One of the major problems encountered during recycling is the gradual increase in the concentration of inert or impurities in the system. A stage may reach when the concentration of these components may cross permissible levels. By bleeding off a fraction of the recycle stream, this problem can be overcome. This operation is known as purging. This is quite common in the synthesis of ammonia and electrolytic refining of copper.
The above (8.1, 8.2, and 8.3) definitions have been shown in Figure 8.1.
Recycle and Bypass 8
WORKED EXAMPLES
8.1 A distillation column separates 10,000 kg/h of a 50% benzene and 50% toluene. The product recovered from the top contains 95%
benzene while the bottom product contains 96% toluene. The stream entering the condenser from the top of the column is 8,000 kg/h. A portion of the product is returned to the column as reflux and the remaining is withdrawn as top product. Find the ratio of the amount refluxed to the product taken out.
Mixing Feed unit
Fresh feed
Recycle
Purge
Process
Sepa-rator Gross product
Net product
Figure 8.1 A scheme indicating recycle, bypass and purge.
Overall balance F = D + W or, 10,000 = D + W Benzene balance gives, 5000 = 0.95D + 0.04W Solving, D = 5,050 kg/h W = 4,950 kg/h
Distillation column 10,000 kg/h ‘F’
50% B, 50% T
96% Toluene
Benzene 95% B Condenser
V
8,000 kg/h
R D
Figure 8.2
Bypass
W
Basis: One hour. KNO3 entering = 2,000 kg/h
\ Crystal leaving crystallizer, C = 2000 0.96
8.3 Metallic silver may be obtained from sulphide ores by roasting to sulphates and leaching with water and subsequently precipitating silver with copper. In the Figure 8.4 shown below, the material leaving the second separator was found to contain 90% silver and 10%
copper. What percentage excess copper was used? If the reaction goes to 75% completion based on the limiting agent Ag2SO4, what is the recycle stream in kg/tonne of product?
Balance around condenser gives, V = D + R
or, 8,000 = 5,050 + R
\ R = 2,950 kg/h Reflux ratio = Refluxed quantity
Actual product R
D = 0.584
8.2 What is the flow rate in recycle stream in Figure 8.3 shown below?
Figure 8.3
2 4 4
312 63.5 (2 107.9) 159.5
Ag SO + Cu 2Ag + CuSO
→ ×
Basis: 1 tonne of product = 1000 kg product (i.e.) 900 kg silver; 100 kg of copper
Ag2SO4 needed = 900 312
Total copper supplied = 265 + 100 = 365 kg
% Excess copper used = 100 ´ 100
265 = 37.7%
We know the reaction is 75% complete. 25% of the limiting reactant (Ag2SO4) was unconverted which goes in the recycle steam.
Ag2SO4 balance: F = R + 1,300 or, F = 1,733.3 kg 0.25F = R
or, 0.25 (R + 1,300) = R \ R = 433 kg
8.4 In the diagram shown in Figure 8.5, what fraction of dry air leaving is recycled?
Water removed by drying = (1.562 – 0.099) = 1.463 g Water removed/g of dry air = 0.0525 – 0.0152 = 0.0373 g
\ Dry air needed = 1.463
0.0373 = 39.22 g
Dry air passing through drier (e) = 52.5 g (between B and C)
\ Dry air recycled (x) = 52.5 – 39.22 = 13.28 g
\ Fraction recycled = 13.28 52.5 x
e = 0.253
8.5 What is recycle, feed and waste for the system shown in Figure 8.6?
Basis: 100 units of feed.
Material balance for A at = 20 + x = (100 + x)0.4 Solving, x = 33.3 units.
Figure 8.6
Feed 40% A 20% A;
80% B
Recycle A only
x
Aw waste
Product P A 5% and B 95%
Recycle 33.3
Feed 100 = 0.333
Material balance for A at = (100 + 33.3)0.4 = Aw + 33.3 + 0.05P Overall balance: 100 = Aw + P
Solving: Aw = 15.81 units Product, P = 84.19 units
8.6 Methanol is produced by the reaction of CO with H2 according to the equation CO + 2H2 ® CH3OH. Only 15% of the CO entering the reactor is converted to methanol. The methanol product is condensed and separated from the unreacted gases, which are recycled. The feed to the reactor contains 2 kmoles of H2 for every kmoles of CO. The fresh feed enters at 35 °C and 300 atm. To produce 6,600 kg/h of methanol calculate:
(a) Volume of fresh feed gas, and (b) The recycle ratio.
