Vector and Tensor+Note

8/8/2019 Vector and Tensor+Note

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i c s Advanced Fluid e c h

a

Mechanics

F l u i d

Presented by:

Prof. Rashtchian

a n c e d

Designed by:

F. Bayati

A d T. Hamzeloueian

A. Noorjahan E. Vafa

(Summer 2007)



بردار چيست؟

i c s

يزي ي: ف ياه تيم

جرم ،حجم ،درجه حرارت ،فشار ،حرارت مخصوص-

e c h a

ع د م ص مي و ه بوسي ه ي

-سرعت ،نيرو ،مومنتوم، گراديان درجه حرارت

) vectorبوسيله هم مقدار و هم جهت مشخص ميشوند(•

F l u i d

δ1V1 δ2V2 δ3V3

تنسور چيست؟

a n c e d

جزء از قبيل: 9بوسيلهτيك تنسور درجه دوم

A d

بصورت زير نشان داده ميشود.bookkeepingمشخص ميشود ،كه فقط به خاطر

11 12 13 21 22 2331 32 33

Chemical & Petroleum Engineering Department

Sharif University of Technology



i c s

م ه ي

11 و

22 و

33 بي ر

م ه ي دي و ي دي و

م ه ر

12 و

13 و

21ر ... و

المانهاي غير ديگاونال گويند.

e c h a

س ير ر ر

ي س

τ12اينصورت تنسور را آنتي سيمتريك گويند. ( τ21و τ13 τ31و τ23 τ32.(

F l u i d scalarو vector.دنتسه روسنت زا يصاخ تلاح

د ه د. 3دا ا nنس اهد خ ناملا

a n c e d

Scalar ~ tensor of order 0 (

30 1)

A d

Vector ~ tensor of order 1 (31 3)

Tensor of order 2 (32 9)

3





Properties

i c s

‐nd nd

e c h a

.

2‐ If m = scalar, Aij = 2nd order tensor then mAij is a 2

nd order tensor.

3‐ Symmetric tensor:

A = ‐ A

F l u i d 4‐ Antisymmetric tensor:

Aij = ‐ Aij

5‐ Aij + A ji 2nd order symmetric tensor

a n c e d 6‐ Unit tensor (2

nd order)

1 0 00 1 0

or KRONECKER DELTA

A d

هستند. δ3و δ2وδ1در واقع ستون ها و رديف هاي يك تنسور درجه دو واحد اجزاء سه بردار واحد يعني





Vector operations from an analytical viewpointاحد د ه د تتنس ان ا استفاده ا ت م ا ت م ا ا ل م ف ا خ ل

)Kronecher deltaه د تنس (

i c s

،دحاو هس)The alternative unit tensor or Levi Civita tensorفيرعت ريز تروص هب هك (

ميشوند ،بشكل خالصه نوشت:

e c h a

0

1 123,231,312

F l u i d

1 321,213,132

0 any two indices are alike a n c e d ميتوان بصورت زير نوشت: زير را با استفاده از به عنوان مثال دترمينان

11 12 1321 22 2331 32 33 123 123

3

k

33

i

A d بردارهاي واحد را ميتوان بصورت زير نوشت: crossوscalarحاصلضرب

. 3

k1 or Chemical & Petroleum Engineering Department




i c s

در يك جمله يا ترم تكرار شود ،آن جمله يا ترم بايد به اندازه تكرار ممكن آن)i,j,kمثلsuffixوقتي يك (

e c h a

su x.دوش را ر ت

.. 3 ; j 1,2,3

F l u i d كه ميتوانم چنين بنويسيم:

; Repeated suffix Free suffix

a n c e d

A d





Definition of a vector and its ma nitude the unit vector

i c s

δ V δ V δ V δ V3 e c h a

i1 δVi

F l u i d

ن هستند. , ,x,y,zبرد ره ي و حد در جهت محوره ي 3و 2و 1ه در

2 2 2 3 a n c e d i1

بردار

ودو

:

گار

گويند

مساوي

را

A d 1 1 , 2 2 , 3 3





در ارتباط با بردارهاي واحد ميتوان نوشت:

i c s δ1. δ1 δ2. δ2 δ3. δ3 1

δ1. δ2 δ2. δ3 δ3. δ1 0

e c h a

δ1 δ2 δ3 و δ2 δ3 δ1 و δ3 δ1 δ2

F l u i d

2 1 3 3 2 1 1 3 2

a n c e d ommen

1‐ A repeated suffix is known as a DUMMY suffix it may be replaced by any

other suitable symbol

A d e. g.

