WO2017018764A1 - Method for concentrating solute-containing aqueous solution at high concentration by reverse osmosis method in non-osmotic pressure difference state - Google Patents

Method for concentrating solute-containing aqueous solution at high concentration by reverse osmosis method in non-osmotic pressure difference state Download PDF

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WO2017018764A1
WO2017018764A1 PCT/KR2016/008104 KR2016008104W WO2017018764A1 WO 2017018764 A1 WO2017018764 A1 WO 2017018764A1 KR 2016008104 W KR2016008104 W KR 2016008104W WO 2017018764 A1 WO2017018764 A1 WO 2017018764A1
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solute
osmotic pressure
aqueous solution
chamber
water
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PCT/KR2016/008104
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French (fr)
Korean (ko)
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장호남
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장호남
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Priority to CN201680049525.1A priority Critical patent/CN108137351A/en
Priority to JP2018523724A priority patent/JP2018520874A/en
Priority to EP16830797.3A priority patent/EP3326977A4/en
Priority to US15/744,945 priority patent/US10953367B2/en
Priority claimed from KR1020160094090A external-priority patent/KR101865342B1/en
Publication of WO2017018764A1 publication Critical patent/WO2017018764A1/en
Priority to SA518390792A priority patent/SA518390792B1/en

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    • BPERFORMING OPERATIONS; TRANSPORTING
    • B01PHYSICAL OR CHEMICAL PROCESSES OR APPARATUS IN GENERAL
    • B01DSEPARATION
    • B01D61/00Processes of separation using semi-permeable membranes, e.g. dialysis, osmosis or ultrafiltration; Apparatus, accessories or auxiliary operations specially adapted therefor
    • BPERFORMING OPERATIONS; TRANSPORTING
    • B01PHYSICAL OR CHEMICAL PROCESSES OR APPARATUS IN GENERAL
    • B01DSEPARATION
    • B01D61/00Processes of separation using semi-permeable membranes, e.g. dialysis, osmosis or ultrafiltration; Apparatus, accessories or auxiliary operations specially adapted therefor
    • B01D61/02Reverse osmosis; Hyperfiltration ; Nanofiltration
    • CCHEMISTRY; METALLURGY
    • C02TREATMENT OF WATER, WASTE WATER, SEWAGE, OR SLUDGE
    • C02FTREATMENT OF WATER, WASTE WATER, SEWAGE, OR SLUDGE
    • C02F1/00Treatment of water, waste water, or sewage
    • C02F1/44Treatment of water, waste water, or sewage by dialysis, osmosis or reverse osmosis

Definitions

  • the present invention relates to a method of concentrating a solute-containing aqueous solution at high concentration, and more particularly, to a method of concentrating a solute-containing aqueous solution to be concentrated at a high concentration by removing water by a hydraulic-membrane process in the absence of an osmotic pressure difference.
  • RO Reverse osmosis
  • Table 1 compares the advantages and disadvantages of the forward osmosis (FO) regeneration method and the method of concentrating the solute-containing aqueous solution at a high concentration by the hydraulic-membrane process in the absence of an osmotic pressure difference.
  • RO has only 50% solvent recovery and most of all, it is difficult to operate at 343.070 bar of 3% -NaCl saturated solution (26.47%) to recover 100% solvent and solute.
  • FO has the advantage of operating at atmospheric pressure, but it is necessary to regenerate the draw solution and it is not easy to recover the solute when the solute in the feed crosses to the draw solution (Jung et al. Process Biochemistry (2015) 50 (4) 669-677 ).
  • the flux of the solvent (water) and the movement of the solutes (salt, VFA, ethanol) are as follows.
  • Jw is the water flux
  • Lp is the water permeation coefficient
  • ⁇ P is the hydraulic pressure difference between the feed chamber and the draw chamber
  • is the osmotic pressure difference between the feed chamber and the draw chamber.
  • the Js is divided into one due to the difference in osmotic pressure and one due to the flux of the solute.
  • the solute When there is no Jw in Equation (2), the solute may move to the feed chamber by the draw chamber due to the osmotic pressure difference.
  • C means concentration
  • R means gas constant
  • T means temperature
  • Vsp is the volume of 1 mole of solvent when the solute concentration is 0
  • is the activity coefficient of the solvent
  • X is the mole fraction of the solute.
  • the salt When 30 g / L of solute is dissolved in water, the salt has an osmotic pressure of 25.4 bar, albumin 0.01 bar and particles 1.2x10 -12 bar.
  • Reverse osmosis and forward osmosis have the advantage of saving energy by using membranes, but as the concentration progresses, the osmotic pressure in the feed chamber increases, making it impossible to concentrate the feed solution or increase the utilization of the feed solution (Loeb, S, Loeb-Sourirajan Membrane, How it Came About Synthetic Membranes, ACS Symposium Series, 153, 1, 1-9, 1981; Loeb, S., J. Membr. Sci, 1, 49, 1976).
