • Figure 2-1 page 27 from book

 If the two plates are separated by a distance y then we define the dynamic viscosity η as:

units of dynamic viscosity
UNITS OF DYNAMIC VISCOSITY
  • From the previous equation:

 [η] = N/m2. m/ms-1= N.s/m2

  • Since Pa= N/m2 [η] = Pa.s
  • Since 1N= 1kg.m/s2[η]= kg/m.s
  • Table 2-1 page 28 gives more units in different systems
kinematic viscosity
KINEMATIC VISCOSITY
  • Kinematic viscosity is defined by the relation:
units of kinematic viscosity
UNITS OF KINEMATIC VISCOSITY
  • From the previous equation

[ν] = kg/m.s x m3/kg = m2/s

  • Table 2-2 page 29 gives you other units in different systems
variation of viscosity with temperature
VARIATION OF VISCOSITY WITH TEMPERATURE
  • EXAMPLE: Engine oil is quite difficult to pour when it is cold, indicating high viscosity. As the temperature is increased, the viscosity decreases
  • A measure of how greatly the viscosity of a fluid changes with temperature is given by the

VISCOSITY INDEX OR VI

viscosity index
VISCOSITY INDEX
  • The viscosity index is especially important for lubricating oils and hydraulic fluids.
  • The general form of the equation for calculating the viscosity index is
  • All kinematic viscosities values are in mm2/s
slide28
U= Kinematic viscosity at 400C of the test oil
  • L= Kinematic viscosity at 400C of a standard oil of 0 VI having the same viscosity at 1000C as the test oil
  • H= Kinematic viscosity at 400C of a standard oil of 100 VI having the same viscosity at 1000C as the test oil
viscosity index1
VISCOSITY INDEX
  • A FLUID WITH A HIGH VISCOSITY INDEX EXHIBITS A SMALL CHANGE IN VISCOSITY WITH TEMPERATURE WHILE A FLUID WITH LOW VISCOSITY INDEX EXHIBITS A LARGE CHANGE IN VISCOSITY WITH TEMPERATURE
viscosity index2
VISCOSITY INDEX
  • Table page 35 and figure 2-3 show the viscosity index.
  • Lubricants and hydraulic fluids with a high VI should be used in engines , machinery and construction equipments used outdoors where temperature vary over wide ranges
  • The high VI can be obtained by blending selected oils with high parrafins contents or adding special polymers that increase VI while maintaining goodlubricating properties
use of figure 2 3
USE OF FIGURE 2.3
  • FOR A GIVEN VI, FIND THE KINEMATIC VISCOSITY AT A GIVEN TEMPERATURE
  • CLASS WORK: FIND THE VALUES OF THE TABLE PAGE 35
viscosity measurements
VISCOSITY MEASUREMENTS
  • ROTATING DRUM VISCOSIMETER:

 APPARATUS SHOWN IN FIGURE 2.4 (A) MEASURES THE ABSOLUTE VISCOSITY

  • CAPILLARY VISCOSIMETER:

 APPARATUS SHOWN IN FIGURES 2.6 AND 2.7 MEASURES THE KINEMATIC VISCOSITY

saybolt universal viscosimeter
SAYBOLT UNIVERSAL VISCOSIMETER
  • SHOWN IN FIGURE 2-11
  • FIGURE 2.12 SHOWS A GRAPH OF SUS ( SAYBOLT UNIVERSAL SECONDS) VERSUS KINEMATIC VISCOSITY ν IN MM2/S FOR A FLUID AT 1000F
  • ABOVE ν = 75 MM2/S USE EQUATION  SUS= 4.632 ν
  • FOR A FLUID AT 2100F, THE EQUATION IS

