Chapter 3: Equations of Change
In the last chapter, we presented examples of microscopic balances in one or
two dimensions for various elementary examples. In this chapter we present the general
balance equations in multidimensional case. The balances, also called equations of
changes can be written in cartesian, cylindrical or spherical coordinates. We will
explicitly derive the balance equations in cartesian coordinates and present the
corresponding equations in cylindrical and spherical coordinates. The reader can consult
the books in reference for more details. Once the equations are presented we show
through various examples how they can be used in a systematic way to model
distributed parameter models.
3.1 Total Mass balance
Our control volume is the elementary volume xyz shown in Figure 3.1. The
volume is assumed to be fixed in space. To write the mass balance around the volume
we need to consider the mass entering in the three directions x,y, and z.
vz)z+z
vy)y+y
y
x
vx)x+x
vx)x
z
vy)y
z
y
vz)z
x
Figure 0-1 Total Mass balance in Cartesian coordinates
102
Mass in:
The mass entering in the x-direction at the cross sectional area (yz) is
(vx)|x yzt
(3.1)
The mass entering in the y-direction at the cross sectional area (xz) is
(vy)|y xzt
(3.2)
The mass entering in the z-direction at the cross sectional area (xy) is
(vz)|z xyt
(3.3)
Mass out:
The mass exiting in the x-direction is:
(vx)|x+x yzt
(3.4)
The mass exiting in the y-direction is:
(vy)|y+y xzt
(3.5)
The mass exiting in the z-direction is:
(vz)|z+z xyt
(3.6)
Rate of accumulation:
The rate of accumulation of mass in the elementary volume is:
()|t+t xyz - ()|t xyz
103
(3.7)
Since there is no generation of mass, applying the general balance equation Eq. 1.2 and
rearranging gives:
(|t+t − |t) xyz = vx|x −vx|x+x)yzt + vy|y −vy|y+y)xzt
+ vz|z −vz|z+z)xyt
(3.8)
Dividing the equation by xyzt results in:
 |t t  |t vx |x vx |xx v y | y v y | yy vz |z vz |zz



t
x
y
z
(3.9)
By taking the limits as y,x,z and t goes to zero, we obtain the following equation
of change:

vx v y vz



t
x
y
z
(3.10)
Expanding the partial derivative of each term yields after some rearrangement:




v v v
 vx
 vy
 vz
 ( x  y  z )
t
x
y
z
x y z
(3.11)
This is the general form of the mass balance in cartesian coordinates. The equation is
also known as the continuity equation. If the fluid is incompressible then the density is
assumed constant, both in time and position. That means the partial derivatives of  are
all zero. The total continuity equation (Eq. 3.11) is equivalent to:
0  (
vx v y vz


)
x y z
or simply:
104
(3.12)
0
vx v y vz


x y z
(3.13)
3.2 Component Balance Equation
We consider a fluid consisting of species A, B … , and where a chemical
reaction is generating the species A at a rate rA (kg/m3s). The fluid is in motion with
mass-average velocity v = nt/ (m/s) where nt = nA + nB + … (kg/m2s) is the total mass
flux and  (kg/m3s) is the density of the mixture. Our objective is to establish the
component balance equation of A as it diffuses in all directions x,y,z (Figure 3.2).
nAy|y+y
z
nAz|z
y
nAx|x+x
nAx|x
z
y
x
nAy|y
nAz|z+z
x
Figure 0-2 Mass balance of component A
Mass of A in:
The mass of species A entering the x-direction at the cross sectional (yz) is:
(nAx)|x yzt
(3.14)
where nAx kg/m2 is the flux transferred in the x-direction
Similarly the mass of A entering the y and z direction are respectively:
(nAy)|y xzt
105
(3.15)
(nAz)|z xyt
(3.16)
Mass of A out:
The mass of species A exiting the x, y and z direction are respectively
(nAx)|x+x yzt
(3.17)
(nAy)|y+y xzt
(3.18)
(nAz)|z+z xyt
(3.19)
A|t+t xyz − A|t xyz
(3.20)
The rate of accumulation is:
The rate of generation is:
-rAxyzt
(3.21)
Applying the general balance equation (Eq. 1.3) yields:
A|t+t − A|t) xyz = (nAx|x+x − nAx|xyzt + (nAy|y+y − nAy|yxzt
+ (nAz|z+z − nAz|zxyt +rAxyzt
(3.22)
Dividing each term by xyzt and letting each of these terms goes to zero yields:
 A n Ax n Ay n Az



