 2000, W. E. Haisler
Conservation of Mass
ENGR 211
Principles of Engineering I
(Conservation Principles in Engineering Mechanics)
Conservation of Mass
 Multicomponent Systems
 Rate Equation
 Multicomponent Rate Expressions
 Multiple Inlets and Outlets
 Steady State
 Multiple Unit Processes
 Recycle and Bypass
1
 2000, W. E. Haisler
2
Conservation of Mass
We consider conservation of mass in the general conservation
statement and write:








Mass accumulation   Mass entering  Mass leaving 
  system during  - system during
within system













during time period   time period   time period 













Mass generated   Mass consumed 



 within system
 -  within system



during time period  during time period 

where








Mass contained 
Mass accumulation
Mass contained
in system at the 
within system
 in system at the
beginning of 
during time period
end of time period


time period




































 2000, W. E. Haisler
Conservation of Mass
We can write a more compact mass conservation statement as
(msys)
 (msys)
 m  mout  mgen  mcon
end
beg in
We can make the statement even more compact by defining the
following:
m
 m  mout and m
 mgen  mcon
in/ out in
gen/ con
so that conservation becomes
(msys)
 (msys)
m
m
end
beg in / out gen/ con
3
 2000, W. E. Haisler
Conservation of Mass
4
Multicomponent Systems
Conservation of mass applies to the total mass of a
multicomponent system and for each chemical component (or
species).
Consequently, for a system that contains n species we can write n
species conservation statements that must sum to give total mass
conservation.
For species i mass conservation, we write
(m
)
 (m
)
 (m )  (m )out  (m )gen  (m )con
i,sys end
i,sys beg
i in
i
i
i
We can write n of the above equations (i=1,…,n). Summing up
the individual conservation statements for all n species gives:
 2000, W. E. Haisler
5
Conservation of Mass
n
n

 (mi,sys )end   (mi, sys )
beg
i 1
i 1
n
n
n
n
 (mi )in   (mi )out   (mi ) gen   (mi )con
i 1
i 1
i 1
i 1
The summation of the mass components (species) must equal the
total mass. Thus, we write
n
(msys)
  (mi,sys )
end
end i 1
,
n
m   (mi )in
in
i 1
, etc
and the top equation becomes
(msys)
 (msys)
 m  mout  mgen  mcon
end
beg in
 2000, W. E. Haisler
Conservation of Mass
6
For the case where there is no massenergy conversion, then
there is no net generation of total mass so that
m
 mgen  mcon  0
gen/ con
and the total mass conservation equation becomes
(msys)
 (msys)
 m  mout
end
beg in
In summary, for a mass system with n components (species), we
have n species conservation equations:
(m
)
 (m
)
 (m )  (m )out  (m )gen  (m )con
i,sys end
i,sys beg
i in
i
i
i
i = 1, …, n
and 1 total mass conservation equation:
(msys)
 (msys)
 m  mout
end
beg in
 2000, W. E. Haisler
Conservation of Mass
Example: Ten pounds mass (lbm) of air is placed in a cylinder.
Air consists of a mixture of 77% nitrogen (by mass) and 23%
oxygen (by mass). If 5 lbm of air is removed from the cylinder
and then 3 lbm of oxygen is added, what is the resultant
composition of the gas mixture in the cylinder?
The system is considered to be the contents of the cylinder. The
time period is whatever time is required to remove the air and
add the pure Oxygen (O2).
7
 2000, W. E. Haisler
3 lbm
Oxygen (O2 )
Added
8
Conservation of Mass
initially
10 lbm Air
(77% N,
23% O2 )
5 lbm Air
Removed
At the beginning of the time period, we have 10 lbm of air
which is composed of 7.7 lbm of N2 (0.77 x 10) and
2.3 lbm of O2 (0.23 x 10).
At the end of the time period, we have an unknown amount of air
and unknown amounts of N2 and O2 .
 2000, W. E. Haisler
9
Conservation of Mass
Applying conservation to the oxygen and nitrogen components,
we can tabulate the following known and unknown information:
Species (m
)
 (m
)
 (m )  (m )out  (m )gen  (m )con
i,sys end
i,sys beg
i in
i
i
i
lbm
lbm
lbm
lbm
lbm
lbm
_______________________________________________________________________________________________________________________________________________________________
Oxygen
X
- 0.23(10)
=
Nitrogen
Y
____
Z
- 0.77(10)
_______
10
=
Total
3
- 0.23(5) + 0
0 - 0.77(5) + 0
___ ______ ___
= 3 - 5
+ 0
-
0
-
0
____
- 0
Solving the O2 equation gives X= (mO ,sys )end = 4.15 lbm O2
2
Solving the N2 equation gives Y= (mN ,sys )end = 3.85 lbm N2
2
Solving the total mass equation gives Z= (msys )end = 8 lbm air
 2000, W. E. Haisler
10
Conservation of Mass
Consequently we have complete total mass and species mass
balance (conservation).
Species (m
)
 (m
)
 (m )  (m )out
i,sys end
i,sys beg
i in
i
lbm
lbm
lbm
lbm
_________________________________________________________________________________________________________________
Oxygen
4.15
-
Nitrogen
3.85
_____
8
-
Total
2.3
7.7
_____
10
= 3
- 1.15
= 0 - 3.85
___
____
= 3 - 5
Notice that the total mass at the end (8 lbm) consists of a mixture
of 4.15/8 = 52% oxygen and 48% nitrogen. It is no longer
“standard” air.
 2000, W. E. Haisler
Conservation of Mass
11
Rate Equation
Rather than consider finite time periods (tbeg to tend), it is useful
to consider rate equations and then integrate over the time period.
Suppose we have the rate of mass entering: m . Then the total
in
amount of mass entering is the integral of the mass rate from tbeg
to tend:
t
end m
 dt  m .
in
in
t
beg

