ABSTRACT
The mass conservation equation in a differential form can be obtained by applying the mass conservation principle on a differential control volume as shown in Figure 6.1. Considering the control volume, the net mass ow rate in the x-, y-, and z-directions can be expressed as
( )∂∂ ρx-direction: x u dx dy dz (6.1)
( )∂∂ ρy-direction: y v dx dy dz (6.2)
( )∂∂ ρz-direction: z w dx dy dz (6.3) The rate of change of mass inside the control volume can be obtained
from the Reynolds transport theory as follows:
t
dV t
dx dy dz CV ∫ ∂ρ∂ = ∂ρ∂ (6.4)
The net mass ux into the control volume should be equal to the rate of change of mass inside the control volume. The mass conservation in differential form can be expressed as
t x
u y
v z
w 0( ) ( ) ( )∂ρ∂ + ∂
∂ ρ + ∂
∂ ρ + ∂ ∂ ρ = (6.5)
Newton’s second law on a differential control volume, as shown in Figure 6.1, can be used to obtain the conservation of momentum equation. The net forces on the control volume should be balanced with the acceleration of the control volume times its mass as follows:
a dm dF∑= (6.6)
The acceleration vector eld a is obtained from the total time deriva-
tive of the velocity vector
V
i j ka d dt
du dt
dv dt
dw dt
= = + + (6.7)
Each component of the velocity elds is a function of the space and time. Using the chain rule, the scalar time derivative can be obtained
du x,y,z, t
dt u t
u x
dx dt
u y
dy dt
u z
dz dt
( ) =
∂ ∂ +
∂ ∂ +
∂ ∂ +
∂ ∂
(6.8)
where u = dx/dt is the local velocity component in the x-direction, v = dy/dt is the local velocity component in the y-direction, and w = dw/dt is the local velocity component in the z-direction. The total derivative of u is the acceleration in the x-direction
a du x,y,z, t
dt u t
u du dx
v du dy
w du dz
x ( )
= =
∂ ∂ + + + (6.9)
The acceleration in the y-and z-directions can be expressed as, respectively,
a dv x,y,z, t
dt v t
u dv dx
v dv dy
w dv dz
y ( )
= =
∂ ∂ + + +
(6.10)
a du x,y,z, t
dt w t
u dw dx
v dw dy
w dw dz
z ( )
= =
∂ ∂ + + + (6.11)
The mass of the control volume must be equal to volume times the density as follows:
dm = ρ dx dy dz (6.12)
The forces on the control volume are of two types: body and surface. The body force is due to the gravity:
dF g dx dy dzb = ρ
(6.13)
while the surface forces are due to the surface stresses, including normal and parallel stresses. The surface stresses in the x-, y-, and z-directions are as follows:
dF x y z
=
∂σ ∂ +
∂τ ∂ +
∂τ ∂
(6.14)
dF y y z
=
∂σ ∂ +
∂τ ∂ +
∂τ ∂
(6.15)
dF x y z
=
∂σ ∂ +
∂τ ∂ +
∂τ ∂
(6.16)
The equations of motion in the x-, y-, and z-directions can be expressed as, respectively,
x-direction: u t
u u x
v u y
w u z
g x y z
ρ ∂∂ + ∂ ∂ +
∂ ∂ +
∂ ∂
= ρ + ∂σ∂ + ∂τ ∂ +
∂τ ∂
(6.17)
y-direction: v t
u v x
v v y
w v z
g x y z
ρ ∂∂ + ∂ ∂ +
∂ ∂ +
∂ ∂
= ρ + ∂σ∂ + ∂τ ∂ +
∂τ ∂
(6.18)
z-direction: w t
u w x
v w y
w w z
g x y z
ρ ∂∂ + ∂ ∂ +
∂ ∂ +
∂ ∂
= ρ + ∂σ∂ + ∂τ ∂ +
∂τ ∂
(6.19)
For Newtonian uid, the stress components are obtained from the theory of elasticity, and they are
P 2 u x
xxσ = − + µ ∂ ∂ (6.20)
P 2 v y
yyσ = − + µ ∂ ∂ (6.21)
P 2 w z
zzσ = − + µ ∂ ∂ (6.22)
u y
v x
xy yxτ = τ = µ ∂ ∂ +
∂ ∂
(6.23)
v z
w y
yz zyτ = τ = µ ∂ ∂ +
∂ ∂
(6.24)
w y
v z
zy yzτ = τ = µ ∂ ∂ +
∂ ∂
(6.25)
Substituting the stress equations into the equations of motion, we have
u t
u u x
v u y
w u z
P x
u x
u y
u z
g 2
2 xρ ∂ ∂ +
∂ ∂ +
∂ ∂ +
∂ ∂
= −
∂ ∂ + µ
∂ ∂ +
∂ ∂ +
∂ ∂
+ ρ
(6.26)
v t
u v x
v v y
w v z
P y
v x
v y
v z
g 2
2 yρ ∂ ∂ +
∂ ∂ +
∂ ∂ +
∂ ∂
= −
∂ ∂ + µ
∂ ∂ +
∂ ∂ +
∂ ∂
+ ρ
(6.27)
w t
u w x
v w y
w w z
P z
w x
w y
w z
g 2
2 zρ ∂ ∂ +
∂ ∂ +
∂ ∂ +
∂ ∂
= −
∂ ∂ +µ
∂ ∂ +
∂ ∂ +
∂ ∂
+ρ
(6.28)
Equations 6.26-6.28 are called the Navier-Stokes equations. They are nonlinear and nonhomogenous partial differential equations.