2‐ Don’t use any suffix more than twise

. . is meanin less3‐ 11 22 33 12 22 32





i c s

Applications e c h a

1‐

Equations of motion of a viscous fluid

F l u i d

1 1 11 2 12 3 13 1 112 122 132 1 2 2 2 2 1 22 22 22

a n c e d 1 1 2 2 3 3 2 12 22 32 2

3 1 3 2 3 3 3 1 2

3 2 2

3 2 2

3 2 3

A d





i c s

1‐ 1 1 2 2 3 3

e c h a

2‐ 1 1 2 2 3 3

F l u i d

3‐ 12 22 32 2

a n c e d . 1,2, … .

4‐ . 1 . 1 2 . 2 3 . 3

A d

5‐ . 1 . 1 2 . 2 3 . 3





i c s

e vec or eren a opera on

e c h a known as “nobla” or “del”

1 1 2 2 3 3

F l u i d

a n c e d

S 11 2

2 3

3 A

d

S





i c s

The divergence of a vector field

e c h a

. .

F l u i d

. . a n c e d

A d





i c s ro uct

nvo v ng

e c h a

F l u i d

111 212 313 1

121 222 323 2 a n c e d

A d





i c s

e c h a

The gradient of a vector field is a tensor

1 2 3 F l u i d

1 1 112 22 32

a n c e d 13 23 33 A d





i c s

e ap ac an o a sca ar e

e c h a

.S .

F l

u i d

2

a n c e d 2

2

A d





i c s

line is given the symbol 2,

hence in the rectangular coordinate:

e c h a

2 212 222 232

F l

u i d

2 1 12 22 22

a n c e d In the spherical coordinate:

2 1 2 1 sin 1 2

A d





Curvilinear Coordinates

i c s

e c h a

F l

u i d

a n c e d

A d



i c s

escr p on o a u mo on

e c h a The development of an analytical description of fluid flow is based upon the

expression of the physical laws related to fluid flow in a suitable mathematical

form.

F l

u i d

To discuss the motion of a fluid we need to look at its prosperities. If S represents

some scalar property of the fluid (concentration, temperature, etc.) then it will be

a function of its co‐ordinates x1, x2, x3 written as xi (i=1, 2, 3) and time t.

a n c e d 1 , 2 , 3 , ,

A d





Time derivatives

i c s

, ,

1 1 2 2 3 3

e c h a

being the time rate of change of S.

The derivatives

i =1 2 3 have no meanin without definition since both x and t are inde endent

F l

u i d

variable.

If taken to represent the velocity of the fluid,

and the time rate of change of S relative to the

moving fluid is

a n c e d written as

Where the operator

is described as differentiation following the fluid motion or

A d “substantial derivative”‐ imagine an observer riding on the fluid particle and continuously measuring S.

however if S is temperature it is not practical to measure the temperature of the fluid with some device,

which moves with the fluid.

, i .

the local change measured .





Three kinds of time derivatives

i c s

1‐

The partial time derivatives; 2‐ Total time derivatives;

e c h a

‐ u s an a me er va ve;

.

F l

u i d

.

a n c e d .

A d

terms; equat ons, : repeate su x, : ree su x





i c s يك ميدان جريان بوسيله معادله زير ارائه شده است:مثال -

e c h a

, , 22

4

بدست آوريد؟1,2,3سرعت و شتاب يك ذره را در مختصات

F l

u i d 1,2,3 2 24

, , a n c e d 4 224 4 4 4

1,2,3 8 252 A d





The differential momentum balance

i c s

Newton’s second law applied to an element of fluid

1 ; , , components of force vector e c h a نوشت:rate of change of momentumرابطه فوق را ميتوان بصورت

1 F l

u i d

;

a n c e d 1

A d نرا به دو نيرو به xنشان دهنده منتج تمام نيروهاي موثر در جهتا ر ه باشد ،مناسب خواهد بود ان سيا م ا رب

صورت زير تقسيم كنيم:

‐ 2‐ Mechanical stresses





i c s

e c h a بر كل اعمال ميشود و آنرا ميتوان بصورت زير نشان داد: (body force)نيروي بدني

F l

u i d

. و جهت اثر نيروي ثقل است xزاويه بين جهت βكه

a n c e d

A d





Stress

i c s

Linear momentum principle

The time rate of change of momentum of a body = force acting on the body

e c h a = body forces + surface forces

The forces acting on a body are either internal or external. To discuss the internal forces

consider a small element of area . Define the outward pointing normal to be positive.

