Maahflmery: TR (Uf, D

 

Guf,deboo[<

tr[uf,d Maahflmery Jose R. Francisco

Table of Comtemts

 

Lesson

I

 

"""

Problems Studentb Self-test Lesson 2

43

"""""'47

HEAD LOSSES IN THE PIPELINE..."""""""""""" 51' """""""""""' 51 Introduction ................

"""""52

Obiectives Classifications of Losses 1. Major Head Losses.."""""""' 2. Minor Head Losses...""""""" Lesson

""""""""""'52 """""""""52 """""""""52

Due to Friction """ 53 Equations for the computation of Major Head Losses Lines ....-.....""""' 58 Equations for the computation of Minor Losses in Pipe Determination of Major Loss and Minor Loss Combined """""""""""""' 60

flf"

Sir" Specifications

Problems Student

""""67 """'71 """"'-""""73

s Self-test (Set A)

Student's Seif-test (Set

Lesson 3

...'...'.""'

B)

"""""""""78

CENTRIFUGAL PUMPS .....................""""""""""""' 583I """""""""""' Introduction ...'............ """"" 83 Lesson Obiectives Definitions of Terms and Useful Information """""""':"' Kinds of Pumps 1. Dynamic Pumps 2. Displacement Pumps Classes of iynamic Pumps

Pumps '...'........... Categories ofltuia Coupling TypilalPumpinglnstallation .."""""" iut.go.i", of PumPs

nr,riJcoupfirg

Classificati,on of Pumps Based on Suction

PumPs

""""

84 84 84 84 86 87

87 87

Lift """""""""""""""'

88

"""

89

Centrifugal Basic Parts of Centrifugal Pump Brake Power urd Po*.' Power

Pumps Losses

Losses Pump EfficiencY Ct #"t".i.tics tf Centrifugal Pumps Types of Centrifugal Pumps SuctionTypes Pump

""""""""""' """"""" """"""""

""""' """"""""""' """""""' """"""""""

Classes of Displacement

ip".Lf

"""" M """""""' M

Single-stage'

"""""' 89

"""""""" 89 """""""""m """"""" 90 """""""""" 90 """""""' 90 """""""""' 91

""""' Volute-type Centrifugal ClassificationofCentr'ifugalPumpsBasedonImpellerType.........'...,,.,.,,,92 """"""""""92 Specific Speed of Centrifugal Pumps " 93 Derivation of Specific SpeJa Equation (English Unit) """"""""""""""""' """"""""""" 95 Derivation of Specific Speed Equation (SI Characteristic """""" 98 Pump Basic """""""' 98 Relation of Pump Characteristics for Similar

Types of tmpeller for a Single-suction,

Curves Laws

'

::""""""" Unit)

Pumps

91

"""n

 

Laws.......... Pump Combinations andArrangements l. Pumps in Series 2. Pumps in Parallel Key krms and Concepts to Remember ,............ Problems Student's Self-test Affinity

Lesson 4

.........99 ............... 104 ...................... 104 .....,.............. lM .................. 107

..... 111 ........ 115

CAVITATION AI\D NET POSTTIVE .................................. .................................... ....................... ..... 119 sucTIoN HEAD ................ ..................... I 19 Introduction .........-.-.... ....... 119 Lesson Objectives ..................... 120 ,zDefinition of Cavitation .........120 Causes of Cavitation in Centrifugal Pumps .................... 120 Bad Effects of Cavitation in Centrifugal Pumps ...............20 Factors Affecting Cavitation ......................121 Net Positive Suction Head (NPSH) ................. ....121 Net Positive Suction Head Available (NPSHA).... . .. ....122 Net Positive Suction Head Required (NPSHR) ............. ... ........123 Solution for Cavitation Problem .....123 Cavitation Parameter

Cavitation Limits of Centrifugal Pumps in Terms of SA ......... ..................124 .......,.......... 131 Key Terms and Concepts to Remember .............

Problems

..... 133 ................ 135 ................ 137

Studentb Self-test (Set A) Student s Self-test (Set B)

Lesmn 5

RECIPROCATING PUMPS....................................... 143 ............,........143 Introduction................ ........143 .................:. Lesson Objectives ...............144 Definitions of Terms and Other Useful Information ........144 Types of Reciprocating Pumps

Classification of Reciprocating Pumps in Terms of the Method of ..................lM Driving the Water Piston or Plunger.. Classification of Reciprocating Pumps in Terms of the Number of 146 Water Cylinders Formulas Advantages and Disadvantages of Direct-acting Reciprocating Pumps over Centrifugal Pumps Key Terms and Concepts to Remember .............

