Transcript Momentum Conservation - Chemical Engineering IIT Madras

Momentum Conservation


Newton’s Second Law
 Force = Mass * Acceleration
 Alternate method: From Reynold’s theorem
Fluid Flow
 Force = Momentum flux + Momentum
Accumulation rate

 F   V  V . n dA t V  d (Vol)
s
Vol
Flux
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Accumulation Rate
Example

Straight Pipe

Steady State
A2
V2
A1
V1
0

 F   V  V . n dA t V  d (Vol)
s
Vol
F  P1 A1  P2 A2
Flux  1 A1V1 V1  2 A2V2 V2
Assumption: No friction

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1
2
Example
Same flow rate, same length of
pipe
Pressure Drop for various fluids
Pressure Drop (Pa) (Log Scale)
100000.00
38 times
10000.00
1000.00
16 times
100.00
10" pipe
2.5" pipe
10.00
5" pipe
1.00
Acetone

Water 80 C
Ammonia
(30%)
Water
Flow in a straight pipe. Realistic case
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Sulfuric acid
(98%)
Phosphoric
acid (85%)
1=1g/cm3 d1=8cm
V1=5m/s d2=5cm
Example


Steady State
0

 F   V  V . n dA t V  d (Vol)
s
Vol
Problem: 5.32
d2,V2
d1,V1
1
2
h=58 cm
F  P1 A1  P2 A2  Friction
P1  P2  Hg g h  water g h
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1 A1V1  2 A2V2
Flux  1 A1V1 V1  2 A2V2 V2
Example

Example 1 (bent pipe), page 47
y
1,V1
By
Area=A1
x
Area=A2=A1
Q
Bx
2,V2
Fx  P1 A1  P2 A2 Cos( )  Bx
Fy  0  P2 A2 Sin( ) W  By
Weight of Fluid
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
FluxX  V2 Cos( ) 2V2 A2   V1 1V1 A1 
out
FluxY  0   V2 Sin( ) 2V2 A2 
in


Steady State
out
 d/dt =0
From Eqn of Conservation of Mass
1V1 A1  2V2 A2  m
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
in
Example

 F   V  V . n dA t V  d (Vol)
s
Vol
V1
By
Area=A1
Area=A2=A1
Q
Bx
V2
 V1  V2Cos( )
 Bx  P1 A1  P2 A2 Cos( )  m
 V2 sin( )
 By  P2 A2 Sin( ) W  m
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example

P1,V1
Pipe with U turn
F
P2,V2
Total Force F  P1 A1  P2 A2
Flux  V A  V A2
2
1 1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
2
Use Gage Pressure!
In case of gas, use
absolute pressure to
calculate density
Example
N

P1
“L” bend
F
Total Force x  Fx  P1 A1
Total Force y  Fy  P2 A2
Assume the force by the pipe on the
fluid is in the positive direction
P2
Fluxx   V A
2
1 1
2
2 2
Fluxy  V A
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
What will the force be, if the flow is
reversed (a) in a straight pipe? (b) in
a L bend?
E

 F   V  V . n dA t V  d (Vol)
s
Vol
Example
M  2 AX  1 AY
Pbm. 5.13 & 5.19
X
P2
2 V2
dx
dy
Vw 

dt
dt
Conservation of Mass
 1 AVw  V1   2 AVw  V2   0
Y
Vw
1
V1
P1
Stationary CV
Momentumin CV  V2 2 A2 x  V11 A1 y

dx
dy
V

d
(
Vol
)

V

A

V

A
2 2 2
1 1 1
t 
dt
dt
Vol
MomentumFlux  1 A1V12  2 A2V22
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Force P2 A2  P1 A1
Example
Pressure difference?
Pbm. 5.19
V2=0 m/s
V1=-3 m/s
X
P2
2
V2
Y
Vw
Stationary CV
1
V1
Vw=Velocity of Sound in water
P1
P2  P1  1 Vw  V1 V2  V1 
~ 41 atm
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
1   2
A1  A2
Drag on object 800N / m depth
V2=?
Pbm. 5.1,5,3
V1=6 m/s
0.6 m 1.2 m

