class notes 1.pdf

Fluid MechanicsHydraulic Engineering

ZEIT 2503ZEIT 2602ZEIT 2602

Harald KleineRobert Niven

S h l f E i i d I f ti T h lSchool of Engineering and Information TechnologyUNSW@ADFACanberra, ACT

1ZEIT 2503 Fluid Mechanics

OverviewI. Fundamental equations

a) differential formb) integral formb) integral form

II. Dimensional analysisIII. Internal flow of real fluids

a) review of fluid propertiesb) Couette flowsc) pipe flowsc) pipe flows

----------------------------------------------------------------------------------------------------------------------------------------

IV. External flow of real fluidsa) boundary layers and the momentum integral methodb) pressure drag

V TurbomachineryV. Turbomachinerya) design calculations for pumps and turbinesb) pump performance

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VI. Introduction to compressible flows

ZEIT 2503 Fluid Mechanics

I. Fundamental Equations

I.1 Fundamentals – what we want to determine and howI.2 Differential form of the conservation equations

I.2a Mass conservationI.2b Momentum conservationI.2c Energy conservation

I.3 Integral form of the conservation equationsI.3a Mass conservationI.3b Momentum conservationI.3c The Bernoulli equationI.3d Energy conservation

Aims:- introduce main parameters that describe a fluid- introduce the concept of control volume analysis- introduce fundamental conservation equations- introduce the concepts of local and convective change (Reynolds Transport Theorem)

- use the integral form of the conservation equations to calculate


“what a fluid does”


I.1 Fundamentals – what we want to determine and howneed to know: velocity vector V

pressure pdensity

consider an infinitesimallysmall volume of fluid1)

density temperature T

six unknowns need six equations

in a flowingmedium

need six equations

conservation of mass 1conservation of momentum 3

ti f 1conservation of energy 1equation of state 1

6

- also: need “material properties” such as viscosity and heat conductivity as functions of pressure and temperature.

often made assumption: these values are constant- often made assumption: these values are constant

- also require information on work input/output and heat transfer (if present)

1) this volume is infinitesimally small compared to any macroscopic dimension of the fluid but it contains


1) this volume is infinitesimally small compared to any macroscopic dimension of the fluid but it contains a large number of molecules so that it can be considered as a “piece of continuous matter”.


Derivation of the conservation equationsDerivation of the conservation equationsby Control Volume Analysis

Choose a control volume (CV) around the area of interest (bounded by the control surface) and determine the changes that this volume undergoes (how does it change with time, is there matter or energy crossing the control surface …?)

conservation equations (mass, momentum, energy)

main advantage: we do not have to know anything about the processes inside the CV but only havewe do not have to know anything about the processes inside the CV, but only have to do an accounting job of what happens at its boundaries

at first: h i fi it i ll ll l d d i th ti i- choose an infinitesimally small volume and derive these equations in

differential form- the resulting equations can be applied to any flow but have to be integrated (with appropriate boundary conditions) to obtain useful results(with appropriate boundary conditions) to obtain useful results

then:- if it is known that the flow properties are constant over certain parts of the CV,

5

one can directly apply the integrated form of these equations


I. Fundamental Equations: Continuity Equation

I.2 Differential form of the fundamental equations

I.2a Mass conservation (continuity equation)

Mass can neither be created nor destroyed.

q

y

mass entering the control volume = mass leaving the control volume

l t d i th t l lplus mass stored in the control volume

control volume (CV):d VolCV = dx dy dz

velocity components:u, v, w

mass stored in CV:

ddd)Vold( CV

6

zyxtt

ddd)Vold( CV



zyxuzyuzyxuzyu ddd)(ddddd)(dd

net mass flux in x - direction: (out minus in)

xx

ddd)(

zxyvy

ddd)(

yxw ddd)(

net mass flux in y - direction:

t fl i di ti

7

yxzwz

ddd)(

net mass flux in z - direction:



summing up:

0dddddd)()()(

zyxt

zyxwz

vy

ux

summing up:

0)()()(

wz

vy

uxt

(1.1)

The inner vector product (a scalar quantity) is known as the

… a bit of Math …

p ( q y)divergence of vector A

AAAA zyx

kji

(If A is a scalar and not a vector, the combination A is known as

zyxA

k

zj

yi

x

gradient of A)

ti it ti i t f 0)(

Vwith V = (u, v, w)

(1 2)

8

continuity equation in vector form: 0)(

Vt

(1.2)



in words:the overall net mass flux

(= sum of the net fluxes in each direction) must be equal to the rate of change of mass

within the infinitesimally small volumewithin the infinitesimally small volume.

in pictures:

drawing from:J. Strybny, O. Romberg:

9

y y, gOhne Panik StrömungsmechanikVieweg, 2007



Summary:Summary:

net mass flux (per volume) into the control volume

= mass (per volume) stored in the control volume

0)(

Vt

(1.2)

special case: incompressible fluid ( = const )

0 V

special case: incompressible fluid ( const.)

(1.3)


I. Fundamental Equations: Compressibility Effects

photo by J. Amann

video byG.S. Settles

11

Penn State University


I 2b Momentum conservation

I. Fundamental Equations: Conservation of Momentum

I.2b Momentum conservation

The overall sum of external forces acting on a system is equal to the time rate of change of the linear momentum of the system.g y

(Newton’s Second Law)

linear momentum: mV

Newton’s Second Law: FVm )(d (1.4)e o s Seco d a FVmt

)(d

i l d

(1.4)

special case m = const.: FamVt

m )(d

d

a : acceleration

(1.5)

- now we have to evaluate each side of the equation for the case of fluid motion, i.e., we have to find expressions for F and a

12

, p


I. Fundamental Equations: Newton’s Laws

Newton’s First LawUnless there is a net force acting on a body, its velocity (both magnitude and direction) remains unchangedmagnitude and direction) remains unchanged.

Newton’s Third LawNewton s Third LawTo every action (force) there is always an equal reaction (force).

drawing from:L.W. Dubeck,S.E. Moshier,J E Boss:

13

J.E. Boss:Fantastic VoyagesSpringer, 2004


I. Fundamental Equations: Newton’s Laws



first: LHS of equation: find expression for acceleration ap of a fluid particle

at time t :particle is at locationr = (x y z)r = (x, y, z)

at time t + dt :particle is at positionparticle is at positionr + dr = (x+dx, y+dy, z+dz)

Th l it f th ti l h h d f V ( t) V( t)The velocity of the particle has changed from Vp(r, t) = V(x, y, z, t) to Vp(r + dr, t + dt) = V(x + dx, y + dy, z + dz, t + dt)

change in velocity dV when moving from r to r + drchange in velocity dVp when moving from r to r + dr chain rule

tt

Vz

z

Vy

y

Vx

x

VVd PPPP dddd

15

tzyx



tt

Vz

Vy

Vx

VVd PPPP dddd

- change in velocity dVp when moving from r to r + dr

tzyxP

tVzVyVxVV ddddd

- then the acceleration is

t

t

t

V

t

z

z

V

t

y

y

V

t

x

x

V

t

Va PPPP

P d

d

d

d

d

d

d

d

d

d

h

wt

zv

t

yu

t

x ppp d

d,

d

d,

d

dwhere

VVVw

Vv

Vu

Va P Dd

so

(1 6)ttz

wy

vx

ut

a P Dd

this shows that the acceleration has two main parts:

(1.6)

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- this shows that the acceleration has two main parts:one due to a change in position and one due to change in time



acceleration ap of a fluid particle

VVVw

Vv

Vu

Va P

P

Dd

(1.6)

ttzyxtP Dd ( )

convective acceleration local accelerationconvective acceleration local acceleration

total or substantialacceleration of a particleaka acceleration of a particleaka

Reynolds Transport Theorem

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Theorem



rewriting the convective acceleration in vector form

VVVVV

)(

VVz

wy

vx

u )(

(1.7)

VVVa

VP

)(

Dleads to

(1.8)

tVVa

t P )(

D( )

Even for a steady flow (= all derivatives wrt t are zero), the acceleration of a fluid particle can be (and most often is) non zero because of the convective termnon-zero because of the convective term.