  • the substances that humanity needs are in the form of solids, liquids, and gases in the ocean, land, and atmosphere, and exist as independent molecules or compounds.
  • the desired material can be obtained through catalysts, chemical reactions, bioreactions and the like.
  • the inventors have developed a method for concentrating a solute-containing aqueous solution using a non-osmotic pressure concentrator including a feed chamber composed of a forward osmosis and / or reverse osmosis membrane and a ⁇ -echolizer chamber to obtain the above materials (International Patent PCT / KR2014 / 000952).
  • the above technique uses a part of the concentrate as an induction solution.
  • the high osmotic pressure difference formed between the feed chamber and the ⁇ -echolizer chamber has to be overcome.
  • the above patent used a method of reducing the residence time of the ⁇ -echolizer solution in the ⁇ -echolizer chamber, but found a problem with the design of the ⁇ -echolizer and required a new method.
  • the present invention (a) concentrating the solute-containing aqueous solution using an osmotic pressure difference concentrator comprising a feed chamber and a ⁇ -echolizer chamber partitioned with a reverse osmosis membrane, one of the following processes Forming a ⁇ reduction condition between the feed chamber and the ⁇ -echolizer chamber using the above process: (i) (Feed) Input-split cascade process; (ii) (Feed) Output-split counter-current process; And (iii) applying the nano filtration membrane and (b) recovering the concentrated solute-containing aqueous solution using another osmotic pressure difference concentrator or reverse osmosis.
  • an osmotic pressure difference concentrator comprising a feed chamber and a ⁇ -echolizer chamber partitioned with a reverse osmosis membrane
  • 1 is a diagram illustrating a process of separating 970 g of water and 30 g of salt in a NaCl solution of 3% (w / w) in a zero / low osmotic pressure difference of the present invention
  • 1 and 2 indicate a normal reverse osmosis process
  • 3 Means an osmotic pressure difference concentrator.
  • FIG. 2 is a schematic diagram of three methods used to achieve the osmotic pressure / low osmotic pressure difference of the present invention
  • the osmotic pressure difference is zero and the feed chamber is concentrated, the feed chamber is concentrated and the ⁇ -Equalizer chamber is diluted so that osmotic pressure difference can occur.
  • B is the Feed Output-Split CC (counter-current) method.
  • both streams flow counter-current.
  • the starting concentration of the feed chamber is 6%
  • the final concentration of the ⁇ -Equalizer stream can be made 3%.
  • C indicates that the use of a membrane with a low solute reflection coefficient allows some of the solute in the feed stream to flow into the ⁇ -Equalizer stream to reduce the osmotic pressure difference.
  • the graph on the left shows the difference between the two chambers according to these three methods. Osmotic pressure difference is shown.
  • (a) is a high pressure feed chamber (A) and an atmospheric pressure draw chamber ( It is an explanatory drawing of a BPS system including B), and
  • (B) is a specific design drawing of A and B.
  • FIG. 4 is a system diagram that can separate 3% NaCl solution into water and 26.47% saturated solution (or salt) using an input-split cascade system.
  • the water is discharged to the outside of the various types of solute-containing aqueous solution to be concentrated using a reverse osmosis separator, and the concentrated aqueous solution includes a feed chamber and a draw chamber partitioned by a reverse osmosis membrane or an forward osmosis membrane.
  • the osmotic pressure difference between the feed chamber and the draw chamber is a low osmotic pressure / no osmotic pressure state It was confirmed that it can be maintained as.
  • the question of whether the feed should be limited to the solute-containing solution after recovering the water by reverse osmosis in a non-osmotic pressure difference concentrator confirms that the total energy consumption is reduced while reducing the amount of water drawn from the solute-containing solution (3%) by reverse osmosis, and pure 3% brine (sea water), which has not undergone reverse osmosis-pure manufacturing, Even when added, it was confirmed that the solvent and the solute can be separated with high efficiency.
  • the present invention in one aspect, (a) the solute-containing aqueous solution is concentrated using an osmotic pressure concentrator comprising a feed chamber and a ⁇ -echolizer chamber partitioned with a reverse osmosis membrane, wherein at least one of Forming a ⁇ reduction condition between the feed chamber and the ⁇ -echolizer chamber using the process: (i) (Feed) Input-split cascade process; (ii) (Feed) Output-split counter-current process; And (iii) applying a nano filtration membrane; And (b) recovering the concentrated solute-containing aqueous solution using another osmotic pressure difference concentrator or reverse osmosis pressure.