Questions ond Topics Problems Studentb Self-test

for Library

Research

146 148

154 157 159 161

Lesmn 6

vtt

 

Classification ofHydraulic

Turbines

.................... 164

Turbine .............1A .......... ..................... 165 ....... 165 Runner ............... 165 Specific Speed Ranges of Specific Speed ................. 166 ..... 166 Hydroelectric Power Plant .......... Types of Hydroelectric Power Plants .................167 Definitions of Terms and Other Useful Information ....-.....,..... 168 Determination of Effective Head for lmpulse Turbine ........... 169 Determination of Effective Head for Reaction Turbine ......... l7l Design Equations ................... ........,..173 Hydraulic Turbine Effrciencies ..........175 Key Tbrms and Concepts to Remember ............. ..................185 Problems .....191 ......... 193 Studentb Self-test

Impulse Reaction Turbine Types of Water Turbine

Lesson 7

GAS COMPRESSORS.....................r.......................... I 95

Introduction .................;.. .................;.........r............ .......r.......................... .............. ..................... 195 Lesson Objectives ........ 195

DefinitionofGasCompressor Practical Uses of Compressed Air ............. Types of Gas Compressors.................

.......... l% ......... 196 ................. 196

Compressor............... ................... 198 Curves ........200 Volumetric Efficiency of a Reciprocating Compressor ..........201 Displacement Volume of Reciprocating Compressor Compressor Efficiency ..................203 Reciprocating Preferred Compression

Multi-stage Reciprocating Compressors............ ....,..,.............214 Three-stage Reciprocating Compressor ...................222 Four-stage Reciprocating Compressor ......................224 Five-stage Reciprocating Compressor.............. ......................226 .....233 (For Library ........,...........235 Questions Studentb .........235

.............. .............

Problems

Self-test

Design (Project) Problem References

nswers to the Student's Self-test

vtrt

Research)

239 243 245

 

',,..,.#.

PRTNCIPLE,S Otr trLUID trLO\M lxrRooucrtoil Lesson 1 reviews the principles of fluid flow including the basic fluid properties such as density, specific volume, specific gravity, specific weight, and viscosity. lt also deals with the Reynolds number, types of flows, continuity equation, Bernoulli's equation, and the three forms of head of fluid. lt discusses as well the derivation of the three forms of head and the total dynamic head (TDH) or external head of pumps, and presents a schematic diagram of a typical pump installation. Sample problems accompany the discussion of each topic. To further enhance the learning process, additional problems involving the principles

of fluid flow are provided for students to solve.

Lessox Oe.recrues At the end of this lesson, the students should be able to: define the basic properties of fluid; determine the types of flow of fluid using the Reynolds number;

o o

o o .

derive the three forms of head of fluid and the total dynamic head (TDH) or external head of pumps using Bernoulli's equation; determine the total dynamic head of a pump; and solve problems involving the principles of fluid flow.

 

DEFINITIONS OF TERMS AND OTHER USEFUL INFORMATION

o o o o o o o o o o

gas form. Fluid machinery refers to machines that handle fluids in either liquid or Fluid mechaniis is the study of the behavior of fluids whether at rest or in motion. Ftuid sfaflcs is the study of fluids at rest. Ftuid dynamics is the study of fluids in motion' in motion' Hydraulics is the branch of science concerned with water or other fluids particles that easily Ftuidsare substances which are capable of flowing, and have move and change their relative position without separation of mass' to an Ftuidsfatics is the study of fluids at rest orwhich have no velocity with respect observer in a gravitationalfield. Mass is the absolute quantity of matter'

weight is the force of gravity in a fluid or body which could be measured by

a

weighing scale. Properties of water at 4oC and 1 atmosphere r Density, P = 1 000 kg/m3 = 1 kg/l = 1'94 slug/ft3 I Specific weight, T= 9 810 N/m3 = 62'4lbltt3 r Specific volume, v = 0.001 m3/kg = 0'51546 ft3/slug r Specific gravitY, SG = 1.0

PROPERTIES OF FLUID

1.

