Energy Balance

1. Degree of Freedom Analysis
• It is the process used to determine if a material
balance problem has sufficient specifications to
be solved.
a) draw and completely label the flowchart
b) count the unknown variables on the chart
c) count the independent equations relating these
variables
d) calculate degrees of freedom by subtracting step (b)
from step (c)
ndf = nunknowns – nindep_eqns
1 ChE 201 Fall 2011

2. Degree of Freedom Analysis
ndf = nunknowns – nindep_eqns
• If ndf = 0, problem can be solved (in principle).
• if ndf > 0, problem is underspecified and at
least ndf additional variables must be specified
before the remaining variable values can be
determined.
• if ndf < 0, the problem is overspecified with
redundant and possibly inconsistent relations.
2 ChE 201 Fall 2011

3. Degree of Freedom Analysis
• Sources of equations relating unknown process
stream variables include:
– Material balances. For a nonreactive process, no more
than nms (number of molecular species or components )
independent material balances may be written.
– Energy balance. An energy balance provides a
relationship between inlet and outlet material flows
and temperatures. (We will see this in ChE 301)
– Process specifications.
– Physical properties and laws.
– Physical constraints.
– Stoichiometric relations. (for reacting systems)
3 ChE 201 Fall 2011

4. • A stream of humid air enters a condenser in which 95%
of the water vapor in the air is condensed.
• The flow rate of the condensate (liquid leaving the
condenser) is measured and found to be 225 L/h.
• Calculate the flow
rate of the gas
stream leaving the
condenser and the
mole fractions of
O2, N2, and H2O.
4 ChE 201 Fall 2011
Degree of Freedom Analysis
Example 1
condenser

5. • 6 unknowns
– 3 material balances (1 each for O2, N2, H2O)
– condensate volumetric to molar flow relation (MW and ρ)
– process specification: 95% of the water is condensed
• ndf = 6 – (3 + 1 + 1) = 1
Underspecified
cannot solve
5 ChE 201 Fall 2011
Degree of Freedom Analysis
Example 1
condenser

6. Degree of Freedom Analysis
• 5 unknowns
– 3 material balances (1 each for O2, N2, H2O)
– condensate volumetric to molar flow relation (MW and ρ)
– process specification: 95% of the water is condensed
• ndf = 5 – (3 + 1 + 1) = 0
Solvable
6 ChE 201 Fall 2011
Example 2
condenser

7. • Density relationship
• 95% condensation specification
• O2 Balance
• N2 Balance
• H2O Balance
• outlet gas composition
• total outlet gas flow rate
   
 
  
  
 
5
4
3
total
total
5
O
H
total
4
N
total
3
O
5
2
1
4
1
3
1
1
2
kg
10
0
.
18
O
H
ol
m
1
L
O
H
kg
h
O
H
L
2
n
n
n
n
n
n
y
;
n
n
y
;
n
n
y
n
n
100
.
0
n
n
79
.
0
900
.
0
n
n
21
.
0
900
.
0
n
n
100
.
0
95
.
0
n
00
.
1
225
n
2
2
2
3
2
)
l
(
2
)
l
(
2































 

7 ChE 201 Fall 2011
Degree of Freedom Analysis
condenser

8. 20%
CB
CC
CD
100%
CA
CC
100%
CA
CB
CC
100%
CA
CB
30%
CC
CD
100%
F1 =10 mole/hr
F3
F2
F4
Find the number of degrees of freedom
Let us count
unknown variables
Degree of Freedom Analysis
Example 3

9. 20%
CB
CC
CD
100%
CA
CC
100%
CA
CB
CC
100%
CA
CB
30%
CC
CD
100%
F1 =10 mole/hr
F3
F2
F4
1
2
3
4
5
6
7
8
9
Find the number of degrees of freedom
Degree of Freedom Analysis
Example 3

10. • 9 unknowns
– 4 material balances (1 each for C1, C2, C3, C4)
• ndf = 9 – (4) = 5
Degree of Freedom Analysis
Example 3

11. Material Balance
General Procedure – Single Unit operation
1. Choose as a basis of calculation an amount or flow
rate of one of the process streams.
– If an amount or flow of a stream is given, it is usually
convenient to use it as the basis of calculation.
Subsequently calculated quantities will be correctly
scaled.
– If several stream amounts or flows are given, always
use them collectively as the basis.
– If no stream amount or flow rate is specified, take as a
basis an arbitrariy amount or flow rate of a stream
with a known composition.
11 ChE 201 Fall 2011

12. General Procedure – Single Unit Op
2. Draw flowchart and fill in all variable values,
including the basis. Label unknown stream variables.
– Flowchart is completely labeled if you can express the
mass / mass flow rate (moles / molar flow rate) of each
component of each stream in terms of labeled quantities.
– Labeled variables for each stream should include 1 of:
a. total mass (or flow), and mass fractions of all stream components
b. total moles (or flow), and mole fractions of all stream components
c. mass, moles (or flow) of each component in each stream
• use (c) if no steam information is known
– incorporate given relationships into flowchart
– label volumetric quantities only if necessary
12 ChE 201 Fall 2011

13. General Procedure – Single Unit Op
3. Express what the problem statement ask you
to do in terms of the labeled variables.
4. If given mixed mass and mole units, convert.
5. Do a degree-of-freedom analysis.
6. If ndf = 0, write equations relating unknowns.
7. Solve the equations in (6).
8. Calculate requested quantities.
9. Scale results if necessary.
13 ChE 201 Fall 2011

