Reynolds Theorem and Applications(Simplified)

8/9/2019 Reynolds Theorem and Applications(Simplified)

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Chapter 4: Fluid KinematicsME33 : Fluid Flow 1

Reynolds—Transport Theorem (RTT

! system is a "uantity o# matter o# #i$ed identity% Nomass can cross a system boundary.

! control volume is a re&ion in space chosen #or study%Mass can cross a control sur#ace%

The #undamental conser'ation laws (conser'ation o#

mass ener&y and momentum apply directly to systems%)owe'er in most #luid mechanics pro*lems control'olume analysis is pre#erred o'er system analysis

There#ore we need to trans#orm the conser'ation laws#rom a system to a control 'olume% This is accomplished

with the Reynolds transport theorem (RTT%



Chapter 4: Fluid KinematicsME33 : Fluid Flow +

!'era&e ,elocity and ,olume Flow Rate

1

c

avg n c

c A

V V dA A= ∫

avg cm V A ρ =&




Mass and ,olume Flow Rates

c c

n c

A A

m m V dAδ ρ = =∫ ∫ &




Reynolds—Transport Theorem (RTT

-eneral RTT non#i$ed C, (inte&ral analysis:

( ) sys

CV CS

dBb dV bV ndA

dt t ρ ρ

∂= +

∂∫ ∫ r rg

Mass Momentum Energy Angularmomentum

. E$tensi'e properties m E* /ntensi'e properties 1 e

mV V

H ( )r V ×




Conser'ation o# Mass rinciple

( ) 0CV CS

d dV V n dAdt ρ ρ + =∫ ∫

r rg

( )

sys

CV CS

dB

b d

V bV ndAdt t ρ ρ

∂

= +∂∫ ∫

r r

g




teady—Flow rocesses



Chapter 4: Fluid KinematicsME33 : Fluid Flow

5ewton6s 7aws

Newton’s laws are relations between motions of bodiesand the forces acting on them.

First law: a body at rest remains at rest, and a body in motion

remains in motion at the same velocity in a straight path whenthe net force acting on it is zero.

Second law: the acceleration of a body is proportional to thenet force acting on it and is inversely proportional to its mass.

Third law: when a *ody e$erts a #orce on a second *ody the

second *ody e$erts an e"ual and opposite #orce on the #irst%




Choosing a Control Volume

Selection of CV can either simplify orcomplicate analysis.

Fixed, moving, and deforming controlvolumes: for mass flow calculation userelative velocity ;

but use absolute velocity for Newton’s Law!




Forces Acting on a CV

Forces acting on CV consist ofbody forcesthat actthroughout the entire body of the CV (such as gravity,electric, and magnetic forces) andsurface forcesthatact on the control surface (such as pressure and viscousforces, and reaction forces at points of contact).

•Body forces act on eachvolumetric portiondVof the CV.

•Surface forces act on each

portiondAof the CS.




.ody Forces

The most common *ody #orceis &ra'ity which e$erts a

downward #orce on e'ery

di##erential element o# the C,

The di##erent *ody #orce

Typical con'ention is that

acts in the ne&ati'e z ;direction

Total *ody #orce actin& on C,




ur#ace Forces

ur#ace #orces are not as simple toanaly<e since they include *oth normal

and tan&ential components

=ia&onal components σ xx , σ yy , σ zz are

called normal stresses and are due to

pressure and 'iscous stresses>##;dia&onal components σ xy , σ xz , etc%

are called shear stresses and are due

solely to 'iscous stresses

Total sur#ace #orce actin& on C



Chapter 4: Fluid KinematicsME33 : Fluid Flow 1+

7inear Momentum E"uation

5ewton6s second law #or a system:

