Solution of continuity and momentum equations in polar form

1. 1

2. Equations
Continuity Equation:
(𝒂)
Which shows conservation of Mass
  0. 


V
t


2

3. Momentum Equation
• Which shows conservation of
Momentum
bT
Dt
VD
 
3

4. • These are the equations
which we use only in
𝒙 𝒚 𝒛 𝒄𝒂𝒓𝒕𝒆𝒔𝒊𝒂𝒏 coordinate
system
4

5. 5

6. Question
• Can we use these equations , when
we have fluid in cylindrical channel
𝒊. 𝒆. In pipe or in Veins??
6

7. Topic of presentation
• Derivation of Continuity equation in
𝒓, 𝜽, 𝒛 𝑪𝒚𝒍𝒊𝒏𝒅𝒓𝒊𝒄𝒂𝒍 coordinate system
• Derivation of Momentum Equation in
𝒓, 𝜽, 𝒛 𝑪𝒚𝒍𝒊𝒏𝒅𝒓𝒊𝒄𝒂𝒍 coordinate system
7

8. Cartesian cylindrical
8

9. 9

10. Results from graph











x
y
yxr
ry
rx
1
22
tan
sin
cos



10

11. Direction of 𝑟 & 𝜃




ji
jir


cossin
sincos
11

12. Conversion of unit vectors in terms
of (r 𝜃 𝑧)






zk
rj
ri


cossin
sincos
12

13. velocity components in (r 𝜃 𝑧)
• As we know that velocity field in
𝒙 𝒚 𝒛 system is
• And the velocity field in cylindrical system is
13

 kwjviuV

 zrV vvv zr


14. Conversion of Del operator in
(r 𝜃 𝑧)
• As we know that Del operator in (𝒙 𝒚 𝒛 )
systems is
14









 k
z
j
y
i
x

15. Solution
• As we know that
15











x
y
yxr
ry
rx
1
22
tan
sin
cos









zk
rj
ri


cossin
sincos

16. • And also we have
• By taking derivatives we have
16


sin
cos






y
r
x
r
ry
ry


cos
sin
















x
y
yxr
1
22
tan

17. As 𝒙 = 𝒓𝒄𝒐𝒔𝜽 𝒂𝒏𝒅 𝒚 = 𝒓𝒔𝒊𝒏𝜽 so that
𝒙 = 𝒙 𝒓, 𝜽 𝒂𝒏𝒅 𝒚 = 𝒚 𝒓, 𝜽 , hence we have
17
yy
r
ry
xx
r
rx




























..
..

18. • Using above values we will get
18

































 






z
z
r
rr
r
rr








cossin
cos
sin
sincos
)sin(
cos

19. • After simplification we will get “Del
operator” in (r 𝜽 𝒛)
19









 z
zr
r
r


1

20. Conversion of
in(𝑟 , 𝜃, 𝑧)
20
V.

21. • For this we will use previous results , we have
• After simplification we will get ,
•
21
V. 












 
z
zr
r
r


1








zr vvv zr

.
V.  
zr
r
rr
vv
v z
r









11

22. Required continuity equation in(𝒓, 𝜽, 𝒛)
22
      0
11












vvv zr
zr
r
rrt





  V.      vvv zr zr
r
rr

  






 11

23. Special case
(Incompressible fluid)
23

24. 𝜌 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
24
      0
11









vvv zr
zr
r
rr 


25. Derivation of Momentum
Equation for any fluid in
Cylindrical Coordinate
System
25

26. • As we know that momentum equation
in Cartesian coordinate system is,
• Where is material derivate is
cacuhy stress tensor and is body
force
26
bT
Dt
VD
 
DT
D
T
b

27. • We will use same procedure as we
did in the derivation of continuity
equation, we will transform every
term in (𝒓 , 𝜽 , 𝒛 ) form
27

28. Material derivative in
(𝑟, 𝜃, 𝑧) form
As we know that ,
28
VV
t
V
Dt
VD



 .

29. As we know from previous results that,
29
V. 












 
z
zr
r
r


1








zr vvv zr

.

