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Unformatted text preview: The Mass Balance Equation Exam #1(Closed Book) Stated simply, a material balance means "what goes in, must come out.”
Matter is neither created nor destroyed in industrial processes. Wednesday, September 21, 2011
7:00  9:00 p.m.
PHSC Room 108 mass in Quiz on Wednesday Sept. 14 mass out Process 2 forms of the equation Mole/mass fractions, temperature, pressure, manometers Input – Output + Generation – Consumption = Accumulation
This is what is used in Felder and Rousseau Chapter 4: Material Balances
SingleUnit Processes Input – Output + Generation = Accumulation MultipleUnit Processes IOGA This is what is used in the CD and some Thermodynamic textbooks. In
this equation the generation term has a positive sign (+) if a compound is
being generated while it has a negative () sign if it is being consumed. Draw the Figure and Label the knowns and unknowns Skim milk is prepared by the removal of some fat from whole milk. Skim
milk has been found to contain 90.5% water, 3.5% protein, 5.1%
carbohydrate, 0.1% fat, and 0.8% ash. Given:
Whole Milk, W If whole milk (the original milk) has 4.5% fat, calculate the composition of
whole milk if only fat is removed in the separation process. Assume that
there are no other losses. Skim Milk, S 4.5% fat
Fat, F 90.5% water
3.5% protein
5.1% carbohydrate
0.1% fat
0.8% ash Find:
(a) Find the composition of the whole milk
Assume:
 steady state
 no chemical reaction
 there is no accumulation in the process
 wt%
Plan:
 use conservation of mass equation
 Input  output + generation  consumption = accumulation Whole Milk, W Whole Milk, W Skim Milk, S 4.5% fat
Fat, F 90.5% water
3.5% protein
5.1% carbohydrate
0.1% fat
0.8% ash Skim Milk, S 4.5% fat
Fat, F Overall Balance
What balances can we write? Fat Balance
W (0.045) = S(0.001) + F W=S+F fat balance
protein balance
carbohydrate balance
water balance
ash balance
overall balance 90.5% water
3.5% protein
5.1% carbohydrate
0.1% fat
0.8% ash Protein Balance 0.035 kg protein Wx protein, w = S kg € 0.051 kg carbs Wx carbs,w = S kg Water Balance
€ € 0.905 kg water Wx water, w = S kg Carbohydrate Balance Ash Balance
€ 0.008 kg ash Wx ash, w = S kg € 1 Whole Milk, W Skim Milk, S 4.5% fat
Fat, F 90.5% water
3.5% protein
5.1% carbohydrate
0.1% fat
0.8% ash
Overall Balance Overall Balance Fat Balance
W(4.5%) = S(0.1%) + F W=S+F Fat Balance
W(4.5%) = S(0.1%) + F W=S+F If one assumes a basis of 100 kg of skim milk
W = 100 kg +F (100kg+F)(4.5%) = 100kg(0.1%) + F We have 3 unknowns but only two equations, can we
Solve this problem?
4.5 kg + 0.045F = 0.1kg + F
We need to choose a basis. 4.4 kg = (10.045)F Let’s choose a basis of 100 kg of skim milk W = S+F = 100 kg + 4.6 kg = 104.6 kg 4.6 kg = F Protein Balance
Carbohydrate Balance 0.035 kg protein Wx protein, w = S kg 0.051 kg carbs Wx carbs,w = S kg Water Balance
€ Ash Balance 0.905 kg water Wx water, w = S kg € € 0.008 kg ash Wx ash, w = S kg Three aqueous streams are mixed in a tank to create an output stream (D) with
a composition of 25% salt and 25% inert. Input streams A, B, and C have the
following compositions:
A
B
C
Salt
15%
40%
30%
Inert
20%
20%
40%
If the flow rate of stream A is 100 kg/hr. Find the flow rates of streams B, C, and
D in kg/hr. € 86.5 % water
3.3 % protein
4.9 % carbohydrate
4.5 % fat
0.8 % ash F= 4.6 kg Step 1: Read problem until you understand it.
What do I need to find? Steps 4: Obtain any data/information that is missing. Unknowns: B, C, D Steps 2 & 3: Draw a sketch, place labels on diagram for
known and unknown variables.
