can you please upload with full calculation. Example 6 Purge Considerable intere

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can you please upload with full calculation.

Example 6 Purge Considerable interest exists in the conversion of coal into more convenient liquid products for subsequent production of chemicals. of that can be generated under suitable conditions from in situ (in the ground) coal combustion in the presence of steam (as occurs naturally in the presence of groundwater) are H2 After cleanup, these gases c be combined to yield methanol according to the following equation: CO 2H2 CH3OH Figure E6.9a illustrates a steady state, open process for the production of methanol. All of the compositions are in mole fractions or percent. The stream flows will be in moles. Overall System Boundary 67.3 H Mix orator CH30H Feed F 32.5 co Reactor 100% 0.2 CH Purge P Recycle R Split H2 y co CH4 Figure E6.9a (Continues)

Explanation / Answer

feed to the process : H2= 67.3 moles, CO=32.5 moles and CH4= 0.2

Under steady state, moles of CH4 in the feed= moles of CH4 in the purge

P= Purge

Writing overall CH4 balance

P*3.2/100= 0.2, P=0.2*100/3.2=6.25 moles

CO and H2 in the purge= 6.25*(1-3.2/100)= 6.05

The reaction is CO+2H2------àCH3OH

Molar ratio of H2:CO (Theoretical) = 2:1

Actual ratio of H2:CO= 67.3:32.5=2.1:1

Excess reactant is H2. Hence all the CO gets reacted.

total of CO and H2 entering =67.3+32.5= 99.8

and x= moles of CO the limiting reactant converted

2x moles of H2 get converted

Hence 3x moles of mixture give x moles of CH3OH

Hence

CO and H2 entering = moles of CO and H2 consumed+CO and H2 in the purge

99.8= 3x+6.05

Hence x=(99.8-6.05)/3= 31.25 moles

So CH3OH produced, E = 31.25

CO= 31.25 moles of CO and moles of H2= 62.50

Hence moles of CO in the purge= 32.5-31.25= 1.25 moles of CO and 67.3-62.5=4.8 moles of H2

Composition of purge :   CO= moles of CO/total moles of purge= 1.25/6.25= 0.2, H2= 4.8/6.25= 0.768

CH4=3.2/100 =0.032

This is also the composition of recycle

Writing balance across the reactor. Let R= recycle

CO entering the reactor = 32.5+R*0.2

CO reacted= (32.5+R*0.2)*0.18

For 1 mole of CO entering, 1 mole of CH3OH is formed

Hence (32.5+R*0.2)*0.18=31.25

32.5+R*0.2= 31.25/0.18= 173.6

R*0.2= 173.6-32.5

R*0.2= 141.1, R= 141.1/0.2= 705.5 moles

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