Can you think of a simple everyday-life example for this ?



LHS of Newton’s Second Law was

VV

wV

vV

umV

m dD

d

this must be equal to the force dF acting on the fluid element

tz

wy

vx

umt

m dD

d

- this must be equal to the force dF acting on the fluid element find the force and its components dFx , dFy , dFz .

… and this leads to nothing else but …

a free-body-diagram !

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here:only forces in x - direction



forces acting on a fluid element (excluding esoteric applications):body forces – gravitysurface forces – stresses

- nomenclature of indices:first index indicates the plane in which the stress actssecond index indicates the direction in which the stress actssecond index indicates the direction in which the stress acts

zx

acts inx - direction

acts on x-y plane,

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normal to z - direction here:only forces in x - direction



summing up in x direction:summing up in x - direction:

- canceling equal terms:g q

net surface force in x - direction

zyxF zxyxxx dddd S

21

zyxzyx

F ddddxS

here:only forces in x - direction



FFF zxyxxx dddddd

net overall force in x - direction

(1 9a)zyxzyx

gFFF zxyxxx ddddd d

xx SBx

(1.9a)

analogously, in y- and z-direction

yxgFFF zyyyxy dddddd

(1 9b)

zyxgFFF

zyxzyx

gFFF

zzyzxz

yyyyy

dddddd

ddddd dyy SBy

(1.9b)

(1 9c)zyxzyx

gFFF z ddddddzz SBz

(1.9c)

- now each side of Newton’s second law has been rewritten for the caseof a fluid element / infinitesimally small control volume



putting it all together gives

(1.10)

f f l tisum of forces = mass acceleration

the differential equations of motion f fl id l t

23

for a fluid element



these relations are not yet useful as they introduce new variables for the stresses need expressions for and to relate them to the given unknowns

St k ’ H th iStokes’ Hypothesis:

ep xxx ~2

and are material-specific proportionality constants

ep yyy ~

~2 V

z

w

y

v

x

ue

ep zzz 2

wvu

velocity divergence

z

w

y

v

x

uzyx

,,

dilatation velocity

hydrostatic pressure

dilatation velocity

terms that describe how thefl id i ld t

24

y pfluid yields to pressure



~observation:

solid elastic body: fluid: ~

dt

d

t~

xzxzyzyzxyxy

etc. are the shear velocities

is a proportionality constant

xy

is a proportionality constantknown as viscosity (more on this in section III.)

x

v

y

u

txy

d

dd

25

y



Combining all the above and assuming that leads to

the NAVIER-STOKES equations

3/2~

the NAVIER-STOKES equations

(1.11)



NAVIER-STOKES equations

m a

gravityforce viscous forces

pressure force

Fam 27ZEIT 2503 Fluid Mechanics


The Reynolds Transport Theorem (review)

from solid mechanics: object of interest is a (closed) systemfrom solid mechanics: object of interest is a (closed) system= quantity of matter of fixed identity

Approach: follow this system and analyse the forces etc. acting on itand determine the system’s reactionand determine the system s reaction (aka Lagrangian approach)

in fluid mechanics: very often highly impractical to use the system descriptiony g y p y p

- more practical to use a control volume (CV) approach= define region in space chosen for study and determine what

h i thi i (i dditi t ti f d ith ithappens in this region (in addition to acting forces, mass and with it momentum and energy may pass through the CV boundaries)

(aka Eulerian approach)

The Reynolds Transport Theorem is the link between these two approaches:Use relations from solid mechanics,

but express change as the sum of local and convective change


but express change as the sum of local and convective change.