Specific

Weight is the force of gravity in a unit volume of a substance'

r-v-

--- W, - m9

v

where T = specific weight of the substance, lb/m3, V = volume of the substance, In3, ft3

kgr/m3' N/m3

m = mass of the substance, kg, lb g = acceleration due to gravity, m/s2, fps2 = 9.8066 m/s2 = 32.2fpsz

2.

Densityis the mass per unit volume of a substance' m

P=v where P

= density of a substance, kg/m3, lb/ft3 m = mass of the substance, kg, lb V = volume of the substance, m3, ft3

3.

reciprocal Specific volume is the volume of a unit mass of a substance or the of densitY.

V1 v=-=mp

 

where

4.

v

= specific volume of the substance, m3/kg, ft3/lb \ V = volume of the substance, m3, ft3 m = mass of the substance, kg, lb

Specific Gravity (SG) is the ratio of the specific weight of any substance to that of water or the ratio of density of any substance to that of water.

SG= Y =

where P p*

5.

.

Y*

P

P* = densitY of anY substance = density of water = 1 000 kg/m3

= specific weight of any substance y T* = specific weight of water

Temperature is the measure of random motion of molecules of a fluid or system. It is the thermal condition of fluid with reference to its ability to communicate heat from one body to another body or fluid. lt is the measure of hotness and coldness of a fluid. Absolute temperature is the temperature of a fluid measured with respect to an absolute zero, which is -460oF or -273oC.

t

t

Common Temperafure Sca/es

a)

Scale - scale used in the metric or Sl system of units Celsius point of Basis: Freezing water 00c Boiling point of water 100"c

b)

Fahrenheit Scale - scale used in the English system of units Basis: Freezing point of 32oF Boiling point of water 212.F

water

Equations of Temperature

a) Fahrenheit

to Celsius temperature

5 _32) t"u =:(tF g,

b)

Celsius to Fahrenheit temperature

t.'

=

9t^ *32

5"

where t, = temperature of fluid,

t" = temperature of fluid,

oF oC

 

.

Common Sca/es Used in the Absolute Temperatures

a)

Kelvin

-

an absolute scale used in the metric or Sl system of units

T*=t"+273

b) Degrees Rankine - an absolute scale used in the English system of units

T*=tr+460 = absolute temperature of fluid, K where Tx Tn obsolute temperature of fluid, oR

=

6.

Pressure is basically defined as the normal force per unit area.

o=AF Gauge pressure - the pressure of a fluid or system measured by pressure measuring-instruments like a pressure gauge or manometer

a)

r r

This pressure may be higher than or lower than the atmospheric pressure. Gauge pressure that is lowerthan the atmospheric pressure is known as vacuum pressure.

b) Absolute pressure - the pressure

of a fluid or system with reference to

an absolute zero pressure

It is the pressure of a fluid or system including the atmospheric pressure.

P"u,

=Pr,rtPn

where p,o" = absolute pressure, kPaa, psia

P"t, = atmosPheric Pressure, kPa, Psi

pe

r

c)

= gauge pressure, kPag, psig

r

Negative (-) if pn of the fluid or system is greater than or higher than the atmospheric pressure Positive (+) if ps of the fluid or system is lower than the atmospheric

.r

pressure Negative pressure (Rn) is also known as vacuum pressure.

Fluid pressure or hydrostatic pressure of fluid

r r

lt is the force exerted by the column of fluid per unit area. As shown in Figure 1.1 on the next page,

p=psh=yh or h=*=i

Flotd'M*{hlntry,

,Iste R, l'ransirnu:":

where  

p = fluid pressure, kPag, psig

p

= density of the fluid, kg/m3, lb/fts y = specific weight of the fluid, kN/m3, lb/ft3 g = 9.81 m/s2 = 32.2fps2 h = depth of the fluid, or head of fluid, m, ft

T h

I

Mi

Column of fluid

Figure 1.'1. Fluid Pressure

r

Absolute pressure at the base of the fluid Po

= P"t,

d) Atmospheric

*

Pn

= Put,

pressure

*

-

P9h = Pr,*

* yh-+ kPaa, psia

the force exerted by the column of atmosphere

per unit area at the surface of the earth

It is the intensity of force per unit area due to the weight of the

atmosphere.

e)

Standard atmospheric pressure - the pressure of the atmosphere measured at the surface of the earth near sea level p",, = 1 atm = 760 mm Hg = 29.92 in. Hg P,,, = 14.69 psi = 101.325 kPa = 1.03 kg/cm2 = 1.01 bar p",, = 1.34 m of HrO = 33.91 ft of HrO

r

When a fluid is at rest, the pressure at any boundary exerted by the fluid (and on the fluid) will be perpendicular to the boundary.

