Dr. Belal Al Zaitone Zuriqat

Transcription

1 King Abdulaziz University Faculty of Engineering Department of Chemical and Materials Engineering Introduction to Chemical Engineering Chapter 4 MATERIAL BALANCES WITHOUT RXN ChE 201 Dr. Belal Al Zaitone Zuriqat balzaitone@kau.edu.sa Office: Building No. 40, Room No. 24G58

2 Chapter 4: Mixing Process Two methanol-water mixtures are flowing into mixing unit. The first mixture contains 40.0 wt% methanol, and the second contains 70.0 wt% methanol. If 200 kg/hr of the first mixture is combined with 150 kg/ of the second, what are the mass and composition of the product? Solution: Input (2) Input (1) m 1 = 200 (kg/hr) x 1,m = 0.4 Kg M/kg x 1,w = 0.6 Kg w/kg Mixing Unit m 2 = 150 (kg/hr) x 2,m = 0.70 x2, w = 0.30 Output m 3 (kg/hr) m 3,w Kg w/hr m 3,m Kg M/hr 2

3 Chapter 4: Mixing Process 3

4 Chapter 4: Air Humidification and Oxyg. Process 4 An experiment on the growth rate of certain organisms requires an environment of humid air enriched in oxygen. Three input streams are fed into an evaporation chamber to produce an output stream. The input streams are: 1: Air (21 mole% O2, the balance N2) 2: Pure oxygen, with a molar flow rate one-fifth of the molar flow rate of stream 1. 3: Liquid water, fed at a rate of 20.0 cm 3 /min The output stream gas is analyzed and is found to contain 1.5 mole% water. Draw and label a flowchart of the process, and calculate all unknown stream variables. Solution:

5 Chapter 4: Air Humidification and oxyg. Process 5 Input,1 Input,2 2 n 2,O2 = 0.20 n 1,air (mol O2 /min) n 1,air (mol air/min) y 1,N2 = 0.79 mol N2/mol y 1,O2 = 0.21 mol O2/mol 1 Air Humid. Unit Output n 4 (mol/min) y 4,w = mol W/mol y 4,O2 (mol O2/mol) y 4,N2 (mol N2/mol) 3 Input,3 V w,3 = 20 cm 3 W/min n w,3 (mol W/min)

6 Chapter 4: Air Humidification and oxyg. Process 6

7 Chapter 4: Extraction of Streptomycin 7 Example 4.1: Streptomycin is used as an antibiotic to fight bacterial diseases, and is produced by the fermentation of a bacterium in a biological reactor with a nutrient of glucose and amino acids. After the fermentation process. Streptomycin is recovered by contacting the fermentation broth with an organic solvent in an extraction process. The extraction process is able to recover the Streptomycin because Streptomycin has a greater affinity for dissolving in the organic solution than in the aqueous solution. Determine the mass fraction of Streptomycin in the exit organic solvent assuming that no water exits with the solvent and no solvent exits with the aqueous solution. Assume that the density of the aqueous solution is 1 g/cm 3 and the density of the organic solvent is 0.6 g/cm 3. Solution:

8 Chapter 4: Extraction of Streptomycin 8 Organic Solvent Input,2 Input,1 Aqueous Solution V 1,Aq = 200 L/min (Aq) Conc. Strep 1 = 10g Strep/L Aq = 1.0 g/cm 3 m 1,Aq (kg Aq/min) V 2,S =10 L/min (Organic Solvent) S = 0.60 g/cm 3 m 2,S (kg S/min) Extraction Process Output 4 Aqueous Solution V 4,Aq L/min (Aq) Conc. Strep 4 = 0.2g Sterp/L m 4,Aq (kg Aq/min) Output 3 Organic Solvent Extracted Strep m 3,S (kg S/min) m 3,Strep (kg Strep/min)