I. Fundamental Equations: Fluid Kinematics


I 2c Energy conservation

I. Fundamental Equations: Conservation of Energy

I.2c Energy conservationEnergy may be changed from one form to the other (e.g., from kinetic to potential or to internal or vice versa), but the overall amount of energy associated with a volume of fluid is constant(if there is no heat transfer or work done by/on the volume).

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videos by G.S. Settles, Penn State University



Three forms of energy per volume:internal energy ekinetic energy ½ V 2

t ti lpotential energy g z

These energies are related to each other via the

First Law of Thermodynamics:If heat ( ) is added to or taken from a closed system (= a fixed amount of matter contained within a closed boundary) and if work (W ) is done on or by th t th V l f th t h di tthe system, the energy E = Vol. e of the system changes according to:

EW dδδ Sign convention:

(1.12)g

heat is added to the system: positive work is done by the systemto its surroundings: W positiveg

“ ” for and W indicates that these increments are dependent on the process (= how they are accomplished); symbol “d ” is used for dE because increment is independent of the process E is a thermodynamic property or state variable of the fluid

31

independent of the process. E is a thermodynamic property or state variable of the fluid, and W are not – they are related to processes.



two forms of heat addition :two forms of heat addition :volumetric (e.g., by combustion processes within the system) across the surface of the control volume (e.g., by thermal conduction and

mass diffusion):mass diffusion):

surfvol

work on or by the system is related to eithervolume (body) forces and/orsurface forces (pressure and shear stress)( )

shearpressbody WWWW

any change of energy must be associated with either work or heat added or taken from the system (the minus sign for the work terms indicate the different sign conventions for work and heat transfer): )

bodyshearpresssurfvol WWWt

E D

D(1.13)



the total change in energy has again two components – a local and a convective contribution

(derivation and rationale analogous to discussion of acceleration in previous section)

VV 22

(derivation and rationale analogous to discussion of acceleration in previous section)

VV

gzeV

gzet

22

22

(1.14)

bodyshearpresssurfvol WWW

sum of all heat flux and work

total (substantial) change of energy

sum of all heat flux and work passing through the system



- in the energy equation a new variable appeared – internal energy- in the energy equation a new variable appeared – internal energy- related to temperature T via

Tce vcv : specific heat capacity

(1.15)

for gases of “moderate” temperatures (< 1000 K) constant, but for higher temperatures cv = cv(T, p)

- still need another equation that links T to the already introducedstill need another equation that links T to the already introduced thermodynamic properties p and :

Equation of State

f f t TRp (1 16)for perfect gases: TRp R : another material constant (specific gas constant)

(1.16)

for liquids: only empirical relations, depending on range of p and T

(1.14) will be needed when heat transfer or work input/losses have to be considered

important for all fluid machinery (pumps, turbines, propellers …)

i t t f ll fl id i hi h th t t h ( important for all fluid processes in which the temperature may change (e.g. temperature distribution within the lubricant of a journal bearing)


I. Fundamental Equations – Integral Form

I 3 Integral form of the fundamental equationsI.3 Integral form of the fundamental equations

- the previous discussion has yielded differential equations for the motion of fluid elements

- to obtain useful results for flows of practical interest, these equations have to be integrated (with appropriate boundary conditions)

i h fl ti k ( d) t b t t ( t i- in cases where flow properties are known (or assumed) to be constant (or to vary in a known functional dependence) over parts of a macroscopic control volume, the conservation equations can be used in their integrated form

- control-volume approach - choose a control volume around the area of interest (bounded by the control surface) and determine the changes that this volume undergoes (how does it change with time, is there matter or energy crossing the control surface …?)



Control Volume RulesRule # 1:Draw the control volume !!

