Question

Modified from Problem 3-33 in Elementary Principles of Chemical Processes, 4th Edition (Felder & Rousseau 2015):

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 0.64 containing 45 wt% water and 47 wt% (dry basis) cellulose are treated to produce 1400 tons/day of dry wood pulp containing 85 wt% cellulose.

You are asked to:

a) draw and label a flowchart for the process

b) estimate the feed rate of logs (logs/min), assuming that the logs have an average diameter of 8 inches and an average length of 7 feet.

Answer 1

Flow rate of cellullose= 1400 tons/day= 1400*1000 kg/day =1.4*10⁶ kg/day

it contains 85% cellulose, cellulose in the product= 1.4*106 kg/day*0.85=119000 kg/day

Cellulose doe not undego any change and is the tie substance

Let F= feed entering the process F*0.47= 119000 kg/day

F= 119000/0.47=2531915 kg/day

Specific gravity =0.64

Density of chips / density of water=0.64

Density of chips =0.64*1000= 640 kg/m3 ( Water density is 1000 kg/m3)

Volumetric flow rate of chips =2531915/640 m3/day=3956.117 m3/day= 3956.117 m3/(24*60)=2.747 m3/min

Volume of each chip = PI*r2*L where r= radius = 8/2 inches= 4 inches=4/12ft =0.3 ft =0.3*0.3048m (1ft=0.3048m)=0.09144m, L= 7ft= 7*0.3048 =2.1336m

Volume = (22/7)*(0.09144)2*2.1336=0.056067 ft3

number of logs/min= volumetric flow rate of logs/ volume of each log= 2.74/0.056067=48.99 logs/min