Continuity Equation and Reynolds Transport Theorem Chapter 4

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Continuity Equation and Reynolds

Transport Theorem

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4.2 The Continuity Equation for One-Dimensional Steady Flow 4.3 The Continuity Equation for Two-Dimensional Steady Flow 4.4 The Reynolds Transport Theorem

Objectives

- Apply the concept of the control volume to derive equations for the conservation of mass for steady one- and two-dimensional flows

- Derive the Reynolds transport theorem for three-dimensional flow - Show that continuity equation can recovered by simplification of the Reynolds transport theorem

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Physical system

~ is defined as a particular collection of matter or a region of space chosen for study

~ is identified as being separated from everything external to the system by closed boundary

• The boundary of a system: fixed vs. movable (moving) boundary

•Two types of system:

- closed system (control mass) ~ consists of a fixed mass, no mass can cross its boundary

- open system (control volume) ~ mass and energy can cross the boundary of a control volume

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(5)

A system-based analysis of fluid flow leads to the Lagrangian equations of motion in which particles of fluid are tracked.

However, a fluid system is mobile and very deformable.

A large number of engineering problems involve mass flow in and out of a system.

→ This suggests the need to define a convenient object for analysis.

→ control volume

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~ a volume which is fixed in space and through whose boundary matter, mass, momentum, energy can flow

~ The boundary of control volume is a control surface.

~ The control volume can be any size (finite or infinitesimal), any space.

~ The control volume can be fixed in size and shape.

→ This approach is consistent with the Eulerian view of fluid motion, in which attention is focused on particular points in the space filled by the fluid rather than on the fluid particles.

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• Principle of conservation of mass

The application of principle of conservation of mass to a steady flow in a streamtube results in the continuity equation.

• Continuity equation

~ describes the continuity of flow from section to section of the streamtube

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Consider the element of a finite streamtube - no net velocity normal to a streamline

- no fluid can leave or enter the stream tube except at the ends

- Now, define the control volume as marked by the control surface that bounds the region between sections 1 and 2.

→ To be consistent with the assumption of one-dimensional steady flow, the velocities at sections 1 and 2 are assumed to be uniform.

→ The control volume comprises volumes I and R.

→ The control volume is fixed in space, but in the system moves downstream.

dt

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(10)

( m

_I

+ m

_{R t}

) = ( m

_R

+ m

_{O t}

)

_+∆_t (1)

For steady flow, the fluid properties at points in space are not functions of time,

m 0

t

∂ =

∂

( m

_R

)

_t

( m

_R

)

_t_+∆_t

→ =

₍₂₎

Substituting (2) into (1) yields

( m

_I

)

_t

= ( m

_O

)

_t_+∆_t

Inflow Outflow

(3)

(11)

Express inflow and outflow in terms of the mass of fluid moving across the control surface in time

dt

1 1 1

( m

_I

)

_t

= ρ A ds

2 2 2

( m

_{O t}

)

_+∆_t

= ρ A ds

₍₄₎

Substituting (4) into (3) yields

1

A ds

1 1 2

A ds

2 2

ρ = ρ

Dividing by

dt

gives

1 1 1 2 2 2

m  = ρ AV = ρ A V

1

ds V dt =

(4.1)

→ Continuity equation

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passing all sections of a stream tube is constant.

m  = ρ AV

= constant (kg/sec)

( ) 0

d ρ AV =

→

d ρ ( AV ) + dA ( ρ V ) + dV ( ρ A ) = 0

(4.2b) (4.2a)

(5) Dividing (a) by

ρ AV

results in

d dA dV 0 A V ρ

ρ ⁺ ⁺ ⁼

→ 1-D steady compressible fluid flow

(13)

→

Set

For incompressible fluid flow; constant density

0 d t

ρ = 0, ^∂ ρ =

∂

^(4.3)

From Eq. (4.2a)

( ) 0 ρ d AV =

( ) 0 d AV =

Q = volume flowrate (m³/s, cms) Then (6) becomes

1 1 2 2

const.

