Chemical Calculations for Chemical Engineering. PROBLEMS in MATERIAL BALANCE, RECYCLE, ByPASS & PURGE

1 Hi folks. I’m Ijotika and I am offering you a compilation of solved Chemical Calculations Problems. However, I omitted the solutions for educational purposes (you can ask for it. Like my mentor, I also hate spoon feeding the student. Originally, this compilation was a work of 5 person, I and 4 of my classmates. We failed this subject. I don’t care why they failed. (OMG this is so simple. If you want to know why I also failed, just contact me in my e-mail.) Well, as part of the requirement to complete the unit, we were asked by our professor to research and compile these problems with their solution. After we submitted the drafts for her to check, I thought our task was complete. After 3 months of checking the solution (can you imagine that? She burned 3 months of our college life waiting for her approval.), she returned the papers noting that we have to change some of us have the same problems solved. (Take note, when she returned the paper it was only 10 days before closing the curtain.) So there we were digging again the net hoping to find another set of problems that will save our grades. It’s not an easy task, but we managed to collect the needed problems. After so much effort in editing, I managed to print this and pass it to her. She refused to accept this because we’re late. Indeed, we’re 6 days late but you can’t blame us for it really is a tiresome task—searching for problems that we don’t have idea if it really exist, unless we make our own problems (that is our last resort.) After so much persuading, she accepted the pile and said that she’ll think about giving our grade. (F*CK! Ka-pasamba mo!) After a couple of weeks, she asked again for our drafts for her to check (WHAT THE F*CK)... well to cut this crap short, three of my classmates passed her subject in the end while I and my best bud didn’t. The problems here, like I said, are compiled, came from different sources. You can find the reference books used at the end of every section. I also provided some definition of every section. I didn’t put extra lecture because I assume that you have acquired that knowledge from your professor. If you have any questions, suggestions, comments or anything and everything, feel free to contact me @ [email protected]. MABUHAY!!!

P.S. I uploaded this document because I want to help you guys that like me suffer/suffers/suffered from the wrath of our prof. this is for you guys. P.P.S. By the way I came from the BICOL UNIVERSITY COLLEGE OF ENGINEERING here in Legaspi City, Albay, Philippines. So if ever you, BU CEngians, came across this page, you already knew which professor I’m talking about. P.P.P.S. Please excuse my grammar. I am not goooooooooood in English. aaenlay tsaltoX aiyeuXX fukremr cmirieh (#$%!@)

ijotika Chemical Calculations 1

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MATERIAL BALANCE A material balance is an application of conservation of mass to the analysis of physical systems. By accounting for material entering and leaving a system, mass flows can be identified which might have been unknown, or difficult to measure without this technique. The exact conservation law used in the analysis of the system depends on the context of the problem but all revolve around mass conservation, i.e. that matter cannot disappear or be created spontaneously. Therefore, mass balances are used widely in engineering and environmental analyses. For example mass balance theory is used to design chemical reactors, analyze alternative processes to produce chemicals as well as in pollution dispersion models and other models of physical systems. Closely related and complementary analysis techniques include the population balance, energy balance and the somewhat more complex entropy balance. These techniques are required for thorough design and analysis of systems such as the refrigeration cycle. The general form quoted for a mass balance is The mass that enters a system must, by conservation of mass, either leave the system or accumulate within the system . Mathematically the mass balance for a system without a chemical reaction is as follows:

Strictly speaking the above equation holds also for systems with chemical reactions if the terms in the balance equation are taken to refer to total mass i.e. the sum of all the chemical species of the system. In the absence of a chemical reaction the amount of any chemical species flowing in and out will be the same. This gives rise to an equation for each species in the system. However if this is not the case then the mass balance equation must be amended to allow for the generation or depletion (consumption) of each chemical species. Some use one term in this equation to account for chemical reactions, which will be negative for depletion and positive for generation. However, the conventional form of this equation is written to account for both a positive generation term (i.e. product of reaction) and a negative consumption term (the reactants used to produce the products). Although overall one term will account for the total balance on the system, if this balance equation is to be applied to an individual Chemical Calculations 1

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species and then the entire process, both terms are necessary. This modified equation can be used not only for reactive systems, but for population balances such as occur in particle mechanics problems. The amended equation is given below. Note that it simplifies to the earlier equation in the case that the generation term is zero.

