Intro to Process Simulation

8/22/2019 Intro to Process Simulation

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University of Adelaide

i A. Bhandari , C.Y. Yeo, E. M.Mah,

Disclaimer

The contents and ideas contained in this report are the properties of the authors involved in this

group project, except where there is a clear acknowledgement and reference of the work of other

authors

Aakriti Bhandari

Chen Ying Yeo

Euu Min Mah


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ii A. Bhandari , C.Y. Yeo, E. M.Mah,


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iii A. Bhandari , C.Y. Yeo, E. M.Mah,

Executive summary

The aim of this project was to design a production facility to produce 5 million litters of ethanol per

year. The ethanol was produced by reacting ethylene with steam over a catalyst. The main objective

was to design a production unit minimizing the operating costs and maximizing the profit. This wasdone using Microsoft Excel using the various capabilities of it such as the iteration and solver

functions.

There were flexibilities available to reach the goal of maximum profit and minimum costs. The

reactor pressure could be varied from 10 to 40 bars. The steam to ethylene feed ratio could be

adjusted from 1:1 to 5:1 and the recycle to purge ratio can be changed from 20:1 to 80:1.

The ideal reactant pressure obtained after solving the problem and optimizing all of the above

conditions was 40 bars. High pressure caused the increase in rate of reaction which in turn increased

the production level and also contributed to the costs and profit.

The ideal steam to feed ratio was calculated to be 2.7:1. This directly affected the costs as blah blah

blah. The model recycle to purge ratio that results in maximum ammonia production and in turn

maximum profit was 80:1. The recycle to purge ratio directly impacted the costs and profits as the

recycling material reduced the overall costs of the reactants. This played a vital role in reducing the

feed in to the process as a higher recycle meant less stress on the feed.

The ideal value parameters that obtained a maximum profit and minimum costs are listed in the

table below:

Total operating cost $ 2,745,485.50

Total Product value $ 4,076,265.58

Overall Profit $ 1,330,780.07Table 1

The above figures show that the main objective of the task was reached. Utilizing the given

variability this production process was successful in production the required amount of ammonia of

5 million litters per year. The maximum profit made was $ 1.3 million per year and the minimum cost

price was calculated to be $2.7 million per year.


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iv A. Bhandari , C.Y. Yeo, E. M.Mah,


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v A. Bhandari , C.Y. Yeo, E. M.Mah,

Table of Contents

1. Introduction

1.1. Problem Statement

1.1.1.Problem Specification

1.2. Background Information

1.3. Key Constraints

1.4. How the Problem was Attacked

2. Results

2.1.1._____________Revenues and Operating Costs at 3:1 Steam to Feed Ratio

2.1.2.Revenues and Operating Costs at 80:1 Recycle to Purge Ratio

2.1.3.Final Decision

2.2. Results for the Optimum Conditions

2.2.1.Process Flow Diagram

2.2.2.Flow Summary Table

2.2.3.System Balances

2.2.3.1. Material Balance

2.2.3.2.

Energy Balance2.2.4. Raw Materials and Operating costs

2.2.5.Revenues

2.2.6.Profit

2.2.7.Investment Summary

3. Discussion

3.1. Key Assumptions

3.2. Key Decisions

3.2.1.Reasons for making Choices

3.2.2.Reasons for Discarding Alternatives

3.3. Computation of Results

3.4. Results Analysis and Effects of the variables on the Profit

3.4.1.Pressure

3.4.2.Steam to Feed Ratio

3.4.3.Recycle to Purge Ratio

4. Conclusion


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vi A. Bhandari , C.Y. Yeo, E. M.Mah,

4.1. The Optimum State

4.2. Further Study and safety considerations

5. Appendix

5.1. Blah

5.1.1.Blah

5.1.2.Blah

5.1.3.Blah

5.2. Blah

5.2.1.Blah

5.2.2.Blah

5.2.3.Blah

5.3. Blah

5.4. Blah

6. References


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1 A.Bhandari , C.Y. Yeo, E.M.Mah,

1. Introduction

1.1. Problem StatementThe key purpose of this design project was to produce a preliminary design for an ethanol

production unit in order to produce 5 million litres per year. The design should be such that the

cost of the production is kept to a minimum and the profit collected from the overall sales

should be maximum.

1.1.1. Problem Specification

The objectives for this problem can be reached by varying the reactor pressure, the ratio of

steam to ethylene feed and the recycle to purge ratio. According to Le Chateliers Principle, if the

pressure is increased in a high rate, the system will respond by shifting the equilibrium to the

right. On one hand, this will increase the production of ethanol, whereas on the other it will also

add to the cost of the equipment as the structure has to be extremely robust1. Therefore a cost

analysis is done on the effects of increasing the pressure in order to identify the most

economical state.

There is flexibility in altering the steam to feed between 1:1 to 5:1. The price of the saturated

steam is significantly low compared to the price of the ethylene feed. So, an excess of steam can

be used in order to move the position of equilibrium to the right. However this could alsodecrease the concentration of ethylene which can affect the production rate and ultimately the

cost.

The recycle if to purge ratio can also be altered to reach the goal of this production. This ratio

has a direct impact on the costs and the profits as the recycling material will cut costs for the

amount of reactants required. However, there needs to be a balance between the two to obtain

the optimum results required.