SUS= 4.664ν

  • FOR ANY OTHER TEMPERATURE T IN 0F IS: SUS ( 1000F) X A ( SHOWN IN FIGURE 2.13)
  • A= 6.061.10-5 T + 0.994
  • WORK EXAMPLES 2.1 to 2.3 PAGE 43-44
sae viscosity grade
SAE VISCOSITY GRADE
  • A system for rating the viscosity of oils was established many years ago by the SAE.
  • An oil's viscosity rating is often referred to as its 'grade' (or from times past, its 'weight').
  • An oil that flows more quickly has a lower viscosity and, consequently, is given a lower rating.
  • Grade numbers are assigned to certain ranges of viscosity ratings.
  • For example, an SAE 30 grade covers a lower range of viscosities than an SAE 40 grade.
sae viscosity grade1
SAE VISCOSITY GRADE
  • Multigrade oils, such as a 10W30 oil, are formulated to flow rapidly to areas requiring lubrication when the engine is cold but also to maintain enough viscosity to protect the engine at higher temperatures and operating loads.
  • The number before the 'W' (for Winter) is the oil's viscosity when cold. The number after the 'W' is its viscosity at operating temperatures
sae viscosity grades table 2 4
SAE VISCOSITY GRADESTABLE 2.4
  • OILS WITH SUFFIXE W ( EXAMPLE SAE 10 W)
  • BASED ON MAXIMUM DYNAMIC VISCOSITY AT LOW TEMPERATURES
  • WICH SIMULATE THE CRANKING OF A ENGINE OR PUMPING CONDITIONS
  • THEY MUST ALSO HAVE A MINIMUM KINEMATIC VISCOSITY AT 1000C
slide37
OIL GRADES WITHOUT SUFFIXE W

 RATED AT HIGH TEMPERATURES BY TWO METHODS:

 KINEMATIC VISCOSITY UNDER LOW SHEAR RATE CONDITIONS AT 1000C

 DYNAMIC VISCOSITY UNDER HIGH SHEAR RATE CONDITIONS AT 1500C

summary
SUMMARY
  • The SAE grades 0W through 25W, where W stands for Winter, have a maximum viscosity specified at low temperatures (—5 through —35°C), to ensure easy starting under low temperature conditions, and a minimum viscosity requirement at 100°C to ensure satisfactory lubrication at the final operating temperature. 
  • The SAE grades 20 through 60 only have limits set at 100°C as these grades are not intended for use under low temperature conditions.
example
EXAMPLE
  • for example SAE-15W40 for a multigrade oil and SAE-40 for a monograde oil.
  • The first number (15W) refers to the viscosity grade at low temperatures (W from winter),
  • whereas the second number (40) refers to the viscosity grade at high temperature.
class work2
CLASS WORK
  • WORK THE FOLLOWING PROBLEMS:

2-19, 2.23, 2.25, 2.29, 2.37, 2.38, 2.46, 2.55, 2.57, 2.65 TO 2.76

what is pressure
WHAT IS PRESSURE?
  • BY DEFINITION :
units of pressure
UNITS OF PRESSURE
  • USING THE DEFINITION:
  • SI UNIT : N/M2 CALLED PASCAL (Pa)
  • US UNIT : LB/FT2 WE ALSO HAVE PSI= LB/IN2
what is atmospheric pressure
WHAT IS ATMOSPHERIC PRESSURE?
  • The atmospheric pressure is the pressure in the surrounding air. It varies with temperature and altitude above sea level.
standard atmospheric pressure
STANDARD ATMOSPHERIC PRESSURE
  • The Standard Atmospheric Pressure (atm) is used as a reference for gas densities and volumes.
  • The Standard Atmospheric Pressure is defined at sea-level at 273oK (0oC) and is 1.01325 bar or 101.325 kPa (absolute). The temperature of 293oK (20oC) is also used.
  • In imperial units the Standard Atmospheric Pressure is 14.696 psi.
gage pressure
GAGE PRESSURE
  • GAGE PRESSURE IS MEASURED WITH ATMOSPHERIC PRESSURE AS ITS BASE AND CAN BE READ IN MANOMETERS
absolute pressure
ABSOLUTE PRESSURE
  • ABSOLUTE PRESSURE USES THE PERFECT VACCUM AS ITS BASE
absolute and gage pressure relationship
ABSOLUTE AND GAGE PRESSURE RELATIONSHIP
  • PABSOLUTE = PGAGE + PATMOSPHERIC
example1
EXAMPLE
  • IF A FLUID PRESSURE IS 5.5 kPa ABOVE STANDARD ATMOSPHERICE PRESSURE

 ITS GAGE PRESSURE AS READ IN MANOMETER IS: 5.5 kPa

  • ITS ABSOLUTE PRESSURE IS:

5.5 + 101.3 = 106.8 kPa

class work3
CLASS WORK
  • WORK EXAMPLES 3-1 TO 3.4
definition
DEFINITION
  • The pressure at any given point of a non-moving (static) fluid is called the hydrostatic pressure

 EXAMPLE: As you go deeper in a fluid, such as swimming pool, the pressure increases

pressure height relationship
PRESSURE- HEIGHT RELATIONSHIP
  • γ is the specific weight
  • h is the height or elevation
class work4
CLASS WORK
  • WORK EXAMPLES 3-5 TO 3.7
class work5
CLASS WORK
  • WORK PROBLEMS :
  • 3.2 TO 3.5
  • 3.15, 3.17,3.19,3.25,3.27,3.31
  • 3.35, 3.37,3.39,3.41,3.45,3.47,
definition1
DEFINITION
  • FORCE DUE TO STATIC PRESSURE

REMEMBER 

STATIC FLUID IS FLUID AT REST)

surfaces under gas pressure figure 4 1 a
SURFACES UNDER GAS PRESSURE ( FIGURE 4.1 A)
  • SINCE GASES HAVE LOW SPECIFIC WEIGHT, THE PRESSURE DOES NOT DEPEND ON THE HEIGHT
  • SINCE WE HAVE UNIFORM PRESURE:
example2
EXAMPLE
  • WORK EXAMPLE 4.1 PAGE 85
horizontal surfaces under liquids figure 4 1b
HORIZONTAL SURFACES UNDER LIQUIDSFIGURE 4.1B
  • WE WILL HAVE THE HYDROSTATIC PRESSURE IN THE BOTTOM:

AND THE HYDROSTATIC FORCE:

example3
EXAMPLE
  • WORK EXAMPLE 4.2 AND 4.3

PAGE 86

rectangular walls figure 4 1 e
RECTANGULAR WALLSFIGURE 4.1 E
  • PRESSURE ACTING ON A VERTICAL WALL INCREASES AS THE DEPTH INCREASES ( FIGURE 4.6 PAGE 88)
  • AVERAGE PRESSURE IS LOCATED AT h/2:
  • BECAUSE THE PRESSURE VARIES LINEARLY WITH h, THE RESULTANT FORCE:
  • THE CENTER OF THE PRESSURE IS LOCATED AT h/3 FROM THE BOTTOM
example4
EXAMPLE
  • WORK EXAMPLE 4.4 PAGE 89
inclined walls figure 4 7
INCLINED WALLS ( FIGURE 4.7)
  • LENGHT OF THE FACE OF THE WALL:
  • AREA OF THE WALL:
  • THE RESULTANT FORCE:
  • THE CENTER OF THE PRESSURE IS AT:

L/3 FROM BOTTOM

class work6
CLASS WORK
  • WORK PROBLEMS:

4.1,4.3,4.5,4.9,4.11,4.13,4.15,4.17

experiment
EXPERIMENT

If one places a copper ball in a pail of water it will sink, whereas a wooden ball will float

WHY?

what is buoyancy
WHAT IS BUOYANCY?

A BODY IN A FLUID , WHETHER FLOATING OR SUBMERGED, IS BUOYED UP BY A FORCE FB EQUAL TO THE WEIGHT OF THE FLUID DISPLACED

buoyant force f b
BUOYANT FORCE FB ?
  • The upward force exerted by the fluid is known as buoyant force FB .
  • Law of Archimedes:The buoyant force FB is equal to the weight of the replaced liquid or gas
the three situations
THE THREE SITUATIONS
  • SITUATION #1:

A) FE IS POSITIVE

 DENSITY OF SOLID > DENSITY OF FLUID

THE SOLID WILL SINK

B) FE IS NEGATIVE

 DENSITY OF SOLID < DENSITY OF FLUID

 THE SOLID WILL FLOAT

C) FE =0

 DENSITY OF SOLID = DENSITY OF FLUID

 PARTIALLY SUBMERGED

class work7
CLASS WORK
  • WORK EXAMPLES 5-1 TO 5-3
stability of submerged bodies figure 5 9
STABILITY OF SUBMERGED BODIES ( FIGURE 5.9)
  • THE CONDITION FOR STABILITY OF BODIES COMPLETELY SUBMERGED IN A FLUID IS:

THE CENTER OF GRAVITY OF BODY SHOULD BE LOWER THAN THE CENTER OF BUOYANCY

center of gravity of the body
CENTER OF GRAVITY OF THE BODY
  • DEFINITION:

the centre of gravity is related to the distribution of weight within the object. 

center of buoyancy
CENTER OF BUOYANCY
  • DEFINITION :

IS THE CENTER OT GRAVITY OF THE VOLUME OF THE DISPLACED LIQUID

stability of floating bodies figure 5 10
STABILITY OF FLOATING BODIES ( FIGURE 5.10)
  • THE CONDITION FOR STABILITY OF FLOATING BODIES IS:

THE CENTER OF GRAVITY SHOULD BE BELOW THE METACENTER

metacenter
METACENTER
  • DEFINITION:

THE INTERSECTION OF THE VERTICAL AXIS OF THE BODY WHEN IT IS AT EQUILIBRIUM WITH THE NEW POSITION OF THE BOUYANCY WHEN THE BODY IS ROTATED SLIGHTLY

metacenter1
METACENTER
  • CALCULATION:

MB= I/VD

WHERE:

 MB= THE DISTANCE FROM THE CENTER OF BUOYANCY TO THE METACENTER

I IS THE LEAST MOMENT OF INERTIA OF A HORIZONTAL SECTION OF THE BODY TAKEN AT THE SURFACE OF THE FLUID

 VD IS THE VOLUME OF DISPLACED FLUID

class work8
CLASS WORK
  • WORK EXAMPLES 5.5 AND 5.6
class work9
CLASS WORK
  • WORK PROBLEMS AT THE END OF THE CHAPTER
what is flow
WHAT IS FLOW?

THE QUANTITY OF FLUID FLOWING IN A SYSTEM PER UNIT TIME

what is volume fow rate q
WHAT IS VOLUME FOW RATE Q?

THE VOLUME OF FLUID FLOWING PAST A SECTION PER UNIT TIME

formula unit of q
FORMULA & UNIT OF Q
  • A= AREA OF THE SECTION
  • v= AVERAGE VELOCITY OF FLOW
  • UNIT ( TABLE 6.1):

M3/S OR FT3/S

what is weight flow rate w
WHAT IS WEIGHT FLOW RATE W?

THE WEIGHT OF FLUID FLOWING PAST A SECTION PER UNIT TIME

formula and unit of w
FORMULA AND UNIT OF W
  • γ= SPECIFIC WEIGHT OF THE FLUID
  • Q= VOLUME FLOW RATE
  • UNIT ( TABLE 6.1)

N/S (SI) OR LB/S ( US)

what is mass flow rate m
WHAT IS MASS FLOW RATE M?

THE MASS OF FLUID FLOWING PAST A SECTION PER UNIT TIME

formula and unit of m
FORMULA AND UNIT OF M
  • ρ =THE DENSITY OF FLUID
  • Q= VOLUME FLOWRATE
  • UNIT KG/s ( SI) OR SLUGS/S ( US)

( Table 6.1)

class work10
CLASS WORK
  • WORK PROBLESM 6.29,6-31 AND 6.33
continuity equation
CONTINUITY EQUATION

When a fluid is in motion, it must move in such a way that mass is conserved

continuity equation relationship
CONTINUITY EQUATION RELATIONSHIP
  • BETWEEN POINT 1 AND POINT 2 WE HAVE  M1=M2
  • SINCE

Q=A.v AND M=ρ.Q

THE CONTINUITY EQUATION IS:

ρ1A1v1= ρ2A2v2

class work11
CLASS WORK
  • WORK EXAMPLE 6.4 PAGE 157
commercially available pipes and tubing
COMMERCIALLY AVAILABLE PIPES AND TUBING
  • STEEL PIPES: DESIGNATED BY A NOMINAL SIZE AND SCHEDULE NUMBER ( STEEL PIPES SCHEDULE 40 AND 80 ARE IN APPENDIX F
  • STEEL TUBING: USED IN POWER SYSTEMS, CONDENSERS, HEAT EXCHANGERS. THEY ARE DESIGNATED BY OUTSIDE DIAMETER AND WAAL TICKNESS

