1320 mol per second of a liquid mix containing 50.0 mol % benzene and the balanc
ID: 483702 • Letter: 1
Question
1320 mol per second of a liquid mix containing 50.0 mol % benzene and the balance toluene placed in an evacuated sealed container initially at 25 degrees C. It's then heated to 95 degrees C. At this temp an analysis of the container contents reveals the liquid to contain 42.5 mol% benzene and the vapor to contain 73.5 mol% benzene. Calculate heating requirements for this process. 1320 mol per second of a liquid mix containing 50.0 mol % benzene and the balance toluene placed in an evacuated sealed container initially at 25 degrees C. It's then heated to 95 degrees C. At this temp an analysis of the container contents reveals the liquid to contain 42.5 mol% benzene and the vapor to contain 73.5 mol% benzene. Calculate heating requirements for this process.Explanation / Answer
Writing the overall mass balance equation
F(feed)= V(Vapor )+ L(Liquid)
1320 = V+L, V= 1320-L (1)
Writing Benzene balance
1320*0.5 = V*0.735+L*0.425
From Eq.1, 660 = (1320-L)*0.735+L*0.425, 660 = 1320*0.735-0.735L+0.425L=970.2-0.31L
0.31L= 970.2-660, L=997 and V= 1320-997=323 mol/sec
The vapor contains 0.735 mole fraction and 1-0.735= 0.265 mole fraction toluene
Vapor contains 323*0.735 moles/sec Benzene . i.e 237.4 mole/sec benzene 323-237.4 moles/sec toluene ,i.e 85.6 mole/sec toluene
Liquid contains 997*0.425 =423.725 moles/sec and 997-423.725 = 573.275 moles/sec toluene
The liquid gains sensible heat from 25 deg.c to 95 deg.c
Sensible heat = moles/sec* specific heat* temperature difference
Average Specific heats : Benzene =135.69 j/mole.K and toluene =156 J/mole.K
Sensible heat of benzene = 660*135.69*(95-25)= 6268878 J/sec
Sensible heat of toluene = 660*156*(95-25)= 7207200 J/sec
Total sensible heat = 6268878+7207200=13476078 J/sec= 13476 Kj/sec
At 95 deg.c, 237.4 mole/sec benzene and 85.6 moles/sec toluene needs to be vaporized by supplying latent heat
Latent heat : Benzene :34.4 Kj/mole and toluene 38 Kj/mole
Heat to be supplied : flow rate* latent heat
Heat to be supplied to benzene : 34.4*237.4 Kj/sec=8167 Kj/sec
And for toluene = 38*85.6 Kj/sec=3253 Kj/sec
Total heat to be supplied = 13476+8167+3253= 24896 Kj/sec
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