Equations (1)

Continuity, Energy, and Momentum

Equations (1)

Lecture 5 Continuity, Energy, and Momentum Equations (1)

Contents

5.1 Control-Volume Concepts

5.2 Conservation of Matter in Homogeneous Fluids

Objectives

- Apply finite control volume to get integral form of the continuity equation - Compare integral and point form equations

- Derive the simplified equations for continuity equation

may be formulated in the Eulerian sense of focusing attention on a fixed point in space.

 Control volume method:

 Finite control volume - 유한검사체적

 Differential (infinitesimal) control volume - 미소검사체적 [Re] Control volume

- fixed volume which consists of the same fluid particles and whose bounding surface moves with the fluid

5.1 Control-Volume Concepts

① Finite control volume – arbitrary control volume

② Differential (infinitesimal) control volume – parallelepiped control volume

- Frequently used for 1D analysis (Ch. 4) - Gross descriptions of flow

- Analytical formulation is easier than differential control volume method - Integral form of equations for conservation of mass, momentum, and energy

[Ex] Continuity equation: conservation of mass

CV dV CS q dA 0

t

 

   





5.1 Control-Volume Concepts

2) Differential control volume method

- Concerned with a fixed differential control volume (=DxDyDz) of fluid

- 2D or 3D analysis (Ch. 5~6)

   

d d

F mq x y zq

dt dt 

D  D  D D D

- Dx, Dy, Dz become vanishingly small

- Point (differential) form of equations for conservation of mass, momentum, and energy

• Consider an arbitrary control volume

• Although control volume remains fixed, mass of fluid originally enclosed regions (A+B) occupies the volume within the dashed line regions (B+C) after dt.

• Since mass m is conserved:

 

 

 

 

 

 

 

 

dt dt

 - - 



(5.1)

(5.2)



5.2 Conservation of Matter in Homogeneous Fluids



volume in the limit

 

 

 

_{ }

m m m

dt t t



 -  

 

 



where dV = volume element

• RHS of Eq. (5.2)

= net flux of matter through the control surface

= flux in – flux out

1 2

n n

q dA q dA

 







5.2 Conservation of Matter in Homogeneous Fluids

where = component of velocity vector normal to the surface of CV  q cos



 

CV

dV

q dA

q dA

t   

   -

   

※ Flux (= mass/time) is due to velocity of the flow.

• Vector form is

 

CV dV CS q dA

t

 

  - 



 

qn



an enclosed control surface cos

q dA q dA 

  

positive for an outflow from cv negative for inflow into cv,

, 90

90 180





 

     

If fluid continues to occupy the entire control volume at subsequent times

→ time independent

LHS:

 

CV dV CV dV

t t

 

  



 

5.2 Conservation of Matter in Homogeneous Fluids

CV dV CS q dA 0

t

 

   





Eq. (5.4) becomes

(5.6)

→ General form of continuity equation→ Integral form [Re] Differential form

Use Gauss divergence theorem

V A i

i

F dV FdA x

 

  

CS

 q dA  

     q dV

 

Then, Eq. (5.6) becomes

 

CV q dV

t

 

     

  

 



Eq. (5.6a) holds for any volume only if the integrand vanishes at every point.

(5.6b) (5.6a)

 

t

 

    



→ Differential (point) form

5.2 Conservation of Matter in Homogeneous Fluids

 Simplified form of continuity equation

1) Steady flow of a compressible fluid → 0

CV dV

t



 



CS





2) Incompressible fluid (for both steady and unsteady conditions)

const. 0, d 0

t dt

 





CSq dA  0



Therefore, Eq. (5.6) becomes

c c 0 uc D

t x x

    -   

c uc c

t x x D x

     

        

• Conservation of mass equations for the individual species

→ Advection-diffusion equation

= conservation of mass equation + mass flux equation due to advection and diffusion

c q 0

t x

   

 

q uc D c x

 - 



5.2 Conservation of Matter in Homogeneous Fluids

5.2.2 Stream - tube control volume analysis for steady flow

• Steady flow: There is no flow across the longitudinal boundary of the stream tube.

• Eq. (5.7) becomes

(5.9)

1 1 1 2 2 2

0

q dA q dA q dA

   -    



const.

 qdA 

Flux = 0

where dQ = volume rate of flow

1 1 2 2

q dA  q dA  dQ

 For flow in conduit with variable density V qdA





→ average density dQ

Q

 



1V A1 1 2V A2 2

   (5.11)

5.2 Conservation of Matter in Homogeneous Fluids

 For a branching conduit 0

 q dA  



1 2 3

1 1 1 2 2 2 3 3 3 0

A

 q dA

 q dA

 q dA

-      

1

V A

V A

V A

    

 At the centroid of the control volume,

 rate of mass flux across the surface perpendicular to x is

Use Infinitesimal (differential) control volume method

, u v w , ,



5.2 Conservation of Matter in Homogeneous Fluids

net mass flux across the surface perpendicular to y flux in

 

2

u dx dydz

u x

 



  

 - 

  

flux out

 

2

u dx dydz

u x

 



  

  



 

net flux = flux in flux out

  u

dxdydz x

-

 

 - 

 

dydxdz y



 - 



net mass flux across the surface perpendicular to z

 

dzdxdy z



 - 

 Time rate of change of mass inside the c.v.





t



 



(A1)



  

 

 

dxdydz

t x y z

   

     

 -   

     

By taking limit dV = dxdydz

  ^u   ^v   ^{w q div}   ^q

t x y z

  

 ^{  }  

-        

   

0

t q

 

    



→ point (differential) form of Continuity Equation (the same as Eq. 5.6b)

5.2 Conservation of Matter in Homogeneous Fluids

[Re]

 

 

 

 

x y z

    

  

     

  

By the way,

q q q

  

    

Thus, (A1) becomes

0

q q

t

  

      

 (A2)

0 ( const.)

d

dt

   

d 0

t q dt

  

     



Therefore Eq. (A2) becomes

0 0

q q

       

u v w 0

x y z

  

  

  

In scalar form,

(A3)

(A4)

→ Continuity Eq. for 3D incompressible fluid

5.2 Conservation of Matter in Homogeneous Fluids

For 2D incompressible fluid,

u v

x y

 

 

 

2) For steady flow, t 0



 



Thus, (A1) becomes

0

q q q

  

      