Chemical Calculations 1

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MATERIAL BALANCE PROBLEMS WITHOUT REACTION

1. A lacquer plant must deliver 1000 lb of an 8% nitrocellulose solution. They have in stock a 5.5% solution. How much dry nitrocellulose must be dissolved in the solution to fill the order?1 M (lb) N= 100% F (lb) N= 0.055 S= 0.945 1.00

P (1000lb) N= 0.08 S= 0.92 1.00

2. A liquid adhesive, which is used to make laminated boards, consists of a polymer dissolved in a solvent. The amount of polymer in the solution has to be controlled for this application. When the supplier of the adhesive receives an order for 3000kg of an adhesive solution containing 13% by weight polymer, all it has on hand is (1) 500kg of a 10% by weight solution, (2) a very large quantity of a 20 wt% solution, and (3) pure solvent. Calculate the weight of each of the three stocks that must be blended together to fill the order. Use all of the 10% by weight solution. 2


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3. Hydrogen pre-carbon in the form of core is burned. a. With complete combustion using theoretical air. b. With complete combustion using 50% excess air. c. Using 50% excess air but with10% of the carbon burning to CO only IN EACH CASE CALCULATE THE GAS ANALYSIS THAT WILL BE FOUND BY TESTING THE FLUE GASES WITH AN ORSAT APPARATUS. 3 4. An evaporator is concentrating solutions coming from two different sources. The solutions from the first source containing 10% NaCl and 10% NaOH flows at the rate of 50 kg/min. the other solution containing 8% NaCl and 12% NaOH flows at the rate of 70 kg/min. The two streams are fed directly to the evaporator. If 50% of the total water is to be evaporated, calculate the composition and the flow rate of the product. 4 50% H2O evaporated

Solution 1 50 kg/min 10% NaOH 10% NaCl

Solution 2 70 kg/min 8% NaCl 12% NaOH

E V A P O R A T O R

? Compositions and flow rate

5. The waste acid from a nitrating process contains 30% H 2SO4, 35% HNO3 and 35% H2O by weight. The acid is to be concentrated to contain 39% H 2SO4 and 42% HNO3 by addition of concentrated sulfuric acid containing 98% H 2SO4 and concentrated nitric acid containing 72% HNO3 (wt). Calculate the quantities of 3 acids to be mixed to get 1000 kg of desired mixed acid. Waste acid 30%H2SO4, 35% HNO3

Concentrated HNO3 72% HNO3

BLENDING Concentrated H2SO4

Desired mixed acid 1000kg 39% H2SO4, 42% HNO3

(98%) Chemical Calculations 1

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REFERENCES: 1. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6th edition. 2. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 3. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 4. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering 5. K.A. Gavhane, Introduction to Process Calculations (www.google.com/books)


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MATERIAL BALANCE WITH CHEMICAL REACTION 1. A natural gas consisting entirely of Methane (CH4) is burned with an oxygen enriched air of composition 40% O2 and 60% N2. The Orsat analysis of the product gas is reported by the laboratory is CO2 : 20.2%, O2 : 4.1%, and N2 : 75.7%. Can the reported analysis be correct? Show all calculations.1

W H4 100%

CO2= 20.2 O2= 4.1 N2= 75.7

P

A enriched air O2 = 40 N2= 60 2. Plants in Europe sometimes use the mineral pyrites (the desired compound in the pyrites is FeS2) as a source of SO2 for the production of sulfite pulping liquor. Pyrite rock containing 48% sulfur is burned completely by flash combustion. All of the iron turns Fe3O4 in the cinder (the solid product), and a negligible amount of SO 3 occurs in either the cinder or the product gas. The gas from such as furnace is passed through milk of lime (CaO in water) absorbers to produce bisulfate pulping liquor. The exit gas from the absorber analyzes: 0.07% SO2, 2.9% O2, 96.4% N2. (Refer to the figure) Calculate the kg of air supplied to the burner per kg of the pyrite burned. 2 3FeS2

+

8 O2



Fe3O4

+

6 SO2


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3. A gas containing 80% CH4 and 20% He is sent through a quartz diffusion tube to recover the Helium. 20% by weight of the original gas is recovered, and its composition is 50% He. Calculate the composition of the waste gas if 1ooKmol of gas are processed per minute. The initial gas pressure is 120kPa, and the final gas pressure is 115kPa. The barometer reads 740mmHg. The temperature of the process is 22°C.