  • an osmotic pressure concentrator comprising a feed chamber and a ⁇ -echolizer chamber partitioned with a reverse osmosis membrane, wherein at least one of Forming a ⁇ reduction condition between the feed chamber and the ⁇ -echolizer chamber using the process: (i)
  • step (c) another non-intrusive pressure difference concentrator of step (b) further comprises the step of maximizing the recovery of the solute using one or more of the processes (i) to (iii). can do.
  • the step (a) may be characterized in that it further comprises the step of concentrating the solute-containing aqueous solution to be concentrated before using the reverse osmosis.
  • the step (b) of concentrating the solute-containing aqueous solution using another osmotic pressure difference concentrator transfers the concentrated aqueous solution to the feed chamber of the osmotic pressure difference concentrator, and is transferred to the feed chamber. Injecting a solution having the same osmotic pressure as the concentrated aqueous solution into the draw chamber to form an osmotic pressure state between the feed chamber and the draw chamber; And (ii) a pressure of up to 0 to 5 atm for forward osmosis (FO) and 10 to 200 atm for reverse osmosis is applied to the feed chamber at zero osmotic pressure, and the water in the concentrated aqueous solution is drawn. And further concentrating the concentrated aqueous solution by transferring to the chamber.
  • a pressure of up to 0 to 5 atm for forward osmosis (FO) and 10 to 200 atm for reverse osmosis is applied to the feed chamber at zero osmotic pressure, and the water in the concentrated a
  • the osmotic pressure difference between the feed stream supplied to the feed chamber and the ⁇ -echoizer supplied to the ⁇ -echolizer chamber is increased (i) an input-split cascade process; (ii) an output-split counter-current process; Or (iii) using the process of applying the nano filtration membrane to reduce the difference.
  • the terms “(feed) input-split cascade process” and “input-split cascade process” refer to half of the feed solution fed to the osmotic pressure concentrator and half to the feed chamber, and half to the ⁇ -echoiser chamber. After forming the osmotic pressure difference, the aqueous solution of the feed chamber is concentrated at low pressure, and the concentrated solute-containing aqueous solution is transferred to another osmotic pressure concentrator.
  • ⁇ reduction condition means a condition in which there is no osmotic pressure difference or a very small value between the feed chamber and the ⁇ -echolizer chamber. That is, the ⁇ reduction condition of the present invention means a case where ⁇ is 0 or 1 to 100 bar or less.
  • (feed) output-split counter-current process” or “output-split counter-current process” of the present invention is characterized by sending a portion of the ⁇ -echolizer stream to the feed stream in a concentrated state in an osmotic pressure concentrator. You can do
  • ⁇ (Pai) -echolizer chamber of the present invention is used in the same sense as the draw chamber, and the dilution of the filter liquid (water) in the feed stream is the same, but the maneuverability in the orthostatic method is concentrated in the draw chamber. Compared to the induction solution, the difference here is the hydraulic pressure ( ⁇ P).
  • ⁇ P hydraulic pressure
  • step (a) may apply steps (i) to (iii), respectively, and (i) and (ii), (i) and (iii), (ii) and (iii) or (i) )) To (iii) may be applied to form the ⁇ reduction conditions.
  • 3% brine is completely converted into crystalline form (solid content) by reverse osmosis in 3% NaCl solution in 26.47% saturated solution by input-split cascade process and output-split counter-current process.
  • the process of concentration was simulated and consulted with a crystallization expert to get advice.
  • the input-split cascade process (Figure 2A) is divided into feed stream and ⁇ (osmotic pressure) -echolizer stream for each concentration section, solute remains in the feed chamber and only solvent, water, is moved to the ⁇ -echoiser to increase the solution concentration in the feed chamber.
  • ⁇ -echolizer decreases.
  • When recovering the salt in the present invention may be characterized by applying a method using thermal energy or electrical energy well known to those skilled in the art in a saturated solution, or by applying a feed input-split cascade process and a feed output-split counter-current process It is not limited to this.
  • the present invention may be characterized by using a nanofiltration membrane.
  • the step (iii) of the step (a) is selected from the beginning when applied to the process (i), the initial selectivity in consideration of the degree of concentrated water recovery and osmotic pressure difference when applied to the process (ii)
  • the higher the late membrane, the greater the osmotic pressure difference may be characterized in that the osmotic pressure difference can be reduced by selecting a membrane with low selectivity, but is not limited thereto.
  • the reverse osmosis method may be used when producing pure water, and the RO process may be used as a system when producing water.
  • the saturated solution concentrated by the method of the present invention can produce power using PRO (pressure retarded osmosis) method using river water or seawater.
  • the osmotic pressure difference concentrator according to the present invention may be composed of a plurality. That is, the feed chamber and the ⁇ -echolizer chamber constituting the osmotic pressure difference concentrator are characterized in that it is composed of a multi-stage.