Q = AV = = AV = A V

(4.4)

(4.5) (6)

(14)

→

plane of the flow,

b

q =

flowrate per unit distance normal to the plane of flow

(

^{m s m}³ ^⋅

)

Q AV

q hV

b b

= = =

1 1 2 2

hV = h V

(4.6) (4.7) [Re] For unsteady flow

+ inflow outflow

t t t

mass

_+∆

= mass −

( m

_{R t}

)

_+∆_t

− ( m

_{R t}

) = ( m

_{I t}

) − ( m

_{O t}

)

_+∆_t

(15)

Divide by

dt

( ) ( )

R t t R t

I t O t t

m m

dt

+∆ +∆

− = −

Define

(

_{R t}

)

_t

(

_{R t}

) ( )

m m m vol

t dt t

+∆

ρ

∂ = − = ∂

∂ ∂

Then

( )

(

_{I t}

) (

_{O t}

)

_t

vol m m

t ρ

+∆

∂ = −

∂

(16)

No slip

(17)

Eq. (4.5) can be applied. However, velocity in Eq. (4.5) should be the mean velocity.

V Q

= A

A A

Q = ∫ dQ = ∫ vdA

1 V

A

vdA

∴ = A ∫

•The product remains constant along a streamline in a fluid of constant density.^AV

→ As the cross-sectional area of stream tube increases, the velocity must decrease.

→ Streamlines widely spaced indicate regions of low velocity, streamlines closely spaced indicate regions of high velocity.

1 1 2 2

:

1 2 1 2

AV = A V A > → A V < V

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The velocity in a cylindrical pipe of radius

R

is represented by an axisymmetric parabolic distribution (laminar flow).

What is ^V in terms of maximum velocity, ?

v

_c

(19)

r

← equation of paraboloid

dr

dA=2πrdr

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2 0 2

1 1

1 2

A c

Q r

V v dA v r dr

A A R R π

π

 

= = =  − 

 

∫ ∫

3 2 4 2 2

2 0 2 2 2 2

0

2 2 2

2 4 2 4 2

R

c R c c c

v r v r r v R R v

r dr

R R R R R

     

=  −  =  −  =  −  =

     

∫

→ Laminar flow

[Cf] Turbulent flow

→ logarithmic velocity distribution

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(22)

(dscos )θ dA

(23)

Consider a general control volume, and apply conservation of mass

( m

_I

+ m

_{R t}

) = ( m

_R

+ m

_{O t}

)

_+∆_t (a) For steady flow:

( m

_{R t}

) + ( m

_{R t}

)

_+∆_t

Then (a) becomes

( m

_{I t}

) = ( m

_{O t}

)

_+∆_t ^(b)

ⅰ) Mass in O moving out through control surface

(

_{O t}

)

_t . .

( cos )

C S out

m

_+∆

= ∫ ρ ds θ dA

mass = ρ vol = ×ρ area 1× = ρ ds dAcosθ

(24)

ds = vdt

Substituting (c) into (b) gives

(

_{O t}

)

_t . .

( cos )

C S out

m

_+∆

= ∫ ρ v θ dAdt

(c)

(d)

By the way,

v cos θ

= normal velocity component normal to C.S. at

dA n 

dA

(

ⁿ^ ⁼¹

)

Set = outward unit normal vector at

n

cos

v v n v θ

∴ = ⋅ =  

← scalar or dot product (e) Substitute (e) into (d)

. . . .

(

_{O t}

)

_t

C S out C S out

m

_+∆

= dt ∫ ρ v n dA ^{ } ⋅ = dt ∫ ρ v dA ^{ } ⋅

dA = n dA

where =directed area element

(4.8)

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[Cf] Tangential component of velocity does not contribute to flow through the C.S.

→ It contributes to circulation

ⅱ) Mass flow into CV

(

_I

)

_t . .

( cos )

C S in

m = ∫ ρ ds θ dA

. . ( cos ) . . ( )

C S inρ v θ dAdt = dt C S inρv⋅ −n dA

∫ ∫

^ ^

90 cos 0 θ > ^ → θ <

{

^{C S in}^{. .}

} {

^{C S in}^{. .}

}

dt ρv ndA dt ρv dA

= −

∫

^{ }⋅ = −

∫

^{ }⋅ For steady flow, mass in = mass out

{ }

. . . .

C S out C S in

dt

∫

ρv dA^{ }⋅ = dt −

∫

ρv dA^{ }⋅ Divide by

. . . .

