For the butane to be separated from the other waste gases, it will have to be liquefied. This will mean it will have to be cooled to below 26 ºC. This would then mean it would then require vaporising before feeding back to the reactor inlet.
Heaters and Compressors
Since the reactor is operating at 2.5 bara, the air taken from the atmosphere at 1 bara will need to be compressed, which would require the need for a compressor. The recycled butane stream would require heating and vaporising as discussed above, which means a heater would also be needed. The modified process flow diagram is shown in Figure 3.
Figure 3: Process Flow sheet
Compressor Calculations
The compressor increases the pressure of the air stream from atmospheric (1 bara) to the reactor operating pressure 2.5 bara. The compressor is assumed to have an isentropic efficiency of 0.75, which is a typical value for compressors. The calculations for the energy requirements for the compressor can be found in Appendix C. The calculations do show that in compressing the air, it reaches a temperature of 137 ºC, which is hot enough to enter the reactor directly without being heated further.
Heater Calculations
The purpose of the heater is to vaporise and heat the butane fed to the reactor. The temperature in the reactor must be kept above 100 ºC to prevent condensation on the catalyst. The heater will use steam to vaporise and heat the butane to just over 100 ºC so that it can be fed to the reactor. The calculations for the heater energy requirements can be found in Appendix D
Cooler Calculations
The product stream leaving the reactor has a temperature of around 400 ºC, which is far too hot for a separation to take place. By cooling the stream to around 250 ºC, the excess heat can be used to generate steam, which can be sold. The calculations can be found in Appendix D.
Process Economics
Raw Materials and Product Value
The butane feed costs $124 per tonne, where as the MA produced has a value of $1000 per tonne. These costs are assumed to be in US dollars, so all the other costs are going to be converted in to US dollars so that a comparison can be made. For an annual production of 30,000 te/yr, the annual income from the sale of the product is therefore $30,000,000.
Since the required feed rate of butane depends on the conversion, the annual cost of raw material will also depend on the conversion. These costs are summarised in Table 3 below.
Table 3: Costs of butane feed
Utility Costs
Cooling Water
Cooling water will be used to cool the product stream down from 400 ºC to 250 ºC. Water may also be required for the reactor, to keep its temperature constant. The cost of the cooling water is 5 pence per tonne, which is equivalent to $0.079 after multiplying by an exchange rate of 1.58 $US/GBP.
A sample calculation finding the required flowrate of cooling water can be found in Appendix E, with summaries shown in Tables 4 and 5.
Water requirements for Cooler
Table 4: Cooling costs for cooler
Water requirements for Reactor
Table 5: Cooling costs for reactor
Note that for the low conversions, the reactor requires net heating to maintain a constant temperature. The heating would be provided by 10 bar g steam, which is available at £6.50 per tonne. The costs for the relevant conversions have therefore been calculated using this figure.
Electricity Costs
The simplest method of driving the compressor is using an electric motor. The motor is assumed to have a typical efficiency of 80%. Electricity costs 5p per kWh, giving the annual electricity costs shown in Table 6.
Table 6: Electricity costs for the compressor
As the table shows, electricity is probably the most expensive utility, and it is highly likely an on-site power generation unit should be used to keep the costs low. The costs are also very high at low conversions because the air flowrate is so large.
Steam Costs
The steam is used for the heater, in vaporising the butane feed and raising its temperature to 100 ºC. Steam is available at 10 bar g at a cost of £6.50 per tonne. The calculations showing how the annual cost of steam is calculated are shown in Appendix E. The results are summarised in Table 7.
Table 7: Costs of steam required for heating
Pumping Costs
The energy requirements for the pump are assumed to be negligible compared to other costs. They are also difficult to calculate since the pressure of n-butane stream as it leaves the separation system is unknown.
Overall Economic Potential
Summing all the values for the various components gives the following values
Table 8: Overall Economic Potential
The following chart is produced.
An inspection of the chart and figures suggests that there is no real optimum conversion where the economical gain is maximised. The general trend suggests that the profit increases as the conversion increases. A conversion of 1 is not really possible, so a comprise value of 0.9 is chosen. The process flow diagram on the following page shows the process for a conversion of 0.9.
Discussion
Since the conditions in the separation unit are unknown, the temperatures and pressures of the various product and output streams are also unknown. This is why some of the boxes are marked with “?” on the PFD. More information would be needed about the separation, so that energy requirements can be found and the costs estimated. Once the pressure drop across the separator is known, the pumping requirements for the butane recycle can also be found. Although the separation was assumed to be perfect, so that mass balances could be made, this may in fact not be the case. Again, more information about how the separation system behaves would help obtain a more accurate estimate for the process economics.
