Problem 6: The specific gravity (SG) of a liquid mixture of benzene (CsHs) and t
ID: 700886 • Letter: P
Question
Problem 6: The specific gravity (SG) of a liquid mixture of benzene (CsHs) and toluene (C,Hg) is 0.874. The composition (by mass) is 35.0 wt% benzene and 65.0 wt% toluene. Determine the respective masses, (in units of lbs) of toluene and benzene present in a 1250 L amount. (i.e. report mass in lbs for each component) what is the composition of benzene and toluene (in mol%) for this material? Using the mass composition and SG provided, determine the mass flow rate (in kg/s) of a stream which has a volumetric flow rate of 1250 L/h. Determine the stream velocity (m/s) if you pump a volumetric flow rate of 1250 L/h of this liquid through a 2 inch Schedule 40 pipe. Hint - Standard pipe sizes are often reported in this way. The 2 inches is a "nominal diameter" and the schedule indicates the wall thickness. 2 inches is not the actual diameter of the pipe that is needed to correctly solve this problem - do a little web research to find actual the diameter for this type of pipe (inner diameter) to solve the problem. a. b. c. d.Explanation / Answer
Part a
Volume of mixture = 1250 L
Specific gravity of mixture = 0.874
Density of mixture = (0.874 x 1000 kg/m3) x (1m3/1000L)
= (0.874 kg/L) x (1lb/0.454kg)
= 1.925 lb/L
Mass of mixture = volume x density
= 1250 L x 1.925 lb/L
= 2406.38 lb
Mass of benzene = 0.35 x 2406.38 = 842.24 lb
Mass of toluene = 0.65 x 2406.38 = 1564.14 lb
Part b
Moles of benzene = mass /molecular weight
= 842.24 lb / 78.11lb/mol
= 10.78 mol
Moles of toluene = mass /molecular weight
= 1564.14 lb / 92.14 lb/mol
= 16.98 mol
Total Moles = 27.76 mol
Mol% of benzene = moles of benzene x 100/total moles
= 10.78*100/27.76 = 38.83%
Mol% of toluene = moles of toluene x 100/total moles
= 16.98*100/27.76 = 61.17%
Part C
Volumetric flow rate of mixture = 1250 L/h
Density of mixture = (0.874 x 1000 kg/m3) x (1m3/1000L)
= 0.874 kg/L
Mass flow rate of stream = volumetric flow x density
= 1250 L/h x 0.874 kg/L x 1h/3600s
= 0.3034 kg/s
Part d
Volumetric flow rate of mixture
= (1250 L/h) x (1m3/1000L) x (1h/3600s)
= 0.000347 m3/s
Inside diameter of 2 inch schedule 40 pipe
= 2.067 inch x 0.0254 m/inch = 0.0525 m
Inside Area = (3.14/4) x (0.0525 m)^2 = 0.0412 m2
Stream velocity = Volumetric flow/Inside area
= (0.000347 m3/s) / (0.0412 m2)
= 0.008422 m/s