Short Notes: Laminar and Turbulent Flow | Short Notes for Mechanical Engineering PDF Download

 Page 1


Laminar and Turbulent Flow 
Laminar Flow
• Laminar flow also known as viscous flow
• In laminar flow, viscous force is highly is highly predominant.
Case-1
Laminar flow between 2- parallel plates
Consider a fluid element in flow field. Element have thickness dy, length dx and y 
distance from bottom plate.
A
dx ¦
]d y
Assumption:- width of flow perpendicular to paper = unity 
Free body diagram of element
Page 2


Laminar and Turbulent Flow 
Laminar Flow
• Laminar flow also known as viscous flow
• In laminar flow, viscous force is highly is highly predominant.
Case-1
Laminar flow between 2- parallel plates
Consider a fluid element in flow field. Element have thickness dy, length dx and y 
distance from bottom plate.
A
dx ¦
]d y
Assumption:- width of flow perpendicular to paper = unity 
Free body diagram of element
(T +3 T/3 X,dx)dy
Pdy
(P+3P/3X,dx)dy
___
Apply equilibrium condition
Z F =0
cp
-t.d x -| p + -^.dx jdy = 0
f &i
=>pdy-j^T +— xiy 
= > pdv - r dx - — .dxdy - rdx - pdy - — dxdv = 0
a y a x
^ — = Q
a. ax
But we know that
a u
i= u —
cy
So,
n i - M .
a y [a y j a x
— and u
G s .
are independent form y 
By integrating equation (1)
c t i 11 a p
— = — I — .y+c,
a y u l a x
= > Again integrate with respect to y
y: j cp ,
u = — + c,y + c,
2 p la x i
Apply boundary condition
> No slip condition
> Due to friction velocity of fluid at 
UDDer and bottom plate is zero.
1. At y = 0, u = 0 
C2 = 0
2. At y=H, u=0
c, -
2u l^axj
So,
Page 3


Laminar and Turbulent Flow 
Laminar Flow
• Laminar flow also known as viscous flow
• In laminar flow, viscous force is highly is highly predominant.
Case-1
Laminar flow between 2- parallel plates
Consider a fluid element in flow field. Element have thickness dy, length dx and y 
distance from bottom plate.
A
dx ¦
]d y
Assumption:- width of flow perpendicular to paper = unity 
Free body diagram of element
(T +3 T/3 X,dx)dy
Pdy
(P+3P/3X,dx)dy
___
Apply equilibrium condition
Z F =0
cp
-t.d x -| p + -^.dx jdy = 0
f &i
=>pdy-j^T +— xiy 
= > pdv - r dx - — .dxdy - rdx - pdy - — dxdv = 0
a y a x
^ — = Q
a. ax
But we know that
a u
i= u —
cy
So,
n i - M .
a y [a y j a x
— and u
G s .
are independent form y 
By integrating equation (1)
c t i 11 a p
— = — I — .y+c,
a y u l a x
= > Again integrate with respect to y
y: j cp ,
u = — + c,y + c,
2 p la x i
Apply boundary condition
> No slip condition
> Due to friction velocity of fluid at 
UDDer and bottom plate is zero.
1. At y = 0, u = 0 
C2 = 0
2. At y=H, u=0
c, -
2u l^axj
So,
u _ y : f'cp; Hy/'cp 
2 |J . I c*x 1 2(i ( dx
Maximum velocity
A t y - %
u
Sja {d x
Mean velocity 
Mass flow rate =
H
fpudy---------(a)
u = —
1 ( dp
2u l dx
Mass flow rate, when consider average velocity.
= pu x H -----(b)
By equating both term {from eq. (a) and (b)} and putting expression of
|pudy = pu x H
u =¦ 1 r 1
12u { d x )
From expression of
U*ux and u 
u j - H : f c’ p ) |
1 2 p l < 2 x J
^ - U 2 r
Su
cp
dx
So. u = - u .
Shear stress distribution:
By Newton’s Law of viscosity
CM
1= ji —
-1 dp. _ .
=> t = px — x— ( H -2 y ) 
2u dx
d p (H } 
x = — - — y 
dx I 2 7
Page 4