1.2. Background InformationEthylene, also known as ethene, is produced in petro-chemical industry by converting large

organic hydrocarbons into smaller compounds and obtaining a distillate. (more on ethylene?)

The hydration of this ethylene produced ethanol. Ethanol also known as ethyl alcohol is volatile

liquid which serves for various purposes.2It has a chemical formula of C2H5OH. A vapour-phase

hydration of ethylene is the process through with ethanol as a pure substance can be produced.

The hydration reaction shown below produces ethanol but in the presence of a catalyst:


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The above reaction is reversible and the process of formation of ethanol is exothermic. 1In order

to produce the maximum possible amount of ethanol, the equilibrium position has to be shifted

as far to the right as possible. As the hydration reaction is exothermic, if the temperature is

lowered, the forward reaction is favoured according to Le Chateliers Principle. The system will

shift to the right and will produce a higher amount of ethanol. This seems theoretically possible,

but in reality, lowering the temperate will lower the rate of the reaction. This is not economical

as it takes longer time and energy. So, in order to conserve time, a compromise has to be made.

This plant uses a compromise temperature of 250C. This produces an acceptable proportion of

ethanol but in a very short time. Pressure is another variable that will have an effect on the rate

of this reaction. According to Le Chateliers Principle, an increase in pressure will cause the

system to respond by favouring the reaction which produces fewer molecules in order to

decrease the pressure. That will be the right hand side on this reaction. The pressure has to be

as high as possible in order to increase the rate of production of ethanol. However, high

pressure cause number of problems such as the increase in cost and polymerisation of ethylene

to make polyethylene

The ethanol produced from this production plant is used hugely worldwide as an alternative to

gasoline to internal combustion engines. It is renewable and has various environmentaladvantages over fossil fuel.

3Ethanol is also sold commercially as alcoholic beverages. However,

most of the ethanol in alcoholic beverages is produced through fermentation. Ethanol is not very

harmful and is used in cosmetic industry for preservation reasons. It can be denatured and

changed into methylated spirits and can be used as commercial disinfectant, spirit burners,

general industrial solvents and for cleaning purposes.3

It can also be sold to pharmaceutical and

research companies that utilize ethanol to preserve liquid drugs as well as for other purposes.

1.3. Key ConstraintsThe optimal condition depends on the production rate and the feed rate as well as various other

constraints. The key constraints for this process were presented in terms of the flexibility in the

operation of this unit. The flexibility include variation in reactor pressure from 10 to 40 bar, the

variation in steam to feed ratio from 1:1 to 5:1 and the variation in the recycle to purge ratio

from 20:1 to 80:1.

1.4. How the Problem was AttackedFirstly, a block flow diagram was drawn of the process given in order to visualise the problem

better. After the block flow diagram was drawn, material balance was performed on a randomly


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chosen set of parameter(pressure:10bar, recycle to purge ratio: 20:1 and steam to feed ratio of

1:1) by taking a basis of 1000mols. Microsoft Excel was used to perform these calculations. Cost

was then optimized for the obtained results of the material balance. Various graphs were drawn

in order to identify the optimum state. Energy balance was then conducted on the optimum

state. Lastly, process flow diagram was then drawn and labelled to present the overall design of

the production facility.


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2. Results

The first section of this part presents the economic analysis of the optimum condition and how it

was obtained and the most profitable operating conditions are presented in the second section.

The maximum profit was $1.33 million, and the optimum conditions were 40 bar operating pressure,

2.7:1 steam to feed ratio and 80:1 recycle to purge ratio. Ethanol was sold for $0.99 per kg which

made annual revenue of $4.04 million. The fuel gas was also sold annually for $38 thousand that

contributed towards the total revenue. The annual operating cost cleared at $2.8 million. The

difference between the revenues and operating costs resulted in the obtained profit.

2.2.1 Economic Analysis

2.2.1.1 Optimum Steam to Feed Ratio

The optimum steam to feed ratio was calculated to be approximately 2.7:1 using Solverin Microsoft

excel. This can be found in more detail in the appendix section of this report.

2.2.1.2 Recycle to Purge Ratio Determination

The steam to feed ratio was determined to be 2.7:1. So, in this section, all the net profits were

graphed for all the 3:1 steam to feed ratio against pressure to identify the optimum Recycle to purge

ratio. This graph shows that there is almost a logistic relationship between the profit at 3:1 S/F and

0

20

40

60

80

100

120

140

160

180

0 10 20 30 40 50

Net

Profit($/hour)

Pressure (bar)

3:1 Steam to feed ratio graph to find the

recycle to purge ratio

20:01

30:01

40:01

50:01

60:01

70:01

80:1

Recycle

to Purge

Figure 2.2.1.2: 3:1 Steam to feed ratio graph


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pressure. The profit gradually increases until it reaches a maximum point. The maximum profit for

the graph is when the recycle to purge ratio is 20:1.

2.2.1.3 Pressure optimization

The Graph above shows net profit for all pressure range for 80:1 recycle to purge ratio (which was

determined to be one of the optimum conditions from 2.2.1.) and steam to feed ratio of 3:1. This

graph shows that the profit is greater for a greater pressure. The flexibility in pressure was between

10-40 bar. It was predicted that increasing pressure will increase the revenue but will also add to the

cost causing a decrease in profit. However, in this case, we can see that the pressure of 40 bar isnt

high enough to add to the cost. So, the optimum operating pressure for the reactor is 40 bar.