  • the reverse osmosis membrane or the forward osmosis membrane partitioning the feed chamber and the draw chamber of the osmotic pressure difference concentrator may be used without particular limitation as long as it does not pass the solute and mainly passes the solvent.
  • the solute means a liquid or solid substance dissolved in water as a solvent.
  • the solute-containing aqueous solution to be concentrated may include sea water, brackish water, cell metabolites, reaction solutions, and the like, and cell metabolites may be cultured cells of animal cells, plant cells, or microorganisms.
  • the concept includes a primary product, a secondary product, an in vitro secreted protein, a biotransformation, and the like, but may be a suitable process when the molecular weight is small and the osmotic pressure is high.
  • reaction solution examples include a reaction solution through a chemical reaction and a reaction solution through an enzyme reaction.
  • the primary products of the microorganisms include organic acids (acetic acid, propionic acid, butyric acid, lactic acid, citric acid, lactic acid, succinic acid, etc.), alcohols (ethanol, butanol, etc.), nucleic acids, amino acids (lysine, tryptophan, etc.), vitamins, polysaccharides, and the like. It may be illustrated, but is not limited thereto.
  • the secondary products of the microorganisms include antibiotics (such as lung nicillin), enzyme inhibitors, physiologically active substances (taxols, etc.), and the in vitro secreted proteins of the microorganisms include enzymes such as amylase and cellulase, insulin, interferon, and single group antibodies.
  • antibiotics such as lung nicillin
  • enzyme inhibitors such as a carboxyl-containing carboxyl-containing carboxyls, etc.
  • the in vitro secreted proteins of the microorganisms include enzymes such as amylase and cellulase, insulin, interferon, and single group antibodies.
  • the biotransformation of the microorganism is a substance produced by using a microorganism or an enzyme, and examples thereof may include steroids, but are not limited thereto.
  • the concentration of ethanol can be concentrated in a reverse osmosis concentrator (RO-1) is about 20%, theoretically 20 ⁇ by the osmotic pressure difference method It is known to be able to concentrate up to 100%.
  • RO-1 reverse osmosis concentrator
  • the saturation degree is about 50 to 60 wt%, so it is theoretically possible to concentrate 100%, and the high solute rejection rate is also 100% concentrated in the absence of osmotic pressure difference.
  • the solute low selectivity membrane is a membrane having a selectivity lower than 1 and higher than 0, which generally corresponds to a nanofiltration membrane, but is not limited thereto.
  • the target material when the target material is a solid, it is easy to crystallize according to temperature and pH, and is suitable for a material having high viscosity at high concentration.
  • alcohol for fuel may be a good application but is not limited to this.
  • the pH of the aqueous solution is 2-13
  • the temperature is the temperature at which water maintains the liquid (usually 0 to 100 °C, preferably 15 to 50 °C, more preferably 20 to 40 °C) or more Or can be For example, mixtures with other solutes / solvents may deviate from the above temperatures.
  • a concentrated aqueous solution transferred to the feed chamber a solution that can be easily separated after use, and the like may be used. It is preferable to use an aqueous solution of the same composition as the concentrated aqueous solution transferred to the feed chamber.
  • the concentration using the osmotic pressure difference concentrator may be performed in a batch or continuous manner to maximize the effect.
  • the batch may be performed when there is no flow with both chambers and the external system, and the continuous may be performed when there is a flow with the external system.
  • the feed chamber and the ⁇ -equalizer chamber is characterized in that it is composed of a multi-stage.
  • the method of recovering the solute and the water from the aqueous solution further concentrated in the osmotic pressure difference concentrator is independently a commonly known multistage evaporation method, dialysis evaporation, pyrolysis method, sulfuric acid method, calcium method and input-split. cascade may be used, but is not limited thereto.
  • the concentration method further comprises the step of (d) maximizing the recovery of either solute or water using any one of the steps (i) to (iii) of step (c). It can be characterized.
  • the pressure-added forward osmosis (PRO) power generation, resource utilization and rare earth recovery process may be further included to increase the added value of the process.
  • it may further comprise a step of optimizing the material balance and the energy balance.
  • the present invention also relates to a method for separating a solvent and a solute from a solute-containing solution using the above concentration method.
  • the solute is a salt
  • the solvent may be characterized in that the water.
  • Table 4 calculates the osmotic pressure of 3% NaCl, the description of each column of the table is as follows.
  • # 1: w / w% is the weight ratio of water and salt.
  • 3% brine consists of 970kg water and 30kg salt
  • # 5 The amount of energy required to produce one ton of water with an unchanged percentage of raw water at each concentration.
  • volume 1 is the amount of water contained in each% solution based on 30 kg of salt.
  • a sample (NaCl aqueous solution) is put in the apparatus as shown in FIG.