?se RTT with b = V and B = mV:




pecial Cases

teady Flow !'era&e 'elocities

!ppro$imate momentum #low rate

To account #or error use momentum;#lu$

correction #actor β




ro*lem

For the el*ow duct !E3 oil at +@C (speci#ic wei&ht is 8+ 5Am3enters section 1 at 30 5As where the #low is laminar and e$its at

section + where trhe #low is tur*ulent: u1B,ma$1(1;(rAR1+

u+B,ma$+(1;(rAR+(1A% !ssumin& steady incompressi*le #low

compute the #orce and its direction o# the oil on the el*ow due to

momentum chan&e only (no pressure chan&es or #riction e##ects%=1B1 cmD =+B2 cmD an&leB3@%




ro*lem

-ra'el is dumped #rom a hopper at a rate o# 20 5As onto a mo'in&*elt as in Fi&% The &ra'el then passes o## the end o# the *elt% The

dri'e wheels are 8 cm in diameter and rotate clocwise at 10

rAmin% 5e&lectin& system #riction and air dra& estimate the power

re"uired to dri'e this *elt%




!n&ular Momentum

Motion of a rigid body can be considered to be thecombination of

the translational motion of its center of mass (U x, U y, U z)

the rotational motion about its center of mass (ω x,ω y,ω z)

Translational motion can be analyzed with linearmomentum equation.

Rotational motion is analyzed with angular momentumequation.

Together, the body motion can be described as a 6–degree–of–freedom (6DOF) system.




Re'iew o# Rotational Motion

!n&ular 'elocity ω is thean&ular distance θ tra'eled per unit time and

an&ular acceleration α is

the rate o# chan&e o#an&ular 'elocity%




Re'iew o# !n&ular Momentum

Moment o# a #orce:Moment o# momentum:

For a system:

There#ore the an&ular momentum e"uation can

*e written as:To deri'e an&ular momentum #or a C, use RTT

with and




!n&ular Momentum E"uation #or a C,

-eneral #orm

!ppro$imate #orm usin& a'era&e

properties at inlets and outlets

teady #low




ro*lem

The hori<ontal lawn sprinler in Fi& has a water #low rateo# 4% &alAmin introduced 'ertically throu&h the center%

Estimate (a the retardin& tor"ue re"uired to eep the

arms #rom rotatin& and (* the rotation rate (rAmin i#

there is no retardin& tor"ue%

d B in

R B 2 in



Chapter 4: Fluid KinematicsME33 : Fluid Flow +1

Mechanical Ener&y

Mechanical ener&y chan&e o# a #luid durin&incompressi*le #low

( )2 2

2 1 2 1

2 1

2

mech

P P V V e g z z

ρ

− −∆ = + + −



Chapter 4: Fluid KinematicsME33 : Fluid Flow ++

-eneral Ener&y E"uation

Recall &eneral RTT

G=eri'eH ener&y e"uation usin& B=E and b=e

.rea power into rate o# sha#t and pressure wor

( ) sys

r CV CS

dB d bdV b V n dA

dt dt ρ ρ = +∫ ∫

r rg

( ), ,

sys

net in net in r CV CS

dE d Q W edV e V n dA

dt dt ρ ρ = + = +∫ ∫

r r& & g

( ), , , , , , ,net in shaft net in pressure net in shaft net inW W W W P V n dA= + = − ∫ r r& & & & g





Ihere does e$pression #or pressure worcome #romJ

Ihen piston mo'es down ds under thein#luence o# F=PA the wor done on thesystem is δ W boundary =PAds.

/# we di'ide *oth sides *y dt we ha'e

For &enerali<ed control 'olumes:

5ote si&n con'entions:

is outward pointin& normal

5e&ati'e si&n ensures that wor done ispositi'e when is done on the system%

pressure boundary piston

dsW W PA PAV

dt δ δ = = =& &

( ) pressure nW PdAV PdA V nδ = − = − ×

r&

n





Mo'in& inte&ral #or rate o# pressure worto R) o# ener&y e"uation results in:

Recall that P/ ρ is the flow work, which isthe wor associated with pushin& a #luidinto or out o# a C, per unit mass%

( ), , ,net in shaft net in r

CV CS

d P

Q W edV e e V n dAdt ρ ρ

+ = + + × ÷ ∫ ∫

r r&





!s with the mass e"uation practical analysis iso#ten #acilitated as a'era&es across inlets and

e$its

ince e=u+ke+e = u+V ! /!+"z

( )