4. Pyrites ore is used for the production of the sulfuric acid. The pyrites is 80% FeS 2 and 20% inert materials. The FeS2 is burned with 30% excess air based on the reaction, 4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2 Only 90% of the FeS2 follows this reaction. The rest goes as follows: 4 FeS2 + 15O2 2 Fe2O3 + 8 SO3 What is the composition of the gases resulting in the combustion?

B 80% FeS2 U 20% inert materials 30% excess airR N E R

SO2 SO3 O2 N2 Inerts

5. A fed containing A, B and inerts enters a reactor. The reaction taking place is: 2A + B C The product stream leaving the reactor is having the following composition by mole: A=23.08%, B=11.54%, C=46.15, inerts= 19.23%. Find the analysis of feed on mole basis. Basis: 100 kmol of product stream

Feed A,B, inerts

REACTOR

Product stream 100 kmol


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REFERENCES: 1. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 2. David Himmelblau, Supplementary Problems For Basic Principles and Calculations in Chemical Engineering 3. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6th edition. 4. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering 5. K.A. Gavhane, Introduction to Process Calculations (www.google.com/books)


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RECYCLE In a reactive process, there is generally some unreacted feed material found in the product. In order to reduce cost and increase efficiency, the unreacted material is often separated and reused in a recycle loop.

Recycle may also be used to recover an expensive catalyst, maintain feed concentration below certain levels, or in a totally closed loop to pump a cooling or heating fluid. Keep in mind that while flow rates of the process effluent, process product, and recycle vary, all three have identical composition.


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RECYCLE PROBLEMS WITHOUT CHEMICAL REACTION 1. Examine the given figure. What is the quantity of the recycle stream in kg/hr? 1

C

Crystals carry off 4% H2O (4 kg H2O per kg total crystals +H2O) 2. Based on the process drawn in the diagram, what is the kg recycle per kg feed if the amount of W waste is 100 kg? The known compositions are inserted on the process diagram.2

X


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3. A planting plant has a waste stream containing Zn and Ni in quantities in excess of that allowed to be discharged into the sewer. The proposed processes to be used as a first step in reducing the concentration of several of the streams are listed in the table, what is the flow ( ℎ ) of the recycle stream R32 if the feed is 1( ℎ ).3 TABLE CONCENTRATION : g/L Stream Zn F 100 Po 190.1 P2 3.50 R32 4.35 W 0 D 0.10

Ni 10.0 17.02 2.19 2.36 0 1.00

4. A fresh feed (contains 5,000 kg of wet material) with 60% moisture is needed to be dried. To facilitate the operation, a part of dried product contains 5% water is recycled and mixed with the feed. The mixing stream of the recycle and the fresh feed contains 30% water. Calculate the water removed and the recycle to feed ratio. 4


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(R) Fresh feed 60% moisture

water evaporated (E) 30% water

Drier

5,000 kg

5% (P)

5. Sea water is to be desalinized by reverse osmosis using the scheme R (S ) 40% salt Seawater 1000lb/hr 3.1 % Salt (T)

Reverse Osmosis Cell

(B) Brine waste

(M) x

5.25% salt

(P) Desalinized water 500 ppm salt Given the data in the figure, determine; (a) the rate of waste brine removal, (b) the rate of desalinized (potable water) production; (c) the fraction of the brine leaving the reverse osmosis cell (which acts in essence as a separator) that is recycled. 5

REFERENCE:

1. David Himmelblau, Basic Principles and Calculations in Chemical edition. 2. David Himmelblau, Basic Principles and Calculations in Chemical edition. 3. David Himmelblau, Basic Principles and Calculations in Chemical edition. 4. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering 5. David Himmelblau, Basic Principles and Calculations in Chemical edition.