2.2.1.4 Final Decision

The results found from the graphs above support the values obtained using solver which can be

found in more detail in the Microsoft Excel Spreadsheets.

The optimum condition determined from this project was a pressure of 40 bar, steam to feed ratio

of 2.7:1 and recycle to purge ratio of 80:1.

0

20

40

60

80

100

120

140

160

180

0 10 20 30 40 50

N

etProfit($/hour)

Pressure(bars)

80:1 Recycle to Purge

3:01

Figure 2.2.1.2: 3:1 Recycle to purge ratio

Steam

to feed


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2.2. Results for the Optimum Conditions

2.2.1. Process Flow Diagram

Figure 2.2.1- Process flow diagram for the production of ethanol


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2.2.4. System Balances

Material Balance on various parameters as well as energy balance in required sector were used to

find the optimum conditions for this system. Microsoft excel was used in order to solve the problem.

All detailed calculations are available in the appendix section of this report.

2.2.4.1 Material Balance

Material balance was conducted with on the system using Microsoft Excel to find the optimum

condition. A basis of 1000 mol/h of ethylene feed is taken in order to find the composition and flow

rates of various other streams. Detailed material balance calculations can be found in appendix.

The tables below consists the material balance calculation for the optimum condition, with pressure

40, recycle to purge ratio of 80:1 and steam to feed ratio of 2.7:1:


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2.2.4.2 Energy Balance

Energy balance was done on the optimum condition of 40 bar, 80:1 recycle to purge ratio and 2.7:1

steam to feed ratio. The detailed calculation on each system is included in the appendix. This section

consists a summary of results of the energy balances determined.

Sysytem Q (kW) Ws (kw)

Heater -0.0757 -

Reactor -0.758 -

Heat Exchanger -0.245 -

Compressor 0 93

Condensor -1739.6 -

Reboilers 192.5 -

Table 2.2.4.2: Energy balance Summary Table

2.2.5. Raw Materials and Operating costs

The raw materials for the optimum state are calculated according to the flow rates presented in

tables 2.2.4.1 (a) and (b). Raw materials included Ethylene and steam and the process included

cooling after reactor, distillation, recycle compressor and waste water treatment.

2.2.5.1. Raw Material Costs

Material Cost($/kg) Total Cost($)

Ethylene 0.77 1,876,160

Steam 0.00245 10,379

Table 2.2.5.1 Raw Materials Costs

2.2.5.2. Operating Costs

Process Cost($) Total Cost($)

Distillation 0.1/kmol 732,916

Cooling after reactor 0.47/kmol 50,438

Reycle compressor 0.008(p in bar)/kmol 75327

Wastewater 100/1000m3 261.34

Table: 2.2.5.2 Operating Costs

The total operating costs is $2.7million


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2.2.6. Revenues

The table below shows the revenues generated per year:

Component Value ($/kg) Total Value($)

Ethanol 0.99 4038299

Fuel Gas 0.4 37966

Table 2.2.8: Revenues

The total revenues is $4.07 million

2.2.7. Profit

The total profit made by subtracting the revenue from the total cost was approximately $1.3 million.

2.2.8 Investment Summary

Careful research must be done before making investment in the project. Capital cost has to be

calculated before constructing the plant.


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3. Discussion

3.1 Key Assumptions

There were various assumptions made throughout this project. When solving the problem, it was

assumed that the process unit operates for 8000 hours a year. It is a realistic as this is about 330

days, which allows for periodic shutdown and maintenance. When calculating the operating costs,

employee wages, utility costs and energy costs were neglected. These costs werent part of the

preliminary design and can be deducted from the profit depending on the number of employees and

consumption rate. So, the profit calculated was just an estimate and investment shouldnt be based

solely on the findings of this report. Also, in this report the effect of side reactions are neglected. The

question suggests that the side reactions should be considered, but it was over looked in the

material balance and energy balance calculations and due to lack of time, its effects were assumed

to be negligible. So, the profit again, is just an estimate as the side reactions could have an impact on

the profit.

3.2 Key Decisions

This reports contains decisions about that the pressure, the steam to feed and recycle to purge ratio

to use in the ethanol production facility. The aim was to minimize costs and mazimize profits in

order to produce 5 million litters of ethanol per year. Decisions were also made on operating

conditions for various systems such as the compressor, heat exchanger, etc which is all summarised

in the process flow diagrams (Figure 2.2.1).

3.2.1. Reasons for making choices

Microsoft excel was used to solve this problem as it was the program the group members were most

familiar with and it also was the easiest. It was also the most accurate. The solver capability of excel

was most helpful when it came to making an estimate mostly about the steam to feed ratio. The

decisions made about the pressure and the ratios of steam to feed and recycle to purge were all

obtained after an extensive work on the material balance, cost optimization and energy balance. The

obtained result seems all reasonable with respect to the assumptions made. The profit calculated

per annum also seems reasonable. This profit could fluctuate depending on the world trade and also

the advances in technology.


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3.2.2. Reason for discarding alternatives

While calculating the optimization, trial and error methods or algebraic methods couldve been

used. These methods were however discarded as they would not provide accurate values. The only

suitable method was to use was computational method and Microsoft Excel.