We can similarly write the total system mass in rate form: m sys
and integrate it over the time period to obtain the obtain the mass
accumulation in the system (total change over the time period):
 2000, W. E. Haisler
t
end m
 sysdt 
t
beg

12
Conservation of Mass
dmsys
t
end
dt 
t
dt
beg





t




end dm

m


m

sys  sys 


t
sys



beg
end
beg

With these ideas, consider the mass conservation statement
written previously:
(msys)
 (msys)
 m  mout  mgen  mcon
end
beg in
In the above equation, we now replace the accumulation term on
the left and all the other terms by their integral forms to obtain:
dm 
t
t
t
t

sys
end 
dt  end m dt  end m dt  end m
dt  end m
dt
gen
con
out
in
t  dt 
t
t
t
t

beg 
beg
beg
beg
beg

t




 2000, W. E. Haisler
Conservation of Mass
13
Removing the integrals provides the mass conservation statement
for a differential time period:







dmsys 
 dt  m dt  m gendt  m condt
dt  m
out

in
dt 

So the rate form of the mass conservation equation becomes
(divide by dt):







dmsys 
  m  m gen  m con
m

dt  in out

In words: Accumulation rate = input rate - output rate
+ generation rate - consumption rate
 2000, W. E. Haisler
Conservation of Mass
14
We could have obtained the integral form in a different way.
Start with the mass conservation statement written previously:
(msys)
 (msys)
 m  mout  mgen  mcon
end
beg in
Write the terms on the right hand side in terms of rates:
(msys)
 (msys)
 m t  m outt  m gent  m cont
end
beg in
where t  t
. Now divide the above equation by t to
t
end beg
(msys)
 (msys)
end
beg  m  m  m
 con
obtain

m
gen
in out
t
(msys )
 (msys )
dmsys
beg


t
beg
Now:
= accumulation rate

t
dt
 2000, W. E. Haisler
15
Conservation of Mass
Substitute above into the mass conservation statement to get:
dmsys
 m  m out  m gen  m con
in
dt
Multiplying the above by dt gives the mass conservation law for
a differential period of time