P re ssu re-me a su ri ng I n stru me

a)

nts

Pressure gauge Bourdon gauge is the device most commonly used to measure pressure commercially. The reference pressure in measuring the pressure of the system is the atmospheric.

o o

Lesson

1

PRINCTPLES OF FLUID FLOW

15

 

Pressure gauge (Bourdon tube)

SYSTEM

Figure 1.2. Pressure Gauge

b)

Manometer - a Utube containing a liquid of known specific gravity, the surface of which moves proportionally to changes in pressure

Types of Manometers

Opentype manometer is a manometer with an atmospheric surface in one leg and capable of measuring gauge pressure. Piezometer - the simplest form of manometer, it is a tube tapped into a wall of a container or conduit for the purpose of measuring the pressure

inside Pressure the system is

T

higher than

atmospheric Working substance of the system

+Ps

Pressure inside is the system higher than atmospheric

Working substance of the system

Mercury

Figure 1.3. Open-type Manometer Differential-type manometer is a manometer without an atmospheric surface and capable of measuring only differences of pressure.

System B

Figure 1.4. Differential-type Manometer   | Guidebook in Fluid Machinerl'

Jose R. Francisco

7.

 

Surface tension is the force of molecular attraction per unit length of free surface. lt is a function of both the liquid and the surface in contact with the liquid. Surface tension of liquid decreases as the temperature increases. It is always tangent to the inter-face. Fluid film

FL_*l

Figure 1.5. Surface Tension

.F oo-_dF or d=_ dLL

where 6 = surface tension, N/m

a) Wetting liquid. A liquid is said to wet a surface in contact with it if the

attraction of the molecules to the sudace exceeds the attraction of the molecules to each other.

Figure 1.6. Wetting Liquid

b)

Non-wetting liquid. A liquid is said to be non-wetting if the attraction of the other liquid molecules to each other is greater than their attraction to the surface. Tube

0 I

Figure 1.7. Non-wetting Liquid Lesson

1

PRINCTPLES OF FLUID FLOW

17

8'  

Compressibilityis the resistance of fluid to change its volume in a confined space. Compressibility of water usually affects the solution of practical probrems in hydraurics onry by changing its unit weight. Modulus of elasticity or butk modulus of the ftuid is the ratio of the stress (change of pressure) to the strain (change in volume divided by the original volume).

'

B

where

AP =- /av\ t-l ('v./

= bulk modurus or modurus F = 300 000 psi (for water) of erasticity of the fluid V = originalvolume AV = change in volume Ap = change in pressure

The negative sign (-) accounts for the fact that as the pressure increases, the volume decreases. Ex' 1'11 Determine the percentage change in the volume of water if its pressure is increased by 30 000 psi. Solution: For water, F = 300 000 psi From the equation of modurus of erasticity of the fruid, ( so ooo \ t'av or 10% decrease

) ap=-|ffi lv,l=- B

)=-o'tO

Note: The negative sign means decrease in volume.

9.

Viscosity is a measure of the resistance to flow of a fluid. lt may be defined as the ratio of the shearing stress or force between adjacent layers of fluid to the rate of change of velocity perpendicular to the direction of motion. lt may also be defined as the property of a fluid that determines the amount of its resistance to a shearing stress. Fluid layer

Shearing

_t   | Guidebook in Fluid Machinery

area,A

Stationary layer

Figure 1.8. Absolute Viscosity Jo19 R,.f.renciscq

 

Absolute viscosity is defined as the unit force required to move one iayer of a fluid at a unit velocity to another layer of the fluid which is at unit distance from the first. ,1,

lt =

a\t' /dv

Absolute viscosity or dynamic viscosity is the ratio of the fluid layer shearing stress to that of the shear stress rate. Shear strain rate is the ratio the velocity of a particular layer to its perpendicular distance from theof stationary surface, and is constant for each

layer.