In the final Process Flow Diagram, (Figure 2.1.1), it was decided that the hot recycle stream, ethanol

feed stream and the steam stream were all mixed at a single point before entering the heat

exchanger. The alternative was to have different mixing point, so exact composition of each material

at a certain stage could be calculated. This was discarded it was easier to have a single mixing point

to calculate the energy balance and material balance.

Finally, the by-product obtained due to side reaction was discarded. It was assumed that the by-

product was minimum and it did not make any significant difference in the profit. (This mightve not

been the case, but due to lack of time, it was simply discarded in the calculations)

3.3. Computation of results

A basis of 1000 moles/h of feed stream was taken in order to solve this problem. The initial basis did

not affect any of the optimization as it was scaled later using a scaling factor. The scaled basis was

10850 moles/h. Moles of individual reactants were determined using this basis. Using Microsoft

excel, the mass flow rate of ethanol production was deduced and the process was scaled to produce

5 million litters per year. This is given in more detail in the appendix. The total cost to produce therequired amount of ethanol was calculated by using the amount of feed required and the prices

given. The overall revenue was calculated likewise. The total cost was then subtracted from the

revenue to calculate the total profit. The range of profit values for varying parameters was obtained.

Solver was used in Microsoft Excel to calculate the steam to feed ratio. The profits were graphed

against the pressure to determine the optimum pressure and a similar method was conducted to

obtain the optimum recycle to purge ratio. After analysing the graph, it was decided the optimum

pressure was 40 bar, the optimum recycle to purge ratio was 80:1 and the optimum steam to feed

ratio was 2.7:1.

3.4. Result Analysis and Effects of Variables on the Profit

3.4.1. Effect of Pressure

According to Le Chateliers Principle, for the given equilibrium reaction, an increase in pressure will

cause an equilibrium shift to the right, producing a higher amount of ethanol. This can be implied for

this reaction. Figures 2.2.1.2and 2.2.1 show that the with the change in pressure the profit also

increases. This shows that with a higher pressure, the production of ethanol is also high which will


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increase the profit. This is generally not the case. If the pressure is extremely high, the rate of

production of ethanol is also high but this increase in pressure will also add to the cost as system has

to extremely robust in order to operate in high pressure. This means that the equipments will be

expensive than usual. This however is not the case for this report. The flexibility in pressure is

between 10-40 bar. These pressures are not high enough to contribute to the increase in the

equipment costs.

3.4.2 Steam to Feed Ratio

As steam is cheaper than the feed, ethylene, is generally used in excess. Increasing the concentration

of steam in the system will lead to shift in the equilibrium to the right, according to the Le Chateliers

Principle. So, we can see from the calculated profit, (refer to the appendix), that there is an increase

in profit from steam to feed ratio of 1:1 to 3:1. After this, the profit starts declining. This is because,

as the steam concentration increases, the ethanol being produced also increases which cause a

decrease in ethylene. In order to overcome this decrease, the equilibrium position shifts to the right

to produce more ethylene which decreases the production of ethanol and also decreases the profit.

3.4.3. Recycle to Purge Ratio

With the increase in recycle to purge ratio, the profit increases. This will decrease the stress on the

feed which will decrease the overall cost. This is because by increasing the recycle to purge ratio,more ethanol can be produced which will add to the profit.


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4. Conclusion

4.1 Optimum State

The production of ethanol is widely influences and manipulated by several parameter. Those include

pressure, ratio of feed streams and the ratio of recycle to purge. Each of them parameters have an

impact on the economics of the production plant. To maximise the profit, the react should run at a

pressure of 40 bars, with steam to feed ratio of 2.7:1 and recycle to purge ratio of 80:1. If run under

these circumstances, the plant is capable of producing 5 million litters of ethanol per annum. This

will have a total cost of $2.75 million per year. The total revenue collected will add up to $4.08

million which will lead to a net profit of 1.33 million.

4.2 Further Study and Safety Considerations

Before making any major financial decisions regarding this production facility, an extensive further

study has to be done. The factors that need study are: the current market, employee figures,

location of the plant, maintenance and energy costs, environmental impacts, variation in the costs of

raw materials.

The demand of ethanol in the current and future market has to be closely analysed. The plant should

be able to make profit for its entire life time.

Costs involved in the operation of employees are fluctuating. In this report, such costs are not

considered. These could alter the profit.

Location of the plant and its environmental impacts are also of concern. As it produces by-products

and they could potentially be toxic, this plant should not be built in the urban setting. And the by-

products and the wastes have to be completely treated before releasing them into the surrounding

as they could pollute the aquifers and affect a large number of population.

Various Safety Considerations include:

The pressure can range from 10-40 bar. These are really high pressures. So appropriatespending has to be made to make this plant a safe work place.

The reactor is operating at a very high temperature, so protection has to employed for the

safety.

Risk assessment, also has to be conducted in the design phase to determine whether it is actually

logical to implement the design.


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5. Appendix

5.1


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5.1 Block Flow Diagram


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5.2 Material Balance

A basis of 1000 moles/h fresh feed of ethylene was taken in order to solve the material balance.

There were three parameter that could be changed are: pressure between 10-40 bar, recycle to

purge ratio between 20:1 to 80:1 and steam to feed ratio between 1:1 to 5:1. Initially, the material

balance was solved taking a set values for each flexible. They were steam to feed ratio of 1:1, recycle

to purge ratio of 20:1 and the pressure at the reactor as 10.