F l

u i d

Assume that the outside material touching the

surface exerts a forces on the inside material across

an element of area, where is a vector which is a

a n c e d

function of position, time and orientation of the

element. is called the stress force/unit area or

simply the stress. is a vector so called stresses act

A d on an area that may in general have components in

each direction. has components normal to (normal

stresses) and parallel to (shear stresses).





STSTIC STRESS IN A FLUID IS ISOTROPIC

i c s

Consider a small tetrahedron with three of it’s faces through P and normal to the co‐

ordinate axes, and the fourth face normal to a given direction n i

e c h a

STSTIC STRESS ON A TETRAHEDRON

F l u i d

a n c e d

A d

Momentum Balance

Rate of change of momentum=Body Forces + Surface Forces

0 = 333322221111nnn δAτeδAτeδAτenδδτρgδV −−−+





Areas of co‐ordinate lanes are ro ection of the obli ue lane onto co‐ordinate lanes

i c s

so

111. A An δ δ =

and

ini A An δ δ =.

Also, 0 ⎯→ ⎯ n A

V δ

δ as ⎯→ ⎯ n Aδ 0

e c h a

=333322221111

eeennn

−−−

But n=−−−

++ 332211 enenen

F l u i d

Hence 0= ...)( 1111 +−−

τ τ nnne

332211τ τ τ τ ===

nn

a n c e d i.e. the normal stress acting on a surface is independent of the orientation of the plane

for fluid at rest.

We associate this stress with the static pressure. As the stress acting the element in the

direction of the outward normal i.e. tension we write

A d

P −===332211

τ τ τ

i.e. ijij P δ τ −= for fluids at rest.





FLUID STATIC’S

i c s

e n on

A fluid deforms continuously under the action of a shear stress.

Thus no shear stress can act normally to any surface within the fluid at rest.

e c h a All stresses act normally to any surface within the fluid.

Thus

)(0 jiij ≠=τ (shear

stress

zero)

It can be shown (see handout) that, ”the normal stress acting on a surface at a point is

”

F l u i d

.

i.e. P −=== 332211 τ τ τ

δ τ −= 01 ii ≠=== δ δ

a n c e d

Definition

A d The pressure at the point in a fluid at rest

is stress that acts on a surface in a direction

opposite to the normal vector.





i c s Newton s

aw

(applied to a parallelepiped of sides 321 ,, x x x δ δ δ ).

e c h a {rate of change of momentum}={body forces}

In 1 direction

F l u i d

321132110 x x x x x g δ δ τ δ δ δ ρ += |11 x x δ + 3211 x x δ δ τ − |

1 x

−

a n c e d Hence in the limit 01 → xδ

11

111

x x g

∂=

∂= ρ

j

p g

∂= ρ

A d

j

This is a vector equation. It summaries 3 equations, 1 for each coordinate drin.





Fluid Dynamics

i c s This is concerned with fluids in motion, in particular the application of Newton’s

Second Law to an element of fluid. Any element will deform, in general, and

e c h a

c ange s ape. e ra e o c ange o momen um mus e or a spec c e emen ,

no matter how it deforms subsequently. The appropriate derivative must follow

the fluid motion‐substantial time derivative.

F l u i d

Local + Convective = body force + pressure force + viscous force

Mathematically

a n c e d

i

i

dt t

udx

x

udU

t x x xU U

∂

∂+

∂

∂=

=

111

32111 ),,,(

A d

i

i x

uu

t

u

dt

du

∂∂

+∂∂

= 111





Consider a parallelepiped δx,δx,δx are stress forces acting in 1 direction of

i c s

this fluid element in a fluid motion are shown in figure Apply Newton’s Second law to the element in the 1 direction

+= δ xδ xδ x.uδ xδ xδ x D

Surface force

e c h a Dt

The mass considered in the element = .δ xδ x ρδ x321

is constant.