V.V zy

According to Newton, the force F required to maintain a constant velocity Vo of the upper (layer) surface is proportional to the area and the shear strain rate.

F*At+) or u=

(to) (n= )

=

F=FAt))

shear stress (N/m'z; shear strain rate (1/s)

where t

= shearing force or stress V = velocity of fluid y = distance perpendicular to flow p = proportionality constant, known as absolute viscosity z = distance between layers or plates

Other Derivation of Viscosity,

trt

to consider: a) .Points lf the distance between the plates and the relative velocity of the upper plate to the lower plate are both kept constant, the force F required will be directly proportional to the area of the moving plate.

F*A

eq.

where A=shearingarea

F = shearirrg force

Lesson

1

PRINCIPLES OF I.LUID ['LO\\'19

1

 

lf both the distance between the plates and the area of the moving plate are kept constant, the force required will be directly proportional to the relative velocity of the plates.

F*V

eq.2

lf the relative velocity and the area of the moving plate are kept constant, but the distance between the plates is varied, the force required will be inversely

proportional to the distance between the plates. This arises from the fact that as the distance between the plates decreases, the shear stress increases.

r*1z

eq. 3

Combiningeq. 1,2,&3, l-x-

AV

z lntroducing the proportionality constant, p,

- '\z/nv \ l--ul-l

)

Dividing both sides byA,

-r = ulTAV\

I

) \z where f : )= t = shearing stress between plates or layers

\A /

/V\ I : l= rate of shearing strain or linear velocity distribution \,)p proportionality constant, known as viscosity

in the fluid

=

Taking the value of p,

Lr=ti]

=

t:) t:)

b)

shearing stress rate of shearing strain

Units of Viscosity or Absolute Viscosity (Sl, mks, & cgs)

N.s m'

cgs English -+

-)

Dvne. cm-

Pa.s

s

lb,. s + -= tn'

10 | Guidebook in Fluid MachinerY

--+ Kg/m. s

Keyn Jose R. Francisco

 

Derivation'.

l=-=

t:l

m/s

tYl

u=[i]

m

['o j = rr')

_ _, N.s _

N/m2

=

m/s

t:)

Lr

_ _, N.s = Pa.s m-

/m2

(kg

.

mis') . s

_kg

m.s

m'

m-

m

Dyne/cm2 m/s

=

t:) 1=[i]=

Dvne.s -r=POlSe cm-

-

cm

Pylg&d _ cm/s

(:)

(9.cm/s') .s _

Dyne .s cm'

cm'

cm

Conversion: 1 Poise=1g/cm-s 1 Poise = 1 Dyne-s/cm2 '1 Reyn = 6.9 x '104 Poise

g

cm.s

1 Dyne=19-cm/s2 1 Poise = 0.10 Pa.s

.

Derivation of 1 Poise = 0.10 Pa s N.S_, (kg . m/s') . s _ (1000 g) . (t OO cm/s') . s 1Pa.s=1 ) -l m' m(100 cm)'?

_1000g.cm/s2.s 100 cm'

=

Then,

1

1

c)

.10 DYn",'

cm-

t

= 1o

(o. cm/s').

=10t"

;'

s

cm-

poise

Poise = 0.10 Pa . s CentiPoise = 0.0'1 Poise = 0.001 Pa . s

Units of Viscosity (English system of units)

rF

)

_[n )_tb,ft'

_tb,.s -" =\------J=----L (v\ fps ftz t-

\=)

rt

rF )

t n I tb,/in2 ' - tb,. :' ;-s =Reyn + afterOsborne =+-'-,'" ' fV) ipn in' u

l,)

Lesson

1

Reynolds

in

PRINCIPLES OF FLUID FLOW

111

 

.