The problem states that:

The feed stream is 99 mol% pure with 1 mol% ethane that was non-reactive

throughout

The equilibrium constant is k=0.0425 bar-1

The reactants reacted stoichiometrically according to the reaction:

(1)The individual molar flow rates:

Using 1:1 ratio for steam to feed,

At this stage, the molar flow rates of fresh feed and reactor inlet streams are:

The equilibrium constant at 250C is 0.0425 bar-1

The extend of reaction is assumed to be moles, ie the amount of reactants reacted is moles.

So, the molar flow rates, out of the reactor are:

mol

mol

For this situation, the pressure is taken as 10 bar so,

Now, finding the extend using the equilibrium constant.

(Where x is the extend of reaction)


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Substituting in the numbers gives,

This quadratic equation can be solved using the quadratic formula,


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The table below shows the reactor-in and out balance

Into Reactor(mol/hr) Out of Reactor(mol/hr)

Ethylene 990 829.2

Ethane 10 10

Water 1000 839.2

Ethanol 0 160.8

Total 2000 1839.2

Table 121

The above table is for a 100% conversion. For a 90% conversion, the adjusted values are:

Into Reactor(mol) Out of Reactor(mol) Out of Reactor

90%(mol)

Ethylene 990 829.2 845.3

Ethane 210 210 210

Water 1000 839.2 855.3

Ethanol 0 160.8 144.7Total 2000 1839.2 1855.53

Into the Flash Separator,

Into Flash Separator(mol) Flash separator, over head (mol)

Ethylene 845.3 845.3

Ethane 210 210

Water 855.3 -

Ethanol 144.7 -Total 1855.53

Now, Calculating the Purge,

mol

Purge(mol)

Ethylene 40.3

Ethane 10

Water -


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Ethanol -

Total 50.3

Calculating the Recycle,

mol

Into Recycle

Recycle(mol)

Ethylene 805.0

Ethane 200

Water -

Ethanol -

Total 1005.0

Now, Iteration is performed in Excel to iterate and solve the problem. The obtained solution is:

The equilibrium reactions reaches a state of near completion due to a higher rate of recycle.


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Distillation

The facts given in the question sheet about the distillation columns include:

Two percent of the total number of moles entering the distillation column is refluxed in the

bottoms.

The re-boiler is assumed to be a splitter.

Two percent of the total number of moles of ethanol entering the distillation column exits in

the wastewater stream.

The condenser is assumed to be a splitter

The ratio in the overhead stream for product to reflux is 12:1.

Feed:

The feed into the distillation column

So, the overall feed stream for the distillation column is listed below in mass:

H20 C2H5OH

Flow rates

kg/h

342108.5581 480853.9581

Mass fractions 0.415703694 0.584296306

Refer to the Microsoft Excel Document for more details.

After calculating the Overhead, Bottoms, Product, Wastewater, Reflux, the overall flow into and out

of the distillation column is listed in the table below:


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H20 C2H5OH

Flow Rates (kg/h) Mass fraction Flow Rates (kg/h) Mass Fraction

Into Distillation 342108.56 0.456 480854 0.584

Into Condenser 502441.9179 0.0758 6126079.427 0.9249

Out of Condenser 38649.4 0.0758 471236.879 0.9249

Reflux 463792.5396 0.0758 5654843 0.9249

Waste water 303459.2 0.9693 9617.079 0.0307

Into Reboiler 606918.4 0.9693 19234.16 0.0307

Out of Reboiler 303459.2 0.9693 9617.079 0.0307

Scaling up

The target for the production unit is to produce 5 million litters per year. The results using a basis of

1000 mol is not able to produce this required target.

So, in order o obtain the required target, the values have to be scaled up. This scaled up value of10850 moles feed is used to determine the cost.

Assuming 8000 hours add up to a year, the scaling up value cam be determined hence.

Scaling

Minimum value 5000000 L per year

=Minimum

value/hours

=625

L per hour

=Minimum

value /density

=4079100

kg per year

Ethanol production target 509.8875 kg per hour

Ethanol We Calculated 509886.3 kg per hour

After Before

Scaling factor 0.001 0.01085

scale basis 1.000002 10.85036

Basis Scaled 1000.002 10850.36


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Cost Analysis

Raw Materials and Operating Cost

The provided valued for the raw materials and the operating costs are listed below as follows:

Ethylene Feed : $0.77/kg

Saturated Steam : $0.00245/kg

Cooling after reactor : $0.1/kmol

Distillation : $0.47/kmol

Recycle Compressor : $0.008 (p in bar)/kmol, pin = 5 bar, pout = 40 bar

Wastewater treatment : $100/1000m3

Assume 1 year = 8000hours

Raw Materials ( per year )

$1,876,162.48

$10,379.42

The total cost of the raw materials was: $1886541.9

Operating costs (per year)

= $50,438.69 =$732,916.36 () =$75,327.22 =$261.34

The overall operating costs was $858943.61

Revenue

The overall revenue calculated is the sum of the gain from selling all the ethanol produced as well as

the fuel gas, which was not the primary product of this production facility.