 F   V  V . n dA t V  d (Vol)
s
Vol
1.2

0
y
x
1.2
2
A2 V2  y  dy   1 A1 V1  y  dy
0
F  P1 A1  P2 A2  Drag
1.2
2
1.2
2
Flux    2 V2  y  dy   1 V1  y  dy
0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
0
Momentum Conservation
• Angular Momentum
• In a moving system
• Torque = Angular Momentum flux + Angular
Momentum Accumulation rate

    r V  V . n dA t  r V  d (Vol)
s
Vol
Flux
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Accumulation Rate
Example
    r V  V . n dA
s
• Example 4 in book
• Find the torque on the shaft
• In a moving system
• Torque = Angular Momentum flux + Angular
Momentum Accumulation rate
Q
IIT-Madras, Momentum Transfer: July 2005-Dec 2005

r V  d (Vol )

t Vol
Example

r V  d (Vol )

t Vol
s

 F   V  V . n dA t V  d (Vol)
s
Vol
    r V  V . n dA
• Approach-1. Find effective Force in X direction
• Find the moment of Force
• Assume: No frictional loss, ignore gravity, steady state,
atmospheric pressure everywhere
ControlVol moving@ r
Vin V0  r
Cons.Mass. Vout V0  r
X directionmomentumout   QV0  r cos
X directionmomentumin QV0  r 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Q
    r V  V . n dA
Example
s

r V  d (Vol )

t Vol
Fon Fluid   QV0  r 1  cos 
FonWheel  QV0  r 1  cos 
Ton wheel  rQV0  r 1  cos 
• Approach-2. Using conservation of angular momentum
• Stationary CV
 r  V  V .n dA
S
In  rQV0
Out   rQ r  V0  r cos 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
• Consider a jet hitting a moving plate
VPlateVPlate
VNoz
•
•
•
•

After 1 second
Vnoz water has entered into the CV
Plate has moved by Vplate
In a control volume which moves with the plate, Vnoz-Vplate water has
entered the CV (and exited at the bottom)
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
VinMO VINGREFERENCE  Vnozzle VPlate
Vout MOVING REFERENCE
Vout MOVING REFERENCE
Vout MOVING REFERENCE
 Vnozzle  VPlate
X Component
Y Component
 Vnozzle  VPlate cos( )
 Vnozzle  VPlate sin( )
Vout MO VINGREFERENCE  Vnozzle VPlate cos( ) 

Vnozzle VPlate sin( ) 

i
j
Vout Stationary REFERENCE  V plate 

Vnozzle  VPlate  cos( ) 


i
i
Vnozzle  VPlate sin( ) 
j
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example

Consider plate moving @ half Vnoz, alpha
= 180 degrees
Vnozzle
Sin( )  0, Cos( )  1
VPlate 
2
Vout Stationary REFERENCE  V plate 

Vnozzle  VPlate  cos( ) 


i
i
Vnozzle  VPlate sin( ) 
j
Vout Stationary REFERENCE  Vplate  Vnozzle  VPlate 
0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
VPlate
Example
• Pbm 5.24
• Thickness of slit =t, vol
flow rate =Q, dia of pipe=d,
density given
• Ignore gravity effects
    r V  V . n dA
s
3ft
6ft

r V  d (Vol )

t Vol
Flux
Accumulation Rate
In  0
9
9
Out    r V V t dr   V t  r dr
0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
3
Example
1
• Pbm 5.31
• P1, P2, density, dia, vol flow
rate given

    r V  V . n dA t  r V  d (Vol)
s
Vol
• Calculate velocity at 1 (=2)
In  0

T  rP2 A2  r A2 V22

IIT-Madras, Momentum Transfer: July 2005-Dec 2005
3ft
2
Energy Conservation
DE

  e V . n dA  e  d (Vol )
Dt
t Vol
s
Friction Loss (Viscous)
dW
DE dQ dW
P


   (V .n)dA 
Dt
dt
dt

dt
s
Heat
Mechanical
Work
done by
the system
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Work done
by pressure force
Energy Conservation

