calculate the momentum deficit correction factor.

Lecture2_MomentumConsC.pdf

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MOMENTUM CONSERVATION

(REVISITED)  eeiiCVCV vmvmFdt )vm(d 



( )e e i i e e i i d

F vdV m v m v v dxdydz m v m v dt t

  

      

              

We apply a similar approach as in mass conservation using the same infinitesimally small control volume with the x‐faces for example and with similar y‐face and z‐face expressions:

( )i i x e e x xm v v vdydz m v v v v v dx dydzx   

      

    

( ) ( ) ( ) ( ) ( )x y z x y z v v v v

v v v v v v v v v v v v t x y z t t x y z

      

                                

         

0 (continuity)

Substitution x y z v v v v

F v v v dxdydz t x y z

     

         

    If we define the Total derivative operator

x y z D

v v v Dt t x y z

           

Dv F dxdydz

Dt 

 :

x x

y y

z z

Dv x dir F dxdydz

Dt Dv

y dir F dxdydz Dt

Dv z dir F dxdydz



 



Unlike Mass Conservation (continuity equation) Momentum Conservation is a VECTOR equation!!

Recall that we have two types of forces; Surface Forces and Body Forces. Neglecting electromagnetic body forces we only need to account for gravitational forces for our control volume:

Body Forces: Gravity gravF mg gdxdydz     gravF

 CG

pxF pdydz

( )pxF p dp dydz  

Surface Forces Net Pressure Force in the x‐direction: ( )Netpx

p p F pdydz p dp dydz pdydz p dx dydz dxdydz

x x   

           Similarly, the net viscous force in the x‐direction:

yxNet xx zx xVISF dxdydzx y z

          

If we repeat the process for the other faces and substitute the sum of all forces to the expression we previously developed for our CV we arrive at the differential form of conservation of momentum which is applicable to any fluid: (No assumptions except for the absence of EM forces.)

DIFFERENTIAL FORM OF CONSERVATION OF MOMENTUM (CARTESIAN COORDINATES)

yxx x x x xx zx x y z x

y y y y xy yy zy x y z y

yzxzz z z z zz x y z z

v v v v p v v v g

t x y z x x y z

v v v v p v v v g

t x y z y x y z

v v v v p v v v g

t x y z z x y z

   

    

   

                       

                 

               

                

COMPACT FORM Dv

g Dt

        



xx yx zx

ij xy yy zy

xz yz zz

p p

       

  

   





Where the stress tensor is

Most common fluids (e.g. water, oil, air) behave as Newtonian fluids which are fluids that exhibit a linear relation between applied shear and strain rate. The constant of proportionality is the viscosity coefficient,  a thermodynamic property or more specifically a transport property. For incompressible flow

2 xxx

yx xy yx

v x

vv y x

 

  

 

  

      

THE NAVIER‐STOKES EQUATIONS OF CONSERVATION OF MOMENTUM FOR LAMINAR INCOMPRESSIBLE FLOW*

Substitution of these expressions and similar ones for the other components of the viscous tensor and assuming uniform viscosity leads to the very‐well known and widely used Nonlinear Partial Differential NAVIER‐STOKES equations for incompressible flow:

2 2 2

x x x x x

y y y y y

z z z z z

Dv v v vp g

Dt x x y z

Dv v v vp g

Dt y x y z

Dv v v vp g

Dt z x y z

  

         

        

              

          

2Dv g p v Dt

         COMPACT FORM

The set is quite formidable and only a limited number of analytic solutions exist. However, numerical solutions are quite accessible through Computational Fluid Dynamics (CFD) computer codes, e.g. ANSYS‐FLUENT, EES, etc.. They are commercially available and some are available within academia for student use. (*& uniform/constant .)

Example: The steady‐state laminar incompressible flow between two long horizontal parallel fixed plates of separation d that are very wide (width>>d) is parabolic and of the form (Example 3.6)

max 2

4 ( ) 1 ,

2 2x y d d

v y v y d

       

 

Is this true? If so, what is vmax if the flow is created by a pressure difference?

CARTESIAN COORDINATES DIFFERENTIAL VOLUME AND OPERATORS

Density 8.89 (kg/dm3) Melting point (Liquidus) 1083oC Melting point (Solidus, Eutectic) 1065oC Coefficient of thermal expansion 17.7‐10‐ 6m/(m.K) Thermal conductivity 388‐W/(m.K) Thermal capacity 385‐J/(kg.K) Electrical resistivity 1.724‐Ohm.mm2/m Electrical conductivity 0.58‐0.59 Ohm.mm2/m Modulus of Elasticity (tension) 115GPa Modulus of Elasticity (shear) 44GPa Poisons ratio 0.33

î

ĵ k̂

Unit vectors

ˆˆdA n dA dxdy k  Differential surface area vector

ˆdA dydz i 

Arbitrary vector notation:

ˆˆ ˆ x y zV V i V j V k  



Differential Volume: dV dxdydz

Gradient (del) operator:

ˆˆ ˆi j k x y z   

      



Laplacian operator: 2 2 2

2 2 2 2x y z

      

  

̂

r̂ k̂

î

ĵ



rd

ˆˆdA n dA rdrd k  Differential cross‐sectional area vector

ˆdA rd dz r 

Differential surface area vector

Unit vectors

Arbitrary vector notation:

ˆˆˆr zV V r V V k   



Differential Volume: dV rdrd dz

Gradient (del) operator: 1 ˆˆr̂ k

r r z 

   

      



Laplacian operator: 2 2

2 2 2 2

1 1 r

r r r r z     

        

CYLINDRICAL COORDINATES DIFFERENTIAL VOLUME AND OPERATORS

z 

CONTINUITY AND NAVIER‐STOKES EQUATIONS FOR LAMINAR INCOMPRESSIBLE FLOW IN CYLINDRICAL COORDINATES

For a velocity vector field: ˆˆ ˆ , ( , , , )r zv v r v v z v v r z t       

Continuity: 1 1

( ) 0zr v v

rv r r r z



  

     

2 2 2

2 2 2 2 2

1 1 2 : r r r r r r r rr z r

v v vv v v v v v v vp r dir v v g r

t r r z r r r r r r z r r     

               

                           

Navier-Stokes:

2 2

2 2 2 2 2

1 1 1 2 : r rr z

v v v v v v v v v v v vp dir v v g r

t r r z r r r r r r z r r          

                   

                             2 2

2 2 2

1 1 : z z z z z z zr z z

vv v v v v v vp z dir v v g r

t r r z z r r r r z   

             

                       

For 2-D axisymmetric flow: ˆ ˆ 0 , ( , , )r zv v r v z v v r z t 

    

AND   

Continuity: 1

( ) 0zr v

rv r r z

  

 

Navier-Stokes:

2 2

1 : r r r r r rr z r

v v v v v vp r dir v v g r

t r z r r r r z r   

                               

1 : z z z z zr z z

v v v v vp z dir v v g r

t r z z r r r z   

                              

2Dv g p v Dt

        

0v    