Engineering 6 th Engineering 6 th Engineering 6 th

Engineering 6 th


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RECYCLE PROBLEMS WITH CHEMICAL REACTION

1. Boron trichloride (BCl3) gas can be fed into a gas stream and for doping silicon. The simplest reaction (not the only one) is 4 BCl3 + Si

3 SiCl4 + 4 B

If all the BCl3 not reacted is recycled, what is the mole ratio of recycle to SiCl 4 exiting the separator? The conversion on one pass through the reactor is 87% and 1 mole per hour of BCl3 is fed to the reactor.1 Unreacted BCl3 BCl3 1mol/hr

87% REACTOR

SiCl4 BCl3 SiCl4

SEPARATOR

2. Many chemicals generate emissions of volatile compounds that needed to be controlled. In the process shown in the accompanying figure, the CO in the exhaust is substantially reduced by separating it from the reactor effluent and recycling the unreacted CO together with the reactant. Although the product is proprietary, the information is provided that the fresh feed stream contains 40% reactant, 50% inert and 10% CO, and that on reaction 2 moles of reactant yield 2.5moles of product. Conversion of the reactant to product is 73% on one pass through the reactor and 90% for the overall process. The recycle stream contains 80% CO and 20% reactant. Calculate the ratio of moles of the recycle stream to moles of the product stream. 2


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X

REACTOR

SEPARATOR

3. Nitroglycerine, a widely used high explosive, when mixed with wood flour is called “dynamite’. It is made by mixing high-purity glycerine (99.9+% pure) with nitration acid, which contains 50% H2SO4, 43% HNO3 and 7% H2O by weight. The reaction is: C3H8O3 + 3HNO3 + (N2SO4)

C3H5O3(NO2)3 + 3H2O + (H2SO4)

The H2SO4 does not take part in the reaction, but it is present to “catch” the H2O formed. Conversion of Glycerine is complete in the nitrator, and there are no side reactions, so all of the Glycerine fed into the nitrator forms into Nitroglycerine. The mixed acid entering the nitrator (stream G) contains 20% excess HNO 3 to assure that all the Glycerine reacts. After nitration, the mixture of Nitroglycerine and spent acid (HNO3, H2SO4, and H2O) goes to a separator (a settling tank). The Nitroglycerine is insoluble in the spent acid and its density is less, so it rises to the top. It is carefully drawn off as product stream P and sent to wash tanks for purifications. The spent acid is withdrawn from the bottom of the separator and sent to an acid recovery tank, where the HNO 3 and H2SO4 are separated. The H2SO4- H2O mixture is stream W, and is concentrated and sold for industrial purposes, the recycle stream to the nitrator is a 70% by weight solution of HNO3 in H2O. In the diagram, product stream P is 96.5% Nitroglycerine and 3.5% H 2O by weight. To summarize: Stream F = 50% by weight H2SO4, 43% HNO3, 7% H2O Stream G = 20% excess HNO3 Stream P = 96.5% by weight Nitroglycerine, 3.5% by weight H 2O Stream R = 70% by weight HNO3, 30% by weight H2O a.) If 1x103 kg of Glycerine/hr is fed to the Nitrator, How many kg/hr of Stream F will be the result? b.) How many kg/hr are in the recycle stream? Chemical Calculations 1

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c.) How many kg of fresh feed, Stream F are fed per hour? d.) Stream W is how many kg/hr? What is its analysis in weight %?

4. Propane is dehydrogenated to form propylene in a catalyctic reactor: C3 H8

C3 H6

+ H2

The process is to be designed for 85% overall conversion of propane. The reaction products are separated into two streams: the first, which contains H 2, C3H6 and 0.656% of the propane that leaves the reactor, is taken off as product, the second stream, which contains the balance of unreacted propane and 5% of the propylene in the product stream, is recycled to the rector. Calculate the composition of the product, the ration (moles recycled/ moles fresh feed) and the single-pass conversion. Illustration: Basis

100 mole of fresh feed

Fresh Feed

A

100 mol C3 H8

Reactor

1 100 + Xr Yr

2 Xp Yp Zp

Separation unit

3 X Y

Xr Yr

85% conversion 0.656%

Z 5%


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5. Reactors that involve biological materials (bioreactors) use living organisms to produce variety of products. Bioreactors are used for producing ethanol, antibiotics and proteins for dietary supplements and medical diagnosis. The overall conversion of the proprietary component in the fresh feed to product is 100% in recycle bioreactor. The conversion of the proprietary component to product per pass in the reactor is 40%. Determine the amount of recycle and the mass percent of component in the recycle stream if product contains 90% product, and the feed to the reactor contains 3% wt of component. (PS) Product Stream 10% water, 90% product

