Lecture+2+MB+and+degrees+of+freedom++calculations (5).pdf

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)

n
df = n
unknowns – n
indep_eqns
1 ChE 201 Fall 2011

2. Degree of Freedom Analysis
n
df = n
unknowns – n
indep_eqns
•If n
df = 0, problem can be solved (in principle).
•if n
df > 0, problem is underspecified and at
least n
df additional variables must be specified
before the remaining variable values can be
determined.
•if n
df < 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 n
ms (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
O
2, N
2, and H
2O.
4 ChE 201 Fall 2011
Degree of Freedom Analysis
Example 1
condenser

5. •6 unknowns
–3 material balances (1 each for O
2, N
2, H
2O)
–condensate volumetric to molar flow relation (MW and ρ)
–process specification: 95% of the water is condensed
•n
df = 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 O
2, N
2, H
2O)
–condensate volumetric to molar flow relation (MW and ρ)
–process specification: 95% of the water is condensed
•n
df = 5 – (3 + 1 + 1) = 0
Solvable
6 ChE 201 Fall 2011
Example 2
condenser

7. •Density relationship
•95% condensation specification
•O
2 Balance
•N
2 Balance
•H
2O Balance
•outlet gas composition
•total outlet gas flow rate    
 
 
 
 
543total
total5OHtotal4Ntotal3O
521
41
31
12
kg100.18
OH olm 1
L
OH kg
h
OH L
2
nnnn
nny ;nny ;nny
nn100.0n
n79.0900.0n
n21.0900.0n
n100.095.0n
00.1225n
222
3
2)l(2)l(2
















7 ChE 201 Fall 2011
Degree of Freedom Analysis
condenser

8. 20% C
B
C
C
C
D
100%
C
A
C
C
100%
C
A
C
B
C
C
100%
C
A
C
B
30% C
C
C
D
100%
F
1 =10 mole/hr

F
3
F
2
F
4
Find the number of degrees of freedom
Let us count
unknown variables
Degree of Freedom Analysis
Example 3

9. 20% C
B
C
C
C
D
100%
C
A
C
C
100%
C
A
C
B
C
C
100%
C
A
C
B
30% C
C
C
D
100%
F
1 =10 mole/hr

F
3
F
2
F
4
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 C
1, C
2, C
3, C
4)
•n
df = 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 n
df = 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 BT33BB x1x ;mmx   3T3B3 mmm   312 mmm  
16 ChE 201 Fall 2011

17. Distillation Column example
•Ex. 4.3-5
4. Convert mixed units in overhead product stream   
  
  
  
kg
T kg
2T
kg
B kg
2B
T kmol
T kg
B kmol
B kg
058.0942.01y
942.0mixture kg 7881B kg 7420y
mixture kg 7881T kg 461B kg 7420
T kg 46113.92T kmol 0.5
B kg 742011.78B 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 1744872.02000m 
i. volumetric flow conversion 
h
B kg
13B 8.62m45.008.0m  
ii. benzene split fraction 
h
B kg
2
3B2B21
766m
mymm45.0




iii. benzene balance   
h
T kg
3T
3T2B21
915m
my1mm55.0



 iv. toluene balance h
kg
h
kg
3T3B21
17441744
mmmm

 
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 9789158.62m  kg
T kg
3T
kg
B kg
h
kg
h
B kg
33B3B
936.0064.01y
064.09788.62mmy

 
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 m
3, x
3)
–mixer: 4 unknowns – 2 balances = 2
–Unit 1: 2 unknowns – 2 balances = 0 (find m
1, x
1)
–mixer: 2 unknowns – 2 balances = 0 (find m
2, x
2)
27 ChE 201 Fall 2011

28. Extraction-Distillation Process
28 ChE 201 Fall 2011

29. Simultaneously solve total mass
and acetone balances to
determine m
1 and m
3.

Solve MIBK balance to
determine x
M1.
Extraction-Distillation Process
29 ChE 201 Fall 2011

30. Solve acetone, MIBK, and water
balances to determine m
A4,
m
M4, and m
W4.
Extraction-Distillation Process
30 ChE 201 Fall 2011

31. For either (just 1) extractor
unit, solve acetone, MIBK, and
water balances to determine
m
A2, m
M2, and m
W2.
Extraction-Distillation Process
31 ChE 201 Fall 2011

32. n
df = 4 unknowns (m
A6, m
M6,
m
W6, and m
5) – 3 balances = 1

underspecified
Extraction-Distillation Process
32 ChE 201 Fall 2011

33. n
df = 4 unknowns (m
A6, m
M6,
m
W6, and m
5) – 3 balances = 1

underspecified
Extraction-Distillation Process
33 ChE 201 Fall 2011