1 ReYn = '1 lbr-s/in2

fF)

-iP- - slus

stug.fpsz

" [y) -- tp_r ft.s rt ,, =

ln

,)

l''l

Conversion: 1 Poise = 0'10 Pa-s 1 cP = 0.001 Pa-s 1 lb,-s/ft2 = 478.8 Poise = 47'88 Pa-s 1 Poise = 0.002089 lb,-s/ft2 1 Pa-s = 2.089 x 10-2 lb,-slft2 1 cP = 2.089 x 105 lb,-s/ft2

d)

Kinematic viscosity is defined as the absolute viscosity divided by density' It is the ratio of the dynamic viscosity of a fluid to its mass density' u

v=i-

o

Units of Kinematic Viscosity (sl or metric system of units)

;r N's/m2 .,-=-=

ko.

m/s2

" a' 'S - m' " p kg/m3 kg/m3 s

Di,ng.* grcm/s2..

c*' -'o a*' ." ,-[= " - p g/cmt= g/cm3 ="*' s

=stoke

lstoke=1cm2/s Units of Kinematic Viscosity (English system of units)

u

V-----

p

lb,'s ft2

slug/ft'

12 | Guidebook in Fluid MachinerY

sluo . fps2 og t2 -+ il'

fl' - slugltt'- -;

Jose R. Francisco

 

Sysfem of Units for Viscosity

1.

Absolute Metric System

lPoise=1g/cm-s lPoise=1Dyne-s/cm2

2.

Absolute British System 1 poundal-s1ft2 = 1 lb./ft-s

Conversion: 1 Poise = 0.0672 poundal-s/ft2 l Poise =0.0672|b /ft-s

3. Absolute "gravitational"

British System, or the Engineers' System

1lb-s/ft2=1slug/ft-s

Conversion: 1 Poise = 0.0020885 lb-s/ft2

1 Poise = 0.0020885 slug/ft-s

Note: 0.0020885 = 0.0672t32.17

4. Units of Kinematic Viscosity Metric System: 1 stoke = 1 cm2ls

Conversion:

1 stoke = 0.001076 ft2ls

Methods of Obtaining Viscosity

1.

Universal Saybolt Viscosimeter (kinematic viscosity)

v=0.0022t'-1'8

where

t' t' = Saybolt second v = kinematic viscosity, stoke

Saybolt second is the time in seconds required for 60 cm2 of liquid to pass through an opening of standard size. u

=(o.rrut'-191)ro"

I

t'J

where v = kinematic viscosity, ft2ls Lesson

1

PRINCIPLES OF FLUID FLOW I13

t

 

Note: For heavy oil, a Saybolt-Furol viscosimeter is used. 1 Saybolt-Furol

= 10 . Saybolt Universal

Engler Viscosimeter

2.

, =(o ,5Bt'-199)',0" tl

t'

where v = kinematic

viscosity, ft2ls t' = Saybolt seconds

Redwood Viscosimeter

3.

u

=(o.rrot'-l t/ 9)ro"

I

where v = kinematic

viscosity, ft2ls t' = Saybolt seconds

Helmholtz Equation for Absolute Viscosity of Water

4.

where

5.

0.01779 1+ 0.03368 r+o.ooo221(

= absolute viscosity, poise t = temperature of water, oC

1r

Grindley and Gibson's Equation forAbsolute Viscosity of Air

where

p = 0.0001 702(1+ 0 00329 t + 0.o0ooo 7 t'z )

= absolute viscosity, poise t = temperature of air, oC

1r

Ex. 1.21Using Helmholtz equation, determine the viscosity of water in Poise and slug/ft-s for a temperature of 15.55"C. lf the density of water at this temperature is 1 gm/cm3, determine the kinematic viscosity in both units.

Given: Water at a given condition t = 15.55"C

P = 1 g/cm3

Required: The absolute viscosity, in Poise, and kinematic viscosity in both units

/ 4 | Guidebook in Fluid Machinery

Jose R. Francisco

 

Solution: Using Helmholtz equation,

p=

0.01779 1 + 0.03368 t+ O.OOO221(

p = 0.01128

0.01779 1

+

0.03368(15.55) + O.OOO221 (15.55)'z

Poise

ANSWER

ln the English unit: 1 Poise = 0.0020885 slugift-s

p

= (0.01128 Poise) (0.0020885 slug/ft-s/Poise)

p = 0.0000236 slug/ft-s For Kinematic Viscosity, 0'0'1128 Poise u=F= = 0.01128 stoke = 0.0000,l 214ft2ls 1glcm3

p

Ex. 1.31 Using Grindley and Gibson's equation, determine the viscosity of air for Specific weight of air is 0.0765 lb/ft3. Find the kinematic viscosity for both units.

60oF.

Given: Air at a given condition

t = 60oF

Y

= 0.0765 lb/ft3

Required: . The absolute viscosity using Grindley and Gibson's equation o The kinematic viscosity for both units Solution. Using Grindley and Gibson's equation,

p = 0.0001 702(1+ 0.00329 t + 0.000007

t'?