, , ,

C

net in shaft net in

out inCV

c

A

d P P Q W edV m e m e

dt

m V n dA

ρ

ρ ρ

ρ

+ = + + − +

÷ ÷ = ×

∑ ∑∫

∫

& & &

r r

2 2

, , ,2 2

net in shaft net in

out inCV

d P V P V Q W edV m u gz m u gz

dt ρ

ρ ρ

+ = + + + + − + + + ÷ ÷

∑ ∑∫ & & &




Ener&y !nalysis o# teady Flows

For steady #low time rate o# chan&e o# the

ener&y content o# the C, is <ero%This e"uation states: t#e net rate o$ ener"ytrans$er to a %V by #eat and &ork trans$ersdur'n" steady $(o& 's e)ua( to t#e d'$$erence

bet&een t#e rates o$ out"o'n" and incomin&ener&y #lows with mass%

2 2

, , ,2 2

net in shaft net in

out in

V V Q W m h gz m h gz + = + + − + + ÷ ÷ ∑ ∑& &





For single-streamdevices mass #low rate

is constant%

( )

( )

2 2

2 1, , , 2 1 2 1

2 2

1 1 2 2

, , 1 2 2 1 ,

1 2

2 2

1 1 2 2

1 2 ,

1 2

2

2 2

2 2

net in shaft net in

shaft net in net in

pump turbine mech loss

V V ! h h g z z

P V P V ! gz gz u u

P V P V gz ! gz ! e

ρ ρ

ρ ρ

−+ = − + + −

+ + + = + + + − −

+ + + = + + + +





=i'ide *y " to &et each term in units o# len&th

Ma&nitude o# each term is now e$pressed as an

e"ui'alent column hei&ht o# #luid i%e% *ead

2 2

1 1 2 2

1 2

1 22 2

pump turbine "

P V P V z h z h h

g g g g ρ ρ + + + = + + + +




The .ernoulli E"uation

/# we ne&lect pipin& losses and ha'e a system withoutpumps or tur*ines

This is the Bernoulli equation

/t can also *e deri'ed usin& 5ewtons second law o#motion (see Cen&el te$t p% 18%

3 terms correspond to: tatic dynamic and hydrostatichead (or pressure%

2 2

1 1 2 2

1 21 22 2

P V P V z z

g g g g ρ ρ + + = + +




)-7 and E-7

/t is o#ten con'enient

to plot mechanical

ener&y &raphically

usin& hei&hts%

)ydraulic -rade 7ine

Ener&y -rade 7ine(or total ener&y

P H#" z

g ρ = +

2

2

P V E#" z

g g ρ = + +





The Bernoulli equation is an arox'mate re(at'on

bet&een ressure,

e(oc'ty and e(eat'on

and 's a('d 'n re"'ons o$

steady, 'ncomress'b(e

$(o& &#ere net $r'ct'ona(

$orces are ne"('"'b(e.

E"uation is use#ul in #low

re&ions outside o#*oundary layers and

waes%



Chapter 4: Fluid KinematicsME33 : Fluid Flow 3+


7imitations on the use o# the .ernoulli E"uationteady #low: d/dt B

Frictionless #low

5o sha#t wor: wpumpBwtur*ineB

/ncompressi*le #low: ρ B constant

5o heat trans#er: )net,'nB

!pplied alon& a streamline (e$cept #or irrotational

#low




ro*lem

Ihen the pump in Fi&% draws ++ m3Ah o# water at +@C #rom the reser'oir the total

#riction head loss is 0 m% The #low dischar&es throu&h a no<<le to the atmosphere%

Estimate the pump power in I deli'ered to the water%




ro*lem

.ernoulli6s 138 treatise *ydrodynam'ca contains many e$cellent setches o# #low

patterns% >ne howe'er redrawn here as Fi&% seems physically misleadin&% Ihat is

wron& with the drawin&J




ro*lem

In the spillway fow o Fig., the fow is assumed uniorm

and hydrostatic at sections 1 and 2. I losses areneglected, compute (a) V 2 and (b) the orce per unit widtho the water on the spillway.

Reynolds Theorem and Applications(Simplified)

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