9 Chapter 4: Extraction of Streptomycin 9

10 Chapter 4: Separation of Gases Using a Membrane 10 Example 4.2: Membranes represent a relatively new technology for the separation of gases. One application that has attracted attention is the separation of nitrogen and oxygen from air. The Diagram illustrates a nanoporous membrane that is made by coating a very thin layer of polymer on a porous graphite supporting layer. What is the composition of the waste stream if the waste stream amounts to 80% of the input stream? Solution:

11 Chapter 4: Separation of Gases Using a Membrane 11 Input,1 n 1,air (mol) y 1,N2 = 0.79 mol N2/mol y 1,O2 = 0.21 mol O2/mol Membrane Process Output 2 Output 3 n 3,air (mol) y 3,N2 = 0.75 mol N2/mol y 3,O2 = 0.25 mol O2/mol n 2,air mol =0.80 n 1,air y 2,N2 = mol N2/mol y 2,O2 = mol O2/mol

12 Chapter 4: Separation of Gases Using a Membrane 12

13 Chapter 4: Analysis of Distillation Column 13 Example 4.3: A new manufacturer of ethyl alcohol (denoted as EtOH) for gasohol is having a bit of difficulty with a distillation column. The process is shown in the flow chart. It appears that too much alcohol is lost in the bottoms (waste). Calculate the composition of the bottom and the mass of the alcohol lost in the bottom based on the data shown in the flow chart that was collected during 1 hour of operation. Solution:

14 Chapter 4: Analysis of Distillation Column 14 Input,1 m 1 = 1000 kg feed x etoh = 0.10 x w = 0.90 m 1,etoH (kg) m 1,w (kg) Distillation Column Output 2(waste) m 2 (kg) m 2,etoH (kg) m 2,w (kg) Output 3(product) m 3 (kg) x 3,etoH = 0.6 x 3,w = 0.40 m 3,etoH (kg) m 3,w (kg) m 3 = 1/10 m 1

15 15 Q 7.14: A drier takes in wet timber (20.1% water) and reduces the water content to 8.6%. You want to determine the kg of water removed per kg of timber that enters the process. Solution: Input,1 m 1 =?? kg x 1,w = x 1,t = m 1,w (kg) m 1,t (kg) Dryer Output 2 Output, 3 m 2,w (kg) m 3 =?? kg x 3,w = x 3,t = m 3,w (kg) m 3,t (kg)

16 Chapter 4: Mixing of Battery (Sulfuric) Acid 16 Example 4.4: You are asked to prepare a batch of 18.63% battery acid as follows: A tank of old weak battery acid (H2S04) solution contains 12.43% H2S04 (the remainder is pure water). If 200 kg of 77.7% H 2 S0 4 is added to the tank, and the final solution is to be 18.63% H2S04, how many kilograms of battery acid have been made? Solution:

17 Chapter 4: Mixing of Battery (Sulfuric) Acid 17 Input,1 m 1 = 200 kg soln x 1,H2SO4 = x 1,w = m 1,H2SO4 (kg) m 1,w (kg) Tank Output 3 Input 2 m 2 (kg) x 2,H2SO4 = x 2,w = m 2,H2SO4 (kg) m 2,w (kg) m 3 (kg) x 3,H2SO4 = x 3,w = m 3,H2SO4 (kg) m 3,w (kg)

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19 Chapter 4: Unsteady state process 19 Time = t1 Then we added 200 kg Time = t2 Accumulation = input - output

20 Chapter 4: Drying 20 Example 4.6: In the processing of the fish, after the oil is extracted, the fish cake isdried in rotary drum dryers, finely ground, and packed. The resulting product contains 60% protein. In a given batch of fish cake that contains 80% water (the remainder is dry cake), 100 kg of water is removed, and it is found that the fish cake is then 40% water. Calculate the weight of the fish cake originally put into the dryer. Solution:

21 21 Input,1 m 1 =?? kg fish x 1,w = 0.80 x 1,BCD = 0.20 Rotary Drum Dryer output 2 m 2,w = 100 kg water Output 3 m 3 =?? (kg) x 3,w = 0.40 x 3,BCD = 0.60

