Linear Momentum Balance (2D) in Cartesian coordinate

For x-component

Figure 1. x- Component of the Linear momentum balance in 2D flow field (Cartesian

coordinate system)
 yx = Viscous flux of x momentum in y direction

In general (for special surfaces where the direction of normal to the surface is z, y, or z
directions)
 ij = ‘i’ is the direction normal to the surface along which the force is acting and ‘j’ is the
direction of the applied force.

For any arbitrary surface, applied notation in means ‘n’ is the direction normal to the surface
and ‘i’ is the direction of applied force.

A general expression for the linear momentum balance:

[Rate of momentum accumulation]1 = [Rate of momentum in]2 – [Rate of momentum out]3 +
[Sum of the forces acting on the system]4 (1)

Momentum transfer to or from the system takes place by 2 mechanisms

i) Convection (by the virtue of bulk fluid flow)
ii) Molecular transfer/Diffusion
1) Rate of momentum accumulation (in ‘x’ direction)
( uxy)
t
2) Rate of momentum in
y[( u)u]  y[ xx ]  x[( u)v]  x[ yx ]

3) Rate of momentum out

(  uu )  (  uv) 
[(  u )u  x]y  [ xx  ( xx )x]y  [(  u )v  y]x  [ yx  ( yx )y]x
x x y y

4) Sum of all forces acting

a) Arising from fluid pressure P

in = Px y

Px
out = [ Px  x ]y
x
b) Body force
[let ‘X’ be the B.F per unit volume] = X xy

Substituting into equation (1)

( uxy)  (  uu )
= y[( u)u]  y[ xx ]  x[( u)v]  y[ yx ] y (  u )u  xy
t x
  (  uv) 
y xx  ( xx )xy  x(  u )v  yx  x yx  ( yx )yx
x y y
P
 Px y  yPx  x xy  X xy
x
After simplification,

( u ) ( uu)  xx ( uv)  yx Px

     X
t x x y y x

( u) ( uu) ( uv)   yx Px

    xx   X (2)
t x y x y x

u  u ( u) v ( u)
LHS =  u  u u  u  v
t t x x y y

u    u  u v   u 
= u  u u      u   u  v u  
t t  x x  x y  y y 
  u u u       u v  
=   u v  u u v     
 t x x   t x y  x y  

Du  D  u v   Du
=  u      = 
Dt  Dt  x y   Dt

D  u v 
[As,       0 , from continuity equation]
Dt  x y 
Then,

Du    yx  Px
    xx   X (3)
Dt  x y  x

Equation 3 is the general linear momentum balance (2D) equation in the x-direction of the
cartesian coordinate system.
Substitution of the relevant expression of  xx and  yx using constitutive relations for Newtonian
fluids will lead to the famous Navier Stokes Equations
Energy Transport
A general expression for energy transport
 Rate accumulation in the CV1 =  Net transfer of energy by fluid flow 2 +
 Net heat transfer by conduction 3 +
 Rate of internal heat generation 4 ‒
 Net work transfer from the CV to its environment 5 (1)

For terms 1 and 2,

Figure 2. Energy transport by fluid flow in a 2D flow field (Cartesian coordinate system)

Define e as the total specific energy of the system,

1
e  uˆ  v 2
2
û = specific internal energy = U/m = û

1 2 1 mv 2 1 2
v = specific Kinetic energy = KE/m = = v
2 2 m 2

1. Rate of energy accumulation

 (  e)
x y (2)
t
2. Net transfer of energy by fluid flow

i) Rate of energy in
y ue  x ve

ii) Rate of energy out

 (  ue) (  ve)
y ue xx  x ve y y = [  ue  x]y  [  ve  y ]x
x y

Net = (in – out) = (i-ii)

(  ue) (  ve)
= [  ]xy (3)
x y

3. Net heat transfer by conduction

Conduction heat flux is the transport element,

Conductive heat
q" 
Area * time

Similarly, Net conduction heat transfer,

  (q"x )  (q"y ) 
   x  y (4)
 x y 

4. Rate of Internal heat generation

Let the volumetric heat generation rate be q"' (heat/volume*time)

Thus, the total internal heat generation is,

q"' xy (5)

5. Net work done by the fluid element against its surrounding
It consists of two parts

i) Work against the body forces (originate from volume)

ii) Work against the surface forces
a) Work against the pressure forces
b) Work against the viscous forces
Recall,
Work = (force) × (distance in the direction of the applied force)
Rate of work = (force) × (velocity in the direction of the applied force)

i) Work against the body forces (originate from volume)

xy [vx g x  vy g y ] 6(i)

g x , g y are the gravitational accelerations in x and y directions respectively ~

Force/mass.

The negative sign appears in Eqn.5(i) as displacement is against the body force.

ii) Work against the surface forces

a) Pressure

Net force (similar to 2 and 3)

 ( pvx ) ( pvy ) 
   xy 6 [ii(a)]
 x y 
 b) Viscous forces

Figure 4. Work done against the viscous force in a 2D flow field (Cartesian coordinate system)

The arrow directions for the  xy and  yx are not physical.

The general representation is  ij , where i is the direction normal to the surface along which
force is acting, and j is the direction of the applied force.