14. Distillation Column example
• Ex. 4.3-5
1. basis is given as a
volumetric quantity
2a. Flowchart drawn from description
2b. Convert mole to mass fractions
2c. no stream information known
write in terms of species flows
14 ChE 201 Fall 2011

15. Distillation Column example
• Ex. 4.3-5
2d. confirm every component mass flow in every process stream
can be expressed in terms of labeled quantities and variables.
2e. process specification
15 ChE 201 Fall 2011

16. Distillation Column example
• Ex. 4.3-5
3. write expressions for quantities requested in problem statement
B
T
3
3
B
B x
1
x
;
m
m
x 

 

3
T
3
B
3 m
m
m 

 

3
1
2 m
m
m 

 

16 ChE 201 Fall 2011

17. Distillation Column example
• Ex. 4.3-5
4. Convert mixed units in overhead product stream
  
  
   
   
kg
T
kg
2
T
kg
B
kg
2
B
T
kmol
T
kg
B
kmol
B
kg
058
.
0
942
.
0
1
y
942
.
0
mixture
kg
7881
B
kg
7420
y
mixture
kg
7881
T
kg
461
B
kg
7420
T
kg
461
13
.
92
T
kmol
0
.
5
B
kg
7420
11
.
78
B
kmol
0
.
95









17 ChE 201 Fall 2011

18. Distillation Column example
• Ex. 4.3-5
5. Perform degree of freedom analysis
=0.942
=0.058
4 unknowns
-2 material balances
-1 density relationship
-1 process specification
0 degrees of freedom
18 ChE 201 Fall 2011

19. Distillation Column example
• Ex. 4.3-5
6. Write system equations
7. Solve
   h
kg
L
kg
h
L
1 1744
872
.
0
2000
m 


i. volumetric flow conversion
   h
B
kg
1
3
B 8
.
62
m
45
.
0
08
.
0
m 
 

ii. benzene split fraction
 
h
B
kg
2
3
B
2
B
2
1
766
m
m
y
m
m
45
.
0







iii. benzene balance
   
h
T
kg
3
T
3
T
2
B
2
1
915
m
m
y
1
m
m
55
.
0








iv. toluene balance
h
kg
h
kg
3
T
3
B
2
1
1744
1744
m
m
m
m



 



iv. total mass balance (check)
19 ChE 201 Fall 2011

20. Distillation Column example
• Ex. 4.3-5
8. Calculate additional quantities
=0.942
=0.058
=1744 kg/h
=915 kg T/h
=62.8 kg B/h
=766 kg/h
h
kg
h
T
kg
h
B
kg
3 978
915
8
.
62
m 



kg
T
kg
3
T
kg
B
kg
h
kg
h
B
kg
3
3
B
3
B
936
.
0
064
.
0
1
y
064
.
0
978
8
.
62
m
m
y





 

20 ChE 201 Fall 2011

21. Balances on Multiple Unit Ops
• A system is any portion of a process that can
be enclosed within a hypothetical box
(boundary). It may be the entire process, a
single unit, or a point where streams converge
or combine.
21 ChE 201 Fall 2011

22. Balances on Multiple Unit Ops
• Boundary A encloses the entire process.
– inputs: Streams 1, 2, and 3
– products: 1, 2, and 3
– Balances on A would be considered overall balances
– internal streams would not be included in balances
22 ChE 201 Fall 2011

23. Balances on Multiple Unit Ops
• B: an internal mixing point (2 inputs, 1 product)
• C: Unit 1 (1 input, 2 products)
• D: an internal splitting point (1 input, 2 products)
• E: Unit 2 (2 inputs, 1 product)
23 ChE 201 Fall 2011

24. Balances on Multiple Unit Ops
• The procedure for solving material balances on
multi-unit processes is the same as for a single unit;
though, it may be necessary to perform balances on
several process subsystems to get enough equations
to determine all unknown stream variables.
24 ChE 201 Fall 2011

25. Two-Unit Process Example
• Variables for Streams 1, 2, and 3 are unknown
25 ChE 201 Fall 2011

26. Two-Unit Process Example
• Variables for Streams 1, 2, and 3 are unknown
• Label unknown stream variables
26 ChE 201 Fall 2011

27. Two-Unit Process Example
• Degree-of-freedom analysis
– overall system: 2 unknowns – 2 balances = 0 (find m3, x3)
– mixer: 4 unknowns – 2 balances = 2
– Unit 1: 2 unknowns – 2 balances = 0 (find m1, x1)
– mixer: 2 unknowns – 2 balances = 0 (find m2, x2)
27 ChE 201 Fall 2011

28. Extraction-Distillation Process
28 ChE 201 Fall 2011

29. Simultaneously solve total mass
and acetone balances to
determine m1 and m3.
Solve MIBK balance to
determine xM1.
Extraction-Distillation Process
29 ChE 201 Fall 2011

30. Solve acetone, MIBK, and water
balances to determine mA4,
mM4, and mW4.
Extraction-Distillation Process
30 ChE 201 Fall 2011

31. For either (just 1) extractor
unit, solve acetone, MIBK, and
water balances to determine
mA2, mM2, and mW2.
Extraction-Distillation Process
31 ChE 201 Fall 2011

32. ndf = 4 unknowns (mA6, mM6,
mW6, and m5) – 3 balances = 1
underspecified
Extraction-Distillation Process
32 ChE 201 Fall 2011

33. ndf = 4 unknowns (mA6, mM6,
mW6, and m5) – 3 balances = 1
underspecified
Extraction-Distillation Process
33 ChE 201 Fall 2011