dQ dW dW
P



   e    V . n dA  e  d (Vol)
dt
dt
dt

t Vol
s 
e  Energy perUnit Mass
v2
 u   gh
2
u  int ernalenergy
Bernoulli' s Eqn:
v2 P
  gh  Const
2 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
•
•
•
•
•
No Frictional losses
Incompressible
Steady
No heat, work
No internal energy change
Example
• Flow from a tank
Dia = d1
2
1
2
v1 P1
v2
P2
  gh1 
  gh2
2 
2

h1
0
• Pressure = atm at the top and at the outlet
h3
Dia = d2
2
3
• Velocity at 1 ~ 0
V3  2 g (h1  h3 )
2 g (h1 )
V2  2gh1  Toricelli’s Law A3  A2
2 g (h1  h3 )
A2V2  A3V3
• Sections 2 and 3
Q  A2 2gh1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Example
• How long does it take to
empty the tank?
• What if you had a pipe all
the way upto level 3?
dh
A2

 V1  V2
dt
A1
A2

2 gh
A1
t  0, h  hinit
t  t, h  0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Dia = d1
1
h1
0
h3
Dia = d2
2
3
• Pressure @ section 2 != atm
• Pressure @ section 3 = atm
Example
• What if you had a pipe all the
way upto level 3?
Dia = d1
V3  2 g (h1  h3 )
h1
0
Q  A3V3  A2V3
h3
w.o. pipe:Q  A2 2gh1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Dia = d2
2
3
• More flow with the pipe
P2  P3  gh3
1
• Turbulence, friction
• Unsteady flow
• Vortex formation
Example
v2 P
  gh  Const
2 
Height is known
• Moving reference; Aircraft
60 km/h
• Find P and r (eg from tables)
150 km/h
P1  atm,V1  0
• Flight as Reference
P1  atm,V1  150
P2  unknown,V2  0
P3  unknown,V3  60
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1
2
3
Flow through a siphon vs constriction
2
2
v1 P1
v2
P2
  gh1 
  gh2
2 
2

h3
V2  V3 , P1  P2  Patm
h1
h2
P3  Patm  g (h3  h2 ) if P3  Pvapor
h1  h2 , A1V1  A2V2
1
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
  A 2 
P2  P1  V12   1   1
  A2 



if P2  Pvapor
2
Example
•
•
•
•
•
v1 P1
v2 P
  gh1    gh  FrictionLoss head
2 
2 
Pbm. 6.4
Steady flow through pipe , with friction
Friction loss head = 10 psi
Area, vol flow rate given
Find temp increase
• Assume no heat transfer
Frictionloss  u2  u1
 Cv (T2  T1 )
1o C
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
Example
•
•
•
•
2
v1 P1
v2
P2
  gh1 
  gh2
2 
2

Pbm. 6.10
Fluid entering from bottom,
exiting at radial direction
Steady, no friction
t
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
P2=atm
h2
D1
• Find Q, F on the top plate
V1 A1  V2 A2
D12
A1  
, A2  D2t
4
D2
P1=10 psig
 F   V  V . n dA
s

V  d (Vol )

t Vol
Example
F
 F   V  V . n dA
s

V  d (Vol )

t Vol
 F  Force from PlateWeight  P A
1 1
y
Momentumin A V
2
1 1
D1
P1=10 psig
Momentumout  0
Hence, Force from Plate  ...
• If the velocity distribution just below the top plate is known,
then P can be found using Bernoulli’s eqn
F   P dA
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
A
Modifications to Eqn
2
v1 P1
v2 P
  gh1    gh  FrictionLoss head
2 
2 
• Unsteady state, for points 1 and 2 along a stream line
v1 P1
v
P
V
  gh1    gh   ds
2 
2 
t
1
2
2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
Draining of a tank
• We can obtain the time it takes to drain a tank
• (i) Assume no friction in the drain pipe
• (ii) Assume you know the relationship
between friction and velocity
• Le us take that the bottom
location is 2 and the top fluid
surface is 1
• Incompressible fluid
V1 A1  V2 A2
R
1
H
L
2
D
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Draining of a tank: Quasi steady state
• Quasi steady state assumption
• Velocity at fluid surface at 1 is very small
• i.e. R >> D
• No friction : L is negligible
• P1 = P2 = Patm
V22 V12