For the AB plane (normal to the x-axis at x=0), forces/unit area acting are  xx and  xy

The velocity in the direction of the force  xx will be u (Since the direction of applied force, the
jth index, is x)
The velocity in the direction of the force  xy will be v (Since the direction of applied force, the
jth index, is y).
Hence, the rate of work done by the viscous forces at the AB plane against the surrounding
(sign not considered) is
[ xxu   xy v]y

Similarly, at the BD plane (perpendicular to the y-axis at y=0)

Work done by the viscous force
[ yy v   yxu]x .
Thus, net work done by the viscous force-

= [ xx u   xy v] x y  [ yy v   yxu ] y x  [ xxu   xy v] x x y  [ yy v   yxu ] y y x

 xxu  xy v  yy v  yxu
 [    ]xy
x x y y

 xx u v   v  u
= [u   xx   xy  v xy  v yy   yy  u yx   yx ]xy
x x x x y y y y

Writing the constitutive equations,

u v
 yx   xy    (
 )
y x
u 2
 xx  2   (.v)
x 3
v 2
 yy  2   (.v)
y 3

For incompressible flow, .v =0

Substituting  yx ,  xx ,  yy with .v =0 in the above equation we get

[  ]xy
6 [ii(b)]

Where,

 u  2  v  2   u v  2
  2           =Viscous dissipation function
 x   y    x y 

Net work done by fluid on the surroundings,

5(i)+ 5 [ii(a)]+ 5 [ii(b)]
 (  u )  (  v)
= [  (ug x  vg y )     ]xy (6)
x y

Substituting (2), (3), (4),(5) and (6) into equation (1)

(  e) (  ue) (  ve)  ( q" )  ( q y )  (  u )  (  v)

"

 [  ] [ x  ]  q '''  [  (ug x  vg y )     ]
t x y x y x y
  ( e)   (  u )  (  v) e e
  e[   ]   u   v  (.q '' )  q '''   (v .g )  .(  v )  
t t x y x y

 (e) 
  e[  .(  v )]  (.q '' )  q '''   (v .g )  .(  v )  
t t

Putting  .(  v )  0 (continuity),
t
 (e)
  (.q '' )  q '''   (v .g )  .(  v )   (7)
t

Now, gravitational force per unit mass g is a vector and can be expressed as a gradient of a
scalar,
g   Where,  is the potential energy

Hence,
D 
 (v .g )    (v . )    
Dt t
Also,
D  
  .(v )   v . ;
Dt t t
D 
(v . )   ;
Dt t

If  is time-independent then, 0
t
D
So,  (v .g )   
Dt
Substituting into equation 7,
D (e   )
  q  .( Pv )    .q ” (8)
Dt

v2
This is an equation of change for total energy, e    uˆ  
2
Now we have to use the laws of thermodynamics to get the transport equation in terms of
Temperature.
According to the First law of thermodynamics
Q  u  w
So, the energy supplied to the system is used to increase the internal energy only. Considering
internal energy only, equation 8 can be rewritten as
De
  q  .( Pv )    .q (9)
Dt
Where e  uˆ only.
Using thermodynamic relation Enthalpy of the system,
H  U  PV
In terms of specific enthalpy,
H / m  h  (U / m  PV / m)  [uˆ  P / (m / V )]  [e  P /  ]

Or,
Dh De 1 DP P D 
   (10)
Dt Dt  Dt  2 Dt

Using Fourier’s law –

q   K T (11)

Substituting equation 10 and 11 in equation 9, we get

Dh DP P D 
  q  P(.v )      .q
Dt Dt  Dt

Dh DP P  D  
  q        (.v )   .( K T )
Dt Dt   Dt 
Dh DP
  q     .( K T ) (12)
Dt Dt

Now, dh  C p dT is only valid for ideal gases (see Table 1)

In general, we know,
dH  TdS  VdP
[As H  U  PV and, dH  (dU  PdV )  VdP  dQ  VdP  TdS  VdP ]
Table 1: Summary of thermodynamic relations and models

In terms of specific property

1
dh  Tds  dP (13)

Now, s  f (T , P)

 s   s 
ds    dT    dP (14)
 T  P  P T

 S   V 
From Maxwell’s relation [      ],
 P T  T  p

we have (for specific properties)

 s   (1/  )  1    
      2  
 P T  T  P   T  P 

Where is  the thermal expansion coefficient

1  v  1  (1/  )      1   
       2    
v  T  P (1/  )  T  P   T  P   T  P

 s  CP
We also have,   
 T  P T
Substituting into equation 14, we have
CP 
ds  dT  dP
T 
Substituting into equation 13, we have
CP  1 1
dh  T [ dT  dP]  dP  CP dT  (1   T )dP
T   
Taking substantial derivative and multiplying by ρ we get,
Dh DT DP
   CP  (1   T ) (15)
Dt Dt Dt
(Relation between h and T)
Using equation 15 and 12, we get the general form of the energy transport equation for
Newtonian fluids
DT DP
 CP  q     T  .( K T ) (16)
Dt Dt
(Common form of  is [.( .v )] )

Special cases,
(A) For ideal gas,   1/ T

So,
DT DP
 CP  q     .( K T ) (17)
Dt Dt
(B) For incompressible liquid,   0

DT
 CP  q    .( K T ) (18)
Dt

In most of the convection problems we have

1) Constant K
2) Zero heat generation
3) Negligible viscous dissipation,   0

Using the above three assumptions, the temperature distribution model will be
DT
C  K  2T (19)
Dt

For an incompressible fluid, (specific heat, C  C p )