  • the membrane used in this example is RE2521-TL (Woongjin Chemical Co, Seoul, Korea; http: //www.csmfilter.com) Ltd, which is a thin-film composite type and the membrane is used in a negatively charged, polyamide, spiral-wound module.
  • RO membrane The permeability is 1.1m 3 / day and the effective area is 1.1m 2 . 99% rejection at 1,500mg / L salt solution, 1.0MPa, maximum pressure is 4.14MPa, maximum flow rate is 1.36m 3 / hr, minimum flow rate (concentrate) is 0.23m 3 / hr.
  • the maximum temperature is 45 ° C. and the pH is 3.0-10.0 and can withstand 2.0-11.0.
  • the 1.5 to 3% portion has a linear relationship where the plus is above the critical pressure, but at subsequent concentrations it is proportional to ( ⁇ P) ⁇ (C) ⁇ where ⁇ is 0 to 1 and ⁇ is -1 to 0 are displayed.
  • concentration the lower the flux and the higher the pressure.
  • the maximum pressure was 40 bar.
  • Example 3 100% Separation of Solute / Solvent Based on Feed Input-Split Cascade Process
  • the process of the present invention is capable of producing an input stream of 3%, a water production system at the bottom, a saturated concentrated solution at the top or a salt thereof.
  • the 3% solution produces 500 L of pure water and enters the 6% chamber with 6%, 470 L water (total 500 kg solution).
  • the input-split method is used to separate 9% concentrate and 3% solution, 6% of which is a 9% chamber at the top, 3% of which produces 235 kg of raw water in the RO-2 unit, and the remaining 235 kg enters the chamber as Recycle. . 9 ⁇ 12 ⁇ 15 ⁇ 18 ⁇ 21 ⁇ 24 ⁇ 26.47% (sat.) And the diluted stream was recovered to 26.47% ⁇ 21% ⁇ 18 ⁇ 15 ⁇ 12 ⁇ 9 ⁇ 6 ⁇ 3 ⁇ 0 (pure) do.
  • Flux movement and energy consumption by solute concentration are 371 kg of total water moved through the membrane in feed input-split cascade.
  • a total pressure of 0.309 Kwh is required at an average pressure difference of 30 bar, and two recycles are required to remove 30 kg of salt.
  • the total energy required is 0.618kwh.
  • 9 steps of mixing and demixing occur from 3% to 26.47%. However, a considerable amount of water can be recovered in steps 1 and 2, which can be considered if not all of the water is recovered.
  • the feed input is 6%, 470L, 30g in the osmotic pressure difference condensation process.
  • the water moves to the chamber by ⁇ -evaporator.
  • the R on the right shows how much salt total flows from the feed chamber end stream to the product (draw) chamber.
  • the concentration is 26.47% at the feed chamber end.
  • the ⁇ -echoarizer is 3.38% and the amount of water recovered is 428.33 kg. This is the amount when the product is a saturated solution.
  • the energy consumption varies greatly depending on the size of the sample entering the target RO apparatus. For example, if you remove 1 ton (100 liters of water and 30 kg of salt) from 100 tons of raw water, the minimum energy will be very small.
  • the energy consumption was only about 10% when a large separator was used at the feed input, but in fact energy consumption is highly related to the recovery rate.
  • the inputs of F250, F485, and F500 RO-1 are 3% brine and the concentration of the output according to the recovery degree of RO-1 entering the recycle stream.
  • the concentration was low, and as in the F500, the recovery of 500L in RO-1 was confirmed to be 6%.
  • the first row is the energy required to recover pure water from 3% brine in the RO-1 unit, and the RO-2 is the energy consumed in the recycle unit.
  • the last one is the desalination process from F250 of Example 2 (3.5S, 3.5Ws). Is the energy required.
  • RO-2 The calculation of RO-2 was performed as follows.
  • RO-2 input (top water) The difference in the output of the bottom water is multiplied by the amount of recycled water divided by 36 to get kwh.
  • the energy difference was obtained by multiplying the pressure difference between feed-output (26.47%) and sub-water input (23.57%) by 41.67.
  • the energy difference between the first and last energy was calculated from the recycling system, the arithmetic mean was calculated, and the total recycled water was multiplied to obtain a simple recycle energy.
  • the cost of water production of RO-2 is the ⁇ process cost, plus 3% of the water production cost from the solution entering the raw water tank.
  • the cumulative total of the three processes shows a low F250 of 1.236 but F485 or The F500 has significantly higher total energy requirements of 1.528 and 1.540, which is not significantly different from “Input-split Cascade” or “Output-split CC.” These values are based on the use of one ton of solute solution.
  • the recovery of salt at low pressure and 100% recovery of water can be significant.
  • Low selectivity membranes are applied to the input-split cascade and output-split CC to significantly improve the membrane flux and to save energy for the recycling process.