F l u i d

a n c e d

A d





i c s

volumeunit

forcesurfaceXDt

Duρ 11 +=

e c h a

volumeunit

forcesurface

=

F l u i d

321

21x31δxx3131x21δxx2132x11δxx11

δxδxδx

333222111−−− +++

( ) j1312111 ττττ ∂=∂+∂+∂=

a n c e d

Du ∂ A d

ji

ji τ

xx

Dtρ

∂+=





i c s Body force

Usually due to gravity ii g X ρ =

e c h a

Stress

force

Stress forces include a contribution from a static pressure force and the rate of strain

‘ ’

F l u i d

.

the following equation.

ji ji jiP σ δ τ +−= where

ji

P δ = the pressure contribution to the stress tensor(normal)

=

a n c e d ji

,

dynamic contribution is given by Newton’s Law of viscosity (the derivation of which is

beyond the scope of this course).

A d





i c s

⎪⎭

⎪⎬⎫⎪⎩

⎪⎨⎧∂∂+

∂∂=

j

i

i

j

ji x

u

x

uμ σ

for constant ρ

e c h a

It

is

interesting

to

calculate

the

trace

of

the

tensor

jiτ

by

the

contraction

j=I.

iiiii

u P

∂+−= μ δ τ 2 and

iu∂ = 0 for constant ρ from continuity

F l u i d

i x i x

Eq. P ii −=3

τ

The trace of a tensor is a scalar, in this case of magnitude ‐3p.

a n c e d Hence substitution in above equation,

⎪

⎬

⎫⎪

⎨

⎧

⎥

⎤

⎢

⎡ ∂+

∂+−

∂+= i j

jii

i uu P g

Duμ δ ρ ρ

A d

ji j





i c s

Now

i

ji

j x P

x ∂−=−

∂δ [ 1= jiδ only if j=i]

2

e c h a Assume constant density ; 0=⎟

⎠

⎜

⎝ ∂∂

=∂∂ j

i

i ji x x x x [see cons. Of mass]

2

F l u i d Hence

j j

i

i

i

j

i j

ii

x x x g

xu

t Dt ∂∂+

∂−=

∂+

∂=

ρ

a n c e d (Local +convective) acceleration =

massunit

A d





i c s

e c h a

. n nown n nown

1 Pressure: P

3 Velocity: iu

9 Stresses: iτ

F l u i d

9− Newtonian equation and viscosity

+ ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐

4

a n c e d

Thus we need four equations for determining P and iu , continuity equation and three

motion equations.

A d





Simple Solution of the N‐S Equations

i c s

Consider laminar flow down a vertical wall,

e c h a

(i) Steady flow 0=∂

∂

t

ui

(ii) Two dimensional 0,0 3 ==∂

uui

F l u i d

3 x

(iii) Suppose the flow is fully developed

00 1 =∂=∂ uui

a n c e d 11 ∂∂

(iv) By continuity 0

1

1

2

2 =∂

∂−=

∂

∂

x

u

x

u

A d Integrating u2 = u2(x1)

At x2 = 0 u2 = 0

u2 = 0 everywhere





i c s

v omen um equa on n rec on s zero.

(vi) Momentum equation in 2 direction

⎪⎫⎪⎧ ∂∂∂−=

∂∂∂ 22

22

222 1 uu puuu

e c h a

⎪⎭⎪⎩ ∂∂∂∂∂∂ 21

22221 x x x x xt ρ

Steady developed u2=0 u

2 = 0 developed

p==

∂

F l u i d

2 x∂

(vii) At liquid surface (still air) 10 x g p p air ρ +=

(viii) Momentum equation in 1 direction.

a n c e d

g x

u

x

u

x

p

x

uu

x

uu

t

u+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∂

∂+

∂

∂+

∂∂

−=∂∂

+∂∂

+∂∂

221

2

21

12

12

12

1

11

1 1υ

ρ

Steady developed u2=0 developed

A d





With g ρ x

pair

1

=∂∂

i c s

⎭⎬⎫

⎩⎨⎧ −−=+−= ρ

ρ1 g g ρ

ρ g dxud υ air air 22

1

2

But ρ ρ ρ liquid air =⟨⟨

e c h a

g dxud 21

2

−=ν

F l u i d

Boundary conditions; at x2 = 0 u1 = 0

at 01 == du x δ No interfacial shear

a n c e d 2 x

Integrating g δcc gxdx

duυ 2

2

1 =+−=

2

A d

⎪⎭⎬

⎪⎩⎨ ⎥⎦⎢⎣

−⎥⎦⎢⎣=

⎪⎭⎬

⎪⎩⎨ −−= 2222

21δ

x

2δ

x g δ

2

xδ x g uυ

δ 3 g δ

Volumetric

flow

rate

==

021 3ν

xu





Flow Between

i c s

ara e ates

e c h a At x2 = ± h u1 = 0

Pressure gradient:

⎤⎡ ∂p

F l u i d

⎦⎣∂ 1x 1 .

Steady flow: 0=

∂

∂

t

ui ;

a n c e d 2 dimensional: 0u0;

x3

3

==∂

Fully developed flow, before)(as0u0x

u0;

x

u0;

x

u2

2

2

1

1

1

i ==∂∂

=∂∂

=∂∂

A d

Continuity is satisfied,

Momentum in 3 direction is satisfied,

Momentum in 2 direction,

ii222 xgρpPwherexρgxρ0 −=∂−=+∂−= P = P(x1)





i c s

. .

Momentum in 1 direction

2

e c h a ii2

2

11

1

x g ρ p P x

υ g x ρ

0 −=∂

++∂

−= where

12 dP 1ud

F l u i d

c}{xdP 1duυ 21 +⎥⎤⎢⎡=

122

dx ρdx=

a n c e d

12

D}cx

2

x{

dx

dP

ρ

1uυ 2

22

1

1 ++⎥⎤

⎢⎡

=

A d

}h{xdx

dP

u2

1u 22

21

1 −⎥⎦

⎤⎢⎣

⎡=





3. Suppose top plate moves with velocity UT

Bottom plate with velocity UB

i c s

Substituting in (1)

⎭⎬⎫

⎩⎨⎧

++⎥⎦

⎤⎢⎣

⎡= DCh

h

dx

dpU T

2

1 2

1 ρ ρ

μ

e c h a

⎭

⎬⎫

⎩

⎨⎧

+−⎥

⎦

⎤⎢

⎣

⎡= DCh

h

dx

dpU B

2

12

1 ρ ρ

μ

{ } Dhdp

U U BT 2)(2 +⎥

⎤⎢⎡

=+μ

F l u i d

x1

{ }Chdx

dpU U BT 2)(

1

⎥⎦

⎤⎢⎣

⎡=−μ

⎫⎧ −⎫⎧ −⎤⎡⎫⎧ + xU U h xdpU U BT BT )(1)( 2

22

2

a n c e d ⎭⎩⎭⎩⎦⎣

=⎭⎩

−hdx

u222 1

1μ

Check when UT = UB=0 simplifies to previous expression

when x2 = + h u1 = UT satisfies B.C.'s

A d x2 = ‐ h u1 = UB

when 0=⎥⎦

⎤⎢⎣

⎡dx

dp

h

xU U U U u BT BT

2

)(

2

21

−+

⎭⎬⎫

⎩⎨⎧ +

=

i.e. linear velocity profile

known as Couette flow





The following velocity profiles are obtained with various (dp/dx)

i c s

e c h a

F l u i d

a n c e d Consider flow relative to lower plate.

If p decreases in direction of flow (normal pressure gradient)

Velocity is always downstream relative to lower plate

A d

If p increases in direction of flow (adverse pressure gradient)

Velocity may become upstream relative to lower plate

s as mpor ance n u r ca on eory.





Plate suddenly brought into motion

i c s

e c h a

F l u i d

Plate

AB

initially

stationary.

Motion

with

velocity

U

starts

at

time

t=0

B.C's t < 0 u1 = 0 at x2 > 0

a n c e d t ≥ 0 u1 = U at x2 = 0

∞→→ 21 xat0u

Momentum equation in 2 directions

⎫⎧∂∂∂∂∂∂

22

uu1uuu

A d ⎪⎭⎪⎩ ∂

+∂

+∂−=∂+∂+∂ 22

2122

21

1xx

νxρx

ux

ut

Assuming an infinite plate and 2D motion

0u0u

0u

221 =∴=

∂∴=

∂

0x

p

2

21

=∂∂∴





i.e. pressure remains hydrostatic

i c s

011

=∂∂=

∂∂

x

P

x

p

e c h a Momentum equation in 1 direction

⎬⎫

⎨⎧ ∂

+∂

+∂

−=∂

+∂

+∂

2

1

2

2

1

2

12

11

1 1 uu puu

uu

uυ

F l u i d

21121 x x x x x

00001

1

12

1

1 =∂∂

=∂∂

==∂∂

x

u

x

P u

x

u

2

a n c e d

22

11

x

u

t

u

∂=

∂υ

c.f. equation in maths course

A d

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−=t

xerf

U

u

υ 21 21

er = . er ∞ = .