dmsys 
dt  m
 dt  m dt  m gendt  m condt

out
in
dt 

The above equation can be integrated from tbeg to tend to obtain
the integral form of the mass conservation law
t  dmsys 
t
t
t
t
end 
dt  end m dt  end m dt  end m
dt  end m
dt
gen
con
out
in
t  dt 
t
t
t
t

beg 
beg
beg
beg
beg





 2000, W. E. Haisler
16
Conservation of Mass
Example: Consider a problem wherein mass is generated and
consumed at a certain rate during a time dependent process. The
m
in
m con
m gen
m out
mass flow rates and generation and consumption rates are given
by the following:
m gen  7eat lbm / hr
m  5eat lbm / hr
in
m out  3eat lbm / hr
m con  4eat lbm / hr
where a= a constant = 1 (hr-1). We want to determine the time
required to accumulate 4 pounds of mass in the system.
 2000, W. E. Haisler
17
Conservation of Mass
Tabulating each term in the mass rate equation gives







dmsys 



dt 
m
 m out  m gen  m con
in

lbm/hr
lbm/hr
lbm/hr
lbm/hr
lbm/hr
_______________________________________________________________________________________________________________________________________________________________







dmsys 
=


dt 
5eat - 3eat
+ 7eat - 4eat

Integrating the rate equation from tbeg to tend gives
 2000, W. E. Haisler
Conservation of Mass
18
dm 

sys dt  tend (5e at  3e at  7e at  4e at )dt
end 


dt
t 
t

beg
beg
t
t

at
accumulation  end (5e  at dt)dt   5e a end
t
t
beg
beg
The left side of the above equation is the accumulation of mass in

t


the system from tbeg to tend . We let the beginning time be 0.
The desidred accumulation is 4 pounds so we write
 at
t
 at end
end 5
5
e
5
e
accumulation  4 pounds  a
 [

]
a
a
0
The above equation gives (recall a= 1 hr-1)
 at
end  1
and solving for tend gives t = 1.61 hr
e
5
 2000, W. E. Haisler
Conservation of Mass
19
Multicomponent Rate Equations
Previously we wrote the conservation equations for
multicompoent systems in integral form as
In summary, for a mass system with n components (species), we
have n species conservation equations:
(m
)
 (m
)
 (m )  (m )out  (m )gen  (m )con
i,sys end
i,sys beg
i in
i
i
i
i = 1, …, n
and 1 total mass conservation equation:
(msys)
 (msys)
 m  mout
end
beg in
We can write the above in rate form following the same ideas
introduced earlier.
 2000, W. E. Haisler
Conservation of Mass
20
In summary, for a mass system with n components (species), we
have n species conservation equations in rate form:
d (m ) sys
i
 (m )  (m )out  (m )gen  (m )con
i in
i
i
i
dt
i = 1, …, n
and 1 total mass conservation equation in rate form:
dmsys
 m  m out
in
dt
 2000, W. E. Haisler
Conservation of Mass
21
Example: Silica gel (a solid) is used as a drying agent to remove
water from processed food. The processed food must be below a
certain water content before it can be shipped.
1) Determine the amount of silica get required per pound mass
of wet food if the food is initially 30% water (by weight) and
must be reduced to 20% water. It is assumed that 3.2 lbm of
silica gel has the capacity to absorb 1 lbm of water.
2) Find the time required to absorb the water if the silica gel
 be at (lbm / min) where t is
absorbs water at the rate m
H 2O
time in minutes, a = 1 min-1 and b=0.13 lbm/min.
 2000, W. E. Haisler
22
Conservation of Mass
Schematically, we have the following for the beginning and
ending states.
Dry Silica Gel
water
Wet Silica Gel
time
Food with
30% Water
Food with
20% Water
system
boundary
Beginning Time
Ending Time
In order to determine the process required on a per pound of food
basis, we assume we have 1 pound of food to begin with. The
mass of the dried food is unknown so we let be an unknown x.
 2000, W. E. Haisler
23
Conservation of Mass
The wet food is considered to be the system. We set up the mass
rate table for the beginning and ending states:
Species (m
)
 (m
)
 (m )  (m )out
i,sys end
i,sys beg
i in
i
lbm
lbm
lbm
lbm
_______________________________________________________________________________________________________________________
Dry Food 0.80x - 0.70(1) =
0
Water
0
?
___
____
0
?
Total
0.20x - 0.30(1) =
_____
_____
x
1
=
-
0
From the dry food equation, we see that x = 0.70/0.8 = 0.875 lbm
(there will be 0.875 lb of food with 20% water content at the
end). Substituting x into the water equation, we see that 0.125
lbm of water must be extracted by the silica gel.
 2000, W. E. Haisler
Conservation of Mass
24
We know that 3.2 lbm of silica gel has the capacity to absorb 1
lbm of water. So the amount of silica gel required to abosrb
0.125 lbm of water is:
lbm of silica gel = 3.2 lbm silica gel x 0.125 lbm water
1 lbm water
= 0.4 lbm silica gel
To determine the time required to absorb 0.125 lb of water, we
write the rate equation for water in the wet food system:


 d (m

)
sys


H 2O

)  (m
)out

  (m


in
H 2O
H 2O
dt




The gen/con term has been omitted. The water input term is
zero. Multiply the above by dt and integrate from 0 to t (where is
the time required to reduce water content in the food to 20%).
 2000, W. E. Haisler
Conservation of Mass
t
[(m
)sys]
[(m
)sys]
  (m
)
H 2O
H 2O
end  t
end  0 0 H 2O out
or
t
accumulation  0.125 lbm   (m
)
0 H 2O out
Note: the negative accumulation means the amount of water in
the food decreases with time. The integral term represents the
rate of absorption of water by the silica gel. We know that the
 be at (lbm / min) .
silica gel absorbs water at the rate of m
H 2O
So we have the equation:
t
  at

t  at

  at

b
b
 0.125 lbm   be dt   a e
 e
1
0




0 a 
Solving for t from the above equation gives
a (0.125)]  1ln[1 1 (0.125)]  3.3min
t  1
ln[
1

a
1
0.13
b
25
 2000, W. E. Haisler
26
Conservation of Mass
Multiple Inlets and Outlets
Stream B
Stream A
System
Stream E
Stream D
Stream C
System
Boundary
 2000, W. E. Haisler
Conservation of Mass
27
Steady State
Steady state means that the state of system is steady; it does not
change with time. Alternately, one can say that the accumulation
of mass within the system is zero. Mathematically, steady state
is described by
(m
)
 (m
)
 accumulation  0
i,sys end
i,sys beg
or
d (m ) sys
i
0
dt
Note that steady state implies the following:
(m )  (m )out  (m )gen  (m )con  0
i in
i
i
i
and for no net generation:
(m )  (m )out
i in
i
 2000, W. E. Haisler
28
Conservation of Mass
Example: The feed to a continuous flash distilliation consists of
25 lbm water, 10 lbm ethanol and 5 lbm methanol per hour.
D
Distillate
unknown composition
and flow rate
F
Input Feed
25 lbm/hr water
10 lbm/hr ethanol
5 lbm/hr methanol
B
24 lbm/hr output
83.3 wgt % water
12.5 wgt % ethanol
4.2 wgt % methanol
The temperature and pressure of the systeam are 95 C and 1 atm.
The total flow rate from the bottom of the column is 24 lbm per
hour. The composition of the bottom stream is 83.3% water (by
weight), 12.5 % ethanol and 4.2 % methanol. The system is
assumed to be operating at steady state and no reactions are
occurring in the system.
 2000, W. E. Haisler
Conservation of Mass
29
a) Set up the species accounting and total mass conservation
equation for the system.
b) Make the proper assumptions, and solve for the unknows.
c) Determine the composition and mass flow rate of the distillate
Since the system is steady state and no reactions are occurring in
the system, mass conservation reduces to: 0  (m )  (m )out .
i in
i
We let the input feed stream be labeled "F", the distillate stream
be "D" and the bottom stream be "B." We can now write
conservation of mass (rate form) in table form:
 2000, W. E. Haisler
30
Conservation of Mass
(m )
i in
Species Stream F
(lbm/hr)
=
(m )out
i
Stream B
Stream D
(lbm/hr)
(lbm/hr)
Water
25
Ethanol
10
Methanol
5
__________
Total
40
=
=
=
0.833(24)
0.125(24)
0.042(24)
________
24
=
+
+
+
+
?
?
?
_____
?
The distillate stream (D) can now be easily solved for:
Water = 5 (lbm/hr), Ethanol = 7 (lbm/hr), Methanol = 4 (lbm/hr)
