Transcript Conservation Principles: Momentum & Energy Conservation

Advanced Transport Phenomena
Module 2 Lecture 5
Conservation Principles: Momentum &
Energy Conservation
Dr. R. Nagarajan
Professor
Dept of Chemical Engineering
IIT Madras
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE
Atmospheric-pressure combustor
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Problem Statement:
Pure methane gas at 300 K, 1 atm, and pure air
at 300K, 1 atm, steadily flow into a combustor
from which a single stream of product gas (CO2,
H2, O2, N2) emerges at 1000 K, 1 atm. Use
appropriate balance equations to determine:
Mass flow rate of product stream out of
combustor
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Chemical composition of product gas mixture
(expressed in mass fractions)
Formulate & defend important assumptions.
Treat air as having nominal composition
wO2 =
0.23, wN2 = 0.73
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Stream 1 Pure CH 4  g 
@ 1 atm
300 K,
m 1 1g s
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Stream 2 Air @ 1 atm
300 K,
m 2  20 g s
Stream 3 Air  Comb Pr oducts
@ 1 atm
1000 K
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
m
3
?
wi , 3  ?  i  1, 2,....,.5 
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Total Mass Balance:

 dV  S  V.n dA  0

V
t
m 3  m 1  m 2 
0 in ss
Conclusion
or
or
m 3   m 1  m 2   0,
m 3 = m1  m 2 ,
m 3 = 1 + 20 = 21 g s ;
i.e., exit stream (1000 K, 1 atm) has mass-flow rate
of 21 g/s (via overall mass balance).
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Chemical Composition of the Exit Stream, i.e.,
wi ,
3
 where i 
1,
2,
O2
3,
4,
N 2 H2 O CO2 CH 4
This can be found via the chemical element
mass
balances, i.e.,

t

V
5
 k  dV    k  v.n dA  0;
S
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
For steady-state, this can be written as
S wk   v.n dA  0
or

w k , 3 m 3 - w k , 1
 
  
or
wk ,3
 
m 1 w k , 2 m
 
w k , 1 m 1 w k ,2 m
 
 

m3
5
where
w k , 3   wk iwi, 3
 
i1
(See following matrix)
Sought
2
,

2

 0.
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Note that we need
w k , 1 for k  1,.....
w k ,
2
for k  1,.....
where w k  
 w
i
We readily find;
k ,i
wi
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Element 1Hydrogen 
w1, 3 
w1 1 m 1  w1 m 2
m3
0.25137 1   0  20 

w1, 3 
 1.197  102
 21
Element 2  Carbon 
w 2, 3 
w 2 1 m 1  w 2 m 2
m3
0.74813 1   0  20 


 3.5625  10 2
 21
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
Similarly,
 N  balance 
 O  balance 
w3, 3 = 0.7333
w 4, 3  0.21905
Calculation of w k ,
3
via w k ,
3
i =1,...,5
In stream 3 for each k = 1,...,4 we have:
N
w k    wk i wi
i1
Just calculated
Matrix Sought
j = 1,...,4
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
k  1 H  balance  gives :
1.1970 102  0.1119w3  w3, 3  0.10697
k  2(C  balance) gives :
3.5625  10 2  0.2727w4  w4, 3  0.13065
k  3( N  balance) gives :
0.7333  1.000w2  w2,
3
 0.7333
k  4(O  balance) gives :
0.21905  1.00w1  0.8881w3  0.7271w4
Therefore :
w1,

3
 0.02905 above

above
MASS CONSERVATION: ILLUSTRATIVE
EXERCISE CONTD…
This completes the composition calculation for the
exit stream (Note:
w5, 3  0
(no unburned methane)
MOMENTUM CONSERVATION
Linear Momentum Conservation
Angular Momentum Conservation
LINEAR MOMENTUM CONSERVATION
Diffusion & Source Terms:
 Net surface 
 Net inflow rate of linear
 


(
contact
)
force

 

momentum by diffusion across CS  acting on CS 


   . n dA
s
LINEAR MOMENTUM CONSERVATION
CONTD…
 Net source of linear   Net body force
 N

 

momentum
within

acting
on
all
species

 
   V i .gi dV
 CV
  within CV
 i 1

 

p = local stress operator
p. n dA = (Vector) element of surface force
gi = local “body” force acting on each unit mass of
species i
LINEAR MOMENTUM CONSERVATION
CONTD…
Integral Conservation Equation for Fixed CV:
N

 vdV    vv.n dA    .n dA    i .g i dV

s
s
V
t V
i 1
Differential form (local PDE):
N

  v   div(  vv)  div  Π    i .gi
t
i 1
Each equation equivalent to 3 scalar equations,
one for each component (direction)
LINEAR MOMENTUM CONSERVATION
CONTD…
 v v = local, instantaneous, convective
momentum flux
Tensor (as is p)
requires 9 local scalar quantities for complete
specification
LINEAR MOMENTUM CONSERVATION
CONTD…
In cylindrical polar coordinates, components
are:
 vr vr
 v vr
 vz vr
 vr v
 v v
 vz v
 vr vz ,
 v vz ,
 vz vz ,
Because of symmetry, only 6 are independent
LINEAR MOMENTUM CONSERVATION
CONTD…
z-component of PDE:
N