Basis: One hour of operation Methanol produced = 6,600
32 = 206.25 kmoles CO + 2H2 ® CH3OH
15% conversion of CO entering gives 206.25 kmoles of methanol
\ CO entering the reactor = 206.25
0.15 = 1,375 kmoles and H2 entering the reactor = 1,375 ´ 2 = 2,750 kmoles Total amount of CO and H2 = 4,125 kmoles CO unconverted (goes into recycle)
= CO entering – CO converted
= (1,375 – 206.25) = 1,168.75 kmoles H2 unconverted (goes into recycle) = (2,750 ´ 0.85) = 2,337.50 kmoles
\ Total moles of feed unconverted = 3,506.25 kmoles Fresh CO needed = 206.25 kmoles
Fresh H2 needed = 412.50 kmoles
\ Total moles of fresh feed = 618.75 kmoles
(a) Volume of feed gas = 618.75 ´ 22.414 ´ 308 1 273 300
¦ µ ¦t µ
§ ¶ § ¶
¨ · ¨ ·
= 52.16 m3
(b) Amount of gas leaving reactor = (3,506.25 + 206.25) = 3,712.50 kmoles Amount of gas recycled = 3,506.25 kmoles
\ Recycle ratio = 3,506.25 3,712.5
¦ µ
§ ¶
¨ · = 0.944 (mole ratio) By weight 1,168.75 kmoles of CO = 32,725 kg;
2,337.5 kmoles of H2 = 4,675 kg
Recycle stream of CO and H2 = 32,725 + 4675 = 37,400 kg
Reactor
Unreacted CO, H2
CH3OH 6,600 kg/h CO, 2H2
35 °C 300 atm
Figure 8.7
Products leaving reactor = (37,400 + 6,600) = 44,000 kg
\ Recycle ratio = 37, 400 44,000
¦ µ
§ ¶
¨ · = 0.85 (weight ratio)
8.7 Limestone containing 95% of CaCO3 and 5% SiO2 is being calcined.
Heat for the reaction is supplied from a furnace burning coke. The hot flue gases analyze 5% CO2. The kiln gas contains 8.65% CO2. In order to conserve some of the sensible heat a portion of the kiln gas is continuously recycled and mixed with fresh hot flue gas. After mixing the gas entering the kiln analyzes 7% CO2
(a) Find the kg of CaO produced/kg of coke burnt
(b) Find the recycle ratio, (i.e.) moles of gas recycled per mole of gas leaving the kiln.
Figure 8.8
Burner
Limestone Kiln K
P
Air
Coke
F
X R
Basis: 12 kg of coke
represent F kmole of flue gas (5% CO2) K, kmole of kiln gas (8.65% CO2) R, kmole of gas recycled (8.65% CO2) P, kmole of product gas (8.65% CO2) X, kmole of gas entering the kiln (7% CO2)
Figure 8.9
F
X R
X = F + R (overall balance) (a)
0.07X = 0.05F + 0.0865 R (CO2 balance) (b)
3 2
56 44
100
CaCO → CaO + CO
Assuming complete combustion
0.05F = 1 kmole CO2\ F = 20 kmoles
(Q 1 kmole of CO2 comes from 12 kg of Carbon) C + O2 Æ CO2
Solving (a) and (b), we have R = 24.2 kmoles; X = 44.2 kmoles Let m kmole of CO2 be added in kiln from the calcinations of limestone
\ K = 44.2 + m
Making CO2 balance, (0.07 X) + m = K ¥ 0.0865 or, 44.2 ¥ 0.07 + m = (44.2 + m)(0.0865)
\ m = 0.8 kmole
\ K = 44.2 + 0.8 = 45.0 kmoles CaO formed = (0.8 ¥ 56) = 44.8 kg (a) CaO formed/kg of coke burnt = 44.8
12 = 3.73 (b) recycle ratio (mole) = 24.2
45 R
K = = 0.538
8.8 Sea water is desalinated by reverse osmosis using the scheme shown in Figure 8.10D stream has 500 ppm salt = 0.05%
Find
(a) rate of B (b) rate of D (c) recycle R
Figure 8.10
Reverse osmosis cell
Desalinated water “D”, 0.05% salt
“B”, Waste 5.25% salt
“R”, Recycle
4%
A 3.1 salt %
sea water 1,000 kg/h
Basis: One hour
Overall balance, 1,000 = B + D
Salt balance = (1,000 ¥ 3.1) = 5.25B + 0.050 Solving, B = 586.5 kg; D = 413.5 kg
R = 5.25%
Overall balance (At entry to cell), 1,000 + R = A Salt balance, 1,000 ´ 0.031 + R ´ 0.0525 = A ´ 0.04 Solving, A = 1,720 kg and R = 720 kg
8.9 In the feed preparation section of an ammonia plant, hydrogen is produced by a combination of steam-reforming/partial oxidation process. Enough air is used in partial oxidation to give a 3:1 H2-N2 molar ratio in the feed to the ammonia unit. The H2-N2 mixture is heated to reaction temperature and fed to a fixed bed reactor where 20% conversion of reactants to NH3 is obtained per pass. The products from the reactor are cooled and NH3 is removed by condensation. The unreacted H2-N2 mixture is recycled and mixed with fresh feed. On the basis of 100 kmoles per hour of fresh feed determine the NH3 produced and recycle rate.
Basis: 100 kmoles of feed Given that H2 : N2 = 3 : 1
hydrogen: 75 kmoles and nitrogen: 25 kmoles N2 + 3H2 ® 2NH3
Overall balance gives:
Figure 8.11
Reactor NH3
Fresh feed
Feed NH3
NH3 formed = 50 kmoles
One mole of N2 gives 2 moles of NH3 Reactor balance gives:
Reactor NH3, N2 and H2
3x, H2
x, N2
Let 3x kmole of H2 and x kmole of N2 enter the reactor Since 20% conversion takes place,
NH3 formed = x ´ 0.2 ´ 2 = 0.4x = 50 kmoles
\ x = 125 kmoles of N2; and 3x = 375 kmoles of H2 Unreacted N2 = 100 kmoles, and H2 = 300 kmoles