Capital costs have also been ignored. These may be significant when considering the compressor, and possibly the catalyst. Other costs not considered include labour costs, losses through shut down, heating, lighting etc.
Conclusions
Though an approximate value for the annual profit for the process was obtained, it is probably not that accurate, since there is no data available about the separation.
Data Sources
Thermodynamic data for the components were obtained from Hysys, using the UNIQUAC fluids package.
The heats of formation for calculating the enthalpy of combustion were obtained from “Physical Chemistry” by Atkins.
The current exchange rate between US$ and GB£ was obtained from an online currency converter,
Appendix A - Balance across reactor
Production of MA
All separations will be assumed to be perfect. This means the required feed to the reactor can be found as follows:
The requirement is to produce 30,000 te/yr.
Molecular Weight of MA =
Required flowrate of MA = kmol/yr
= kmol/hr
This is assuming there are 8760 working hours per year.
Reactor Inlet
For a given conversion, the required flowrate of n-butane in to the reactor, FR, can be found, using the information in the above diagram.
The calculations will be worked through using a conversion 0.5, to demonstrate the typical values obtained.
For X = 0.5:
S = 0.70 – 0.3 x 0.52 = 0.625
The required butane feed to the reactor is therefore
kmol/hr
The amount of n-butane left unreacted at the reactor exit is calculated using
For X = 0.5
kmol/hr
All the unreacted n-butane will be recycled to the reactor inlet, meaning that the make-up fresh butane feed can be found. For a recycle flowrate R, the fresh feed of butane, FF, is given by
The fresh butane feed required for a conversion of 0.5 is therefore
kmol/hr
5% of the liquid feedstock of butane is made up of iso-butane, which is taken to form carbon dioxide instantaneously under the reactor conditions. The flow of iso-butane in to the reactor is given by
Since the n-butane must be kept below its lower explosion limit of 1.7 mol%, this means there is a minimum flowrate of air given by:
For a feed to the reactor of 112 kmol/hr (corresponding to a conversion of 0.5), the minimum flowrate of air is
kmol/hr
The air is taken to be made up of 79% nitrogen and 21% oxygen only. The small amounts of carbon dioxide and other gases have been ignored. The flowrates and compositions of the flows at the reactor inlet are now defined. A summary for different conversions is shown below in Table 9.
Table 9: Feed to Reactor
Reactor Outlet
Since the flows entering the reactor are known, the composition of the stream exiting the reactor can be found.
The amount of CO2 produced in the reactor will come from two sources, the iso-butane, and a proportion of the n-butane. For every mole of butane that reacts, 4 moles of carbon dioxide will be produced. The amount of butane reacting to form CO2 will depend on the selectivity, and hence the conversion.
The rate of CO2 production will be given by
Nitrogen is an inert, which means the amount leaving the reactor will be the same as the amount entering the reactor, i.e.:
The amount of oxygen present in the outlet stream will be dependent on the amount used up in the reactions. For the desired main reaction, 3.5 moles of O2 are used up for every mole of butane. The side reaction uses up 6.5 moles of O2 per mole of butane burnt. The amount of O2 left in the outlet stream will be given by
Both the side and main reactions produce water, giving the total rate of water production as
The rate of MA production is as explained above,
kmol/hr
The amount of n-butane left in the reactor exit stream is
This now defines the component flowrates out of the reactor.
A summary for different conversions is shown in Table 10.
Table 10: Flowrates in the product stream
Appendix B - Reactor Cooling
The enthalpy on heating of the reactants is calculated by first calculating the average specific heat for the stream using the method shown above. The table below shows a summary of the calculations for a conversion of 0.5:
Table 11: Heat capacities for the reactant stream
Average Stream Cp = 32.59 kJ/kmol K
Energy required to heat stream: = kJ/hr = 18.67 MW
Enthalpy of reaction for combustion of butane:
Substituting in values gives an enthalpy of combustion of –2657 kJ/mol
Cooling Reactants
Since the enthalpy of combustion is valid at 298 K, energy will be produced by cooling the reactants to 298 K. This only applies to those reactants that will partake in the reaction.
The extent of the side reaction is given by
For a conversion of 0.5, the extent of reaction = 23.87 kmol/hr
Only the butane and the oxygen will react in the combustion, so that the value of Cp for these reactants is calculated to be 104 kJ/kmol K.
The enthalpy given out when these reactants are theoretically “cooled” to 298 K is calculated to be 930000 kJ/hr = 0.2583 MW.
Enthalpy of Reaction
For a conversion of 0.5, the enthalpy given out from the combustion reaction is calculated by: = 17.61 MW
Heating the Products
Since the products from the reaction are at 400 ºC, they will have to be heated from 25 ºC to 400 ºC.
The enthalpy for this heating process is calculated to be 4688000 kJ/hr = 1.302 MW.
