Mass, kg
c) % recovery of NaCl = 65.58100 % 100 = 65.58%
You can try solving this problem using purely algebraic method. However, the resulting system of simultaneous equations is difficult to evaluate.
RECYCLE STREAMS
Recycling operations refer to processes in which part of the
product is returned and mixed with the feed entering a process equipment. Fig. 3-27 shows a recycle stream.
In any process, some of the objectives are to enhance the yield and to obtain products with minimum impurities. In the manufacture of ammonia, recycle streams are indispensable.
Conditions do not allow 100% conversion to NH3. The unreacted nitrogen and hydrogen are recycled and mixed with the fresh feed. In the petroleum industry, the reactors yield mixtures consisting of complex components, the result of incomplete conversion. Distillation is used to separate these mixtures. After the recovery of the desired products, the other components are recycled back to the appropriate reaction units. In the production of sulfuric acid, the fluid used for absorbing SO3 is commonly part of the product that is recycled. Make-up water or dilute sulfuric acid solution is added to this stream to enhance the absorbing capacity.
Other unit operations employ recycle operations for energy conservation and for better control of product quality. In distil-lation, in which high purity products are desired, the enrich-ment of the product is facilitated by recycling or "refluxing"
part of the product back to the distillation column. In drying operations, we can conserve heat energy by recycling part of the exhaust air and mixing it with the incoming fresh medium.
Since less quantity of incoming fresh air feed has to be condi-tioned (by dehumidification and heating), less amount of heat energy is necessary. In absorption, part of the liquid from the tower bottom is recirculated. This results in an increased
Product Feed
Recycle stream
Fig. 3-27
DRAFT
In Fig. 3-28, stream A is the fresh feed while stream D is the net product obtained from the system. Stream C, on the other hand, is the total product obtained from the process.
This splits to form streams D and E. Stream E is the recycle stream, which is mixed with the fresh feed to form the gross feed, stream B. In some cases, stream E may be recycled to other units located ahead of the unit where it comes from.
A B C D
1 E
2
3 4
Figure 3-28 Mass balances around a process involving recycle stream.
Mass balance calculations involving recycle operations should pose little difficulty if balances are set up properly. As usual, the arithmetic or the algebraic method is suitable depending on the presence of a tie component. Mass balances can be taken around four positions: (See Fig. 3-28.)
1. Taking a balance around the whole system, (boundary represented by 1) the mass in stream A equals the mass in stream D. Note that the recycle loop is within the system so that stream E is not involved.
2. Taking a balance around the process within the recycle loop (represent by line 2), the mass in stream B equals the mass in stream C.
3. Taking a balance around the mixing point of the recycle and the fresh feed, (line 3), the mass in stream E plus the mass in stream A equals the mass in stream B.
4. Taking a balance around the point of splitting of the total product (line 4), the mass of stream C equals the mass of the recycle stream E plus the mass of the net product, stream D.
The above four mass balances are not independent, however. Combinations of any three give the fourth balance.
122 National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
Do not make a mistake of setting up a mass balance around the boundary shown in Fig. 3-29.
Boundary
Figure 3-29
Setting up the balance around the whole system, but cutting through the recycle line is a poor choice if the recycle line is unknown. Note that the effect of the recycle line cancels out.
In setting up an overall mass balance around the whole system (outside the recycle loop), the recycle stream is not involved in the mass balance equation.
Two important terms in recycle operation are the recycle-to-fresh-feed ratio and the total-feed-to-recycle-to-fresh-feed ratio.
Comparing the balance around the whole system and the balance within the recycle loop, the net effect of the recycle is to lower the capacity of an equipment. For instance, the maximum capacity of an equipment is 1000 kg per hour, a once-through flow (without recycle), results in a 1000 kg per hour feed being processed. If, however, a recycle stream is provided, say 500 kg per hour, then only 500 kg product stream can be obtained. The equipment can handle only 1000 kg: 500 kg recycle and 500 kg fresh feed. Capacity is sacrificed for other more important factors. It can be a choice between a product with less impurity and a product with high level of impurity. The use of recycle becomes factor in optimization of the process.
Either the arithmetic or algebraic method can be used for recycle calculations. With the use of the mass balances around other boundaries, we can easily solve the unknown quantities.