)

where 1= 91oo -32)=15.56"c 9' p = o.oo01 70211+ 0.0032e(1 5.56) + 0.000007(1 5.56)'?] = o

OOO'' 8

Poise

ln the English unit: 1 Poise = 0.0020885 slug/ft-s

p = (0.00018 Poise) (0.0020885 slug/ft-s/Porse; = 37.59 x 10-8 slug/ft-s

tesson

1

PRINCIPLES OJ FLUID FLOW IT5''

 

For Kinematic Viscosity,

,=f =

=1.5g22x104 ft2ls

\

v =(1.sgx104

ft2ls,I /

lstoke

-

)=0.1a7 stoke

\0.001076ft,1s )

Ex. 1.41 No. 6 fuel oil has a viscosity of 300 SSU and a specific gravity of approximately Determine the kinematic viscosity and absolute viscosity of the oil for both units.

Given: Lr =

t'=

1

.0.

No. 6 fueloil

300 SSU

SG = 1.0

Required: The kinematic viscosity and absolute viscosity in both units Solution: a) For heavy oil, 1 SSF where SSF SSU

= = =

10SSU Saybolt Second Furol Saybolt Second Universal

t' b)

3oo SSU 10

= 3o SSF

For English unit,

,

/

=10.236t'

I

T}

o'

=

[o'r.u(r.)-(# I,

o"

= 6.13 x 1O

6

ft2ls

c) For Sl unit,

v = 0.0022t'

d)

= o.oo22(30)-1 -19 t'\,30

=

0,006 cmr/s or stokes

Solving for the absolute viscosity, p = vp

where

p

=

(1 000) (SG)

=

(1 000) (1.0;

=

1

000 kg/m3

Then, p = (0.006 cm2/s) (0.001 kg/cm3) = 6 x

10-3

= 0.001

kg/cm3

g/cm-s or Poises

e) For English System, 1 Poise = 0.0020885 slug/ft-s

p = (0.006 Poise) (0.0020885 slugift-siPoise; = 1 .2531 x 10 5 slugift-s 16 | Guidebook

in tr'luid Machinery

Jose R. Francisco

 

DEFINITIONS OF TERMS AND OTHER USEFUL INFORMATION Viscometer or viscosimefer is a device used to measure the viscosity of a fluid. Hydrometer is a device used to measure the specific gravity of a fluid. Saybolt viscometer is the most commonly used device for determining the absolute viscosity of liquids.

a a a

Saybolt Seconds Universal (SSU) is the Saybolt time used to measure the

viscosity of fluid. Relationship between SSU and Kinematic viscosity Let t = time, in seconds

r

For t < '100 seconds, SSU = O.22ot

r

For t > 100 seconds, SSU = O.22Ot

o e Ex.

1

-

-

195

-)

t

in centistokes

135 -+ in centistokes t

Conversion of SSU to centistokes, centistokes = 0.308(SSU Saybolt Seconds Furol (SSF) - another viscosity index r 60 SSF = 600 SSU

.51

-

26)

The absolute viscosity of a fluid at atmospheric condition is 6 x 10-3 kg,-s/m2. Find

this viscosity in

a) Reyn b) Poise c) lb,-s/ft2 d) Pa-s

Solution: a) For the viscosity in Reyn,

u=

[u

x 1o 3

Y= )t-#

=8.535 x 106

b)

=[u *

ro'ks*,

p=0 05886 I

=g.535 x 106 Reyn

For the conversion to Poise, u

Lesson

F

)t-*; tTffi'l

r,

)tru)=o

ourru]

.'[##=)=o

.r=0.05886 pa .s

uuuu Poise

PRII{CIPLES Or FLUID FLOW

117

l

 

c)

For the conversion to lb,-s/ft2

u=|.o x 103

' I d)

lb'll'' | =1.22stx1s, + ft' m' /l lkg J[3.28f1 , 9+)f

2'205

For the conversion to Pa-s

,' =lo x 10', ks'='s l[ggU)=0.05886

m' J[ trs, )

[.

