Transcript rv．ndA

The control surface A can be considered to consist of three regions:
Ain: the surface of the the region where the fluid enters the control volume;
Aout: the surface of the region where the fluid leaves
Awall; the fluid is in contact with a wall
A = Ain + Aout + Awall
As shown in Fig. 1.2-2(b) the outward mass flow
rate through dA is rv．ndA, since n is points outward,
thus the inward mass flow rate through dA is - rv．ndA
For a homogeneous fluid of a pure material
undergoing no chemical reactions, the conservation
law for mass is
Rate of mass accumulation (term 1)
= Rate of mass in (term 2)
- Rate of mass out (term 3)
(b)
Fig. 1.2-2 Volume flow rate through a
differential surface element dA: (a) threedimensional view; (b) two-dimensional view
The mass of the fluid in the differential volume element dΩ is dM= rdΩ
22
The mass of the fluid in the differential volume element dΩ is dM= rdΩ
The total mass in the control volume is M =


The rate of mass change in the control volume is
The net inward mass flow rate into Ω is
That is
rd
dM 
   rd
dt
t
  rV  ndA
A
dM 
   rd   rV  ndA
A
dt
t
[1.2-4]
 ( rvdA)in  ( rvdA)out  min  mout
A
A
Examples 1.2-1 and 1.2-2
23
1.3 Differential mass-balance equation
According to Eq. 1.2-4

 rd    rV  ndA
A
t 

 rd 
t 
r
 t d
[1.2-4]
[1.3-1’]
According to Gauss divergence theorem [A.4-1]

A
rV  ndA     ( rV )d
[1.3-2]
Substituting Eq.[1.3-1’] and [1.3-2] into [1.2-4], we obtained


{
r
   ( rv)}d  0
t
[1.3-3]
The integrand must be zero everywhere since the equation must hold for
any arbitrary Ω. Therefore, we have
r
   ( rv )  0
t
[1.3-4]
24
Equation [1.4-4] is the differential mass-balance equation, that is, the equation
of continuity. For an incompressible fluid, the density r is constant and Eq. 1.3-4
becomes
 v  0
Example 1.3-2 Normal stress due to creeping flow around a sphere
25
1.4 Overall momentum-balance equation
Consider an arbitrary stationary control volume Ω bounded by
surface A through a moving fluid is flowing, as shown in Fig. 1.4-1.
The control surface A can be considered to consist of three parts:
A = Ain + Aout + Awall
Consider conservation law for momentum
of the control volume:
Rate of
momentum
accumulation
=
Rate of
momentum
in
Rate of
- momentum
out
+
Sum of forces
acting on
system
[1.4-2]
This equation is consistent with Newton’s second law of motion
The mass contained in a differential volume element dΩ in the control volume is
rdΩ and its momentum is dP = rvdΩ. The momentum of the fluid in Ω is :
P   dP  rvd

26
The rate of momentum change in Ω is :
dP 
  rvd
dt t 
[1.4-a]
The inward mass flow rate through dA : -r(v．n)dA
The inward momentum flow rate through dA : -rv(v．n)dA
The net inward momentum flow rate into Ω : 
or
 
Ain  Aout  Awall
 rv(v  n)dA
A
rv( v  n)d ( Ain  Aout  Awall )
Since v = 0 at the wall, the equation can be expressed as
 ( rv(v  n)dA)in  ( rv(v  n)dA)out
A
A
Define v =｜v．n｜
 (  rvvdA) in  (  rvvdA) out    rvv  ndA
A
[1.4-b]
A
27
The pressure force acting on dA is dFp = -pndA, The pressure force acting
on the entire control surface A is
Fp   pndA
[1.4-c]
A
The viscous force exerted on the fluid by the surrounding over dA is
dFv = -t‧ndA, the viscous force exerted on the fluid by the surroundings over A is
Fv   t  ndA
[1.4-d]
A
The body force acting on the differential volume element dΩ is dFb = fbdΩ,
the body force acting on the entire control volume is
Fb   f b d
[1.4-e]

Substituting [1.4-a] through [1.4-e] into [1.4-2]