30. • After simplification we will get
30
z
zz
z
z
z
rz
r
z
rrrr
r
r
zr
r
r
r
rrV
vvv
vvvvv
vvv
zr
zrr
zr

























































111
.

31. • Now
31
VV. .







zr vvv zr


































































z
zz
z
z
z
rz
r
z
rrrr
r
r
zr
r
r
r
rr
vvv
vvvvv
vvv
zr
zrr
zr











111

32. • After simplification we will get
• 𝒓 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
• 𝜽 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
• 𝒛 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
32

33. 𝒓 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
𝜽 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
𝒛 − 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 𝒅𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆
33
zrrrt
v
v
vvvv
v
v r
z
rr
r
r










 2


zrr
v
rt
v
v
vvvv
v
v
z
r
r










 

zr
v
rt
v
v
vv
v
v z
z
zz
r
z













34. Stress Tensor
34
 V

35. By putting previous results we will get
After simplification we will get stress
tensor in (𝒓, 𝜽, 𝒛) we have
35
  V
35













 
z
zr
r
r


1

























 
zrz
zr
r
r vvv zr

 
1
.

36. • 𝒓 − 𝒔𝒕𝒓𝒆𝒔𝒔 𝑻𝒆𝒏𝒔𝒐𝒓 𝒊𝒏 "𝒓" 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏
• 𝜽 − 𝒔𝒕𝒓𝒆𝒔𝒔 𝑻𝒆𝒏𝒔𝒐𝒓 𝒊𝒏 "𝜽" 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏
• 𝒛 − 𝒔𝒕𝒓𝒆𝒔𝒔 𝑻𝒆𝒏𝒔𝒐𝒓 𝒊𝒏 "𝒛" 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏
36
  2
2
22
2
2
211
z
v
rrr
r
rrr
vv
v rr
r




















  2
2
22
2
2
211
z
v
rr
r
rrr
vv
v r

















 


2
2
2
2
2
11
zrr
r
rr
vvv zzz



















37. • Put these values in Momentum Equation we
will get in Momentum Equation in
𝒓, 𝜽, 𝒛 𝒇𝒐𝒓𝒎
• We have
37

38. • 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒓 𝒅𝒊𝒓𝒆𝒄𝒊𝒐𝒏
• Component of 𝒎𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝜽 direction
38



















zrrrt
v
v
vvvv
v
v r
z
rr
r
r
2



  b
vv
v r
rr
r
z
v
rrr
r
rrr






















 2
2
22
2
2
211



















zrr
v
rt
v
v
vvvv
v
v
z
r
r



  b
vv
v z
v
rr
r
rrr
r
























 2
2
22
2
2
211

39. • Component of 𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 z 𝒅𝒊𝒓𝒆𝒄𝒊𝒐𝒏
39



















zr
v
rt
v
v
vv
v
v z
z
zz
r
z


b
vvv
z
zzz
zrr
r
rr




















 2
2
2
2
2
11

40. Navier Stock Equation
• As we know that N.S equation in 𝒙 , 𝒚 , 𝒛
40
  bTP
Dt
VD
 

41. • By using previous results we will get
• 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒐𝒇 𝑵𝒂𝒗𝒊𝒆𝒓 𝑺𝒕𝒐𝒄𝒌 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒓 𝒅𝒊𝒓𝒆𝒄𝒊𝒐𝒏
41



















zrrrt
v
v
vvvv
v
v r
z
rr
r
r
2



  b
vv
v r
rr
rr
z
v
rrr
r
rrr
P 

 



























 2
2
22
2
2
211

42. • Component of Navier stock 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝜽 direction
42



















zrr
v
rt
v
v
vvvv
v
v
z
r
r



  b
vv
v z
v
rr
r
rrr
P r


 

 


























 2
2
22
2
2
211

43. • Component of Navier stock 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 z 𝒅𝒊𝒓𝒆𝒄𝒊𝒐𝒏
43



















zr
v
rt
v
v
vv
v
v z
z
zz
r
z


b
vvv
z
zzz
zrr
r
rr
P 

 

























 2
2
2
2
2
11

44. Applications
• Blood flow in veins
• Fluid flow through cylindrical pipe
(house pipe)
44

45. Any Question
45