CH_3OH is synthesized from CO and H_2 in a continuous vapor-phase reactor at 3.0
ID: 1059645 • Letter: C
Question
CH_3OH is synthesized from CO and H_2 in a continuous vapor-phase reactor at 3.00 atm(abs). The feed contains CO and H_2 in stoichiometric proportions and enters the reactor at 25 degree C and 3.00 atm(abs) at a rate of 17.2 m^3/h, measured at the stream T and P. The product stream emerges from the reactor at 127 degree C. The rate of heat transfer from the reactor is 14.2 kW. Calculate the fractional conversion achieved and the volumetric flowrate (m3/h, at the stream T and P) of the product stream The following data is available:Explanation / Answer
Since the feed is at 25 deg.c which is also the reference temperature, the enthalpy of reactants =0
The reaction is CO+2H2-àCH3OH
Moles of feed , n= PV/RT = 3*17.2*1000L/hr./(25+273.15)*0.0821 moles/hr=2108 moles/hr
So moles of CO= moles of H2= 2108/2= 1054 moles/hr
Let x= moles of CO converted
Moles of CH3OH formed = x moles
Products contains
CO= 1054-x, H2= 1054-3x and CH3OH= x
deltaHf for the reaction = enthalpy of products- enthalpy of reactants =
-201.2*103*x -(1054-x)*(-110.52*103 Joules/hr =
-201.2*1000*x -{-1054*110.52*1000+110.52*1000x }
-201.2*1000x+1054*110.52*1000 -110.52*1000x =( 1.164*108-311.72*1000x) Joules/hr
=(1.164*105- 311.72x) Kj/hr =(1.164*105- 311.72x)/3600 Kj/sec = (32.33-0.087x)Kw
Enthalpy of products =
x*49.07*+ (1054-3x)*28.87 +(1054-x)*29.29{127-25)
={x*49.97+30429-86.61x +30872-29.29x}102
=(61301-65.93x)*102 = 6252702- 6725x J/hr = ( 6252702- 6725x)/1000 = (6252.702- 6.725x) Kj/hr
=( 1.74- 0.001868x ) Kw
Heat transferred = 14.2 Kw= enthalpy of products + heat of reaction- enthalpy of reactants
14.2 = (1.74-0.001868x)+(32.33-0.087x)
Hence x*(0.001868+0.087)= 19.87
Hence x = 223.59 moles
Fractional conversion = 223.59/1054 =0.212
Products : CO= 1054-223.59 =830.41 moles. H2= 1054-3*223.59 = 383.23 moles and CH3OH= 223.59 moles
Total moles = 830.41+383.23 +223.59 =1437.23 moles/hr
Volumetric flow rate, V= nRT/P= 1437.23*0.0821*(127+273.15)/3 =15739 L/hr =15.739m3/hr