Rule # 2:The boundaries of the control volume are arbitrary – you can choose any shape you y y y p ylike. However, since you will have to make an accounting balance of forces and fluxes along and across the boundaries, resp., it is advisable to select the control volume in such a way thata) the boundaries should only be in regions of the flow where you know the flow

properties (velocities, pressure …)b) mass fluxes are normal to the boundary (if possible)) if h CV b d i lid b d f h fl fi ld ( h ll)c) if the CV-boundary intersects a solid boundary of the flow field (such as a wall) you

have to introduce a compensating holding force

Th t l l l i i ll li bl t l d b d f i t llThe control volume analysis is a generally applicable tool and can be used for virtually any kind of flow.


general conservation equation for a control volume CV with surface CS:


general conservation equation for a control volume CV with surface CS:

N : a generic flow quantity (mass, momentum or energy) within CV

n : the same flow quantity per unit massn : the same flow quantity per unit mass

N = n dVol (1.17)

: density

Total change of N in CV consists of two parts:1) the amount of N that is stored in the control volume local change1) the amount of N that is stored in the control volume - local change2) the net flux across the surface CS (out minus in) – convective change

D NN )d(D

D

CSCVtotal

AVnt

N

t

N

(1.18)



flux:

Vn : velocity normal to the control surface (tangential velocity Vt need not be considered as fluid with Vt does not cross the

AVAV ndd

t t

control surface)

relevant component for transfer across controltransfer across controlsurface

sign convention:if h l i d hif the velocity component Vn and the vectornormal to the surface dA are pointing in thesame direction, the product Vn dA is positive(t i ll f fl l i th CV)(typically for flow leaving the CV)

+ : leaving CV: entering CV

38

- : entering CV


I. Fundamental Equations – Integral Form: Conservation of Mass

I 3a Mass conservation: mass cannot be generated or destroyedI.3a Mass conservation: mass cannot be generated or destroyed total change must be zero

mass entering the control volume = mass leaving the control volume= mass leaving the control volume

+ mass stored in the control volume

formally:N = m n = m/m = 1N = m, n = m/m = 1

0d1D

D !

CVl

AVt

m

t

m AVnt

N

t

Nd

D

D

0d AV

m

CSCVtotalttDCSCVtotal

(1 19)

special cases:

0dCSCV

AV

t

AV d0

(1.19)

(1 19 )steady flow in = out

i ibl fl ( )

AV d0CS

(1.19a)

39

incompressible flow ( = const.) AV d0CS (1.19b)



average velocity (means that flow can be treated as one dimensional)

AVAV d (1.20)

average velocity (means that flow can be treated as one-dimensional)

CS

AVAVQ dflow rate [m3/s]

CS

(1.21)

Example 1.1:Determine the average velocity of the given pipe flow with the following velocity distribution V(r) = 5 (1 - (r / R)2) [m/s](R i di )(R: pipe radius)


result: Vavg = 2.5 m/s


Example 1 2:Example 1.2:A circular swimming pool ( 5m) is filled with two garden hoses (inside 17 mm), with flow velocities of 2 m/s and 1.5 m/s, resp. Calculate the time required to fill the pool to a depth of 2 mthe pool to a depth of 2 m.

density of water: = 1000 kg/m3

result: t = 13.7 h

Example 1.3:In most metalworking shops plate steel is cut by means of an oxyacetylene torchIn most metalworking shops, plate steel is cut by means of an oxyacetylene torch. Oxygen for cutting is supplied via tanks 30 cm in diameter and 1.3 m tall. These tanks are charged to an internal pressure of 13 782 kPa. A 12.5 mm diameter valve is located at the top of the tank If the tank valve is opened fully oxygenvalve is located at the top of the tank. If the tank valve is opened fully, oxygen escapes at 1.5 m/s. Assuming that this exit velocity is constant, determine the tank pressure after 60 s. Take the temperature in the tank to be unchanging and equal to = 25C and assume ideal gas behaviour

41

equal to 25 C and assume ideal gas behaviour.Oxygen: R = 260 J/(kg K) T [K] = 273.15 + [C]


result: p (60s) = 12 227 kPa

I 3b Momentum conservation

I. Fundamental Equations – Integral Form: Conservation of Momentum

I.3b Momentum conservationchange in momentum must equal an external force

linear momentum VmVmnVmN /,

d

)(

D

)(DAVV

t

Vm

t

Vm total change:

Newton’s second law:

CSCVtotal

D tt

)()(D VmVm

CSCVtotal

d)(

D

)(DAVV

t

Vm

t

VmF

N t ’ d l

(1.22)

Newton’s second law:All forces applied externally to the control volume (gravity, friction, pressure, surface tension, electric & magnetic …) result in and are equal to a corresponding change of (linear) momentumof (linear) momentum.