  • the current use of the RO process is one There needs 1kwh / m 3 to overcome the minimum osmotic pressure difference in the production of water per ton to overcome the osmotic pressure actually is 2kwh is consumed 2kwh, yet other processes in the osmotic process, the RO process, the total 4kwh Is known to be consumed.
  • the cost of electric energy is $ 1 per ton of water, it is very important to reduce it to a little since 40% of the cost of water is assumed to be $ 0.1 / kwh.
  • the difference in concentration is 1%
  • one method may be to input-split at first and then switch to output-split later.
  • Example 6 F500, F485, and F250 first examined 500L, 485L, and 250L recovery through RO in 970L water of 3% raw water. Then, let's consider recirculation of RO-1, which is the extreme condition, all zero.
  • Feed input 3%, 970 L + 30 kg salt and the recycle stream of the ⁇ -echolizer chamber will be 1.52% with 970 L + 15 kg salt water.
  • the osmotic pressure of 3% is 23.743 bar and the osmotic pressure of 1.52% is 11.746 bar.
  • the difference between the two osmotic pressure is 11.997 bar, the average is 17.744 bar.
  • the pressure difference at the input is 11.997 bar and the pressure difference at the input outlet (26.47%) / pi-Equalizer start point is 55.148 bar.
  • High pressure in the middle process is not a problem because there are many solutions such as low ⁇ membrane or input-split cascade.
  • RO-1 uses only an average of 3% and 4%, later using both an average of 4% and 2.04% and pressure difference in recycle energy.
  • the input of the ⁇ -equalizer chamber be 15S, 41.67W (1/2 of the concentrated water component) and 23.57% (Example 6).
  • Ethanol for fuel is currently used in 99.5% or 99.6% purity. There is a thermal method (gas generates steam, electricity is used). According to the Renewable Fuels Association (March 06, 2016) in the United States, the minimum energy is 23,424 BTU / gal from Iowa WDG. If you change this to kwh / kg-fuel ethanol, you get 2.27kwh / kg-ethanol. Here, the concentration of up to 99.5% by applying an osmotic pressure difference RO technology will be considered.
  • Table 10 calculated the osmotic pressure of the alcohol by Lewis equation (eq.-4). Initially it goes up to 28 bar at 5% or 6010 bar at 99.50%. Unlike high pressures, high pressures have very low volume, which results in very low energy content. Like 3% -NaCl, most of the energy content is concentrated at low concentrations.
  • the energy required for concentration can be calculated.
  • the energy required for concentration up to 5% 99.5% is 0.08819 kwh / kg, 10% ⁇ 99.5% is 0.0373kwh / kg and 7% ⁇ 99.5% is 0.04787kwh / kg.
  • the osmotic pressure difference technology can make the pressure level that we can handle (e.g. below 100 bar).
  • the membrane is used, the azeotrope phenomenon between water and ethanol is also achieved. It can be said that it can be solved.
  • the global desalination market is 60 million / d, according to 2016 (google image: desalination market accessed on 07-24-2016).
  • An economic evaluation was conducted on the 65,000 ton / d seawater desalination plant in Gijang-gun, Busan, Korea. Economics compared the price of water at $ 1 per tonne based on the current international price of each element.
  • the method of concentrating an aqueous solution by hydraulic pressure in a non-osmotic pressure difference state consumes less energy and can be concentrated until the maximum saturated aqueous solution concentration or the concentration of the solute is 100% without using an extraction solvent. There is an advantage that does not need to use a separate osmotic induction solution.

Abstract

The present invention relates to a method for concentrating an aqueous solution with low pressure in a non-osmotic pressure difference state and, more specifically, to a method for concentrating a solute-containing aqueous solution to be concentrated with low pressure in a non-osmotic pressure state. When the method of the present invention is used, energy consumption is reduced, and the aqueous solution can be concentrated, without using an extraction solution, until the saturated aqueous solution of the maximum solute is prepared or the concentration of the solute is 100%. In addition, the method has an advantage of not requiring the use of a separate osmotic induction solution.

Description

무삼투압차 상태에서 역삼투압법으로 용질 함유 수용액을 고농도로 농축하는 방법Concentration of solute-containing aqueous solution at high concentration by reverse osmosis in the absence of osmotic pressure difference
본 발명은 용질 함유 수용액을 고농도로 농축하는 방법에 관한 것으로서, 더욱 상세하게는 농축하고자 하는 용질 함유 수용액을 무삼투압차 상태에서 유압-막공정법으로 물을 제거하여 고농도로 농축하는 방법에 관한 것이다.The present invention relates to a method of concentrating a solute-containing aqueous solution at high concentration, and more particularly, to a method of concentrating a solute-containing aqueous solution to be concentrated at a high concentration by removing water by a hydraulic-membrane process in the absence of an osmotic pressure difference.