 rA
t
t
t
t
(3.23)
We know from Section 1.11.1, that the flux nA is the sum of a term due to convection
Av) and a term due to diffusion jA (kg/m2s):
nA = Av + jA
106
(3.24)
Substituting the different flux in Eq. 3.23 gives:
 A (  A v x ) (  A v y ) (  A v z ) j Ax j Ay j Az






 rA
t
x
y
z
x
y
z
(3.25)
For a binary mixture (A,B), Fick’s law gives the flux in the u-direction as :
j Au  DAB
wA
u
(3.26)
where wA = A/Expanding Eq. 3.25 and substituting for the fluxes yield:
v y v z    A
 v
 A

 
   v x
  A  x 

 v y A  v z A 
t
y
z  
x
y
z 
 x
  DAB wA
 DAB wA
 DAB wA 
  (
) (
) (
)   rA
x
y
y
z
z
 x

(3.27)
This is the general component balance or equation of continuity for species A. This
equation can be further reduced according to the nature of properties of the fluid
involved. If the binary mixture is a dilute liquid and can be considered incompressible,
then density  and diffusivity DAB are constant. Substituting the continuity equation (Eq.
3.13) in the last equation gives:
  2 A  2 A  2 A 
 A   A

 
  vx
 v y A  vz A   DAB 


)   rA
2
2 
 x 2
t 
x
y
z 

y

z


(3.28)
This equation can also be written in molar units by dividing it by the molecular weight
MA to yield:
  2C A  2C A  2C A 
C A
C A 
 C A
  RA
  D AB 
  v x
 vy
 vz



 x 2
x
y
z 
y 2
z 2  reaction













accumulation
C A

t


Convection
Diffusion
107
(3.29)
The component balance equation is composed then of a transient term, a convective
term, a diffusive term and a reaction term.
3.3 Momentum Balance
We consider a fluid flowing with a velocity v(t,x,y,z) in the cube of Figure 3.3.
The flow is assumed laminar. We know from Section 1.11.2 that the momentum is
transferred through convection (bulk flow) and by molecular transfer (velocity
gradient).
zx |z
yx |y+y
xx |x+x
xx |x
zx |z+z
yx |y
y
x
z
Figure 0-3 Balance of the x-component of the momentum
Since, unlike the mass or the energy, the momentum is a vector that has three
components, we will present the derivation of the equation for the conservation of the xcomponent of the momentum. The balance equations for the y-component and the zcomponent are obtained in a similar way. To establish the momentum balance for its xcomponent we need to consider its transfer in the x-direction, y-direction, and zdirection.
Momentum in:
108
The x-component of momentum entering the boundary at x-direction, by
convection is:
(vxvx)|x yzt
(3.30)
The x-component of momentum entering the boundary at y-direction, by
convection is:
(vyvx)|y xzt
(3.31)
and it enters the z-direction by convection with a momentum:
(vzvx)|z xyt
(3.32)
The x-component of momentum entering the boundary at x-direction, by
molecular diffusion is:
(xx)|x yzt
(3.33)
The x-component of momentum entering the boundary at y-direction, by
molecular diffusion is:
(yx)|y xzt
(3.34)
and it enters the z-direction by molecular diffusion with a momentum:
(zx)|z xyt
(3.35)
Momentum out:
The rate of momentum leaving the boundary at x+x, by convection is:
(vxvx)|x+x yzt
109
(3.36)
and at boundary y+y,:
(vyvx)|y+y xzt
(3.37)
(vzvx)|z+z xyt
(3.38)
and at boundary z+z:
The x-component of momentum exiting the boundary x+x, by molecular
diffusion is:
(xx)|x+x yzt
(3.39)
(yx)|y+y xzt
(3.40)
(zx)|z+z xyt
(3.41)
and at boundary y+y:
and at boundary z+z::
Forces acting on the volume:
The net fluid pressure force acting on the volume element in the x-direction is:
(P|x – P|x+x)yzt
(3.42)
The net gravitational force in the x-direction is:
g|xxyzt
Accumulation is:
110
(3.43)
(vx|t+t − vx|t)xyz
(3.44)
Substituting all these equations in Eq. 1.5, dividing by xyzt and taking the limit
of each term goes zero gives:

 ( v x )  ( v x v x )  ( v x v y )  ( v x v z )


P



 ( xx  yx  zx ) 
 g x
t
x
y
z
x
y
z
x
(3.45)
Expanding the partial derivative and rearranging:
v y
 
v x
v z
v x 



x
y
z
 t
(

v x
v x
v x
 v x
  
 t  v x x  v y y  v z z




 

(3.46)
 yx
 xx
 zx
P


)
 g x
x
y
z
x
Using the equation of continuity (Eq. 3.10) for incompressible fluid, Equation (3.46) is
reduced to:

 v x
v
v
v 


P
 v x x  v y x  vz x   ( xx  yx  zx ) 
 g x
x
y
z 
x
y
z
x
 t

(3.47)
Using the assumption of Newtonian fluid, i.e.
 xx   
v x
,
x
 yx   
v x
,
y
 zx   
v x
z
(3.48)
Equation 3.47 yields:
  2v
 v
v x
v
v 
 2v
 2 v  P
   v x x  v y x  v z x     2x  2x  2x  
 g x

t
x
y
z 
x
y
z  

x 






 


accumulation
generation

transport by bulk flow
transport by viscousforces
111
(3.49)
The momentum balances in the y-direction and z-direction can be obtained in a similar
fashion:
  2 v y  2 v y  2 v y  P
v y
v y 
 v y

   v x
 vy
 vz
 g y
    2  2  2  

t

x

y

z
x
y
z  

y 

 



 


accumulation
(3.50)
  2 v z  2 v z  2 v z  P
 v z
v z
v z
v z 

   vx
 vy
 vz
 g z
    2  2  2  

t
x
y
z 
x
y
z  

z 






 


accumulation
generation
(3.51)
v y
transport by bulk flow
transport by viscousforces
transport by bulk flow
generation
transport by viscousforces
These equations constitute the Navier-Stock’s equation.
3.4 Energy balance
In deriving the equation for energy balance we will be guided by the analogy
that exists between mass and energy transport mentioned in Section 1.11.3 We will
assume constant density, heat capacity and thermal conductivity for the incompressible
fluid. The fluid is assumed at constant pressure (Fig 3.4).
The total energy flux is the sum of heat flux and bulk flux:
e = q + CpTv
(3.52)
Therefore, the energy coming by convection in the x-direction at boundary x is:
(qx + CpTvx)yzt
(3.53)
Similarly the energy entering the y and z directions are
(qy + CpTvy)xzt
(3.54)
(qz + CpTvz)xyt
(3.55)
112
The energy leaving the x,y and z directions are:
(qx + CpTvx)|x+x yzt
(3.56)
(qy + CpTvy)y+yxzt
(3.57)
(qz + CpTvz)|z+z xyt
(3.58)
The energy accumulated is approximated by:
CpT|t+t −CpT|txyz
(3.59)
q y |y
qz |z+z
y
x
qx |x+x
qx |x
z
qy |y+y
q z |z
Figure 0-4 Energy Balance in Cartesian coordinates
The rate of generation is H where H includes all the sources of heat generation, i.e.
reaction, pressure forces, gravity forces, fluid friction, etc. Substituting all these terms in
the general energy equation (Eq. 1.7) and dividing the equation by the term xyzt
and letting each of these terms approach zero yield:
( CpT )  ( CpTvx )  ( CpTv y )  ( CpTvz ) qx q y qz






 H
t
x
y
z
x
y
z
113
(3.60)
Expanding the partial derivative yields:
  v v y vz  qx q y qz
 T
T
T
T 

Cp  vx  v y  vz   CpT   x 



 H
x
y
z 
x
y
z  x y z
 t
 t
(3.61)
Using the equation of continuity (Eq. 3.10) for incompressible fluids the equation is
reduced to:
 T
T
T
T  qx q y qz

