Lecture 9 9-12-11 - The Mass Balance Equation Exam...

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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 Single-Unit Processes Input – Output + Generation = Accumulation Multiple-Unit 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 = (1-0.045)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 Problem-solving 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.

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