(W) PRODUCT RECOVER Y

(F)

MIXER

90% water

3% (M) component

BIOREACTOR

waste stream Water Dead cell

CELL SEPARATOR

10% component

REFERENCE: 1. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 2. David Himmelblau, Supplementary Problems for Basic Principles and Calculations in Chemical Engineering 6th edition. 3. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6th edition. 4. Felder and Rousseau, Elementary Principle of Chemical Processes (Revised Problem) 5. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. Chemical Calculations 1

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BYPASS

Consider a juice concentration process in which the dehydration process runs most efficiently by removing more water than is desired. A portion of the feed may be directed around the dehydrator in a bypass loop, to be mixed with unprocessed feed. The figure below illustrates a bypass loop.


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BYPASS PROBLEMS WITHOUT CHEMICAL REACTION 1. Air at 310K saturated with water vapor is dehumidified by cooling to 285K and by consequent condensation of water vapor. Air leaving the dehumidifier, saturated at 285K is mixed with a part of the original air which is bypassed. The resulting air stream is reheated to 320K. It is desired that the final air contains water vapor not more than 0.03kg per kg of dry air. Calculate the mass of dry air (in kg) bypassed per kg of dry air sent through the dehumidifier.1

2. The figure shows a 3-stage separation process. The ratio of P 3/D3 is 3, the ratio of P2/D2 is 1, and the ratio of A to B in stream P2 is 4:1. Calculate the composition and percent of each component in stream E.2


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3. A process needs an air supply which should contain 0.12 kmol / kmol da exactly. 1500 m3/ min of air at 25 °C and 101.3 k Pa is to be treated. Part of this air goes to a spray chamber. Where the air picks up water goes out with the humidified air to produce a 1850-kg mixture containing 0.12 kmol H2O/ kmol da. What is the water consumption (kg/min)? What is the ratio of the flowmeter readings of the bypass stream and the stream to the water spray?3 m3 bypass m2

m1 1500 m3 /min 25°C, 101.kPa

spray chamber

m4

m5 1,850 kg air mixer 0.12 kmol H2O/kmol da 0.3 kmol H2O/ kmol d.a

4. In a textile industry, it is desired to make a 24% solution (by mass) of caustic soda for a mercerization process. Due to the very high heat of dissolution of caustic soda in water, the above solution is prepared by two step process. First, in a dissolution tank, caustic soda is dissolved in the correct quantity of water to produce 50% solution. After complete dissolution and cooling, the solution is taken to dilution tank where some more water is added to produce 24% solution. Assuming no evaporation loss in dissolution tank, calculate the mass ratio W 1/W2.

W2

Solid NaOH W1 Water

DISSOLUTION TANK

50% NaOH

DILUTION TANK

24% NaOH(P)


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REFERENCES:

1. K.A. Gavhane, Introduction to Process Calculations Stoichiometry. 2. David Himmelblau, Basic Principles and Calculatons for Chemical Calculations 6 th edition. 3. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering. 4. K.A. Gavhane, Introduction to Process Calculations Stoichiometry.


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BYPASS PROBLEMS WITH CHEMICAL REACTION 1. Perchloric acid (HClO4) can be prepared from Ba(ClO4)2 and HClO4 , as shown in the diagram below. Sulfuric acid is supplied 20% in excess in reactor 1 that converts 80% of the Ba(ClO 4)2. 6125lb of the feed is bypassed to the second reactor where 10% in excess H2SO4 is supplied and converts 60% of Ba(ClO4)2. Between the adjacent reactors is a stream that contains 63% HClO4 by mole

.

. The over-all reaction produces 7400 lb HClO4

over the time period. Calculate;2 a. lbmole of waste H2SO4 per lbmole of BaSO4 that exits the separator b. lb HClO4 in G per lb feed c. over-all conversion of Ba(ClO4)2.