Cp
 vx
 vy
 vz


 H
x
y
z  x
y
z
 t
(3.62)
Using Fourier's law:
qu  k
(3.63)
dT
du
into the last equation gives:
  2T  2T  2T 
 T
T
T
T 


Cp
 Cp vx
 vy
 vz
H
  k  x 2  y 2  z 2   


t

x

y

z






 
generation




accumulation
Transport by bulk flow
(3.64)
Transport by thermal diffusion
The energy balance includes as before a transient term, a convection term, a diffusion
term, and generation term. For solids, the density is constant and with no velocity, i.e. v
= 0, the equation is reduced to:
  2T  2T  2T 
T
Cp
 k  2  2  2    H


t
x
y
z  generation










accumulation
Transport by thermal diffusion
114
(3.65)
3.5 Conversion between the coordinates
So far we have shown how to derive the equation of change in Cartesian
coordinates. In the same way, the equations of change can be written other coordinate
systems such as the cylindrical or spherical coordinates. Alternatively, one can
transform the equation of change written in the Cartesian coordinates to the others
through the following transformation expressions.
The relations between Cartesian coordinates (x,y,z) and cylindrical coordinates
(r,z,) (Figures 1.3 and 1.4) are the following:
x = rcos(),
z=z
(3.66)
y
  tan 1 ( )
x
(3.67)
y = rsin(),
Therefore;
r  x2  y 2 ,
The relations between Cartesian coordinates (x,y,z) and spherical coordinates (r,,)
(Figures 1.3 and 1.5) are:
x  r sin( ) cos(),
y  r sin( ) sin( ),
z  r cos()
(3.68)
Therefore;
r x y z ,
2
2
2
x2  y 2
  tan (
),
z
1
y
  tan 1 ( )
x
(3.69)
Accordingly, we list the following general balance equations in the three coordinates.
These equations are written under the assumptions mentioned in previous sections. For
the more general case, where density is considered variable the reader can consult the
books listed in the references.
115
3.5.1 Balance Equations in Cartesian Coordinates
Mass Balance
(
vx v y vz


)0
x y z
(3.70)
Component balance for component A in binary mixture with chemical reaction rate RA:
  2C A  2C A  2C A 
C A  C A
C A
C A 
  D AB 
  v x
 vy
 vz


)   R A
2
2
2
t

x

y

z

x

y

z




(3.71)
Energy balance
Cp
T
T
T
T
 2T  2T  2T
 Cp(v x
 vy
 vz
)  k( 2  2  2 )  H
t
x
y
z
x
y
z
(3.72)
Momentum balance
 x component
vx
vx
vx
vx
 2vx  2vx  2vx P

 (vx
 vy
 vz
)  ( 2  2  2 ) 
 g x
t
x
y
z
x
y
z
x
(3.73)
 y component:

v y
t
 (vx
v y
x
 vy
v y
y
 vz
v y
z
)  (
 2v y
x
 z component
116
2

 2v y
y
2

 2v y
z
2
)
P
 g y
y
(3.74)
vz
vz
vz
vz
 2vz  2vz  2vz P

 (vx
 vy
 vz
)  ( 2  2  2 ) 
 g z
t
x
y
z
x
y
z
z
(3.75)
3.5.2 Balance Equations in Cylindrical Coordinates
Mass balance
(
1 rvr 1 v vz


)0
r r
r  z
(3.76)
Component balance for component A in binary mixture (A-B)with reaction rate RA:
 1  C A
C A  C A
C A 
1 C A
1  2 C A  2 C A 
  vr
 v
 vz
(r
) 2

)  RA
  D AB 
 r r
t
r
r 
z 
r
r  2
z 2 


(3.77)
Energy balance
Cp
T
T
1 T
T
1  T
1  2 T  2T
 Cp(vr
 v
 vz
)  k(
(r )  2

)  H
t
r
r 
z
r r r
r  2 z 2
(3.78)
Momentum balance
 r component

v v r v2
v r
v
v
  (v r r  

 vz r ) 
t
r
r 
r
z
(
2
2
 1 rv r
1  vr
2 v  v r
P
(
) 2


)
 g r
2
2
2
r r r
r
r 
r 
z
  component:
117
(3.79)

vv
v
v
v v
v
  (vr      r   vz  ) 
t
r
r 
r
z
(
(3.80)
 1 rv
1  2v
2 vr  2v
P
(
) 2