Conservation of Momentum

i c s

Consider parallel piped 321 ,, x x x δδδ fixed in space. Carry out a momentum balance in the i direction. (Say 1 direction)

{ }321i δ xδ xδ xu ρ∂

⎫⎧ momentumldirectionaiof onaccumulatiof Rate

e c h a 321 δ xδ xδ x

=⎭⎩ volumeunit

δ x xi321 xi321

δ xδ xδ x

uδ xδ xu ρuδ xδ xu ρ11i +−

=⎬⎫

⎨⎧

volumeunit

convectionbymomentumldirectionaiof inflowof RateNet

F l u i d + … +

)uu( ρ

x

ji

j∂

∂−=

a n c e d

321 δ xδ xδ x=

⎭⎬

⎩⎨

volumeunit

x321iδ x x321i

δ xδ xδ x

δ xδ xτ δ xδ xτ 11i

−=⎬

⎫⎨⎧ +

volumeunit

stressestoduedirectioniinelementonactingForce

A d + … +

)( ji j x

τ∂∂

−=

∂ ji ji

j

p

x

σ+−

∂

−=

)( ji ji x x

pσ

∂∂

+∂∂

−=





∂∂∂∂

i c s

ence. ji

ji

i ji

j

i

x x

g uu

x

u

t

σ

∂

+

∂

−ρ=ρ

∂

+ρ

∂)()}({ ji

ii jii

ji

i

x x

p g u

xu

t

pu

x

uu

t

uσ

∂∂

+∂∂

−ρ=ρ∂∂

+∂∂

+∂

∂ρ+

∂∂

ρ

e c h a

)( ji j

ii

i

x

g

x

p

Dt

Duσ

∂

∂+ρ+

∂

∂−=ρ

F l u i d

Multiply by ui

i ji jiiiiiiii x

uu xu g x

u P P u xuut

D

∂∂

σ−σ∂∂

+ρ=∂∂

+∂∂

−=ρ )()()2

1(

a n c e d

Equation of mechanical energy (scalar)

A d





Ener E uations

i c s

Momentum Equation

p ∂∂∂∂

e c h a i

jii ji

ji

x x xt ∂∂−=

∂∂

)()}({ ji ji

i j j

ii j

ji

i

x x p g u

xu

t pu

xuu

t u σ

∂∂+

∂∂−ρ=ρ

∂∂+

∂∂+

∂∂ρ+

∂∂ρ

F l u i d

Continuity Equation

u D ∂∂∂

a n c e d

ji

j x Dt eu

xt ∂ρ−==ρ

∂+

∂..

)(ii

i g p Du

σ∂

+ρ+∂

−=ρ

A d

ji x





Multiply by ui:

i c s

j

i ji jii

jii

i

ii

iii

x

uu x

u g x

u p pu x

uu Dt

D

∂∂σ−σ

∂∂+ρ+

∂∂+

∂∂−=ρ )()()

2

1(

e c h a Physical Interpretation

(1) )}({2

1)}

2

1()

2

1({)

2

1( jiiii jiiii u

t

puuuuuuu

t uu

Dt

Dρ

∂∂

+∂∂

+∂∂

+∂∂

ρ=ρ

F l u i d

)2

1()

2

1( jii

jii uuu

xuu

t ρ

∂∂

+ρ∂∂

=

= {Rate of increase of kinetic energy/ unit volume} + {Net rate of kinetic energy/volume}

a n c e d

(2) =∂∂

− )( pu x

ii

{Rate of work done on element by pressure forced per unit volume}

∂

D

u ˆ

)1

(

A d

(3) ==ρ=⎥⎦⎢⎣ρ−=

∂v

Dt v Dt p

Dt x p

i ˆspecific volume (i.e. per unit mass)