Special case:steady flow: d AVVF (1.22a)

42

steady flow: CS

(1.22a)


(1.22) or (1.22a) give one equation in each direction of the coordinate system


( ) ( ) g q y(subscripts denote directions). Note that the sign of the integral term depends on whether the flow enters or leaves the CV and the direction of the velocity itself !

Example 1 4:Example 1.4:Water from a garden hose ( 15 mm) hits a wall with a velocity of 2 m/s. Determine the force exerted by the water on the wall, when the angle between the water jet and the wall is 90 (2)and the wall is 90 .

density of water: = 1000 kg/m3

(1)

(2)

CS

d AVVF

(1)

)(- 121z111z2z222 zzzz VVmVAVVAVF

CS

in z – direction :

(1.23)where 222111 VAVAm

)( VVVAVVAVF

43

)(- 121r111r2r222 rrrr VVmVAVVAVF in r – direction :


result: Fz = 0.707 NFr = 0 N

I. Fundamental Equations: Bernoulli Equation

I.3c The Bernoulli equationq

p + gh + ½ V 2 = const.follows from either the energy conservation equation for steady incompressible flow when there is no heat transfer and no external work except that by pressure

(1.24)

flow when there is no heat transfer and no external work except that by pressure forces, or from the momentum conservation equation when there are no friction forces

- it represents the sum of all mechanical energies in the flow and states that in theit represents the sum of all mechanical energies in the flow and states that in theabsence of irreversible losses, this sum remains constant, i.e., energies can be distributed into different mechanical forms (kinetic, potential, pressure), but the overall amount remains the same

Note: Bernoulli’s equation is only valid as long as no loss- or energy-generating elements are encountered in the flow. It cannot be applied across sharp-edged inlets abrupt area changes or elements that change the energy of the flow (suchinlets, abrupt area changes, or elements that change the energy of the flow (such as fans, pumps …). However, in most cases, useful information can be found by applying Bernoulli’s equation from a reservoir condition up to the loss- or energy-generating element.generating element.

The Bernoulli equation is derived from the momentum conservation equation by integration along a streamline – hence it can only be applied along but in general not across a streamline

44

not across a streamline.


I. Fundamental Equations: Bernoulli Equation

Summary Bernoulli equation:- can only be used when

- viscous effects are negligiblefl i t d d i ibl- flow is steady and incompressible

- flow is along a streamline

The constant of integration in the Bernoulli equation can be evaluated if at oneThe constant of integration in the Bernoulli equation can be evaluated if at one location along the streamline sufficient information about the flow is available.

The Bernoulli equation can be considered as a special case of the general energyThe Bernoulli equation can be considered as a special case of the general energy equation – it represents the balance of mechanical energies.

Example 1 5:Example 1.5:Water flows from a faucet (area 2 cm2) with a velocity of 2 m/s.Further downstream, the column of water narrows downFurther downstream, the column of water narrows down to an area of 1 cm2. At which distance from the faucet does this happen ?


result: z = 0.612 m


Example 1.6:A water jet of velocity Vj = 3.5 m/s and a cross-sectional area of 0.5 m2 strikes a curved vane as shown. The vane is moving at a velocity of Vv = 1 m/s in the positive x-direction, and it deflects the jet through an angle of 60. Assuming no frictional l b t th j t d th f dlosses between the jet and the surface and negligible height differences, determine the reaction forces Fx and Fy .