22 Chapter 4: Hemodialysis Example 4.7:. Dialysis solution, the cleansing solution, is pumped around these fibers. The fibers allow wastes and extra fluids to pass from your blood into the solution that carries them away. This example focuses on the plasma components of the streams: water, uric acid (UR), creatinine (CR), urea (U), P, K, and Na. You can ignore the initial filling of the dialyzer because the treatment lasts for an interval of two or three hours. Given the measurements obtained from one treatment, calculate the grams per liter of each component of the plasma in the outlet solution. Solution: Dialysate in Blood in Blood out Dialysate out 22

23 Chapter 4: Hemodialysis Input 1, Blood V 1,water = 1100 ml/min C 1,UR = 1.16 g/l w C 1,CR = 2.72 g/l w C 1,U = 18.0 g/l w C 1,P = 0.77 g/l w C 1,k = 5.77 g/l w C 1,Na = 13.0 g/l w Input,2 V 2,water =1700 ml/min dialyzer Process Output 3 V 3,water =?? ml/min Output 4, Blood V 4,water = 1200 ml/min C 4,UR = 60 mg/l w C 4,CR = 120 mg/l w C 4,U = 1.51 g/l w C 4,P = 40.0 mg/l w C 4,k = 2.10 mg/l w C 4,Na = 5.21 g/l w C 3,UR =?? g/l w C 3,CR =?? g/l w C 3,U =?? g/l w C 3,P =?? g/l w C 3,k =?? g/l w C 3,Na =?? g/l w 23

24 Chapter 4: Hemodialysis 24

25 Chapter 4: Scale-up Process 1 25 A mixture (by moles) of A and B is separated into two fractions. A flowchart of the process is shown below. It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h. Scale the flowchart accordingly. Solution: Input n 1 = 100 (mol/hr) y 1,A = 0.6 mol A/mol y 1,B = 0.4 mole B/mol Separation Unit Output n 2 = 50 (mol/hr) y 2,A = 0.95 mol A/mol y 2,B = 0.05 mole B/mol Output n A3 = 10.5 mole A/hr n B3 =39.5 mole B/hr

26 Chapter 4: Scale-up Process 1 26

27 Chapter 4: Scale-up Process 2 27 Consider the distillation of Benzene-Toluene mixture. Scale-up to flow rate of 100 Ibm/s. Solution: Input m 1 = 2.5 g/s x 1,T = 0.6 g T/g x 1,B = 0.4 g B/g Distillation Unit Output m 2 = 1.0 g/s x 2,T = 0.9 g T/g x 2,B = 0.1 g B/g Output m 3 = 1.5 g/s x 3,T = 0.3 g T/g x 3,B = 0.7 g B/g

28 Chapter 4: Scale-up Process 2 28

29 Chapter 4: Gas Mixing 29 Q1:. Tank A containing 90% nitrogen is mixed with Tank B containing 30% nitrogen to get Tank C containing 65% nitrogen. You are asked to determine the ratio of the gas used from Tank A to that used from Tank B. Solution: Tank A Tank C Tank B n A =?? mole y N2 = 0.9 y g = 0.1 n C =?? mole y N2 = 0.65 y g = 0.35 n B =?? mole y N2 = 0.3 y g = 0.7

30 Chapter 4: Quartz Diffusion Tube Q2: A gas containing 80% CH 4 and 20% He is sent through a quartz diffusion tube to recover the helium. Twenty percent by weight of the original gas is recovered, and the composition of the recovered stream is 50% He. Calculate the composition (mole fraction) of the waste stream gas if 100 kmoles of gas are processed per minute. Solution: Input,1 n 1 = 100 kmole/min y 1,CH4 = 0.80 y 1,He = 0.20 quartz diffusion Tube Output, 3 waste n 3 =?? kmole/min y 13CH4 =?? y 3,He =?? Output 2 recoverd n 2 =?? kmole/min y 2,CH4 = 0.50 y 2,He =

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