  g  H  L
2
2
 g H 
V2  2  g  H 
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Draining of a tank: Quasi steady state
d ( H )
V1 
dt
V2 A2
dH


dt
A1
2  g  H  D22
4 R12
• Original level of liquid is at H = H0
• Integrating above equation from t=0, H=H0 to t=tfinal, H
=0, we can find the efflux time
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Draining of a tank: Unsteady state
• BSL eg.7.7.1
• At any point of time, the kinetic + potential energy of the fluid
in tank is converted into kinetic energy of the outgoing fluid
• We still neglect friction
4


1
D
1

2
2
2
2
K .E.   R H   V1    R H   V2
4 
2 
 2 16 R 




• Potential Energy of a disk at height z and thickness dz
P.E.    R z   g dz
2
H
Total P.E.    R2 z   g dz
0
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Draining of a tank: Unsteady state
4
2



d 
1
D
H


2
2
2
 R g
  R H    V2

4 
dt 
2 
 2 16 R 


  D2   1 2 

V2    V2 
 4
 2 
• Also, using continuity equation V    dH
2

 dt
2
  4R 
 2 
 D 
• Substituting, you get a 2nd order non linear ODE with
two initial conditions. Please refer to BSL for solution
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Draining of a tank (accounting for friction)
• What if the flow in the tube is laminar and you
want to account for friction?
• Bernoulli’s eqn is not used (friction present)
• Continuity
R
V1 A1  V2 A2  V3 A3
• Hagen-Poiseuille’s eqn
 P R34
Q
8 L
 dH
P   g  H  L
V1 
dt
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
1
H
L
2
D
Draining of a tank (accounting for friction)
• Substituting and re arranging,
8 L R
d ( H  L)
 4
dt
H L
R3  g
2
1
• Integrating with limits
@ t  0, H  H 0
@ t  t final , H  0
t final
128  LR12
H )

l
n(
1

L
 gD34
• Note: The answer is given in terms of diameter of tube, so that
it is easier to compare with the answer given in the book
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
v1 P1
v
P
V
  gh1    gh   ds
2 
2 
t
1
2
Example
2
2
• A1,A2, initial height h1 known
• A1 >> A2
P1  P2  Patm
V1  0
1
• Consider section 3 and 2
V3  V2
V
t
Section3 2
L
h1
3
dV2

dt
V
dV2
1 t ds  dt L
2
• Pseudo Steady state ==> Toricelli’s law
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
Example
dV2
V22
L  gh1 
dt
2
dV2
dt

2
2 gh  V2 2 L
@ t  0, V2  0
• Rearranging and solving, we get
V2
t
 tanh(
2 gh )
2L
2 gh
• As t increases, the solution
approaches the Toricelli’s equation
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
t
lim tanh(
2 gh )  1
t 
2L
v1 P1
v
P
V
  gh1    gh   ds
2 
2 
t
1
2
Appendix:Example
2
2
• Oscillating fluid in a U-tube
P1  P2  Patm
V1  V2  V3
1
• Let h=h1-h2
V dV s

t
dt
1
2
 L3  Const(k )
1
1 dx
V
2 dt
2
h2
2
 ds  h  h
h1
x  h1  h2
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
L3
d 2 x  2g 
  x  0
2
dt  k 
Appendix:Example
• Blood Flow in vessels
• Minimization of ‘work’
• Murray’s Law:
3
3
r

r
 in  out
• Laminar Flow, negligible friction loss (other than
that due to viscous loss in laminar flow) , steady
• Turbulent, pulsating flow
• Assume
8LQ
p 
R 4
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Q2
Work  Q p   4
r
Appendix:Example
rinner
• If the ratio of ‘smaller’ to larger capillary is constant k 
router
• And Metabolic requirement =m=
power/volume
 1  2  2
2

m


1
Lr


r
 

• Work for maintaining blood vessel
 k 

• Total work
• Optimum radius
 Q2 
W    4    r2
 r 1
Q 3
 
k
IIT-Madras, Momentum Transfer: July 2005-Dec 2005