C S inρv dA C S outρv dA

−

∫

^{ }⋅ =

∫

^{ }⋅ dt

(26)

where integral around the control surface in the counterclockwise direction

→ Continuity equation for 2-D steady flow of compressible fluid Combine C.S. in and C.S. out

. . . .

0

C S

ρ v dA ⋅ =

C S

ρ v ndA ⋅ =

∫ ^{ } ∫ ^{ }

 

Integral form

. . C S =

∫

[Cf] For unsteady flow

( mass inside c v . .) t

∂

⁼mass flowrate in – mass flowrate out

. . . .

C S out C S in

∫ ∫

(27)

(2) Infinitesimal control volume

(28)

Expand to first-order accuracy

ABρv ndA⋅ + BC ρv ndA⋅ + CDρv ndA⋅ + DAρv ndA⋅ = 0

∫

^{ }

∫

^{ }

∫

^{ }

∫

^{ } ^(f)

2 2

AB

dy v dy

v ndA v dx

y y

ρ ⋅ ≅ −^ρ − ^∂∂ρ ^ − ∂^∂ ^

∫

^{ }

AB 2

dx x ρ ≈ −ρ ^∂ρ

∂

2 v n v v dy

y

 ∂ 

⋅ = − − ∂ 

 

(g)

2 2

BC

dx u dx

v ndA u dy

x x

ρ ⋅ ≅ ^ρ + ^∂∂ρ ^ + ^∂∂ ^

∫

^{ }

2 2

CD

dy v dy

v ndA v dx

y y

ρ ⋅ ≅ ^ρ + ^∂∂ρ ^ + ^∂∂ ^

∫

^{ }

2 2

DA

dx u dx

v ndA u dy

x x

ρ ⋅ ≅ −^ρ − ^∂∂ρ ^ − ^∂∂ ^

∫

^{ }

(29)

Substitute (g) to (f), and expand products, and then retain only terms of lowest order (largest order of magnitude)

( )2

2 2 4

v dy dy v dy

vdx dx v dx dx

y y y y

ρ ρ

^∂ ^∂ ^{∂ ∂}

− + + −

∂ ∂ ∂ ∂

( )

2

2 2 4

v dy dy v dy

vdx dx v dx dx

y y y y

ρ ρ

ρ ρ ^∂ ^∂ ^{∂ ∂}

+ + + +

∂ ∂ ∂ ∂

( )

2

2 2 4

u dx dx u dx

udy dy u dy dy

x x x x

ρ ρ

ρ ρ ^∂ ^∂ ^{∂ ∂}

+ + + +

∂ ∂ ∂ ∂

( )2

2 2 4 0

u dx dx v dx

udy dy u dy dy

x x x x

ρ ρ

^∂ ^∂ ^{∂ ∂}

− + + − =

∂ ∂ ∂ ∂

v u 0

dxdy v dxdy dxdy u dxdy

y y x x

ρ ρ

ρ ^∂ ^∂ ρ ^∂ ^∂

∴ + + + =

∂ ∂ ∂ ∂

(30)

y y x x

∂ ∂ ∂ ∂

( u ) ( v ) 0

x ρ y ρ

∂ ∂

+ =

∂ ∂

→ Continuity equation for 2-D steady flow of compressible fluid (4.10)

ρ

=

• Continuity equation of incompressible flow for both steady and unsteady flow ( const.)

u v 0

x y

∂ + ∂ =

∂ ∂

^(4.11)

(31)

For steady 3-D flow of incompressible fluid

u v w 0

x y z

∂ ∂ ∂

+ + =

∂ ∂ ∂

Continuity equation for unsteady 3-D flow of compressible fluid

( u ) ( v ) ( w ) 0

t x y z

ρ ρ ρ ρ

∂ + ∂ + ∂ + ∂ =

∂ ∂ ∂ ∂

(32)

(33)

Apply (4.9) to control volume ABCD

AB

ρ v ndA ⋅ +

BC

ρ V ndA ⋅ +

CD

ρ V ndA ⋅ +

DA

ρ V ndA ⋅ = 0

∫ ^{ } ∫ ^{ } ∫ ^{ } ∫ ^{ }

2 2

t AB t

d v d

V ndA ρ θ v θ dr

ρ ρ

θ θ

∂ ∂

  