 2 )
 g
r r r

r  2 r 2 
z
 z component

v v
vz
v
v
1  vz
1  2vz  2vz
P
  (vr z   z  vz z )   (
(r
) 2
 2 )
 g z
t
r
r 
z
r r
r
z
r  2
z
(3.81)
3.5.3 Balance Equations in Spherical Coordinates
Mass Balance
 1  (r 2 v r )
1 (v sin( ))
1 v
 2


r
r sin( )

r sin( ) 
r

0


(3.82)
Component balance for component A in binary mixture (A-B) with reaction rate RA :
C A  C A v
  vr

t
r
r

 1
D AB  2
r

v
C A
C A 



r sin( )  
 2 C A
1

C
1
 2C A 
(r
) 2
(sin( ) A )  2 2
)  RA
r
r

r sin( ) 
r sin ( )  2 
(3.83)
Energy balance
Cp
v T 
 T v T
T

 Cp v r


t
r  r sin(  )  
 r
1 
T
1

T
1
 2T 
  H
k  2 ( r 2
) 2
(sin(  ) )  2 2
r
r sin(  ) 

r sin ( )  2 
 r r
Momentum balance
 r component
118
(3.84)

2
2
 v
v
v v
vr
vr v  v 
   vr r   r 


 r

t
r


r
sin(

)


r


 1 2
 
 r r
2
2
( r 2 vr ) 

v
1
 2 vr
(sin( ) r )  2 2

r sin( ) 
r sin ( )  2
1
2
 P

 g r
 r

(3.85)
  component:

2
 v
v
v
v v
v vr v v cot( ) 
   vr     



 r

t
r  r sin( ) 
r
r


 1  2 v
 sin( )v
 2v 
1 
1
(
r
)

(sin(

)
)

2
r

r 2 
r 2 sin 2 ( )  2 
 r r
2  v
cos( ) v  1 P

 2  r 
 g
r   sin 2 ( )   r 
 
(3.86)
  component

 v v v
v
v vr v v v cot( ) 
   vr
 




 r

t
r  r sin( ) 
r
r


v
 1  2 v
 2v 
1 
1  sin( )v
1
(
r
)

(
)

 r 2 r
r

r 2  sin(θ )
r 2 sin 2 ( )  2 


(3.87)

2
vr
2 cos( ) v 
1
P

   2
 2 2
 r sin( )   g




r
sin(

)
r
sin
(

)


3.6 Examples of Application of Equations of change
Practically all the microscopic balance examples treated in the previous chapter
can be treated using the equations of change presented in this chapter. In this section we
review some of the previous examples and present additional applications.
3.6.1 Liquid flow in a Pipe
To model the one dimensional flow through the pipe of an incompressible fluid
(Example 2.2.1) we may use the continuity balance. For constant density we have the
continuity equation in cylindrical coordinates (Eq. 3.76).
119
1 rvr v vz


0
r r
r z
(3.88)
The plug flow assumptions imply that vr = v = 0, and the continuity balance is reduced
to:
vz
0
z
(3.89)
3.6.2 Diffusion with Chemical Reaction in a Slab Catalyst
To model the steady state diffusion with chemical reaction of species A in a slab
catalyst, (Example 2.2.3) we use the equation of change (Eq. 3.71). The fluid properties
are assumed constant,
C A
C A
C A
C A
 2C
 2C A  2C A
 vx
 vy
 vz
 D A ( 2A 

)  RA
t
x
y
z
x
y 2
z 2
Since the system is at steady state we have
C A
0.
t
(3.90)
If we assume that there is no bulk
flow then vx = vy = vz = 0. For diffusion in the z-direction only, the following holds:



0.
x y
The equation is then reduces to:
d 2C A
 DA
 RA
dz 2
3.6.3 Plug Flow Reactor
120
(3.91)
The isothermal plug flow reactor (Example 2.2.5) can be modeled using the
component balance equation (3.77). The plug flow conditions imply that vr = v = 0 and