해수담수화 공정에서 제일 많이 활용되고 있는 역삼투압법(reverse osmosis, RO), 최근 드로우 챔버에 고삼투압 용액을 사용하여 해수 혹은 하수로부터 물을 회수하여 삼투압유발물질과 물을 분리하여 사용하고 드로우 용액은 다시 재생 사용하는 정삼투압법(forward osmosis, FO) 및 무삼투압차 상태에서 유압-막공정법으로 용질함유 수용액을 고농도로 농축하는 방법의 장단점을 서로 비교하면 아래 표 1과 같다. Reverse osmosis (RO), the most widely used seawater desalination process, recently used high-osmotic solutions in draw chambers to recover water from seawater or sewage, to separate osmotic inducing substances and water. Table 1 below compares the advantages and disadvantages of the forward osmosis (FO) regeneration method and the method of concentrating the solute-containing aqueous solution at a high concentration by the hydraulic-membrane process in the absence of an osmotic pressure difference.
RO는 용매회수율이 50%에 그치고 있고 무엇보다도 100% 용매와 용질을 회수하려면 3%-NaCl 포화용액(26.47%)의 343.070 bar에서 운전해야 하는 어려움이 있다. FO는 상압에서 운전하는 장점은 있으나 드로우 용액을 재생해야하고, 또 피드의 용질이 드로우 용액 쪽으로 넘어 오면 용질을 회수하는 것이 쉽지 않다 (Jung et al. Process Biochemistry (2015) 50(4) 669-677). 반면에 무삼투압차 상태에서 유압-막공정법으로 용질함유 수용액을 고농도로 농축하는 방법(△π=0 공법)은 피드 챔버와 드로우 챔버 사이에서 공정이 진행됨에 따라 발생하는 삼투압차를 극복해야 하는 어려움이 있다.RO has only 50% solvent recovery and most of all, it is difficult to operate at 343.070 bar of 3% -NaCl saturated solution (26.47%) to recover 100% solvent and solute. FO has the advantage of operating at atmospheric pressure, but it is necessary to regenerate the draw solution and it is not easy to recover the solute when the solute in the feed crosses to the draw solution (Jung et al. Process Biochemistry (2015) 50 (4) 669-677 ). On the other hand, the method of concentrating the solute-containing aqueous solution to a high concentration by using the hydraulic membrane process in the absence of osmotic pressure difference (△ π = 0 method) has to overcome the osmotic pressure difference caused by the process between the feed chamber and the draw chamber. There is this.
Figure PCTKR2016008104-appb-T000001
Figure PCTKR2016008104-appb-T000001
현재 막 공법은 역삼투압법 외에 정삼투압법으로 전력을 생산하는 데 연구되고 있어 성공적으로 개발되면 물문제와 에너지 문제를 동시에 해결할 수 있는 가능성까지 가지고 있다. 하기 표 2의 제한이유가 기술적/경제적으로 해결되면 화학 산업/생물산업 및 환경 산업에 큰 미칠것으로 예상되고 있다.Currently, the membrane method is being researched to produce electric power by forward osmosis method in addition to reverse osmosis method, and if successfully developed, it has the possibility of simultaneously solving the water problem and energy problem. If the reason for limitation in Table 2 is solved technically and economically, it is expected to have a big impact on chemical industry / biotechnology industry and environmental industry.
Figure PCTKR2016008104-appb-T000002
Figure PCTKR2016008104-appb-T000002
막 공법에서 용매(물)의 플럭스와 용질(소금, VFA, 에탄올)의 이동에 대해서 살펴보면 다음과 같다.In the membrane method, the flux of the solvent (water) and the movement of the solutes (salt, VFA, ethanol) are as follows.
Jw = Lp (△P -σ△π) ----- (1)Jw = Lp (△ P -σ △ π) ----- (1)
Js = Cs (1-σ)Jw+ω△π ----- (2)Js = Cs (1-σ) Jw + ω △ π ----- (2)
여기서, Jw는 물의 흐름(water flux), Lp는 수분 투과 계수, △P는 피드 챔버와 드로우 챔버의 수력압력 차이, △π는 피드 챔버와 드로우 챔버의 삼투압 차이를 의미한다. 그리고, 상기 Js는 용질의 flux로 Jw에 의한 것과 삼투압 차이에 의한 것으로 나누어진다. Here, Jw is the water flux, Lp is the water permeation coefficient, ΔP is the hydraulic pressure difference between the feed chamber and the draw chamber, Δπ is the osmotic pressure difference between the feed chamber and the draw chamber. In addition, the Js is divided into one due to the difference in osmotic pressure and one due to the flux of the solute.