2. TiCl4 can be formed by reacting Titanium Dioxide with Hydrochloric Acid. 1200Kg feed with 64% TiO2, 3% inerts, 20% HCl, and 13%H2O is the initial feed. 800Kg of it is fed in to unit 1 that removes all the inerts from the original feed but unfortunately, also removes 20% of TiO 2 from the feed. That is why a bypass stream is needed with an amount of 400Kg. The stream leaving unit 1 together with the bypass is sent to unit 2 which converts all TiO 2 into TiCl4. A separator unit isolates the desired product from the rest. 3 Required: A. amount of TiCl4 produced B. amount of W and its composition


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PURGE When a process uses a recycle loop, there can often be a buildup of some undesired material within the system. By using a purge, a fraction of the recycle loop material is removed. This purge fraction is generally only a few percent of the recycle flow rate.


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PURGE PROBLEMS WITHOUT CHEMICAL REACTION 1. Caustic soda, among the largest producing product in Chemical Industry, is produced by electrolysis of common salt, NaCl, using membrane cell electrolyzers. The NaCl solution feed contains 30% by vol. of the solute. The NaOH produced is collected on storage tanks and the unreacted material containing 25% NaCl is being recycled. A purge stream is used to separate/remove the Cl 2 from the recycle stream. Calculate the following: a. Recycle stream flow rate/ recycle stream and product rate. b. Purge rate if it contains 40% Cl2 to produce 32.5 mol of NaOH, equivalent to 65% by vol. in the final product.

2. To save energy, stack gas from a furnace is used to dry ice. The flow sheet and known data are shown in the figure below. What is the amount of recycle gas ( in lb mole) per 100 lb of P if the concentration of water in the gas stream entering the dryer is 5.20%?

B


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3. One half of the high-pressure steam after being utilized for power generation is sent to the plant as process steam. The other half is returned to the boiler as condensate carrying 50 ppm solids.

To keep the solid level in the boiler below 1600 ppm solids, a part of the boiler water is blowdown (purge) continuously. the fresh boiler feed is found to contain 500ppm solids. if steam produced is free of solids, calculate the weight ratio of feed water to the blowdown (purge) water. 4. An evaporator is concentrating solutions from the two different sources. The solution from the first source containing 20% NaCl and 15% NaOH, flows at the rate of 40 kg/min. The other solution containing 8% NaCl and 12% NaOH flows at the rate of 70 kg/min. The two streams are fed directly to the evaporator, in this process, 60% of the water evaporated. In order to obtain the desired concentration,80% NaOH and 65% of NaCl were recycled. However, 20% of H2O was mixed on the recycle stream; to avoid this, purging of H2O must be done. Calculate the composition of NaOH and NaCl in the recycle stream and the product in the evaporator.


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Illustration: 60% H2O 40 kg/min 20% NaCl 15% NaOH

Solution:

EVAPORATOR

70 kg/min 8% NaCl

80% NaOH

12% NaOH

65% NaCl

20% H2O

5. In the production of NH3 from hydrogen and nitrogen, the nitrogen, the NH 3 produced is condensed for the reactor (convertor) product stream and the unreacted material is recycled. If the feed contains 0.2% Argon (from the nitrogen separation process), calculate the purge rate required to hold the argon in the recycle stream below 5.0%. R Purge

Feed 0.2% Argon

5% Argon REACTOR CONDENSER

Liquid NH 3


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REFERENCES: 1. K. Sinnott, John Metcalfe Coulson, and John Francis Richardson, Chemical Engineering Design. 2. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 3. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 4. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering 5. K. Sinnott, John Metcalfe Coulson, and John Francis Richardson, Chemical Engineering Design.


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PURGE PROBLEMS WITH CHEMICAL REACTION 1. Ethylene dioxide is produced by oxidation of ethylene as per the following reaction: C2H4 + ½ O2  C2H4O Fresh feed containing ethylene and air is mixed with recycle feed and mixed feed enters into a reactor. The proportion of C2H4 : O2 : N2 in mixed feed is 1:0.5:5.65 (on mole basis) 50% per pass conversion is achieved in the reactor are feed to the absorber where only all C 2H4O formed removed. The gases from the absorber containing C 2H4, O2 and N2 are recycled each. To avoid built up of N2 in the system, small portions of recycle stream is continuously purged based on 100 mol ethylene is mixed feed, calculate the fresh feed to the process, the purge stream, recycle ratio, combined feed ratio and overall conversion of ethylene.


