In petroleum refineries, a mixture of paraffins and cycloparaffins is commonly r
ID: 533753 • Letter: I
Question
In petroleum refineries, a mixture of paraffins and cycloparaffins is commonly reformed in a fixed-bed catalytic reactor to produce blending stocks for gasoline and aromatic precursors for making petrochemicals. A typical multicomponent product from catalytic reforming is a mixture of ethylbenzene with the three xylene isomers. If this mixture is separated, these four chemicals can then be subsequently processed to make styrene, phthalic anhydride, isophthalic acid, and terephthalic acid. Compute, using the following data, the minimum work of separation in Btuh for T_0 = 560 degree R if the mixture below is separated at 20 psia into three products.Explanation / Answer
The minimum work of separation can fe calculated by the following equation is as follows:
Wmin= (out) nb- (in) nb
Where n = flow rate of the component
b= stream availability and is given by:
b = H - TO. S
Where, H = enthalpy of the given component
To = coolant temperature= 560oR
S= entropy
For feed,
n =( 150+190+430+230)lbmol/h=1000 lbmol/h
b= (29290)-(560 × 15.32)=29290- 8579.2= 20710.8 Btu/lbmol
For product 1:
n= (0.96× 150) + ( 0.005×190) + (0.004×430) + (0.00× 230)
= 144+0.95+1.72+0=146.67 lbmol/h
b = 29750-(560×12.47)=29750-6983.2= 22766.8 Btu/lbmol
For product 2:
n = (0.04×150) + (0.99×190)+(0.99×430)+(0.015×230)=
= 6+ 188.1+425.7+ 3.45=623.25 lbmol/h
b= 29550- (560× 13.6)=29550-7616= 21934 Btu/lbmol
For product 3:
n = (0 × 150) + (0.005× 190) +( 0.006×430) + (0.985 ×230)
= 0+ 0.95 + 2.58 + 226.55= 230.08 lbmol/h
b= 28320 - ( 560× 14.68)=28320- 8220.8=20099.2 Btu/lbmol
Hence,
Wmin= [(146.67×22766.8) + (623.25× 21934) + (230.08×20099.2)]- (1000×20710.8)= (3339206.556 +13670365.5+ 4624423.936) - (20710800)
=21633995.99 -20710800= 923195.992 Btu/h.