0.
r 
Equation (3.77) is reduced to:
C A
C A
 2C A
 v z
 DAB
 RA
t
z
z 2
(3.92)
For the non-isothermal plug flow reactor (Example 2.2.6), the energy balance is
obtained by using Eq. 3.78. For fluid with constant properties and at constant pressure,
we have:
Cp
T
T v T
T
 2T 1 T 1  2T  2T
 Cp(vr

 vz
)  k( 2 


)  H
t
r
r 
z
r r r 2  2 z 2
r
Using the plug-flow assumptions, vr = v = 0 and



0,
r 
(3.93)
and neglecting the viscous
forces, the term H includes the heat generation by reaction rate RA and heat exchanged
with the cooling jacket, htA(T –Tw). Equation (3.93) is reduced to:
Cp
T
T
 2T
D
 Cpv
 k 2  H r ko e E / RT C A  ht
(T  Tw )
t
z
z
A
(3.94)
3.6.4 Energy Transport with Heat Generation
Consider the example of a solid cylinder of radius R in which heat is being
generated due to some reaction at a uniform rate of H (J/m2s). A cooling system is
used to remove heat from the system and maintain its surface temperature at the
constant value Tw (Figure 3.5). Our objective is to derive the temperature variations in
the cylinder. We assume that the solid is of constant density, thermal conductivity and
heat capacity. Clearly this is a distributed parameter system since the temperature can
vary with time and with all positions in the cylinder. We will use then the equation of
change (Eq. 3.78) in cylindrical coordinates for a solid:
121
T
k  2T 1 T 1  2T  2T


( 2 
 2 2  2 ) H
t Cp r
r r r 
z
Cp
(3.95)
T (r = R)
R
-R
T (r = -R)
Figure 0-5 Cylindrical solid rod
A number of assumptions can be made:
 The system is at steady state i.e.
T
0
t
 The variation of temperature is only allowed in radial directions. Therefore, the
terms
 2T
 2T
and
are zero.
z 2
 2
The energy balance is reduced to:
0
k d 2T 1 dT

( 2 
) H
Cp dr
r dr
Cp
(3.96)
Or equivalently:
(
d 2T 1 dT


) H
2
dr
r dr
k
with the following boundary conditions:

The temperature at the wall is constant:
122
(3.97)
T (r  R)  Tw
(3.98)
 The maximum temperature will be reached at the center ( r = 0), therefore:
dT
0
dr
at r  0
(3.99)
Note that Equation (3.97) can also be written as follows:
1 d dT

(r
) H
r dr dr
k
(3.100)
since qr = −k dT/dr, this equation is equivalent to:
1 d
(rqr )   H
r dr
(3.101)
The left hand side is the rate of diffusion of heat per unit volume while the right hand
side is the rate of heat production per unit volume.
3.6.5 Momentum Transport in a Circular Tube
We revisit example 2.2.2 where we derived the steady state equations for the
laminar flow inside a horizontal circular tube. We will see how the model can be
obtained using the momentum equation of change. We assume as previously that the
fluid is incompressible and Newtonian. The momentum equations of change in
cylindrical coordinates are given by Eqs. (3.79-3.81). A number of simplifications are
used:
 The flow has only the direction z, i.e. vr = v = 0.
 The flow is at steady state,
v z
0
t
The momentum equation in cylindrical coordinates Eq. 3.81 is reduced to:
123
vz
  2v 1 vz 1  2vz  2vz  P
vz

  2z 


 g z
z
r r r 2 2 z 2  z
 r
(3.102)
Using the continuity equation (Eq. 3.76), and since vr = v = 0 gives:
dv z
0
dz
(3.103)
We also note that because the flow is symmetrical around the z-axis we have necessarily
no variation of the velocity with , i.e.
  2vz 
 2   0
  
(3.104)
Equation 3.102 is then reduced to
 d 2 v z 1 dv z  dP
 

2
dr
r
dr

 dz
 
Since the left hand side depends only on r, this equation suggests that
(3.105)
dP
is constant.
dz
Therefore:
dp P

dz
L
(3.106)
where P is the pressure drop across the tube. Equation 3.105 is equivalent to:
 d 2 vz 1 dvz  P