A (k g / h r)
15% s alt
20% inert
65% H2O
B (k g / h r)
40% s alt
20% inert
40% H2O
C (k g / h r)
30% s alt
40% inert
30% H2O Are the % in wt% or mol%? Must be wt% What is the MW of inert? Plan: Use Material Balance Equation Mixer D (k g / h r)
25% s alt
25% inert
50% H2O Input  Output + Generation  Consumption = Accumulation
Assumptions: No chemical rxn, steady state
Input = Output 2 Step 5: Choose a basis. Step 8: Write the equations A (k g / h r)
15% s alt
20% inert
65% H2O Basis: 100 kg/hr of A B (k g / h r)
40% s alt
20% inert
40% H2O Overall Balance Step 6: Determine number of unknown variables A+B+C=D A (k g / h r)
15% s alt
20% inert
65% H2O Mixer D (k g / h r)
25% s alt
25% inert
50% H2O C (k g / h r)
30% s alt
40% inert
30% H2O Salt Balance
Unknowns: B, C, D B (k g / h r)
40% s alt
20% inert
40% H2O Mixer 0.15 kg salt 0.40 kg salt 0.30 kg salt 0.25 kg salt A + B
+C = D kg
kg
kg
kg D (k g / h r)
25% s alt
25% inert
50% H2O Inert Balance C (k g / h r)
30% s alt
40% inert
30% H2O 0.20 kg inert 0.20 kg inert 0.40 kg inert 0.25 kg inert A + B
+C = D kg
kg
kg
kg Step 7: Determine number of independent equations
1. Overall Balance
2. Salt Balance H2O Balance 3. Inert Balance 0.65 kg H 2 O 0.40 kg H 2 O 0.30 kg H 2 O 0.50 kg H 2 O + B
+C = D A kg
kg
kg
kg 4. H2O Balance Subtract Salt Equation from Water Equation Rewriting the Inert Balance Equation with the known values 0.65 kg H 2 O 0.40 kg H 2 O 0.30 kg H 2 O 0.50 kg H 2 O + B
+C = D A kg
kg
kg
kg 0.20 kg inert 0.40 kg inert 200 kg 0.25 kg inert 100 kg 0.20 kg inert + B
+C
= kg
kg
kg
kg hr hr 0.15 kg salt 0.40 kg salt 0.30 kg salt 0.25 kg salt + B
+C = D A kg
kg
kg
kg 0.20 kg inert 0.40 kg inert 30 kg inert
+C
=
B kg
kg
hr 0.50 kg H 2 O 0.25 kg H 2 O = D A kg
kg Rewriting the Overall Balance Equation
100 kg/hr + B + C = 200 kg/hr
B + C = 100 kg/hr kg 0.50 kg H 2 O 100 hr kg = 200 kg
D=
hr 0.25 kg H 2 O kg B = 100 kg/hr  C
Substitute this into the Inert Balance equation above 0.40 kg inert 30 kg inert
kg 0.20 kg inert +C
=
 C 100 hr
kg
kg
hr Substitute this into the Inert Balance equation above
A (k g / h r)
15% s alt
20% inert
65% H2O 0.40 kg inert 30 kg inert
kg 0.20 kg inert  C 100
+C
=
hr
kg
kg
hr B (k g / h r)
40% s alt
20% inert
40% H2O 0.20 kg inert 0.40 kg inert 30 kg inert
20 kg inert
+C
=
C hr
kg
kg
hr 100 kg
100 kg 50 kg 50 kg
−C =
−
=
hr
hr
hr
hr B= 50 kg
hr B= D (k g / h r)
25% s alt
25% inert
50% H2O C (k g / h r)
30% s alt
40% inert
30% H2O 0.20 kg inert 10 kg inert
=
C kg
hr C= Mixer € 50 kg
hr C= 50 kg
hr D = 200 kg
hr € 3 Multiple Unit Processes
Problemsolving strategy is essentially the same, but we
Often need to analyze multiple systems:
We can do separate material balances around each
unit, for any combination of units, and for the overall
process.
Chemical processes can be extremely complex, constructed
of many units. For each choice of system, we can write N balance
equations, where N is the number of chemical
components Good organization and careful preparation of a detailed
flowchart are essential to working the problem correctly. Draw the Figure
A liquid stream (900 g/s) of 20%A, the rest B, is sent to a Unit I where a
stream of pure A, 100 g/s is removed. The other stream is combined with
200 g/s of pure C and sent to Unit II. From this Unit II, a top stream is
removed and it has 10% C. A bottom stream is produced with 12.0% A,
55.8% B, and the rest C. Determine All Stream flow rates and their
compositions in mass fractions. F2=100 g/s
100% A
F1=900 g/s
20% A
80% B F6=? Unit 1
?% A
?% B
F3=? ?% A
?% B
?% C
F5=? ?% A
?% B
10% C
Unit 2 F4=200 g/s
100% C F7=?
12% A
55.8% B
32.2% C F2=100 g/s Overall Balance for the Entire Process F1 = F2 + F3 100% A
F1=900 g/s
20% A
80% B F6=? Unit 1
?% A
?% B
F3=? F4=200 g/s
100% C Input = Output F1 + F4 = F2 + F6 + F7 100% A Total Mass Balance Around Unit 1 F2=100 g/s ?% A
?% B
?% C
F5=? ?% A
?% B
10% C
Unit 2 F7=?