density of water: = 1000 kg/m3 result: Fx = 1562.5 N

Example 1.7:A 45 reducing elbow can often be found in domestic water piping systems. Water flows

density of water: 1000 kg/m x

Fy = 2706.3 Ndirections as shown

into the elbow and is deflected through an angle of 45. The inlet diameter is 25 mm and the outlet diameter is 12 mm. The volume

3flow rate of water is 0.0008 m3/s. The inlet and outlet pressures are, respectively, 160 kPa and 150 kPa. If the elbow is located i h i t l l h d t i th result: F = 59 1 N

46

in a horizontal plane as shown, determine the force exerted on it by the moving water.


result: Fx = 59.1 NFy = 12.2 N

directions as shown


Example 1.8:A vertical jet of water is issued from a nozzle (area A0) with velocity V0. It impinges on a horizontal disk as shown. This disk has a mass M and can only move in vertical direction.Obtain an expression for the speed V(h) of the water jet as a function of height habove the nozzle exit, and find the height to which the disk will rise and remain t ti Vi ff t b i d Al i it t tstationary. Viscous effects can be ignored. Also given: gravity constant g .

A0

result:

2

20

1

2)(

M

ghVhV


00

202

1

AV

MgV

gh

I. Fundamental Equations – Integral Form: Conservation of Energy

I.3d Conservation of energy:gychange in energy is caused by work and heat transfer

N = E (energy of the system) (specific energy = E/m)

AVnt

N

t

Nd

D

D

CSCVtotal

EE 1D

gzVem

En 2

2

1

AVgzVet

E

t

Ed

2

1

D

D

CS

2

CVtotal

thi h f h t b l t th ll t f k d h t t f- this change of energy has to be equal to the overall amount of work and heat transferdone on the control volume

- for all further calculations in this course we assume:the flow is steady and incompressible- the flow is steady and incompressible

- there is no heat transfer

- if we consider a CV with an inlet and outlet area and if we assume that all properties

VVE

D 22

- if we consider a CV with an inlet and outlet area and if we assume that all propertiesare constant over these areas, the energy change becomes


QgzV

egzV

et

E

inout

22D

D(1.25)


h Q i th l t i fl t- here Q is the volumetric flow rate

(1.26)outoutinin AVAVVAQ

V : velocity ; A : area

- the RHS of the energy equation becomes, in the absence of heat transfer :

bodyshearpressbodyshearpresssurfvol WWWWWW bodyshearpressbodyshearpresssurfvol

rate of work done by pressure forces

rate of work done by friction forces

rate of work done bybody forcesby pressure forces by friction forces body forces

- the work done by body forces typically represents so-called shaft work i.e. work provided by some (usually moving) mechanical device

- the work done by pressure and friction forces can be expressed in the following way:

work = force displacement rate of work = force displacement / timef l it


= force velocity


t f k d b frate of work done by pressure force:- the magnitude of the pressure force is - the direction of the pressure force is always normal on the surface, pointing towards

the surface

ApFpress dd

- the resulting rate of work is a scalar whose value depends on the orientation that the area vector and the velocity vector have with respect to each other

the surface- the rate of work done by the pressure force is (1.27)VApWpress dd

that the area vector and the velocity vector have with respect to each other.- if the area vector has the same direction as the velocity (= flow enters CV surface

at right angle), then VApWpress dd (1.27a)

where the sign depends on whether the pressure force and the flow velocity are pointing in the same (+) or in opposite (-) direction

rate of work done by friction force:- the magnitude of the friction or shear force is - the direction of the friction force is always parallel to the surface, along a vector t

AFshear dd y p , g

that has the magnitude “1” and that is perpendicular to the area vector dA- the rate of work done by the friction force is

VtAWtAF )d(d)d(d (1 28)


VtAWtAF shearshear )d(d)d(d (1.28)


There are two cases in which the rate of work done by the friction force is zero, even if the friction force itself is non-zero.

by choosing the CV properly, the work contributions of the friction forces canbe removed

- along solid walls, V = 0 and hence regardless of the value of

- at ports (= areas where flow is crossing the boundaries of the CV), the value of