SEE APENDIX G

commercially available pipes and tubing1
COMMERCIALLY AVAILABLE PIPES AND TUBING
  • COPPER TUBING : SIX TYPES ARE AVAILABLE IN THE MARKET
  • DUCTILE IRON PIPE:
  • PLASTIC PIPE AND TUBING
  • HYDRAULIC HOSE
class work12
CLASS WORK
  • WORK EXAMPLES 6.6 TO 6.8 PAGES 164-165
class work13
CLASS WORK
  • WORK PROBLEMS 6.37, 6.39, 6.41, 6-43, 6-45 , 6-49 and 6.54
what is the law of conservation of energy
WHAT IS THE LAW OF CONSERVATION OF ENERGY?

ENERGY CAN NEITHER BE CREATED OR DESTROYED, BUT IT CAN BE TRANSFORMED FROM ONE FORM TO ANOTHER

what are the different energies involved in a pipe flow system
WHAT ARE THE DIFFERENT ENERGIES INVOLVED IN A PIPE FLOW SYSTEM ?
  • POTENTIAL ENERGY
  • KINETIC ENERGY
  • FLOW ENERGY
what is potential energy
WHAT IS POTENTIAL ENERGY
  • DUE TO ITS ELEVATION, THE POTENTIAL ENERGY OF THE FLUID RELATIVE TO SOME REFERENCE IS :
what is kinetic energy
WHAT IS KINETIC ENERGY?
  • DUE TO ITS VELOCITY, THE KINETIC ENERGY OF THE FLUID IS
what is flow energy
WHAT IS FLOW ENERGY?
  • SOMETIMES CALLED PRESSURE ENERGY OR FLOW WORK,IT REPRESENTS THE AMOUNT OF WORK NECESSARY TO MOVE THE FLUID
analysis of figure 6 5 page 167
ANALYSIS OF FIGURE 6.5 PAGE 167
  • IF NO ENERGY IS ADDED TO THE FLUID OR LOST BETWEEN SECTION 1 AND SECTION 2  THE PRINCIPAL OF CONSERVATION OF ENERGY REQUIRES:

E1=E2

bernoulli s equation
BERNOULLI’S EQUATION
  • E1=E2
  • DIVIDING BY w
restrictions of bernoulli s equation
RESTRICTIONS OF BERNOULLI’S EQUATION
  • INCOMPRESSIBLE FLUID

 w= CONSTANT

  • NO WORK ADDED OR REMOVED IN/FROM THE SYSTEM
  • NO HEAT ADDED TO OR REMOVED FROM THE SYSTEM
  • NO ENERGY LOST DUE TO FRICTION
class work14
CLASS WORK
  • WORK EXAMPLE 6.9 PAGE 170 and 6-10 PAGE 172
class work15
CLASS WORK
  • WORK PROBLEMS 6-61, 6-63, 6-65,6-67,6-69 AND 6-71
restrictions of bernouilli equation
RESTRICTIONS OF BERNOUILLI EQUATION
  • INCOMPRESSIBLE FLUID

 w= CONSTANT

  • NO WORK ADDED TO OR REMOVED FROM THE SYSTEM
  • NO HEAT ADDED TO OR REMOVED FROM THE SYSTEM
  • NO ENERGY LOST DUE TO FRICTION
expansion of the bernouilli equation
EXPANSION OF THE BERNOUILLI EQUATION
  • WE WILL NOW:
  • CONSIDER ENERGY LOST FROM SYSTEMS THROUGH FRICTION, VALAVES AND FITTINGS ( hL)
  • CONSIDER ENERGY ADDED TO THE SYSTEM BY A PUMP ( hA)
  • CONSIDER ENERGY REMOVED FROM SYSTEM BY MOTORS OR TURBINES ( hR)
assumptions
ASSUMPTIONS
  • SINCE WE ARE DEALING WITH LIQUIDS ONLY :