and Total = 16.
The composition on a mass basis of the distillate is easily
calculated: Water = (5/16) 100% = 31.2%, Ethanol = 43.8% and
Methanol = 25%.
 2000, W. E. Haisler
31
Conservation of Mass
Multiple Unit Processes
Multiple unit processes involve flow of mass from one system
into another; often as a "chain" of systems. An example
schematic is shown below.
A
B
C
In the above, A, B and C can be treated as individual systems. In
addition, a system boundry can be drawn around the combination
of A, B and C. See p. 50-52 for an example.
 2000, W. E. Haisler
Conservation of Mass
32
Recycle and Bypass Systems
In many processes, such as reactions and separation, some of the
material that leaves the system is introduced back into the system
as a recycle stream. Bypass streams are used to remove material
that requires no further processing. A schematic of a recycle and
bypass system is shown below.
Bypass Stream
Recycle Stream
 2000, W. E. Haisler
33
Conservation of Mass
Example: A membrane system is used to filter waster products
from the blood stream.
R=recycle stream
Filtered
Blood
A
Mem bran
e
Unfiltered Blood U
25 g/min
0.17 wgt % waste
R
F
W
Pure Waste
30 mg/min
B
The membrane can extract 30 mg/min of waste , without
removing any blood. The waste concentration in the entering
blood is 0.17 wgt %, and the flow rate of blood is 25 g/min. A
recyle stream is used to help control the concentration of the
waste before it enters the membrane.
 2000, W. E. Haisler
Conservation of Mass
34
a) If the recycle rate is twice that of the filtered blood rate, what
is the composition at A?
b) If the composition at A is to be 0.1 % waste, what must be the
recycle rate, assuming streams U, W and F remain the same?
We first label the individual streams U, F, W, A, B, and R as
shown on the sketch. U=unfiltered blood, A=input stream to
filter (A=U stream + R stream), B = output stream from filter,
F=filtered blood, W = pure waste, R=recycle stream.
We can draw several “systems.” We will choose the membrane
plus recycle stream as the first system.
 2000, W. E. Haisler
35
Conservation of Mass
R=recycle stream
Filtered
Blood
U
A
Mem brane
Unfiltered Blood
25 g/min
0.17 wgt % waste
R
F
W
Pure Waste
30 mg/min
B
We start with the rate equation
d (m ) sys
i
 (m )  (m )out  (m )gen  (m )con
i in
i
i
i
dt
For steady state and no gen/con in the system, the above reduces
to (m )  (m )out . We construct the mass balance table:
i in
i
 2000, W. E. Haisler
36
Conservation of Mass
Note that the Recycle Stream is inside the system and does not
enter into the mass in and out terms.
Species
Blood
Waste
Total
(m )
i in
Stream U
(g/min)
=
0.9983(25) =
0.0017(25) =
__________
25.0
=
(m )out
i
Stream W
Stream F
(g/min)
(g/min)
0
0.030
______
0.030
+
+
+
?
?
_____
?
From the above, we conclude that for stream F the filtered blood
flow rate is 24.9575 g/min, the waste flow rate is 0.0125 g/min or
12.5 mg/min and the total flow rate is 24.9700 g/min. The
filtered blood contains 0.05 % waste (0.125/24.97x100%)
 2000, W. E. Haisler
37
Conservation of Mass
