Process Engineering Project Design Product 2009 Ammonia Plant

Discussion

The point of the Process engineering project is to design an Ammonia Plant which will produce 2300

tonne/day of ammonia starting from a feed of Carbon Dioxide 14% and Methane 86% and a feed of

steam at a rate of 3.5Kmols per Kmol of carbon. The gasses firstly go through 3 reactors and undergo

2 reactions which are reform and a shift. In the first to reactors it undergoes both reactions, in the

third only the shift reaction is undertaken. The reactions within the reactors are as follows:

CH 4+H 2O→CO+3H 2 ReformReaction ΔHf=2.06E5 kJKmol

of CH 4

CO+H 2O→CO 2+H 2Shift Reaction ΔHf=−4.12E4 kJKmol

of CO

The reforming Reaction does not reach equilibrium in both reactors, so as to get an accurate Kp

value from Kp tables a value called approach temperature is used, this is value that is added or

subtracted from the temperature of the reactor, you then find the Kp for that new temperature. The

shift Reaction reaches equilibrium in the first and the second Reactor, but not in the shift reactor.

Reactor 1 Temp= 835’C Reforming Approach Temperature=+31’C

Reactor 2 Temp 950’C Reforming Approach Temperature=+39’C

Shift Reactor Temp 950’C Shift Reaction Approach Temperature=-31’C

After this reactor Air is added at 120 Kmol/day (78%Nitrogen, 21%Oxygen and 1%Argon), this means

there are some inert's entering the Reactor 2 and onwards.

Using the excel Solver values were found for CO and CO2 values for both Reformer reactors.

Although results were found, the excel solver is a difficult program to use, it is hard to know whether

the values given are the correct ones, many iterations are needed to find the correct values.

Reformer 1 CO= 32.7 CO2=39.8 Kmols/ Day

Reformer 2 CO=52 CO2=42.58 Kmols/Day

As can be seen the values from the Reformer2 are greater than that from Reformer 1, this is to do

with the CO and CO2 from the reactor one adding to the newly formed CO and CO2. In the second

Reformer CO increases far more then CO2, this due to the fact that reaction which generated CO is

endothermic and as Reformer 2 has a greater temperature 950’c as opposed to reformer 1’s 835’C

caused by the oxygen brought in with the air. through le chatliers principle the system will go against

the condition, thus try decreasing the temperature which in turn will induce the CO generating

reaction therefore producing more CO then that of CO2. From mass balances the values for the

other parts of the streams can be generated this is shown in appendix, the inert gas that enter

reformer 2 do not change, i.e. N2 and Ar. For reformer 2 one of the assumptions is that the oxygen

that enters via the air stream undergoes a reaction which the Kp and reaction kinetics are ignored,

this assumption means that the total reactions undergoing in Reformer 2 are not known, although

this makes the calculations easier for a real plant this would not be acceptable, as it causes a issue

when it comes down to total understanding of the system, if this was a real plant the Kp and kinetics

of the reaction of the oxygen with the CH4 should be looked into greater depths.

For the shift reactor the excel solver is not needed, instead simultaneous equations are generated

from the mass balances and the fact that the shift reactor reduces the CO to 0.95% dry (no water).

These equations are used to find all the mass balances within the system, thus find out all products

made and reactants consumed; this is shown in Appendix 1. For the shift Reactor CH4, N2 and Ar are

constant. The CO is reduced to 4.64 Kmols/Day, CO2 increased to 90Kmols/day. A massive increase

in CO2 caused by the fact the reactor is at a of 950’C a high temperature inducing a high yield of the

endothermic reaction, and as the shift reaction is the only one occurring there is no reform reaction

to reduce the efficiency of the shift reaction.

After the shift reaction water is reduced in the gas stream by cooling to 53’C at 39 bar as shown on

the flow sheet. The flow leaving the water extractor has a very small water content due to it being

100 saturated, to work out the value of water still within the flow using steam tables a saturated

vapour pressure of water was used to find the mol fraction, this is done as the saturated vapour

pressure divided by the total pressure is equal to the mole fraction of water. The saturated vapour

pressure found=0.1575 bars. The H2O still in the flow=1.955Kmols/day.