rvd    rvv  ndA   pndA   t  ndA   f b d


A
A
A

t
or

rvd    rvv  ndA   F


A
t
[1.4-3]
28
dP
 (  rvvdA) in  (  rvvdA) out  Fp  Fv  Fb
A
A
dt
[1.4-8]
dP
 (rvav Av av )in  (rvav Av av ) out  Fp  Fv  Fb
dt
 (mv av )in  (mv av ) out  Fp  Fv  Fb
dP x
 (mv av x )in  (mv av x ) out  Fpx  Fvx  Fbx
dt
Where
P = momentum of fluid in control volume
m = mass flow rate at inlet or outlet
Fp = pressure force acting on control volume by surrounding
Fv = viscous force acting on control volume by surrounding
Fb = body force acting on control volume
The correction factor 
 ( vvdA) /( Avav vav )
A
 = 4/3 for laminar flow in round pipes
~ 1 for turbulent flow
29
Example 1.4-1 Force exerted by an impinging jet
30
Example 1.4-2 Back trust of a jet
31
Example 1.4-3 Thrust on a pipe bend
32
Example 1.4-4 Friction force on a pipe wall
33
Example 1.4-5Creeping flow around a sphere: Stoke’s law
34
Example 1.4-5
35
Example 1.4-6 Laminar flow over a flat plate
36
1.5 Differential momentum-balance equation
1.5.1 Derivation
Eq. 1.4-3 represented the integral form of momentum-balance equation

rvd    rvv  ndA   pndA   t  ndA   f b d


A
A
A

t
[1.5-1]
The surface integrals in Eq. 1.5-1 can be converted into their corresponding
volume integrals. From Eq. [A.4-4], we have
 rvv  ndA     (rvv)d
A
A

A
pndA   pd

t  ndA    td
A

Eq. 1.5-1 can be rewritten as follows:


{

( rv)    ( rvv )  p   t  f b }d
t
37
The integrand must be zero everywhere since the equation must hold for any
arbitrary region Ω, therefore
∵

( rv)    ( rvv )  p   t  f b  0
t

v
r
and   ( rvv )  v(  rv)  rv  v
rv  r  v
t
t
t
It can be further simplified with the help of the equation of continuity as following:
v
r  rv  v  p   2 v  f b
t
(for constant r and )
[1.5-8]
If the gravity force is the only body force involved, that is
r
v
 rv  v  p   2 v  rg
t
The equation is called Navier-Stokes equation
[1.5-9]
38
The equation of motion is often expressed in terms of the Stokes derivative D/Dt,
Which is defined as follows:
D 




  vx
 vy
 vz
  v
Dt t
x
y
z t
The equation of motion (Eq. 1.5-9) becomes
Dv
r
 p   2 v  rg
Dt
(1)
(2)
(3)
(4)
Where
term (1): the inertia force per unit volume, namely, the mass per unit volume
times acceleration
(2):pressure force, (3) viscous force, and (4) gravity force per unit volume
39
1.5.2 Dimensionless form
The equation of motion and the continuity equation can be expressed in
the dimensionless form.
In forced convection, a characteristic velocity V, length L, and time L/V are
used to express the equation qualitatively. By substituting V for v, (L/V)-1 for
D/Dt, L-1 for ▽, and L-2 for ▽2.
2
Dv
V
r
V
inertia force: r
r

Dt
L /V
L
V
viscous force:  2 v   2
L
Define:
v
dimensionless velocity v 
pressure force: p 
p
L
gravity force: rg
p - po
rV 2
V
x y z
tV



dimensionless time t  
dimensionless coordinates x , y , z  , ,
L L L
L
Dimensionless operator   L  ex   e y   ez 
x* 2 y * 2 z * 2
Dimensionless operator 2  L2 2  e   e   e 
x
y
z
40
x*2
y*2
z *2

dimensionless pressure p 
Substituting Eqs. [1.5-12] through [1.5-17] into Eqs.[1.5-18] and [1.5-19], we have
Continuity:
Motion:
1 * *
 v V  0
L
V  *
*
* * V
r
( v V )  rVv   v
*
L t
L
1 * * 2
1 *2 *
   p rV   2  v V  rge g
L
L
Multiplying Eq.[1.5-21] by L/V and Eq.[1.5-22] by L/rV2, we have
Continuity:
*  v*  0
Motion:
v*

Lg
*
* *
* *
*2 *
 v   v   p  (
) v  ( 2 )e g
*
t
LV
V
41
Therefore, for forced convection
Continuity:
Motion:
Where
*  v*  0
v*
1 *2 * 1
*
* *
* *