0d shearW

( g )shear stress rate of work depends on the orientation that the shear stress vector and the velocity vector have with respect to each other

- if V is parallel to dA (equivalent to the flow passing through the surface at a right p ( q p g g g

angle), then it is automatically perpendicular to t so that

again regardless of the value of 0d0 shearWVt

again regardless of the value of

in both cases, the RHS of energy equation is simplified



i isummarising

AVpWWAVgzVet

E

t

Eshearbody dd

2

1

D

D 2

- the term of the rate of work done by pressure forces is often taken to the LHS ofthe equation as it also represents an area integral of the same structure as the

tt y2D CSCSCVtotal

the equation as it also represents an area integral of the same structure as theconvective energy term:

(1 29)WWAVVpE

d1 2

- the combination e + p/ is called the enthalpy

(1.29)shearbody WWAVgzVp

et

d

2CS

2

CV

peh (1 30)the combination e + p/ is called the enthalpy

- for the case of steady flow entering and leaving the CV at right angles and all

eh (1.30)

y g g g gproperties being constant over the port areas, (1.29) simplifies to

QgzV

hgzV

hW

22

(1 31)


QgzhgzhWinout

body

22

(1.31)


(1 31) tifi h th t f k b ht i t ( t k t f) th CV h th(1.31) quantifies how the rate of work brought into (or taken out of) the CV changes theenergy contained in the CV. This energy change has two parts:

mechanical energies: (1.32)QgzVp

gzVp

22

mechanical energies:

internal energies:

( )

(1.33)

Qgzgzinout

22

inoutvinout TTcQeeQ

- the change of internal energies is usually related to a temperature change as internalenergy often depends only on temperature (see (1.15))

- if such a change of internal energy occurs, it is at the expense of the mechanical energies

- this is most easily shown (and quantified) for a steady incompressible flow of a fluid in an adiabatic flow section where no shaft work is applied

- choose an adequate CV so that the work rate contribution of the shear force dropschoose an adequate CV so that the work rate contribution of the shear force drops out



(1 31) th b(1.31) then becomes

022

QgzVp

egzVp

e 022

Qgzegzeinout

if th i h i h f i ti th t i th t t th- if there is a mechanism such as friction that increases the temperature, then

inout gzVp

gzVp

ee

22

inout

inout gg

22

- hence any mechanism in the flow that increases the temperature will reduce the mechanical energy of the system(converting “useful” energy into “useless” energy)



Example 1.9:Turbines convert the energy contained within a fluid into mechanical energy or shaft work. Turbines are often used in power plants with generators to produce electricity. O h i t ll ti i h b l W t i itt d t fl th hOne such installation is shown below. Water is permitted to flow through a passagewayto the turbine, after which the water drains downstream. For the data given in the figure, determine the power available to the turbine when the discharge at the outlet is 30 m3/sdischarge at the outlet is 30 m3/s.Assume uniform flow profiles at the outlet, negligible heat transfer, no change in internalenergy of the water, and steady flow. water = 1000 kg/m3

result: MW73Wresult: MW7.3W



Example 1.10:The pump of a water distribution system is powered by a 15 kW electric motor whoseefficiency is 90%. The water flow rate through the pump is 50 l/s. The diameters of thei l t d tl t i th d th l ti diff th iinlet and outlet pipes are the same, and the elevation difference across the pump is negligible. If the pressures at the inlet and the outlet of the pump are measured to be100 kPa and 300 kPa (absolute), resp., determinea) the mechanical efficiency of the pumpa) the mechanical efficiency of the pumpb) the temperature rise of water as it flows through the pump

cv = 4.18 kJ/kg K water = 1000 kg/m3

result: = 74.1% T = 0 017 K


T 0.017 K

Coming up next:g p

Dimensional Analysis

or:

The story of mice and men,

polar bears and dinosaurs,

atomic bombs blast wavesatomic bombs, blast waves,

and some really serious work rationalisation.


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