{m'

..=0.05886 pa.

s

Ex. 1.61 Given a fluid with an absolute viscosity m between 1-m2 surfaces spaced 1 mm apart, find how fast the surfaces will move relative to each other if a 10-N force is applied in the direction of the surfaces when m is a) 0.001 N-s/m2 (water) b) 0.100 N-s/m2 (a thin oil at room temperature) c) 10.0 Pa-s (syrup; cold oil) d) 108 Pa-s (asphalt) Solution:

F=

10N

\/

z=1mm

a)

Solving for the relative velocity if p = g.96', N-s/m2

" b)

)t-m** )=,

=

[i)[;)= (1#)(-#fl+*

)=

o'o

m/s

Solving for the relative velocity if p = 10 Pa-s

" d)

[i)[; )= [i*]

0,,.

Solvin for the relative velocity if p = 6.100 N-s/m2

" c)

=

=

[i I;

)

=

(i# )(#* ,*

]

=o

oo,

*,.

Solving for the relative velocity if p = 108 Pa-s

, =(il;)=

[#)[#H#)=,0

l8 | Guidebook in Fluid Machinery

",,, Jose R. Francisco

 

Ex. 1.7] Mercury at 20oC has a viscosity of 1 .58 x 10-2 Poise. What is the force necessary to maintain a relative velTitV of 2 m/s between two plates that are separated by 10 cm and * 'rt or. pa s - l whose area is 0.10 m2? | 1 Poise = 0.,10

t'm')

A = 0.10 m'

z=0.10m

Solving for the force necessary to maintain the relative velocity,

N '=.Slli[o.to m'(2 ' m/s)-l 'l=0.00316N

'=u[])=[,ru'x103 m' JL 010m

]

REYNOLDS NUMBER The Reynolds number is a dimensionless parameter used to determine the type of flow of fluid. The Reynolds number equation is as follows:

R"=YDV=PDV where

y p p

D

V

g weight of fluid, N/m3, lb/ft3 = specific = density of fluid, kg/m3, lb/ft3 ir

l.r

= absolute viscosity, poise, Pa-s, gm/cm-s, slug/ft-s = internal diameter of the pipe, m, ft = velocity of fluid, m/s, fps

t r 3:r"r1".'1ff:11:,nJ3,"'"'

m/s2'

rPS2

TYPES OF FLOW OF FLUID 1

.

Laminar flow is the type of flow of fluid in which the fluid pafticles move along straight, parallel paths in layers or laminae, in which R" . 2 000 -+ low velocity.

Figure 1.9. Laminar Flow

Lesson

I

PRINCIPLES OF FLUID FLOW

119

 

Turbulent flow is the type of flow of fluid in which the fluid particles move in a haphazard fashion in all directions. lt is impossible to trace the motion of an individual particle because of its high velocity and variable direction.

2.

,4

R"

,

:}

liti 2iitei.:i{,*it i

4 000 high velocitY

Figure 1.10. Turbulent Flow

3.

Criticalflow (transitional flow) is the combination of laminar flow and turbulent flow.

R"=20001o4000 CONTINUITY EQUATION Continuity equation is an equation derived from the first law of thermodynamics flow for a steady flow, open system. lt is used to determine the mass flow rate and volume rate of fluid.

1. Massflowrate, m = PAV= 2. Volume flow rate, Q = A V where

PQ

m = mass flow rate, kg/s p = density of the fluid, kg/m3

A = cross-sectional area, m2 = velocity of the fluid, m/s o = volume flow rate, m3/s

'l Ex. 1.Bl Air having a density of 1 .01 kg/m3 and an absolute viscosity of .79 x 104 Poise (1.7g r10-5 pa-s) flows through a 30.48-cm diameter pipe at the rate of 1814.0 kg/hr. Determine the type of flow existing in the pipe.

Given: Air flowing inside a PiPe

m=1814.0kg/hr

= 1.01 kg/m3 D = 30.48 cm p = 1.79 x 10-a Poise = 1.79 x

P

20 I Guidebqok in, Fluid ftIachinery

10-5

Pa-s

Jose R. Frencisco

 

Required: The type of flow of fluid inside the pipe Solution: For the volume flow rate,

e=[-' p

(1l14ko/hr)f -1 hr ) "'l36oosJ=o.5om3/s 1.01 kg/m'

For the air velocity,

+(o'sm'ls)-

v=9= a9 A nD2 n(O.SO+a m)'

=6.85m/s

For the Reynolds number,

R" =

+

=

(t ot t