Laminar and Turbulent Flow 
Laminar Flow
• Laminar flow also known as viscous flow
• In laminar flow, viscous force is highly is highly predominant.
Case-1
Laminar flow between 2- parallel plates
Consider a fluid element in flow field. Element have thickness dy, length dx and y 
distance from bottom plate.
A
dx ¦
]d y
Assumption:- width of flow perpendicular to paper = unity 
Free body diagram of element
(T +3 T/3 X,dx)dy
Pdy
(P+3P/3X,dx)dy
___
Apply equilibrium condition
Z F =0
cp
-t.d x -| p + -^.dx jdy = 0
f &i
=>pdy-j^T +— xiy 
= > pdv - r dx - — .dxdy - rdx - pdy - — dxdv = 0
a y a x
^ — = Q
a. ax
But we know that
a u
i= u —
cy
So,
n i - M .
a y [a y j a x
— and u
G s .
are independent form y 
By integrating equation (1)
c t i 11 a p
— = — I — .y+c,
a y u l a x
= > Again integrate with respect to y
y: j cp ,
u = — + c,y + c,
2 p la x i
Apply boundary condition
> No slip condition
> Due to friction velocity of fluid at 
UDDer and bottom plate is zero.
1. At y = 0, u = 0 
C2 = 0
2. At y=H, u=0
c, -
2u l^axj
So,
u _ y : f'cp; Hy/'cp 
2 |J . I c*x 1 2(i ( dx
Maximum velocity
A t y - %
u
Sja {d x
Mean velocity 
Mass flow rate =
H
fpudy---------(a)
u = —
1 ( dp
2u l dx
Mass flow rate, when consider average velocity.
= pu x H -----(b)
By equating both term {from eq. (a) and (b)} and putting expression of
|pudy = pu x H
u =¦ 1 r 1
12u { d x )
From expression of
U*ux and u 
u j - H : f c’ p ) |
1 2 p l < 2 x J
^ - U 2 r
Su
cp
dx
So. u = - u .
Shear stress distribution:
By Newton’s Law of viscosity
CM
1= ji —
-1 dp. _ .
=> t = px — x— ( H -2 y ) 
2u dx
d p (H } 
x = — - — y 
dx I 2 7
H dP
V e lo city P ro file ¦ tt;
Pressure difference b/w two points along flow
Consider average velocity expression
1 **---------- ------------- >•!
- H : cp
u = ------- -
12u 5x
n - L
12uu
=>Pi-P:
12uL
H,
Laminar flow through pipe: (circular)
Consider a fluid element having radius r and length dx 
Free body diagram of element
Apply horizontal equilibrium equation 
Internal flow:-
Page 5


Laminar and Turbulent Flow 
Laminar Flow
• Laminar flow also known as viscous flow
• In laminar flow, viscous force is highly is highly predominant.
Case-1
Laminar flow between 2- parallel plates
Consider a fluid element in flow field. Element have thickness dy, length dx and y 
distance from bottom plate.
A
dx ¦
]d y
Assumption:- width of flow perpendicular to paper = unity 
Free body diagram of element
(T +3 T/3 X,dx)dy
Pdy
(P+3P/3X,dx)dy
___
Apply equilibrium condition
Z F =0
cp
-t.d x -| p + -^.dx jdy = 0
f &i
=>pdy-j^T +— xiy 
= > pdv - r dx - — .dxdy - rdx - pdy - — dxdv = 0
a y a x
^ — = Q
a. ax
But we know that
a u
i= u —
cy
So,
n i - M .
a y [a y j a x
— and u
G s .
are independent form y 
By integrating equation (1)
c t i 11 a p
— = — I — .y+c,
a y u l a x
= > Again integrate with respect to y
y: j cp ,
u = — + c,y + c,
2 p la x i
Apply boundary condition
> No slip condition
> Due to friction velocity of fluid at 
UDDer and bottom plate is zero.
1. At y = 0, u = 0 
C2 = 0
2. At y=H, u=0
c, -
2u l^axj
So,
u _ y : f'cp; Hy/'cp 
2 |J . I c*x 1 2(i ( dx
Maximum velocity
A t y - %
u
Sja {d x
Mean velocity 
Mass flow rate =
H
fpudy---------(a)
u = —
1 ( dp
2u l dx
Mass flow rate, when consider average velocity.
= pu x H -----(b)
By equating both term {from eq. (a) and (b)} and putting expression of
|pudy = pu x H
u =¦ 1 r 1
12u { d x )
From expression of
U*ux and u 
u j - H : f c’ p ) |
1 2 p l < 2 x J
^ - U 2 r
Su
cp
dx
So. u = - u .
Shear stress distribution:
By Newton’s Law of viscosity
CM
1= ji —
-1 dp. _ .
=> t = px — x— ( H -2 y ) 
2u dx
d p (H } 
x = — - — y 
dx I 2 7
H dP
V e lo city P ro file ¦ tt;
Pressure difference b/w two points along flow
Consider average velocity expression
1 **---------- ------------- >•!
- H : cp
u = ------- -
12u 5x
n - L
12uu
=>Pi-P:
12uL
H,
Laminar flow through pipe: (circular)
Consider a fluid element having radius r and length dx 
Free body diagram of element
Apply horizontal equilibrium equation 
Internal flow:-
F flow — 0
d u
p . dA — (p + -7 T- • dx)dA — t( 2 7 t ? ‘ . dx) = 0
C /
= = > — — .d fc .d A = — t .'It iv . dfc 
ox
dp
dx
nr, / = — r . 2 7 r/
r = - ^ , d p
2 d x
According to Newton’s Law of Viscosity 
cti
T = U----
' f y
„ r dP dn
So---------= u— ..... (a)
2 fix 5y
from first figure in this section
y + r = R => dy = -c r 
put the value of 
c\- = -cr 
in eq. (a)
r < ? p _ cti 
2 5x Sr
Su
a 7
r Sp 
2u Sx
By integrating it
u - — f — 1 + C 
4pl
At
r = R, u = 0 
---no slip condition
C - -R 4 
4u I^ S x
So
n — L f *
4u\ C X 
Maximum Velocity:
(r 2 - r )
V at r = 0
m n
So from expression of u, put r=0