The values of Ethanol and Fuel gas provided in the question sheet are as follows:

Ethanol : 0.99/kg

Fuel gas :0.4/kg


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The overall revenue was: $4,076,265.58

Profit

The profit calculated was :$1,330,780.07

Profit for Pressure 40 bar

The Overall Cost analysis:

Process Material Annual Costs AnnualRevenue

Annual Profit

Ethylene Feed $1,876,162.48 $0.00 -

$1,876,162.48

Saturated

Steam

$10,379.42 $0.00 -$10,379.42

Cooling after reactor $50,438.69 $0.00 -$50,438.69

Distillation $732,916.36 $0.00 -$732,916.36

Recycle Compressor $75,327.22 $0.00 -$75,327.22

Wastewater

treatment

$261.34 $0.00 -$261.34

Ethanol(95vol%)

$0.00 $4,038,299.16 $4,038,299.16

Fuel Gas $0.00 $37,966.42 $37,966.42

Total $2,745,485.50 $4,076,265.58 $1,330,780.07

Recycle to

Purge Ratio:

30 30 40 50 60 70 80

Steam:Ethylene

Feed Ratio

(Y:1)1 17.82822154 1.880512734 -

15.06207235

-

32.43189545

-

50.02428785

-

67.74734343

-85.5536459

2 141.8819265 148.0413424 151.0631507 152.607949 153.3309078 153.5465283 153.4257696

3 154.1505955 159.7928337 162.5870615 164.1189043 164.9797891 165.4412425 165.6455781

4 148.5132214 153.3583313 155.7001009 156.9578739 157.6481755 158.0042511 158.1468936

5 139.4134511 143.8409261 145.9550074 147.0768196 147.6822261 147.984705 148.0943053


34/44




35/44



ENERGY BALANCE

Heater

Basis : 1hour

Assume Ideal Gas and process is steady -state.

General energy balance equation on a continuous process, open system:

Energy Equation : Ek+ Ep + H = QWs

whereEk: kinetic energy

Ep: potential energy

H: enthalpy change

Q : heat flow

Ws : shaft work

Assume all components in stream 1,2 and 20 are not mixed.Ek,Ep , Ws = 0

Q= H

Stream Stream 1 2 20

Component Component C2H6 C2H4 H2O C2H6 C2H4

Flowrate mol /h Flowrate/h 108.50362 10741.858 29420.135 24532.76877 8680.289579Flowrate mol /s Flowrate/s 0.0301399 2.9838494 8.1722597 6.8146579 2.41119155Stream Stream 3 4

Component Component C2H6 C2H4 H2O C2H6 C2H4 H2O

Flowrate mol /h Flowrate/h 8788.793199 35274.62713 29420.13537 8788.793199 35274.62713 29420.1353

Flowrate mol /s Flowrate/s 2.4413314 9.7985075 8.1722598 2.4413314 9.7985075 8.1722598

Cp (kJ/mol) Table B.2

HC2H6 49.373x10-3

+13.92x10-5

T-5.816 x10-8

T2+7.28 x10

-12T

3

HC2H4 40.75x10-3

+11.47x10-5

T-6.891 x10-8

T2+17.66 x10

-12T

3

HC2H5OH 38.58

H H2O 33.46x10-3

+0.6880x10-5

T+0.7604 x10-8

T2-3.593 x10

-12T

3

Calculation for Input

Treference =25C, Tfresh feed = 25C , Trecycle = 312.4C (as calculated @ Compressor balance)

Tfresh-reference= 287.4 C, Tfeed-reference= 0 C

From Table B.6, Temperature of saturated steam is found to be 250.3C at P=40bar.

Tsteam-reference= 225.3C

Hin=

HeaterStream 4C2H68788.793199 mol

C2H435274.62713 mol

H2O29420.13537 mol

T=250C, P=30barT=50C, P=30bar

Stream 1

C2H6 108.50362 molC2H410741.858 mol

Stream 2

H2O29420.135 mol

Stream 20

C2H4 24532.76877 molC2H68680.289579 mol

3

20

1

2

4


36/44



=0+2.41119155 [40.75x10-3Tfresh-reference +(11.47x10-5

Tfresh-reference2)/2- (6.891 x10

-8Tfresh-reference

3)/3+

(17.66 x10-12

Tfresh-reference4)/4] +0+ 6.8146579 [49.37x10

-3Tfresh-reference +(13.92x10

-5Tfresh-reference

2)/2-(5.816 x10

-8

Tfresh-reference3)/3+(7.28 x10

-12Tfresh-reference

4)/4] +8.1722597 [33.46x10

-3Tsteam-reference +(0.6880x10

-5Tsteam-reference

2)/2+(0.7604 x10

-8Tsteam-reference

3)/3-(3.593 x10

-12Tsteam-reference

4)/4]

= 38.4185477 + 132.8177569 + 63.25186294

= 234.488 kWCalculation for Output

Treference =25C, Treactor= 250C (as calculated @ Compressor balance)

Treactor-reactor= 225 CHout= =9.7985075 [40.75x10

-3Treactor-reactor+(11.47x10

-5Treactor-reactor

2)/2- (6.891 x10

-8Treactor-reactor

3)/3+

(17.66 x10-12

Treactor-reactor4)/4] +2.4413314 [49.37x10


-5Treactor-reactor

2)/2-(5.816 x10

-8

Treactor-reactor3)/3+(7.28 x10

-12Treactor-reactor

4)/4] +8.1722597 [33.46x10


-5Treactor-reactor

2)/2+(0.7604 x10

-8Treactor-reactor

3)/3-(3.593 x10

-12Treactor-reactor

4)/4]

= 115.8356653 + 35.19322815 + 7.729219321= 158.758 kW

Q= HoutHin

= 158.758 - 234.488= -75.73 kW


37/44



Reactor Analysis


Basis: 1hour

Assume Ideal Gas and process is steady -state.

