Number:
Name:
14-10-2016
CHE 205 Materials and Energy Balances
Homework I
1. A process for the production of flake NaOH is given by the following representation
of the flowchart.
Fresh feed to the process is 2000 kg/h with 30% aqueous NaOH solution. Fresh feed is
combined with the recycled filtrate to produce 50% NaOH solution which is then fed to
the filter. The filter produces a filter cake that is composed of 95% NaOH crystals and
5% solution that itself consists of 45% NaOH. The filtrate on the recycle stream contains
45% NaOH. Determine the flow rate of H2O removed by the evaporator and the recycle
rate.
2. 3500 kilograms per hour of a Na2CO3 solution (30 wt%) and a recycle stream are fed
into an evaporator. The concentrated stream leaving the evaporator contains 51%
Na2CO3, this stream is fed into a crystallizer in which it is cooled and then filtered. The
filter cake consists of Na2CO3 crystals and a solution that contains 37% Na2CO3; the
crystals account for 90% of the total mass of the filter cake. The filtrate on the recycle
stream contains 37% Na2CO3. Draw and label the process flowchart. Calculate the rate
of evaporation of water, the rate of production of crystalline Na2CO3 and the recycle
ratio (mass of recycle/mass of fresh feed).
3. Calculate the compositions of Stream 3, 5 and 6 according to the flowchart given
below.
4. At a factory producing concentrated syrup, water is evaporated using the process
below. The fresh feed is comprised of 90.0% (wt) water and 10.0% (wt) solid
particles. A portion of the fresh feed bypasses the evaporator as shown in the figure.
This stream is mixed with concentrated syrup and final product is obtained. The water
content of the final product is 60.0% (wt). In the evaporator, 92.0% (wt) of the water
fed is evaporated. Calculate
a) Amount of the final product
b) Amount of water evaporated
c) Bypass rate
SOLUTIONS
1.
Overall balance
݉ሶ ଵ = ݉ሶ ଶ + ݉ሶ ସ
Overall NaOH balance
(2000)(0.3) = ݉ሶ ସ (0.95) + ݉ସሶ (0.05)(0.45)
݉ሶ ସ = 617 ݇݃/ℎ
Overall water balance
(2000)(0.7) = ݉ሶ ଶ + ݉ସሶ (0.05)(0.55)
(2000)(0.7) = ݉ሶ ଶ + 617(0.05)(0.55)
݉ሶ ଶ = 1383 ݇݃/ℎ
NaOH balance on the filter
݉ሶ ଷ (0.5) = ݉ሶ ସ (0.95) + ݉ସሶ (0.05)(0.45) + ݉ହሶ (0.45)
݉ሶ ଷ (0.5) = 600 + ݉ହሶ (0.45)
Water balance on the filter
݉ሶ ଷ (0.5) = ݉ସሶ (0.05)(0.55) + ݉ହሶ (0.55)
݉ሶ ଷ (0.5) = 17 + ݉ହሶ (0.55)
݉ହሶ = 5830 ݇݃/ℎ
2.
݉ሶ ସ = 0.9 (݉ሶ ସ + ݉ሶ ହ )
݉ሶ ହ = 0.111 ݉ሶ ସ
Overall N (Na2CO3) balance
(3500)(0.3) = ݉ሶ ସ + (0.37) ݉ହሶ
݉ሶ ସ = 1008.6 ݇݃/ℎ
݉ሶ ହ = 111.95 ݇݃/ℎ
Overall total mass balance
(3500) = ݉ሶ ଶ + ݉ሶ ସ + ݉ሶ ହ
(3500) = ݉ሶ ଶ + (1008.6 + 111.95)
݉ሶ ଶ = 2379.45 ݇݃/ℎ
Mass balance around filter
݉ሶ ଷ = ݉ሶ ସ + ݉ሶ ହ + ݉ሶ ଺
݉ሶ ଷ = 1008.6 + 111.95 + ݉ሶ ଺
݉ሶ ଷ = 1120.55 ݇݃/ℎ + ݉ሶ ଺
Water balance around filter
݉ሶ ଷ (0.49) = ݉ሶ ହ (0.63) + ݉ሶ ଺ (0.63)
݉ሶ ଷ (0.49) = 70.53 ݇݃/ℎ + ݉ሶ ଺ (0.63)
݉ሶ ଷ = 4536.21 ݇݃/ℎ
݉ሶ ଺ = 3416.22 ݇݃/ℎ
݉ሶ ଺ ݇݃ ‫݈݁ܿݕܿ݁ݎ‬/ℎ
3416.22
=
= 0.976
3500 ݇݃ ݂‫ݏ݁ݎ‬ℎ ݂݁݁݀/ℎ
3500
Mass balance around recycle-fresh feed mixing point
3500 ݇݃/ℎ + ݉ሶ ଺ = ݉ሶ ଵ
݉ሶ ଵ = 6916.22 ݇݃/ℎ
Check: A mass balance around evaporator
݉ሶ ଵ = ݉ሶ ଶ + ݉ሶ ଷ
݉ሶ ଵ = (2379.45 + 4536.21)݇݃/ℎ
݉ሶ ଵ = 6915.66 ݇݃/ℎ
3.
య
య
௠ ଺଴௠௜௡
௠
SG=0.75 ρ=0.75*1000=750 kg/ m3 ܸሶ = 0,0111 ௠௜௡ ଵ௛ =0.666 ௛
m1= ρ*ܸሶ =750*0.666=499.5 kg/h
m1 =499.5 kg/h; 0.5 kgC/kg, 0.5 kg D/kg
m2 =200 kg/h; 0.9 kgC/kg, 0.1 kg D/kg
m1 = m2+ m3 499.5 =200 + m3 m3 =299.5 kg/h
C balance around Unit 1:
499.5*0.5 =200*0.9+299.5 (wC3) wC3=0.233 kg C/kg ; wD3=0.767 kg C/kg
Mass balance around mixing point:
m5 = m4+ m3 = 299.5 +150 =449.5 kg/h
C balance:
299.5*0.233 + 150*0.3 = 449.5 wC5 wC5 = 0.255kg C/kg
Mass balance around Unit 2:
m5= m6+ m7 449.5 =150 + m6 m6 =299.5 kg/h
C balance:
449.5*0.255 =299.5* wC6 +150*0.6 wC6 =0.082
4.
Basis: 100 kg/min
Overall solid balance: 10=m4(1-0,6) m4 =25 kg/min
Overall mass balance: m1 = m4+ m6 100 -25 =75 kg/min= m6
Water entering the evaporator: 75/0.92=81.52 kg/min
Solid entering the evaporator: 81.52 (10/90)= 9,06 kg/min
m2 = 81.52+9.06=90,58 kg/min
Bypass amount: 100-90,58=9,42 kg/min