To evaluate stream A we can consider the junction of streams U,
R and A as a system:
R=recycle stream
Filtered
Blood
U
A
Mem brane
Unfiltered Blood
25 g/min
0.17 wgt % waste
R
F
W
Pure Waste
30 mg/min
B
We write the mass balance equation for the A, U and R streams:
 2000, W. E. Haisler
38
Conservation of Mass
Note: The factor of 2 in stream R comes from the problem
statement which states the recycle rate (stream R) is twice that of
the filtered blood rate (stream F).
Species
Blood
Waste
Total
=
(m )
i in
Stream U
Stream R
(g/min)
(g/min)
24.9575
+ 2(24.9575)
0.0425
+ 2(0.0125)
__________
______
25.0
+ 2(24.9700)
=
=
=
(m )out
i
Stream A
(g/min)
?
?
_____
?
From the above, stream A consists of 74.8725 g/min blood,
0.0675 g/min of waste, total mass rate of 74.9400 g/min and the
waste in stream A is 0.09% by mass.
 2000, W. E. Haisler
39
Conservation of Mass
We finally take the membrane itself as the system and construct
the following table to evaluate stream B:
Species
Blood
Waste
Total
(m )
i in
Stream A
(g/min)
=
74.8725
=
0.0675
=
__________
74.9400
=
(m )out
i
Stream W
Stream B
(g/min)
(g/min)
0
0.030
______
0.030
+
+
+
?
?
_____
?
From the above, stream B consists of 74.8725 g/min blood,
0.0375 g/min of waste, total mass rate of 74.9100 g/min.
 2000, W. E. Haisler
40
Conservation of Mass
For part b) we wish to determine the recycle rate so that the
composition in stream A is 0.10% waste (assuming that streams
U, W, and F remain the same). Again, we assume steady state
and no net generation. Schematically, we take the membrane as
the system. The flow rates in streams A and B are unknown.
R=recycle stream
A
Mem brane
Unfiltered Blood U
25 g/min
0.17 wgt % waste
R
Filtered Blood
24.9575 g/min blood F
0.0125 g/min waste
W
Pure Waste
30 mg/min
B
We construct our mass conservation table for the membrane:
 2000, W. E. Haisler
Species
Blood
Waste
Total
41
Conservation of Mass
(m )
i in
Stream A
(g/min)
=
0.999A
=
0.001A
=
__________
A
=
(m )out
i
Stream W
Stream B
(g/min)
(g/min)
0
0.03
______
0.03
+ 0.9995B
+ 0.0005B
_____
+
B
Solving the Blood and Waste equations simultaneously, we
obtain stream A Total = 59.97 g/min and stream B total = 59.94.
 2000, W. E. Haisler
42
Conservation of Mass
The recycle rate will have changed since stream B has changed.
Again, we draw a schematic of system and take the juncuture of
the B, F and R streams.
R=recycle stream
A
Mem brane
Unfiltered Blood U
25 g/min
0.17 wgt % waste
R
Filtered Blood
24.9575 g/min blood F
0.0125 g/min waste
W
Pure Waste
30 mg/min
B
We write mass conservation for the B, F and R streams in table
form:
 2000, W. E. Haisler
Species
Blood
Waste
Total
43
Conservation of Mass
(m )
i in
Stream B
(g/min)
=
59.91
=
0.03
=
__________
59.94
=
(m )out
i
Stream F
Stream R
(g/min)
(g/min)
24.9575
0.0125
______
24.97
+
+
+
?
?
_____
?
Solving the Blood and Waste equations simultaneously, we
obtain stream A Total = 59.97 g/min and stream B total = 59.94.
Consequently, the recycle stream R flow rates are: Stream R
blood=34.9525 g/min, stream R waste=0.0175 g/min and stream
R total=34.97 g/min.