⋅ ≅ −   − ∂   − ∂  

∫ ^{ }

( )

2 2

r BC r

dr v dr

V ndA v r dr d

r r

ρ ⋅ ≅ ^   ρ + ^∂ ∂ ρ ^   + ^∂ ∂ ^   + θ

∫ ^{ }

2 2

t CD t

d v d

V ndA ρ θ v θ dr

ρ ρ

θ θ

 ∂  

⋅ ≅   + ∂   + ∂  

∫ ^{ }

2 2

r DA r

dr v dr

V ndA v rd

r r

ρ

⋅ ≅ −^

ρ

− ^∂∂

ρ

^ − ^∂∂ ^

θ

∫

^{ }

(34)

2 2 4

θ θ θ θ

∂ ∂ ∂ ∂

( )2

2 2 4

t t

v d d v d

v dr

θ

dr v

ρ θ

dr

ρ θ

dr

ρ ρ

θ θ θ θ

∂ ∂ ∂ ∂

+ + + +

∂ ∂ ∂ ∂

2 2

r r

v dr v dr

v rd v drd rd drd

r r

ρ θ ρ θ ρ

^∂

θ ρ

^∂

θ

+ + + +

∂ ∂

2 2

2 2 2 2

r r

dr dr dr v dr v

v rd v drd rd drd

r r r r r r

ρ θ ρ θ ρ θ ρ θ

∂ ∂ ∂   ∂ ∂   ∂

+ ∂ + ∂ + ∂   ∂ + ∂   ∂

2

2 2 2 0

r r

v dr dr v dr

v rd rd v rd rd

r r r r

ρ ρ

ρ θ ρ ^∂ θ ^∂ θ ^{∂ ∂ } ^ θ

− + ∂ + ∂ − ∂ ∂    =

t r

v v

d dr v d dr rdrd v rdrd

r r

ρ ρ

ρ θ θ ρ θ θ

θ θ

∂ ∂ ∂ ∂

+ + +

∂ ∂ ∂ ∂

2 2 3

1 1 1

( ) ( ) ( ) 0

2 2 2

r r

v v

v drd dr d v dr d dr d

r r r r

ρ ρ

ρ θ ρ ^∂ θ ^∂ θ ^{∂ ∂} θ

+ + + + =

∂ ∂ ∂ ∂

(35)

Divide by

drd θ

1 1 1

2 2 2 0

t r r r

v v v v

v v r v dr v dr dr

r r r r r r

ρ ρ ρ ρ

θ θ

∂ + ∂ + ∂ + ∂ + + ∂ + ∂ + ∂ ∂ =

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

r t 0

r r t

v v

r v r v v

r r

ρ ρ

ρ ρ ρ

θ θ

∂ ∂ ∂ ∂

∴ + + + + =

∂ ∂ ∂ ∂

Divide by

r

r r t

0

r t

v v v

v v

r r r r r

ρ ρ

ρ ρ ρ

θ θ

∂ + ∂ + + ∂ + ∂ =

∂ ∂ ∂ ∂

( ) ( )

t

0

r r

v

v v

r r r

ρ

ρ ρ

θ

∂

∴ ∂ + + =

∂ ∂

^(4.12)

For incompressible fluid

t

0

r r

v

v v

r r r θ

∂

+ ∂ + =

∂ ∂

^(4.13)

(36)

Qeth

V

eth

A mixture of ethanol and gasoline, called "gasohol," is created by pumping the two liquids into the "wye" pipe junction. Find and

691.1 kg m3

ρ

mix =

1.08 m s V

mix

=

3 3

30 / 30 10 m /s Qgas = l s = × ⁻

680.3 kg m

3

ρ

gas

=

788.6 kg m

3

ρ

eth

=

(37)

(38)

4

3

1 30 10 / 0.031 0.97 m/s

V = × ⁻ =

1

ρ v n dA ⋅ +

2

ρ v n dA ⋅ +

3

ρ v n dA ⋅ = 0

∫ ^{ } ∫ ^{ } ∫ ^{ }

1

ρ

v n dA⋅ = −680.3 0.97 0.031× × = −20.4 kg/s

∫

^{ }

2 2

2

ρ

v n dA⋅ = −788.6× ×V 0.0079 = −6.23V

∫

^{ }

3

ρ

v n dA⋅ = 691.1 1.08 0.031 23.1 kg/s× × =

∫

^{ }

. 20.4 6.23 2 23.1 0

c s

ρ

v ndA V

∴

 ∫

^{ }⋅ = − − + =

2 0.43 m/s V =

3 3

2 2 (0.43)(0.0079) 3.4 10 m /s 3.4 /s

Qeth V A ⁻ l

→ = = = × =

(39)