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2. It is proposed to produce ethylene oxide ((CH 2)2O) by the oxidation of ethane (C2H6) in the gas phase; +

 (CH2)2O + H2O

The ratio of the air to the C2H6 in the gross feed into the reactor is 10 to 1, and conversion of C2H6 on one pass through the reactor is 18%. The unreacted ethane is separated from the reactor products and recycled as shown in the figure below. What is the ratio of the recycle stream to the feed stream and calculate the composition of purge stream and the stream exiting the reactor.

Water

Ethane air fresh feed

REACTOR

Purge O2 N2

SEPARATOR

Ethylene oxide

O2 N2 C2H6 100% Ethane recycle

Figure3.


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3. Methanol maybe produced by the reaction of CO 2 and H2 CO2 + 3H2

CH3OH + H2O

The fresh feed to the process contains CO 2 and H2 in stoichiometric proportions, 0.5mol% Inerts(I). The reactor effluent passes to a condenser which removes essentially all of theCH3OH and water formed, none of the reactants or inerts. The latter substances are recycled back to the reactor to avoid build-up of the inerts in the system; a purge stream is withdrawn fron the recycle. The feed to the reactor contains 2% inerts and the single pass conversion is 60%. Calculate the molar flow rates of the fresh feed, the total feed to the reactor and the purge stream for CH 3OH production rate of 1000mol/hr.


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4. The fresh feed to an ammonia production process contains N2, H2 and 0.125 mole %. The feed is combined with the recycle stream and the combined stream contains: 24.00 mol% nitrogen, 74 mol% hydrogen and 2% inerts, then it is feed to the reactor, in which 25% single-pass conversion of nitrogen is achieved. The product pass through a condenser in which essentially all the ammonia is removed and the remaining gases are recycled. However, to prevent buildup of the inert in the system, a purge stream must be taken off. Calculate the overall conversion of nitrogen, the ratio (moles purge/mole of gas leaving the condenser) and the ratio (moles fresh feed/moles feed to the reactor).

ILLUSTRATION: nr (mol ) np (mol) X4 N2 mol X4 N2 mol X4 H2 mol X4 H2 mol (1-X4N2–X4H2) mol I (1–X4N2–X4H2)mol I n0 X0mol N2 (0.99875–N2)H2 NH3 .00125 I

REACTION:

n 4 (mol) X4 N2 mol X4 H2 mol (1 - X4 N2 -X4 H2)mol I

REACTOR 24.00% 74.00% H2

CONDENSER 25% N 2 conversion n1 mol N2

2.00% I

n3 mol

n 2 mol H2 n3 mol NH3 2 mol I N2

+ 3 H2

NH3


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5. Carbon monoxide and hydrogen reacts to give methanol: CO+2H 2

CH3OH

The conversion of CO entering the reactor is 20%. A feed stream consisting 33% CO, 66.5% H2, and 0.5%CH4 is mixed with a recycle stream and sent to a reactor. The methane leaving the reactor is separated and the unconverted gases are recycled. To prevent the accumulation of CH 4 and keep its concentration, in the recycle stream at 3%, a portion of the recycle stream is blown off. For 100 moles of fresh feed, determine the following: a. recycle stream (moles), b. purge stream (moles) c. composition of purge (moles) and d. moles of methanol.

Purge (P, mol) CO= x CH4= 0.03 Recycle H2= 0.97-x

CO= x CH4= 0.03 H2= 0.97-x

100 mol CO, H2, CH4

Product REACTOR

SEPARATOR

CH3OH (M, mol)


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REFERENCES:

1. Narayanan, B. Lakshnikutty and K.V. Narayanan, Stoichiometry and Process Calculations 2. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6th edition. 3. David Himmelblau, Basic Principles and Calculations in Chemical Engineering 6 th edition. 4. Wilfredo I. Jose, Introductory Concepts in Chemical Engineering. 5. Narayanan, B. Lakshnikutty and K.V. Narayanan, Stoichiometry and Process Calculations


Chemical Calculations for Chemical Engineering. PROBLEMS in MATERIAL BALANCE, RECYCLE, ByPASS & PURGE

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