v


v


p

 v  eg
*
t
Re
Fr
rVL rV 2 / L inertia force
Re 





V / L viscous force
LV
V 2 rV 2 / L inertia force
Fr 


gL
rg
gravity force
42
Given a physical problem in which the dependent parameter is a function of n-1
independent parameters. We may express the relationship among the variables in
functional form as
q1  f (q2 , q3 , q4 ,...., qn )
where q1 is the dependent parameter, and q2, …, qn are n-1 independent parameters.
Mathematically we can express the functional relationship in the equivalent form
g (q1 , q2 , q3 ,...., qn )  0
Where g is an unspecified function, different from f.
The Buckingham Pi theorem states that: Given a relation among n parameters
of the form
g (q , q , q ,...., q )  0
1
2
3
n
then the n parameters may be grouped into n-m independent dimensionless
ratios, or n parameters, expressed in a functional form by
G(1 ,  2 ,....,  nm )  0
or
1  G1 ( 2 , 3 ,....,  nm )
The number m is usually, but not always, equal to the minimum number of
independent dimensions required to specify the dimensions of all the parameters
43
q1, q2, …and qn.
Example
The drag force F, on a sphere depends on the relative velocity, V,
the sphere diameter, D, the fluid density, r, and the fluid viscosity, , Obtain a
set of dimensionless groups that can be used to correlate experimental data.
Given: F=f(r, V, D, ) for a smooth sphere
Find: An appropriate set of dimensionless groups
Solution:
Step 1: Find the parameters: F, r, V, D, and .
n=5
Step 2: Select primary dimensions: M, L, t or F, L, t
r=3
Step 3: Find the dimension of the parameters selected in step 1
F
r
V
D

ML/t2
M/L3
L/t
M/L3
M/(Lt)
Step 4: Select repeating parameters with the number equal to
the primary dimensions r, V, D
m=3
Step 5: n-m=2. Two dimensionless groups will result.
Step 6: Setting up dimensional equations Π1= raVbDcF, and Π2= rdVeDf
Step 7: Summing exponents
1  r aV b D c F  (
M a L b c ML
0 0 0
)
(
)
L
(
)

M
LT
3
2
L
t
t
44
 2  r aV b D c   (
M: a+1 =0
L: -3a+b+c+1=0
T:-b-2=0
Similarly
2 
M a L b c M
0 0 0
)
(
)
L
(
)

M
LT
3
L
t
Lt
=> a = -1
=> c = -2
=> b =-2
1 
Therefore
F
rV 2 D 2

rVD
Step 8: Check using F, L, t dimensions
F
L4 t 2 1
1 
 F 2 ( ) 2  [1]
2 2
rV D
Ft L L
and

Ft L4 t 1
2 
 2 2
 [1]
rVD L Ft L L
The functional relationship is Π1=f(Π2), or
F


f
(
)
2 2
rV D
rVD
As noted before. The form of the function, f, must be determined experimentally.
45
1.5.3 Boundary Conditions
46
47
48
49
50
51
Example 1.5-1 Tangential annual flow (Incompressible, Newtonian, laminar)
Flow is in q direction only, vr=vz=gq=0
Flow is axisymmetric, no pressure variation in the
q direction. Neglect end effects,  vq / z  0
The equation of continuity in cylinder coordinates is
r 1 
1 


( r rv r ) 
( r v q )  ( r vz )  0
t r r
r q
z
vq
0
Which can be reduced to
q
The velocity distribution vq(r), the q component of the equation of motion is
vq vq vq v r vq
vq 
 vq
r
 vr


 vz


t

r
r

q
r

z


   1 rvq  1  2 vq 2 v r  2 vq
1 p

 
 2
 2
 2
2

r q
r q
z
 r  r r  r q

  r gq

d  1 d (rvq ) 
0


dr  r dr 
The boundary conditions are vq =0 at r = r1, vq = r2w2 at r = r2
52
Example 1.5-2 Laminar flow through a vertical tube, find the velocity, volume flow rate,
and shear stress (Steady-state, incompressible, Newtonian)
Flow is in z direction only, vr=vq=0, vz is independent of q
r 1 
1 


( r rv r ) 
( r v q )  ( r vz )  0
t r r
r q
z
v z
0
z
v v v z
v 
 v z
 vr z  q
 vz z  
r
r q
z 
 t
 1   v z  1  2 v z  2 v z 
p
 
 2   r gZ
r
 2
2
z
z 
 r r  r  r q
r
1 d  dv z

r
r dr  dr
 dp
 rg

dz

Subjected to B.Cs dv z  0 at r = 0 and
dr
v z  0 at r = R
Find vav
  ( PL  P0 )  r gL  R 2
vav 
8 L
1 d  dv z
r
r dr  dr
 ( PL  P0 )  r gL

L

 ( PL  P0 )  r gL  2 2
vz  
 (R  r )
4

L


  ( PL  P0 )  r gL  R 4
Q
8 L
53
Example 1.5-3 Flow of a rising film
v x =v y  0, v z  f ( x)
Continuity Eq.
r 


 ( r vx )  ( r v y )  ( r vz )  0
t x
y
z
v z
0
z
Eq. of motion
 v z
v
v
v 
 vx z  v y z  vz z  
x
y
z 
 t
r
 2vz 2vz 2vz 
p
    2  2  2   r gz
z
y
z 
 x
2vz r
 g
2
y

B.Cs
vz
 0 at y=0 and v z  V at y=L
y
vz  V +
rg 2 2
(y  L )
2
54
Example 1.5-5 Adjacent flow of two immiscible fluids
55