12% A
55.8% B
32.2% C F1=900 g/s
20% A
80% B F3 = F1 − F2
g
g
g
F3 = 900 − 100 = 800
s
s
s F6=? Unit 1
?% A
?% B
F3=? ?% A
?% B
?% C
F5=? ?% A
?% B
10% C
Unit 2 F4=200 g/s
100% C F7=?
12% A
55.8% B
32.2% C Mass Balance on ‘A’ Around Unit 1 0.2 g A 1.0 g A F1 x = F2 g + F3 ( 3,A )
g x 3,A = 0.1 g A
g 0.2 g A 1.0 g A F1 − F2 g g F3 x 3,B = 0.9 g B
g x 3,A = 4 F2=100 g/s What balances can I write around
the mixing pt? F1=900 g/s Overall Mass Balance Around Mixing Pt 20% A
80% B F6=? Unit 1
?% A
?% B
F3=? Mass Balance on ‘A’ Around Mixing Pt ?% A
?% B
?% C
F5=? ?% A
?% B
10% C
Unit 2 F4=200 g/s
100% C Mass Balance on ‘B’ Around Mixing Pt F2=100 g/s Overall Mass Balance Around Mixing Pt 100% A F3 + F4 = F5 ⇒ g
g
g
F5 = 800 + 200 = 1000
s
s
s 12% A
55.8% B
32.2% C 0.1 g A
F3 (
) = F5 (x 5,A ) ⇒ x 5,A
g F6=? Unit 1 20% A
80% B ?% A
?% B
F3=? Mass Balance on ‘A’ Around Mixing Pt
F7=? Mass Balance on ‘C’ Around Mixing Pt 100% A
F1=900 g/s ?% A
?% B
?% C
F5=? ?% A
?% B
10% C
Unit 2 F4=200 g/s
100% C F7=?
12% A
55.8% B
32.2% C 800 g 0.1 g A
) (
0.08 g A
s
g
=
=(
)
g 1000 g s Mass Balance on ‘B’ Around Mixing Pt
F3 ( 800 g 0.9 g B
) (
0.9 g B
0.72 g B
s
g
) = F5 (x 5, B ) ⇒ x 5, B = =(
)
g
g 1000 g s Mass Balance on ‘C’ Around Mixing Pt
F4 ( Overall Mass Balance Around Unit 2 F5 = F6 + F7 ⇒ F6 = 1000 g
− F7
s F2=100 g/s 20% A
80% B F6=? Unit 1
?% A
?% B
F3=? Mass Balance on ‘C’ Around Unit 2 F5 (x 5,C ) = F6 (x 6,C ) + F7 (x 7,C )
g 0.2 g C
0.1 g C
0.322 g C ) = F6 (
) + F7 (
)
1000 (
s
g
g
g ?% A
?% B
?% C
F5=? ?% A
?% B
10% C x 6,B = F4=200 g/s
100% C F7=?
12% A
55.8% B
32.2% C 100 gC
0.222 g C
= F7 (
)⇒
s
g F7 = 450.5 g
s 1000 g 0.72 g B 450.5 g 0.558 g B  s 0.85 g B
g
g s =(
)
g 549.5 g s Stream 1
F1=900 g/s
20% A
80% B F6 = 549.5 g
s F1=900 g/s
20% A
80% B F6=? Unit 1
?% A
?% B
F3=? ?% A
?% B
?% C
F5=? F4=200 g/s
100% C ?% A
?% B
10% C
Unit 2 F7=?
12% A
55.8% B
32.2% C Mass fraction of A in Stream F6 x 6,A + x 6,B + x 6,C = 1 g C 1000 g
0.1 g C
0.322 g C
=(
 F7 )(
) + F7 (
)
s
s
g
g 100% A F (x )  F7 (x 7, B )
F5 (x 5, B ) = F6 (x 6, B ) + F7 (x 7, B ) ⇒ x 6, B = 5 5, B
F6 Unit 2 Substitute eq (1) into eq (2) and solve for F7 200 F2=100 g/s Mass Balance on ‘B’ Around Unit 2 100% A
F1=900 g/s 200 g 1.0 g C
) (
1.0 g C
0.20 g C
s
g
) = F5 (x 5,C ) ⇒ x 5,C = =(
)
g
g 1000 g s Stream 2
F2=100 g/s
100% A Stream 5
F5=1000 g/s
8% A
72% B
20% C Stream 6
F6=549.5 g/s
5% A
85% B
10% C ⇒ x 6,A = 1 − 0.85 − 0.1 = 0.05
Stream 3
F3=800 g/s
10% A
90% B Stream 4
F4=200 g/s
100% C Stream 6
F7=450.5 g/s
12% A
55.8% B
32.2% C 5 ...
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This note was uploaded on 09/26/2011 for the course CH E 2033 taught by Professor Staff during the Fall '11 term at The University of Oklahoma.
 Fall '11
 STAFF

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