Remember that with steady state operation, the flow rates are constant. The recycle line is just a mass stream that is circling within a loop, combining with a feed at a certain point (near the entrance of an equipment or a system), and splitting up with the net product at another point (after the exit of an equipment or a system). See Figure 3-30
Recycle loop
Figure 3-30
Equipment or system
Net feed Net product
We must keep in mind that the above mass balances are
DRAFT
calculation is different. In these cases, we would encounter accumulation of mass in the system. Non-chemical engineers become confused with the concept.
Example 3-12
In a pilot process, a sticky material containing 80% water is to be dried. To facilitate the operation, a part of the dried product containing 5% water is recycled and mixed with the feed. If the material entering the drier contains 30% water, calculate (a) the kg water removed per 2000 kg fresh feed
(b) the recycle-to-feed ratio.
Solve using the arithmetic and algebraic methods.
Solution:
Basis: 2000 kg of fresh feed
Part 1: Using the arithmetic method.
To obtain the amount of water evaporated, consider a balance around the entire system (outside the recycle loop).
Using ratios,
kg H2O evaporated = 2,000 kg fresh feed %0.2 kg bone-dry solids kg fresh feed around the drier inside the recycle loop.
Amount of the mixture entering the drier
= 1579 kg H2O evaporated % 1 ( 0.301− 0.30 − 0.05
1− 0.05 )
kg H2O evaporated kg bone-dry solids
% kg mixed feed
(1 − 0.30) kg bone-dry solids in the mixed feed = 6000 kg
124 National Engineering Center CEEC Seminar June 2 -4, 2004 Drier
DRAFT
6000 kg Recycle
Mixer 2000 kg
Figure 3-34
Consider a mass balance around the mixer Recycle = 6000 - 2000 = 4000 kg
Therefore,
the recycle-to-feed ratio = 40002000 = 2.0
Part 2: Using the algebraic method.
Let R = recycle P = product
E = water evaporate
Overall mass balance (outside the recycle line) 2000 = E + P
Bone-dry solid balance:
(1 - 0.8) (2000) = (1 - 0.05)P P= (0.2)(2000)
0.95 = 421 kg Solving for E
E = 2000 - 421 = 1579 kg
Balance around the drier (inside the recycle loop)
Drier Mixe
r
E R
R + P
2000 kg 2000+R
Figure 3-33
P
DRAFT
Bone-dry solid balance
(1 - 0.3) (2000 + R) = (1 - 0.05)(R + 421) 0.7(2000 + R) = (0.95)(R + 421)
Solving for R
R = 4000 kg and
recycle
feed = 4000 2000 = 2
BYPASS
The bypass arrangement is closely related to recycle. A main stream is split into two. One stream goes to the process equip-ment while the other is mixed with the outlet stream from the equipment. Fig. 5-8 shows this.
Feed
Bypass stream
Process
Equipment Product Net Feed
Figure 3-35
The splitter consists of control valves. Bypassing a stream is useful in attaining precise control of concentration. In general, it is easier to make a large change in concentration or property in a small mass rather than make a small change in a large mass. For instance, if we desire to lower the water content of an air-water mixture, which could be done by cooling the mixture to condense part of the water vapor, it is easier to use a bypass stream. A small portion of the air is cooled to the appropriate temperature and is mixed with the bypassed stream after water removal. Aside from attaining accurate control, energy is also conserved.
A dehydrating agent can also be used for H2O removal. A bypass stream can be used as a control to offset the gradually diminishing power of the dehydrating agent.
Another example of the use of a bypass is in lowering the salt content of a saline solution by evaporation. It will not be necessary to evaporate all the water from the entire solution.
Only a portion of the saline water is evaporated. The rest bypasses the evaporator and is mixed with the salt-free water from the evaporator, resulting in the desired lower salt content.
Mass balance calculations involving bypass streams are similar to that for recycle operations. Either the arithmetic or the algebraic method is suitable. Mass balances can be taken
126 National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
around four positions: (See Fig. 3-36.)
1
2 4
3
Process Equipment
Figure 3-36 Mass balances in bypass operation (1) around the whole system, (2) around the process within the bypass loop, (3) around the point of splitting and (4) around the point of mixing.
Comparing the balance around the whole system and that within the bypass loop, we can see that a bypass stream results in the use of a smaller-sized equipment in contrast to that of a recycle stream where the equipment capacity has to be larger.