General energy balance equation on a continuous process, open system:

Energy Equation : Ek+ Ep + H = QWsQ=0 (adiabatic), Assume Ek,Ep ,

General energy balance equation on reactive process:

where

: heat of reaction at reference condition (25C , 1 atm), : enthalpy change of species i/j relative to reference condition

Heat of reaction

Reaction : C2H4 (aq) + H2O (l) C2H5OH (aq)

where

: extent of reaction

: specific heat of reactionComponent Specific heat of formation (kJ/mol) [Table B.1]

C2H5OH -235.31

= -38.58 kJ/mol = 11594.66527 mol/hour= 3.2207 mol/ s

= -235.31x 3.2207 = - 757.872 kW

Reactor

Stream 5

T=250C,P=30bar

Stream 4

T=250C,P=30bar


38/44



Enthalpy of reactant, H

H= Where = number of moles, = specific enthalpy of species

Stream 4

Component C2H6 C2H4 H2O

Flowrate mol /h 8788.793199 35274.62713 29420.13537Flowrate mol /s 2.441331444 9.798507535 8.172259826Stream 5

Component C2H6 C2H4 H2O C2H5OH

Flowrate mol /h 8788.793199 24839.42838 18984.93663 10435.19874Flowrate mol /s 2.441331444 6.899841217 5.273593508 2.898666318


HC2H6 49.37x10-3

+13.92x10-5

T-5.816 x10-8

T2+7.28 x10

-12T

3

HC2H4 40.75x10-3

+11.47x10-5

T-6.891 x10-8

T2+17.66 x10

-12T

3

H H2O 33.46x10-3

+0.6880x10-5

T+0.7604 x10-8

T2-3.593 x10

-12T

3

H C2H5OH 61.34 x10-3

+ 15.72 x10-5

T -8.749 x10-8

T2

+19.83 x10-12

T3

Using 25C as a reference temperature

T=250C -25C = 225C

H = H C2H4

=9.798507535 [40.75x10-3T +(11.47x10-5T2)/2-(6.891 x10-8T3)/3+(17.66 x10-12T4)/4]= 118.29 kW

H C2H6=2.441331444 [49.37x10-3T+(13.92x10-5T2)/2-(5.816 x10-8T3)/3+(7.28 x10-12T4)/4]

= 35.72 kW

H H2O

=8.172259826 [33.46x10-3T+(0.6880x10-5T2)/2+(0.7604 x10-8T3)/3-(3.593 x10-12T4)/4]= 62.95 kW

HTotal = 118.29 kW + 35.72 kW + 62.95 kW = 216.96 kW

Using the same method, but different number of moles to calculate the enthalpy of the product,

Enthalpy of the product

H C2H4 = 83.30 kW

H C2H6 = 35.72 kWH H2O = 40.62 kWH C2H5OH = 51.54 kW

HTotal = 83.30 + 35.72 + 40.62 + 51.54 = 211.18 kW

Hfinal = + HproductHreactant= - 757.872 kW + 211.18 kW216.96 kW

= - 763.65 kW

Q = -763.65 kW


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Heat Exchanger Analysis

The reactor outlet stream is cooled to the operating temperature of the flash separator 50CEnergy Equation : Ek+ Ep + H = QWs

Assume Ek,Ep , Ws = 0

Q= H

, since T=250C - 250C = 0Ethanol Information (Table B.1) :

Tb = 78.5C, Hv =38.58 kJ/molWater Information ( Table B.1) :

Hv = 40.656 kJ/mol

T1=250C -78.5C = 171.5C

T2=78.5C - 50C = 28.5 CT3 = 250C -50C = 200C

T4 = 250C -100C = 150C

T5 = 100C-50C = 50C

Hout= = H 1 +H 2 +H 3 + H4 +H 5

H 1 = 2.898666318 [61.34 x10-3

T2+( 15.72 x10-5

T22)/2+(-8.749 x10

-8T2

3)/3 + (19.83 x10

-12T2

4)/4]

+ 38.58 + 2.898666318 [103.1 x10-3T2]= 48.848 kW

Stream 5


Flowrate mol /h 8788.793199 24839.42838 18984.93663 10435.19874Flowrate mol /s 2.441331444 6.899841217 5.273593508 2.898666318Stream 6


Flowrate mol /h 8788.793199 24839.42838 18984.93663 10435.19874Flowrate mol /s 2.441331444 6.899841217 5.273593508 2.898666318


HC2H6 49.373x10-

+13.92x10-

T-5.816 x10-

T +7.28 x10-

T

HC2H4 40.75x10-3

+11.47x10-5

T-6.891 x10-8

T2+17.66 x10

-12T

3

HC2H5OH(g) 61.34 x10-3

+ 15.72 x10-5

T -8.749 x10-8

T2

+19.83 x10-12

T3

HC2H5OH(l) 103.1 x10-3

H H2O (g) 33.46x10-3

+0.6880x10-5

T+0.7604 x10-8

T2-3.593 x10

-12T

3

H H2O (l) 75.4 x10-

Heat Exchanger Stream 6

T=250C, P=30barT=50C, P=30bar

Stream 5


40/44



H 2 = 6.899841217 [40.75x10-3

T3 +(11.47x10-5

T32)/2-(6.891 x10

-8T3

3)/3+(17.66 x10

-12T3

4)/4]