U D ˆ

ˆ

1= =[ rate of conversion to internal energy / unit volume]





(4) ii u g ρ = [Rate of work done by gravity forces per unit volume]

i c s

(5) =σ∂∂ )( jii j

u x

[Rate of work done by stress forces per unit volume]

e c h a By difference

(6) ii ji ji

u

x

u

x

u

x

u

∂∂

∂∂

+∂

∂μ−=

∂∂

σ− ][ (see tutorial 1)

F l u i d = [Rate of conversion to internal energy due to viscous dissipation / unit volume]

This is the equation of mechanical energy (No.2) It states that

a n c e d

[Rate of accumulation of kinetic energy / unit volume]= [Sum of rates of work done by

pressure, gravity and stresses / unit volume] – [Rate of conversion to internal energy / unit

volume

A d





i c s

e c h a

ons er e para e ep pe o s e 321 x x x

The first law of thermodynamics will be applied

[Rate of accumulation of internal and kinetic

F l u i d energy] = [Net rate of internal energy gained by

convection and conduction] + [Net rate of work

done

on

the

system

by

surrounds]

a n c e d

[Rate of accumulation of internal and kinetic energy / unit volume] =

A d 321

2

321 /}]2

1ˆ{[ x x xV U x x xt

δ δ δ δ δ δ ρ +∂∂

= [net rate of internal and kinetic energy in by convection / unit volume]



[Rate of accumulation of internal and kinetic energy / unit volume] = 3212

321 δxδx}]/δV1

U{δxδxδx[ρ ×+∂ ˆ



[Rate of accumulation of internal and kinetic energy / unit volume] 321321 δxδx}]/δV2

U{δxδxδx[ρt

×+∂

⎤⎡ 11 ˆˆ

i c s

=

⎥⎥⎥⎥

⎥

⎦⎢⎢⎢⎢

⎢

⎣

+−+

=⎥⎥⎥⎦⎢

⎢⎢⎣

+→ 321

δxx321

x321

0δx δxδxδx

V2

UxxρuV2

Uxxρu

Lim

volumeunit/convection

byinenergykinetic

andinternalof ratenet

111

i

e c h a +

⎥⎥

⎦⎢⎢

⎣++

−=⎥

⎦⎢⎣

+

→...

δxδxδxLim

volumeunit/conduction

byinenergyheatof ratenet

321

32δxx1x1

0δx

111

i

+ ⎥⎤

⎢⎡

++=⎥⎦

⎤⎢⎣

⎡

→...ρ

δxδxδx

δxδxδxguLim

volumeunit/forces

gravitybydoneworkof ratenet

321

32111

0δx i

F l u i d +

⎢⎢

⎣

⎡ −=⎥

⎦

⎤⎢⎣

⎡ +

→ 321

x32111δxx32111

0δx δxδxδx

δxδxτuδxδxτuLim

volumeunit/forcesstressby

doneworklof ratenet111

i

x32122δxx32122 δxδxτuδxδxτu111

−+

a n c e d

321 δxδxδx

+321

x231211δx2x231211

δxδxδx

δxδxτuδxδxτu −+

+ … +

A d

+ … + … + ]

)τu{x

guρx

q)}V

2

1U(u{ρ

x}}V

2

1U{ρ

tjii

jii

i

i2i

i

2

∂∂

++∂∂

−=+∂∂

++∂∂ ˆˆ{

)σ(up)u{guρq

)}V1

U(D

ρ{)}u(ρρ

){V1

U( jiiiiii2

i2 ∂

+∂

−+∂

−=++∂

+∂

+ ˆˆ xxxx jiii

)σ(ux

p)u{x

guρx

q)}V

2

1U(

Dt

Dρ{ jii

ji

iii

i

i2

∂∂+

∂∂−+

∂∂−=+ˆ



Equation of total energy



Equation of total energy

i c s

Subtract equation (2) from equation (3) to get

Equation of

Thermal

Energy

iii uσ]

up[

qUDρ

∂+

∂−

∂−=

ˆ

e c h a jii xxxDt ∂∂∂

Physical Interpretation

[Rate of change of internal energy / unit volume] = [Rate of input of internal energy by conduction / vol.] [Rate

F l u i d

. .

