CHEMICAL ENGINEERING PROBLEM In the manufacture of paper, logs are cut into smal
ID: 538147 • Letter: C
Question
CHEMICAL ENGINEERING PROBLEM
In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several of the chemical constituents of wood but not cellulose.
The slurry of undissolved chips in solution is further processed to recover most of the original solution constituents and dried wood pulp.
In one such process, wood chips with a specific gravity of 0.640 containing 45.0 wt% water are treated to produce 1600.0 tons/day of dry wood pulp containing 85.0 wt% cellulose.
The wood chips contain 47.0 wt% cellulose on a dry basis.
Estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 9.00 inches and an average length of 9.00 feet.
31,35 are incorrect
Explanation / Answer
production of dry wood pulp=1600 tons/day
it contains 85 wt% cellulose, hence cellulose in dry wood pulp=1600*0.85 tons/day=1360 tons/day
wood chips contains 45% cellulose. let F= Feed rate of cellulose
writing cellulose balance F*0.45=1360, F= 1360/0.45 tons/day=3022tons/day=3022*1000 kg/24 hr =125917 kg/hr
=125917/60 kg/min =2099 kg/min
dimensions of log : diameter(d)= 9inches (12inches= 1ft) = 9/12=0.75 ft , 1 ft=0.3048m, 0.75ft= 0.75*0.3048m=0.23 m
depth(h) = 9ft= 9*0.3048m= 2.74m
volume of each log= (PI/4)*d2*h= (22/28)*0.23*0.23*2.74 m3 =0.1138 m3
specific gravity of wood chips= 0.64 , density of wood chips= specific gravity*desnity = 0.64*1000 =6400kg/m3
mass of single wood chip=0.1138*6400 kg =728.32 kg
wood chips required= 2099 kg/min
no of logs required/min= 2099/728.32= 3