• Reynolds Transport Theorem (RTT)

~ A general relationship that converts the laws such as mass conservation and Newton’s 2^nd law from the system to the control volume

Most principles of fluid mechanics are adopted from solid mechanics,

where the physical laws dealing with the time rates of change of extensive properties are expressed for systems.

→ There is a need to relate the changes in a control volume to the changes in a system.

Osborne Reynolds (1842-1912);

English engineer

(40)

• Two types of properties

Extensive properties ( E ): total system mass, momentum, energy Intensive properties ( i ): mass, momentum, energy per unit mass

(41)

system momentum,

E i

m

mv  ( )

2

m v 

v 

( ) v 

2

system mass,

system energy,

1

system system

E = ∫∫∫ i dm = ∫∫∫ i ρ dvol

^(4.14)

(42)

(43)

Consider time rate of change of a system property

(a) (b.1)

(b.3) (b.2)

(b.4)

(

0

) ( )

t dt t R t dt R I t

E

₊

− E = E + E

₊

− E + E

0 . .

( )

_{t dt}

c s out

E

₊

= dt ∫∫ i v dA ρ ^{ } ⋅

(

_{. .}

)

(

_I

)

_t

c s in

E = dt − ∫∫ i v dA ρ ^{ } ⋅

( )

(

_{R t dt}

)

R t dt

E

₊

i dvol ρ

= ∫∫∫

+

( )

(

_{R t}

)

R t

E = ∫∫∫ i dvol ρ

(44)

( ) ( )

{ }

t dt t

1

R t dt R t

E E

i dvol i dvol

dt dt ρ ρ

+

∴ − = ∫∫∫ − ∫∫∫

. . . .

c s out

i v dA ρ

c s in

i v dA ρ + ∫∫ ^{ } ⋅ + ∫∫ ^{ } ⋅

(

. .

) (

. .

)

. .

= = ∂ + ⋅

∫∫∫

^{c v} ∂

∫∫∫

^{c v}

 ∫ ∫

^{c s} ^{ }

dE d

i dvol i dvol i v dA

dt dt

ρ

t

ρ ρ

(4.15)

(45)

dE

dt

(

c v. .i dvol

)

t ρ

∂

∫∫∫

. .

c s i v dAρ ⋅

∫ ∫

^{ }



E

① = time rate of change of in the system

② = time rate change within the control volume → unsteady term

③ = fluxes of across the control surface

▪ Application of RTT to conservation of mass

For application of RTT to the conservation of mass,

in Eq. (4.15), , and

E = m _i = ₁ dm 0

because mass is conserved.

dt =

(

c v. . dvol

)

c s. . v dA

(

c s out. . v dA c s in. . v dA

)

t ρ ρ ρ ρ

∴ ∂ = − ⋅ = − ⋅ + ⋅

∂

∫∫∫  ∫ ∫

^{ }

∫∫

^{ }

∫∫

^{ } ^(4.16)

Unsteady flow: mass within the control volume may change if the density changes

(46)

For one-dimensional flow

. . . . 0

c s outρv dA⋅ + c s in ρv dA⋅ =

∫∫

^{ }

∫∫

^{ } ~ same as Eq. (4.9)

2 2 2 . .

c s out

ρ v dA ⋅ = ρ V A

∫∫ ^{ }

1 1 1 . .

c s in

ρ

v dA⋅ = −

ρ

V A

∫∫

^{ }

1

AV

1 1 2

A V

2 2

ρ ρ

∴ =

• In Ch. 5 & 6, RTT is also used to derive the work-energy, impulse-momentum, and moment of momentum principles.

(4.1)

(47)

Homework Assignment # 4 Due: 1 week from today

Prob. 4.9 Prob. 4.12 Prob. 4.14 Prob. 4.20 Prob. 4.31

Continuity Equation and Reynolds Transport Theorem Chapter 4