After that using a CO2 absorption column which used the down flow of liquid MEA

monoethanolamine which is a very good solvent for CO2, the CO2 flow is decreased to 0.06% dry

basis. Reducing the CO2 flow to 0.239Kmol/day. An assumption made here is that only CO2 leaves

the MEA column, in actual terms this would be incorrect, some of the other components would be

carried and taken out via the downward stream, again if this was a real reactor the detail of what

components are lost via the MEA should be looked into greater detail.

After this the flow enters the methanator, this is a device used to remove all the CO2 and CO via

reacting it with Hydrogen to form Methane, all the Carbon Oxides are converted into Methane using

the following reactions:

CO+3H 2→CH 4+H 2O∆Hf=−2.06E5 kJKmol

CO

CO2+4H 2→CH 4+2H 2O∆Hf=−1.65E5 kJKmol

CO2

Both of these reactions are assumed to meet completion all CO and CO2 are removed, the hydrogen

percentage is decreased and methane and water percentages are increased. The reaction reaching

completion is assumed to make the calculations easier however it is very unlikely that a reaction will

ever reach completion in a real reactor there would be CO and CO2 left in the output stream. This

assumption would not be made in real conditions as the CO and CO2 that don’t react will be inerts

and the following reactions may affect and slowed down the following system, this definitely would

be taken into account for the recycle loop. A system should be put in place in this position in the

system to purge the excess reactants. After the methanator water is again removed however there is

a different saturated vapour pressure for water as there are different conditions 39’C and 36bar.

Saturated Vapour pressure=0.07bar. Nothing else is affected by the water that is removed, the water

left in the flow after the absorber is .744Kmols/day. After this the flow is compressed to 145 bar and

temperature is decreased to 17.5’C, at this point it is mixed with an ammonia rich recycle path.

The reactants enter the ammonia recycle loop within the loop the inerts CH4 and Ar remain at a

constant % of flow of 17.5, the components are compressed, then cooled to a temperature of -28’C.

This means all the water in the system freezes, for this particular system it is assumed that it is fine

to leave the ice coating the pipe walls, this would not assumed in a real plant for as long as the

recycle loop runs more water will run in and freeze until a point where the pipe is completely

blocked by ice which would stop the hole process, which would stop reacting and economically

would not be accepted. What would be recommended is before the cooling remove all water via an

absorption column. After cooled the flow enters an Ammonia separator, this is used to collect the

product, the ammonia will be liquid at -28’C whereas the other components will be at gas phase thus

an easy separation. This flow for the project needs to be 2300Tonnes/day, the value generated from

this system is 160.04Kmol/day, A Multiplication factor is generated in the appendix to up all flows to

generate enough to produce 2300Tonnes/day of ammonia.

Once the flow has left the absorption column it is heated to 160’C before entering the Ammonia

reactor, after the reactor the temperature is 290’C, it is then cooled, once cooled a some of the flow

is purged, this is done to keep the inert's at the constant percent 17.5%. The purge is a flow divider;

it does not separate different species. After the purge the flow joins stream 14, the input from the

methanator, this stream, stream 20 is the recycle loop feed.

Conclusion

For there is no experimental data to conclude, this part consists of things which should be done

differently within the system.

Before Ammonia Loop

Excel Solver is a difficult programme to use to increase accuracy of results another

programme should be used

The assumptions made over reactor 2, although make calculations easier, leaves quite a

large unknown process which would generate a greater number of inerts, as this system

ends in recycle loop greater detail is needed to look into purging so the inerts don’t build up.

The assumption made that only CO2 leaves in the CO2 absorption column, would also affect

the general reactor so would need to be looked into in greater detail

The methanator would not reach completion in real life, thus the flow output would be

different to the values generated, there would be some CO and CO2 which would add to the

inerts and generate issues as explained above.

The Methane generated from the methanator could be recycled and used as the input for

flow 1, this would increase yield and decrease costs, once the system had been built.

Ammonia Synthesis Loop

The assumption that the water freezes and is left is not chemically or economically viabale, if

the pipe were to freeze solid it could cause massive amount of damaged which would mean

great cost to replace as well as the money lost via the reaction not being undertaken, the

water should be purged.

The multiplication factor to convert 160.04Kmol/day to a Kmol/day value which would equal

2300Tonnes/day is 35.22.