Example 3-13
We desire to lower the n-hexane content of a nitrogen-hexane mixture at 45oC (the partial pressure of n-hexane = 310 mm Hg) and a total pressure of 757 mm Hg to a mixture contain-ing 15% mole n-hexane. This could be achieved by chillcontain-ing the mixture to 10°C to condense out some of the n-hexane. The gas leaves saturated at 10oC (the partial pressure of n-hexane
= 64 mm Hg) and at a pressure of 750 mm Hg. The bypass gas then mixes with the gas from the chiller to form the 15%
mixture. 300 m3/min of the original gas mixture is to be treated.
Condensed n-hexane Bypass stream
Chiller
15% n-hexane Net Feed
Nitrogen gas T = 45oC PT = 757 mm Hg PnHex = 310 mm Hg
T = 10oC PT = 750 mm Hg PnHex = 64 mm Hg
Figure 3-37
a.How much n-hexane is condensed in the chiller?
b.What is the volume of the gas that bypasses the chiller?
Assume ideal gas behavior.
Solution:
Basis: One minute operation.
The arithmetic method will be used since the n-hexane-free
DRAFT
By a n-hexane balance, the n-hexane condensed is equal to the n-hexane in the N2 entering the system (point 1) minus the n-hexane in the product leaving the system (point 4). See Fig.
3-38.
1 2 3 4
Figure 3-38
Nitrogen entering = 300 mmin %3 757
760 % 0+ 273
45+ 273 % 1 kmol 22.4 m3 % (757 − 310) kmol N2 /min
757 kmol gas mixture = 6.76 kmol n-hexane entering =
6.76 kmol Nmin 2 % 310 kmol n-hexane(757 − 310) kmol N2 = 4.69 kmol/min Nitrogen balance
Nitrogen entering = nitrogen leaving
Therefore the n-hexane in the nitrogen leaving
= 6.76 kmol N2% 0.15 kmol n-hexane0.85 kmol N2 = 1.19 kmol/min The amount of n-hexane that condensed
= 4.69 - 1.19 = 3.50 kmol /min Considering the chiller alone:
Phex= 310 mm Hg PT = 757 mm Hg
Condensed n-hexane
= 3.50 kmol Chiller
Phex= 64 mm Hg PT = 750 mm Hg
Figure 3-39 Using the ratio method:
mole n-hexane condensed
mole n-hexane free nitrogen gas = 310
757− 310 − 64 750− 64 The moles of nitrogen entering the chiller
128 National Engineering Center CEEC Seminar June 2 -4, 2004
DRAFT
= 3.5 kmol n-hexane condensedmin
% 1
( 310
757− 310 − 64
750− 64 )kmol n-hexane condensed kmol nitrogen gas
= 5.83 kmol/min
Volume entering the chiller
= 5.83 kmol N2% 22.4 mkmol %3 757
757− 310 % 760
757 % 45+ 273 0+ 273
= 258.63 m3/min Volume that bypasses the chiller
= 300 - 258.63 = 41.37 m3/min
PURGE
Purge or bleed is a flow arrangement used in recycling opera-tions where a fraction of the recycle stream becomes an output stream. In some processes with recycle stream, a separator removes only the desired product from the stream going out of a process unit. The inerts and impurities, which enter the system in the fresh feed, gradually accumulate in the recycle stream. We can avoid the accumulation by bleeding out or purging a portion of the recycle line. The arrangement is shown in Fig.5-13.
Feed stream
Recycle stream
Process
Equipment Gross Product
product
Purge stream
Separator
Figure 3-40
If the desired product does not contain any inert material or impurity, then by a component balance around the whole system (outside the recycle loop) the inerts and impurities entering in the feed stream is equal to the inerts and impuri-ties in the purge stream.
An example of a process where a purge stream is applied is in the manufacture of ammonia from N2 and H2. The nitrogen comes from the air and contains inert gases such as argon.
The mixture goes to a reactor in which the reaction goes only to about 70% completion. The product gases lead to a cooler where most of the ammonia condenses out and is separated. If
DRAFT
and mixed with the feed, the inerts will gradually accumulate in the system. By providing a purge stream, the concentration of the recycle line is kept constant.
Another example is in the use of recirculating cooling water in refrigeration or air-conditioning systems. The water used usually contains dissolved mineral impurities. As the water recirculates, some is lost by evaporation and make-up water has to be added. Since the impurities are nonvolatile, they will accumulate and cause formation of scale deposits within the system. To prevent the accumulation, a portion of the water is bled-off or purged.
Calculations involving purge are similar to that for recycle problems. They should not pose any problem as long as proper mass balances are set up.