INCOMPRESSIBLE FLUID

 w= CONSTANT

  • IN FLUIDS MECHANICS WE ASSUME NO HEAT EXCHANGE
we will add the following terms to bernouilli equation
WE WILL ADD THE FOLLOWING TERMS TO BERNOUILLI EQUATION
  • hL= ENERGY LOST BY THE SYSTEM BY FRICTION IN PIPES OR MINOR LOSS DUE TO VALVES AND FITTINGS.
  • In general hL= K ( v2/2g) where K is the resistance coefficient. We will study it in the next chapter
  • hR= ENERGY LOST BY THE SYSTEM DUE TO TURBINES OR MOTORS
  • hA= ENERGY ADDED TO THE SYSTEM BY PUMPS
power and efficiency of pumps
POWER AND EFFICIENCY OF PUMPS
  • POWER OF PUMP:

PA= hA.γ.Q

  • EFFICIENCY OF PUMPS:

e= PA/PI WHERE PI IS THE POWER OF THE MOTOR TURNING THE PUMP

power and efficiency of motors
POWER AND EFFICIENCY OF MOTORS:
  • PR= hR.γ.Q
  • Efficiency = Power of motor/ PR
class work16
CLASS WORK
  • WORK EXAMPLES 7.1 PAGE 204 AND 7.2 PAGE 205
class work17
CLASS WORK
  • WORK PROBLEMS : 7.1, 7.3, 7.9,7.11,7.13,7.15
what is reynolds number
WHAT IS REYNOLDS NUMBER?
  • GIVE AN IDEA ABOUT THE BEHAVIOUR OF FLUID
  • TELL US IF THE FLUID IS LAMINAR OR TURBULENT

HOW?

definition of reynolds number
DEFINITION OF REYNOLDS NUMBER
  • THE REYNOLDS NUMBER IS DEFINED BY THE FOLLOWING EQUATION:

NRE= V.D.ρ/η

OR

NRE= V.D/ν

laminar or turbulent flow
LAMINAR OR TURBULENT FLOW?
  • IF NRE < 2000 LAMINAR
  • IF NRE > 4000 TURBULENT
  • 2000
class work18
CLASS WORK
  • WORK EXAMPLES 8.1, 8.2 AND 8.3 PAGES 231-232
darcy s equation for friction loss
DARCY’S EQUATION FOR FRICTION LOSS
  • DARCY’S EQUATION: the energy loss due to friction is

hL= f.(L/D).V2/2g

where f is the friction factor

friction factor f for laminar flow
FRICTION FACTOR f FOR LAMINAR FLOW
  • HAGEN-POISEUILLE EQUATION:

hL= 32.η.L.V/γ.D2

Then:

f=64/NRF

friction factor f for turbulent flow
FRICTION FACTOR f FOR TURBULENT FLOW
  • MODY DIAGRAM ( FIGURE 8.6 PAGE 237 OF BOOK)
class work19
CLASS WORK
  • WORK EXAMPLES 8.3 AND 8.4 PAGES 232 AND 234
  • WORK EXAMPLES 8.5 , 8.6 AND 8.7 PAGES 239-240
homework
HOMEWORK
  • DO PROBLEMS 8.1 , 8.3,8.5 AND 8.9 AND 8.11
what is open channel flow
WHAT IS OPEN CHANNEL FLOW?
  • AN OPEN CHANNEL FLOW IS A FLOW SYSTEM IN WHICH THE TOP SURFACE OF THE FLUID IS EXPOSED TO THE ATMOSPHERE
uniform steady flow
UNIFORM STEADY FLOW

VOLUME FLOW RATE( OR DISCHARGE IN OPEN CHANNEL FLOW) REMAINS CONSTANT IN THE SECTION OF INTEREST AND THE DEPTH OF THE FLUID IN THE CHANNEL DOES NOT VARY ( FIGURE 14.2 PAGE 445)