N.B. the last two terms represent interconversion of thermal and mechanical energy

VDpuiˆ

−=∂

−

a n c e d DtVxi

ˆ∂ . .

This term is reversible because

Dt

VD ˆ may be true or –ve

uuuu i jii ∂∂∂∂

A d xxxx ji j j ∂∂∂∂

and may be written as a sum of squared terms

This term is irreversible because it is always positive.

represen s egra a on o mec an ca energy o n erna energy +ve n

In this course we seek solution of equation (4).





Thermodynamic Relationships

i c s

The simplification of equation (4) required the following

e c

h a

(1) VPdSTdUd ˆˆˆ −=

; dpVSTdHd ˆˆˆ −=

PV

ST

V

U −⎥⎤

⎢⎡∂∂=⎥

⎤⎢⎡∂∂

ˆ

ˆ

ˆ

ˆ ; V

P

ST

P

H ˆˆˆ+⎥

⎤⎢⎡∂∂=⎥

⎤⎢⎡∂∂

F l u i d

P]T

PT[

V−

∂∂

= ˆ

; V]T

VT[ P

ˆˆ

+∂∂

−= … (a)

a n c e d

(2) dT]T

U[Vd]

V

U[Ud

VT ˆ

ˆˆ

ˆ

ˆˆ

∂∂

+∂∂

=

; dT]T

H[dP]

P

H[Hd PT ∂

∂+

∂∂

=ˆˆ

ˆ

Pˆˆˆ

∂−

Vˆˆ

ˆ∂−

A d T V∂ T p∂

...

PVdSTdHd (3)

ˆˆ

ˆˆˆ +=

STddTCTTTC ppp

p

ˆˆˆ

=⇒⎟⎟ ⎠⎜

⎜⎝ ∂=⎟

⎟ ⎠⎜

⎜⎝ ∂=





i c s

Hence at constant pressure

dTCVpdUd pˆˆˆ +−= … (c)

e c

h a

(4) i

ixu

DtDρ

ρ1

Dt

)

ρ

1D(

ρDtVDρ

∂∂=−==ˆ … (d)

F l u i d

(5) dpVVpdUdHd ˆˆˆˆ ++= … (e)

a n c e d

A d





DpuqHD ∂∂ˆ

i c s

Dtxσ

xDtρ

j

ji

i

+∂

+∂

−=

And

ˆ

e c

h a ]dp)

TT(V[dTCHd pp ∂

−+= ˆˆˆ

Or Dp

])V

T(V[DT

CHD ∂

−+=ˆ

ˆˆˆ

F l u i d

ttt

Hence

a n c e d

]Dt

DP})

T

VT(V{

Dt

DTCρ[ pp ∂

∂−+

ˆˆˆ =

Dt

DP

x

uσ)

x

TK(

x j

i ji

ii

+∂∂

+∂∂

−∂∂

−

A d

Dt

DP

x

uσ

xx

TK

Dt

DP])

T

VρT(

Dt

DPVρ

Dt

DTCρ

j

i ji

ii

2

pp +∂∂

+∂∂

∂=

∂∂

−+ˆ

ˆˆ

ˆ



2



Dt

DP)

T

V(

V

T

x

uσ

xx

TK

Dt

DTCρ p

j

i ji

ii

2

p ∂∂

+∂∂

+∂∂

∂=

ˆ

ˆˆ

i c s

Or

ˆ

e c

h a

ˆ

Dt]

ln(T)

n[

x

uσ

xxK

DtCρ p

j

i ji

iip ∂

+∂

+∂∂

=ˆ

F l u i d

در اين ترم ا ر دانسيته ثابت باشد خر را مورد بررسي قرار مي دهيم. صفر مي شود و ا ر شار ثابتVnحال در رابطه فوق ترم

صفر مي شود. DP/Dtباشد

∂)DP)Vlnدر نتيجه كل ترم در فشار يا دانسيته ثابت صفر است.ˆ

a n c e d

رابطه فوق را مي توان بصورت زير نوشت:

i ji

2

jp

x

uσ

xx

TK]

x

TU

t

T[Cρ

∂

∂+

∂∂

∂=

∂

∂+

∂

∂ˆ

A d

برابر صفر مي شوند. jiσوUjگار سيال در حالت سكون باشد2 TKT ∂

=∂

iipxxCρt ∂∂∂


Vector and Tensor+Note

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