Enthalpy of Main Reaction
The enthalpy change for the main reaction is –1230 kJ/mol = -1230000 kJ/kmol
The extent of this reaction is 34.95 kmol/hr, which gives an heat output from the main reaction of:
kJ/hr = 11.96 MW
The overall heat duty for the reactor is then the sum of these parts:
MW
The reactor would therefore require the removal of 9.86 MW of energy to maintain a constant temperature.
Table 3 below summarises the heating/cooling duty for the reactor, operating at different conversions.
Appendix C - Compressor Calculations
The air is compressed from 1 bar to 2.5 bar, giving a pressure ratio of 2.5. Assuming the inlet temperature of the air is 20 ºC (293 K). Assuming the compression of air is isentropic with a value of γ of 1.4, the ‘ideal’ temperature of the compressed gas is given by
Assuming of isentropic efficiency of 1.
Substituting in values gives:
The isentropic efficiency, η, is defined by:
Assuming the compressor is not ideal, and in fact has an isentropic efficiency of 0.75 (typical for a compressor) gives the real temperature of the compressed gas is
ºC
This shows that the temperature of the compressed is high enough to not require additional heating, and can be fed directly to the reactor.
Energy required by compressor
The rate of energy use for the compressor is given by
The Cp for air is 29.48 kJ/kmol K
For a conversion of 0.5, the required air flowrate in to the reactor is 6757 kmol/hr.
The duty of the compressor is therefore calculated to be
kJ/hr = 6.47 MW
The results for other conversions are shown in Table 12 below.
Table 12: Duty for the compressor
Appendix D - Heater Calculations
. The energy required by the heater can be calculated from a simple enthalpy difference as follows
The enthalpy of vaporisation for butane, ΔHvap, is 21114 kJ/kmol
The specific heat of butane is 111.5 kJ/kmol K
Assuming the temperature difference is approximately 80 ºC, the butane is being heated from 26 ºC, the boiling point of butane, to just over 100 ºC. The flowrate of butane in to the reactor, for a conversion of 0.5, is 112 kmol/hr. Substituting in values gives
kJ/hr = 0.934 MW
Cooler Calculations
To calculate an enthalpy change in the stream as it is cooled, an overall value for Cp for the stream has to be calculated. Table 13 shows the values for Cp for the individual components
Table 13: Heat Capacities for the product stream
For a conversion of 0.5, the flowrates of the components in the product stream is a shown in Table 14 below.
Table 14: Product stream composition for X=0.5
Since the flowrates for the product stream are known, the mole fractions for the stream can be calculated, and are shown in Table 15 below.
Table 15: Mole fractions in product stream for X =0.5
The overall stream heat capacity is calculated by taking a weighted average for the components:
The energy released as the stream is cooled is given by
Substituting in values gives
kJ/hr = –9.58 MW
Appendix E - Utility Costs
Cooling Water
Assuming the cooling water is fed to the process at a temperature of 15 ºC, the rate of cooling water, and hence the cost, can be estimated.
The water will be heated from 15 ºC to the boiling temperature of water, before being vaporised to form steam.
The energy required to heat and vaporise one kmol of water is given by:
The specific heat of water is 77.72 kJ/kmol K
The heat of vaporisation for water is 39668 kJ/kmol
Substituting values in to the above equation gives
kJ/kmol
The flowrate of cooling water can then be found by dividing the cooling duty by the energy required to vaporise one kmol of water.
For a cooling duty of -9.58 MW:
kmol/s
Converting this to an annual requirement of tonnes per year, assuming 8760 hours of operation per year, gives the following
te/yr
Cooling water costs 2.5 pence per tonne, which gives an annual cost of approximately £2940. This is equivalent to $4640 using an exchange rate of 1.58 US$/GBP.
The amount of cooling water required to cool the reactor is calculated using the same method.
Steam
Steam is used in the heater to vaporise and heat up the butane feed to the reactor. For a conversion of 0.5, this process has an energy requirement of 0.934 MW.
Assuming the sensible heat effects are negligible in the condensing steam, the energy will come from the enthalpy of condensation. For water, this has a value of 39670 kJ/kmol.
The required flowrate of stream for this duty is given by:
kmol/s
Converting this to an annual steam requirement for the heater gives:
te/yr
10 bar g steam costs £6.50 per tonne, which gives an annual cost of approximately £86,700, equivalent to $137,000 per year.
Electricity
For the case where the conversion is 0.5, the energy requirement for the compressor is 6.47 MW. Assuming a typical motor efficiency of 80 %, the actual electrical requirement is
The annual energy use (in kWh) is therefore:
/yr
At a cost of 5 pence per kWh, the annual cost of the electricity for the compressor is £3,542,000, which is equivalent to $5,597,000.