TO ACHIEVE CONSTANT DISCHARGE , THE CHANNEL SHOULD BE PRISMATIC

varied steady flow
VARIED STEADY FLOW
  • THE DISCHARGED OR VOLUME FLOW RATE REMAINS CONSTANT BUT THE DEPTH OF THE FLUID VARY . THIS WILL HAPPEN WHEN THE CHANNEL IS NOT PRISMATIC
unsteady varied flow
UNSTEADY VARIED FLOW
  • HAPPENS WHEN THE DISCHARGES ( VOLUME FLOW RATE) VARIES BECAUSE OF THE CHANGE OF DEPTH ( APPLIES FOR PRISMATIC ORNON PRISMATIC)
varied flow
VARIED FLOW
  • THEY ARE CLASSIFIED AS RAPIDLY VARYING FLOW OR GRADUALLY VARYING FLOW DEPENDING ON THE RATE OF CHANGE IN DEPTH ( FIGURE 14.3 PAGE 445)
hydraulic radius for open systems
HYDRAULIC RADIUS FOR OPEN SYSTEMS
  • THE CHARACTERISTIC DIMENSION IN OPEN SYSTEMS IS THE :

HYDRAULIC RADIUS

definition of hydraulic radius
DEFINITION OF HYDRAULIC RADIUS
  • THE RATIO OF THE NET CROSS SECTION AREA OF THE FLOW STREAMTO THE WETTED PERIMETER OF THE SECTION:

R= AREA/ WETTED PERIMETER

example5
EXAMPLE
  • IN FIGURE 14.1.C PAGE 444, DETERMINE THE HYDRAULIC RADIUS OF THE TRAPEZOIDAL SECTION ( W=4 FT, X=1FT, D=2FT)
  • YOU WILL FIND PROPERTIES OF AREA OF DIFFERENT SHAPES IN APPENDIX L PAGE 611
solution
SOLUTION
  • AREA = WD+ 2( XD/2)= 10 FT2
  • WETTED PERIMETER = WP
  • WP= W+2L WITH L = ( X2+D2)0.5
  • WP= 8.48 FT
  • R= 1.18 FT
reynolds number for open channels
REYNOLDS NUMBER FOR OPEN CHANNELS
  • IN OPEN CHANNELS, THE REYNOLDS NUMBER IS ( NRE= V.R/v)
  • EXPERIMENTAL EVIDENCE HAS SHOWN THAT, FOR OPEN CHANNELS, THE LAMINAR FLOW OCCURS WHEN NRE < 500 AND THE TURBULENT FLOW OCCURS WHEN NRE > 2000
froude number
FROUDE NUMBER
  • THE REYNOLDS NUMBER AND THE LAMINAR AND TURBULENT TERMS ARE NOTSUFFICIENT TO DESCRIBE FLOWS IN OPEN CHANNELS
  • THE FROUD NUMBER IS INTRODUCED TO TAKE INTO ACCOUNT THE RATIO OF INERTIAL FORCES TO GRAVITY FORCES WHICH IS NEGLIGEABLE IN PIPES.
definition of froud number
DEFINITION OF FROUD NUMBER
  • NF= V/ (g.yh)0.5 where yh= A/T
  • A is the area and T is the width of the free surface of the fluid at the top of the channel
classification of flows in open channels
CLASSIFICATION OF FLOWS IN OPEN CHANNELS
  • SUBCRITICAL LAMINAR

NRE < 500 AND NF< 1

  • SUBCRITICAL TURBULENT

NRE > 2000 AND NF < 1

  • SUPERCRITICAL LAMINAR

NRE < 500 AND NF > 1

  • SUPERCRITICAL TURBULENT

NRE> 2000 AND NF> 1

manning equation for uniform steady flow
MANNING EQUATION FOR UNIFORM STEADY FLOW
  • MANNING EQUATION IS USED TO CALCULATE THE AVERAGE VELOCITY OF UNIFORM FLOW.
  • IN SI UNITS:

V= (1/n).R(2/3).S0.5 ( V IN M/S AND R IN M)

n is the resistance factor ( table 14.1 page 449)

S is the slope of the the channel

Usually in open channels the slope is small then we have S= h/L ( figure 14.5 page 449)

normal discharge q
NORMAL DISCHARGE ( Q)
  • Q= ( 1/n). A.R(2/3).S0.5
manning equation
MANNING EQUATION
  • US UNITS FOR MANNING EQUATION OF STEADY FLOW IN OPEN CHANNELS

V= (1.49/n). R(2/3).S0.5 ( VIN FT/S AND R IN FT)

Q= ( 1.49/n). A.R(2/3).S0.5

class work20
CLASS WORK
  • WORK PROBLEMS 14.2 , 14.3 AND 14.4 PAGES 450 TO 453)
ad