(  vz )   div(  vv)   div  Π   z   i .g i , z
t
i 1
where
1 
1 

(r  vr vz ) 
(  v vz )  (  vz vz ),
 div(  vv)z 
r r
r 
z
(analogous expression can be written for [div p]z
“Jump” condition across surface of discontinuity:
LINEAR MOMENTUM CONSERVATION
CONTD…
 Net outflow rate
  Net surface force 
of linear momentum  due to stresses acting 

 




by convection relative  on both sides of

to the moving CS
  CS

CONSERVATION OF ENERGY (I LAW OF
THERMODYNAMICS)
In
chemically-reacting
systems,
thermal,
chemical & mechanical (kinetic) sources of
energy must be considered
Heat-addition & work must be included
CONSERVATION OF ENERGY (I LAW OF
THERMODYNAMICS) CONTD…
Definitions:

q  = total energy flux in prevailing material
mixture

q =
volumetric energy source for material
mixture
 Typically
derived from interaction with a local
electromagnetic field (“photon phase”)
CONSERVATION OF ENERGY
Definition of Terms:
 Net inflow rate 
 Rate at which work is

of total energy 
done by (surface) stresses 







q
.n
dA






s
by diffusion 
along CS on fluid within 
across CS

CV

 Rate at which work is done 
 Net " source " of 
by each of the local body






total
energy
within

q
dV


 V


CV

 forces g i acting on the corr  



esponding moving species 

CONSERVATION OF ENERGY CONTD…
Work Terms:

S
  .n dA.v
N
and
'' .g dV

i

V mi
i 1
CONSERVATION OF ENERGY CONTD…
Integral Conservation Equation for Fixed CV:



v2 
v2 
  e   dV     e   v.n dA    q.n dA

S
s
t V 
2
2


N
  q dV     .ndA .v    mi .g i dV
V
S
i 1
V
where e = specific “internal” energy of mixture
(function of local thermodynamic state), including
chemical contributions
CONSERVATION OF ENERGY CONTD…
v2/2 = specific kinetic energy possessed by each
unit mass of mixture as a consequence of its
ordered motion
CONSERVATION OF ENERGY CONTD…
Local PDE for Differential CV:
•
 
 
v2 
v2  
   e     div    e   v   div( q )  q'''  div( Πv )
t  
2 
2 
 
N
  mi .gi
i 1
“Jump” condition for surface of discontinuity:
 Rate at which work 
 Net inflow rate of  

is
done
by
stress
 Net outflow rate of
 



 total energy due  
total energy by convection   
  system at both


to
q
at
both
relative to the moving CS  
 


  surfaces of CS   surfaces of CS





CONSERVATION OF ENERGY CONTD…
When body forces gi (per iunit mass) are same
for all chemical species (e.g., gravity):
N

''
'' 
''
m
.
g

m
.
g

m
.g

i i
 i 
i 1
 i 1 
N
With the constraints:
N
w
i 1
i
1
N
j
i 1
''
i
0
CONSERVATION OF ENERGY CONTD…
If gi is associated wioth a time-independent
potential energy field, f, then the total energy
density field becomes:
1 2


 e  v f 
2


(separate body-force term on RHS not required)
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I
Problem Statement:
Numerically, compute and compare the
following energies (after converting all to the
same units, say calories):
a. The kinetic energy of a gram of water moving at
1 m/s
b. The potential energy change associated with
raising one gram of water through a vertical
distance of one meter against gravity (where
g=0.9807*103 cm/s2 )
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
c. The energy required to raise the temperature of
one gram of liquid water from 273.2 K to 373.2 K.
d. The energy required to melt one gram of ice at
273.2 K.
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
e. The energy released when one gram of H2O(g)
condenses at 373 K.
f. The energy released when one gram of liquid
water is formed from a stoichiometric mixture of
hydrogen (H2(g)) and oxygen (O2(g)) at 273.2 K
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
What do these comparisons lead you to expect
regarding the relative importance of changes of
each of the above mentioned types of energy in
applications of law of conservation of energy? Is
H2O “singular,” or are you conclusions likely to
be generally applicable?
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
1 m/s
Solution Procedure:
m=1g
Z=1 m
a) KE/mass for 1 g H2O @ 1 m/s
1 m/s
z  1m
2
KE v 1  2 cm 
1J
1 cal
4 erg
  10
 7 
  0.5 10
mass 2 2 
s 
g 10 erg 4.184 J
KE
 1.19 104 cal g
mass
2
Z=0 m
m=1g
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
b. f  g z

3 cm 
  0.9807 10 2  102 cm 
s 

1J
1cal
5 erg
 0.9807 10
 7

g 10 erg 4.184 J
 234 103 cal g
2
c.  h thermal  c p T  1100  110 cal g
H2O(l)
cal
K
gK
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
kcal
103 cal 1g  mole


d.  h  fusion  1.4363
g  mole 1kcal 18.016 g
 79.7 cal g
 0.797 102 cal g
3
kcal
10
cal
1
g

mole
e.  h 
@373K  9.717


vaporiz .
g  mole 1kcal 18.016 g
 540 cal g
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…
f.
 h comb to form 1 g H 2O(l ) : 3.79 10
3
cal g @ 298 K
kcal
Remark: H f (298) = - 57.798
;
g  mole
this means that for the formation reaction :
1
H 2 (g) + O2  g   H 2O(g), H 298  57.798
2
CONSERVATION OF ENERGY: ILLUSTRATIVE
EXERCISE I CONTD…