= 70.843 kW

H 3 = 2.441331444 [49.37x10-3

T3+(13.92x10-5

T32)/2-(5.816 x10

-8T3

3)/3+(7.28 x10

-12T3

4)/4]

= 30.531 kW

H4 = 5.273593508 [33.46x10-3T4 +(0.6880x10-5T4 2)/2+(0.7604 x10-8T4 3)/3-(3.593 x10-12T4 4)/4] +

40.656 + 5.273593508 [103.1 x10-3T5]= 94.76 kW

Q = -(H 1 +H 2 +H 3 + H4 )= -(48.848 + 70.843 + 30.531 + 94.76 )

= - 244.982 kW

Compressor Analysis

Stream 19

TotalCompressor

Stream 20Total

= 9.225849542 mol/s

T=50C,

P=5bar

T=250C,P=40bar


41/44



= 9.225849542 mol/s


Assume Ek,Ep , Q= 0

Ws =12,500 (kmol/min)[ ] ---(1)

[ ] [ ]---(2)

Using (1),

Ws =12,500 x 9.225849542 x [(40/5)0.286

-1]

=93702.788 W = 93.702 kW

Tout=[(40/5)0.286

]x (50+273)

= 585.4 K=312.4 C

Condenser Analysis

Condenser

T=50C, P=1bar

P=1bar

Tdew = 78.68C

=

11

12

13


42/44



Q= heat of vaporisation of water

Temperature change between bubble point and dew point of ethanol is found to be 0.11, which is relatively

small. Hence, we assume enthalpy change of ethanol to be zero.

Energy Equation : Ek+ Ep + H = QWsAssume Ek,Ep , Ws = 0

Stream 10 12 13

Component H2O C2H5OH H2O C2H5OH H2O C2H5OHFlowrate kg/h 502441.9179 6126079.427 463792.5396 5654843 606918 19234.2Flowrate kg /s 139.5671994 1701.68873 128.831261 1570.789597 10.73593842 130.899133Flowrate mol/s 7745.13 36929 7149.35 34088.3 595.779 2840.69

(kJ/kmol) Table B.1HC2H5OH 38.58

H H2O 40.656

Q= H

niHi= -[38.58(36929) + 40.656(7745.13)]= -1739606.825 W

= -1739.607 kW


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Reboiler

Stream 16 17

Component H2O C2H5OH H2O C2H5OH

Flowrate kg/h 303459 9617.08 303459 9617.08Flowrate kg /s 84.2942 2.67141 84.2942 2.67141Flowrate mol/s 4677.81 57.9733 4677.81 57.9733

Energy Equation : Ek+ Ep + H = QWsAssume Ek,Ep , Ws = 0

Heat of vaporisation of water at 100C and 1atm is found to be

(kJ/kmol) Table B.1H H2O 40.656

Q= H niHi= (4677.81+ 57.9733) x 40.656= 192538.006 W = 192.538 kW

Reboiler

Stream 17

Stream 16

T=100C, P=1bar

T=100C, P=1bar


44/44


References

http://www.sciencedaily.com/articles/e/ethanol_fuel.htm3 31/5

http://www.agar.com.au/site/files/ul/data_text01/1804309.pdf 3-8 31/5

http://cbau2058.wordpress.com/declaration-of-originality-and-disclaimer/4 6/7

Journal of chemical and engineering data, Vol. 20, No.1, 1975, pg 104, Table IV 5

Lecture Notes (7)

http://www.chemguide.co.uk/physical/equilibria/ethanol.html(1) 9/7

http://www.britannica.com/EBchecked/topic/194354/ethyl-alcohol2 11/7
http://www.sciencedaily.com/articles/e/ethanol_fuel.htmhttp://www.sciencedaily.com/articles/e/ethanol_fuel.htmhttp://www.agar.com.au/site/files/ul/data_text01/1804309.pdf%20%093-8%2031/5http://www.agar.com.au/site/files/ul/data_text01/1804309.pdf%20%093-8%2031/5http://cbau2058.wordpress.com/declaration-of-originality-and-disclaimer/http://cbau2058.wordpress.com/declaration-of-originality-and-disclaimer/http://www.chemguide.co.uk/physical/equilibria/ethanol.htmlhttp://www.chemguide.co.uk/physical/equilibria/ethanol.htmlhttp://www.britannica.com/EBchecked/topic/194354/ethyl-alcoholhttp://www.britannica.com/EBchecked/topic/194354/ethyl-alcoholhttp://www.britannica.com/EBchecked/topic/194354/ethyl-alcoholhttp://www.chemguide.co.uk/physical/equilibria/ethanol.htmlhttp://cbau2058.wordpress.com/declaration-of-originality-and-disclaimer/http://www.agar.com.au/site/files/ul/data_text01/1804309.pdf%20%093-8%2031/5http://www.sciencedaily.com/articles/e/ethanol_fuel.htm

Intro to Process Simulation

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