2
r dr  L
 dr
 
(3.107)
with the following conditions identical to those in Example 2.2.2. Note also that Eq.
3.107 can also be written as:
124

d rv z
P
( )
dr dr
L
(3.108)
d ( r rz ) P

rdr
L
(3.109)
which is also equivalent to:
where rz is the shear stress. The left term is the rate of momentum diffusion per unit
volume and the right hand side is in fact the rate of production of momentum (due to
pressure drop). Note then the similarity between Eq. 3.109 for momentum transfer with
Eq. 3.101 for heat transfer.
3.6.6 Unsteady state Heat Generation
We reconsider Example 3.6.4 but we are interested in the variations of the
temperature of the reactor with time as well. This may be needed to compute the heat
transferred during start-up or shut-down operations. Keeping the same assumptions as
Example 3.3.4 (except the steady state assumption), the energy balance in cylindrical
coordinates yields:
Cp

T
1 d dT
k
(r
) H
t
r dr dr
k
(3.110)
with the initial and boundary conditions:
T ( R , t )  Tw
(3.111)
dT
| r 0  0
dr
T ( r ,0)  Tw
(3.112)
3.6.7 Laminar Flow Heat Transfer with Constant Wall Temperature
125
(3.113)
We consider a fluid flowing at constant velocity vz into a horizontal cylindrical
tube. The fluid enters with uniform temperature Ti. The wall is assumed at constant
temperature Tw. We would like to model the variations of the fluid temperature inside
the tube. To apply the energy equation of change (Eq. 3.78) we will assume that the
fluid is incompressible, Newtonian and of constant thermal conductivity. Since the
system is at steady state d/dt = 0 and the flow is one-dimensional
vr = v = 0, the
energy equation in cylindrical coordinates Eq. 3.78 is reduced to:
Cpv z
T
 2T 1 T 1  2T  2T
 k( 2 


)
t
r r r 2  2 z 2
r
(3.114)
Tw
Tb(z)
T(r,z)
r
Tb(0)
Figure 0-6 heat transfer with constant wall temperature
Since the temperature is symmetrical then
the conduction term
 2T
z 2
 2T
T
= 2


= 0. In some cases we can neglect
compared to the convective term v z
described by the following energy and momentum equations:
126
T
. The system is then
z
Cpv z
T
 2T 1 T
 k( 2 
)
t
r r
r

d rv z
P
( )
dr dr
L
(3.115)
(3.116)
These two equations are therefore coupled by vz, with the previous boundary conditions
for vz,
At r = R,
vz = 0
(3.117)
At r = 0,
dvz/dr = 0
(3.118)
At z = 0,
T = Ti
(3.119)
At r = 0,
dT/dr = 0
(3.120)
At r = R,
T = Tw
(3.121)
3.6.8 Laminar Flow and Mass Transfer
We consider the example of a fluid flowing through a horizontal pipe with
constant velocity vz. The pipe wall is made of a solute of constant concentration CAw that
dissolved in the fluid. The concentration of the fluid at the entrance z = 0 is CAo. The
regime is assumed laminar and at steady state. The fluid properties are assumed
constant. We would like to model the variation of the concentration of A along the axis
in the pipe. The component balance for A (Eq. 3.77) is:
vr
C A
C A
C A
 v
 vz
 D AB (
r

z
C A
)
2
2
r   C A   C A )
rr
r 3  2
z 2
 (r
(3.122)
Since vr = v = 0, the balance equation becomes:
C A
vz
 D AB (
z
C A
)
2
r   C A )
rr
z 2
 (r
127
(3.123)
If the diffusion term
vz
C A
z
 2C A
z 2
in the z-direction is negligible compared to convection term
then the last equation reduces to:
C A 

)
  (r
C A
r 
vz
 D AB 
 rr

z




(3.124)
The equation (Eq. 3.116) is unchanged.
The two equations are coupled through vz. The boundary conditions are analogous to
the previous example:
At r = R,
vz = 0
(3.125)
At r = 0,
dvz/dr = 0
(3.126)
At z = 0,
CA= CAo
(3.127)
At r = 0,
dCA/dr =0
(3.128)
At r = R,
CA= CAw
(3.129)
Note the similarity between this example and the heat transfer case of the previous
example.
128
102