상기 (2)식에서 Jw가 없는 경우는 삼투압 차이에 의해 용질이 드로우 챔버에 의해서 피드 챔버로 움직일 수 있다. σ는 용질의 막에 의한 반사계수로 σ=1이면 용질은 완전 불통과이며 두 챔버간의 삼투압 차이도 최대가 된다. When there is no Jw in Equation (2), the solute may move to the feed chamber by the draw chamber due to the osmotic pressure difference. σ is the reflection coefficient of the solute film. If σ = 1, the solute is completely untransmitted, and the osmotic pressure difference between the two chambers is maximum.
삼투압은 π =CRT ------------------(3)로 표시된다. Osmotic pressure is represented by π = CRT ------------------ (3).
여기서, C는 농도, R은 기체상수, T는 온도를 의미한다.Here, C means concentration, R means gas constant, and T means temperature.
또 용질을 많이 포함하는 고농도 용액인 경우 Lewis 근사식은 In the case of a high concentration solution containing a large amount of solute, Lewis approximation
π=RT/vsp ln (1-γX)-----------------(4)π = RT / v sp ln (1-γX) ----------------- (4)
이다(참고문헌: Lewis, G. N., The osmotic pressure of concentrated solutions and the laws of perfect solution. Journal of the American Chemical Society 1908, 30, 668-683.).(Ref .: Lewis, GN, The osmotic pressure of concentrated solutions and the laws of perfect solution.Journal of the American Chemical Society 1908, 30, 668-683.).
여기서 Vsp는 용질의 농도가 0인 경우 용매의 1몰의 부피이고 γ는 용매의 활성화 계수 (activity coefficient), X는 용질의 몰 분율이다.Where Vsp is the volume of 1 mole of solvent when the solute concentration is 0, γ is the activity coefficient of the solvent, and X is the mole fraction of the solute.
30g/L의 용질이 물에 녹아 있을 경우 소금은 25.4bar, 알부민은 0.01bar, 입자는 1.2x10-12bar의 삼투압을 가진다.When 30 g / L of solute is dissolved in water, the salt has an osmotic pressure of 25.4 bar, albumin 0.01 bar and particles 1.2x10 -12 bar.
역삼투압 및 정삼투압 방식은 막을 이용하여 에너지를 절약하는 이점은 있으나 농축이 진행될수록 피드 챔버속의 삼투압은 증가되어 더 이상 피드용액을 농축하거나, 피드용액의 활용도를 높이는 것이 불가능해진다 (Loeb, S, Loeb-Sourirajan Membrane, How it Came About Synthetic Membranes, ACS Symposium Series, 153, 1, 1~9, 1981; Loeb, S., J. Membr. Sci, 1, 49, 1976). Reverse osmosis and forward osmosis have the advantage of saving energy by using membranes, but as the concentration progresses, the osmotic pressure in the feed chamber increases, making it impossible to concentrate the feed solution or increase the utilization of the feed solution (Loeb, S, Loeb-Sourirajan Membrane, How it Came About Synthetic Membranes, ACS Symposium Series, 153, 1, 1-9, 1981; Loeb, S., J. Membr. Sci, 1, 49, 1976).
위의 (4)식에서는 몰 분율이 쓰이고 있지만 현실적으로는 mole (g/L) 혹은 wt%가 많이 쓰이고 있으므로 wt%와 삼투압을 표시하도록 하겠다. 몰농도의 어려운 점은 Lewis의 삼투압은 모든 용질에 적용되나, 몰농도의 경우, 용질과 용매를 혼합하면 질량은 보존이 되지만 부피는 보존이 되지 않기 때문에 현실적으로는 wt%, Lewis식, activity coefficient estimation혹은 측정하는 것이 더 편리하다. 아래 표 3은 △π=0 공법의 모델로 사용할 3% NaCl의 대표적인 %에서 삼투압, 총수분양, 용질 1g당 물의 양, 그리고 각 % 수용액에서 물 1m3을 생산하는 데 소요되는 전력량을 표시하고 있다. In the above formula (4), mole fraction is used, but in reality, mole (g / L) or wt% is used a lot, so wt% and osmotic pressure will be displayed. The difficulty of molarity is that the osmotic pressure of Lewis is applied to all solutes.However, in case of molarity, the mass is preserved when the solute and solvent are mixed, but the volume is not. Or it is more convenient to measure. Below Table 3 and displays the amount of power required to produce water 1m 3 in △ π = 0 Method osmotic In exemplary% of the 3% NaCl used as a model, the total pre-sale, the solute amount of water per 1g, and each% aqueous solution .
Figure PCTKR2016008104-appb-T000003
Figure PCTKR2016008104-appb-T000003
인류가 필요로 하는 물질은 고체, 액체, 기체의 형태로 해양, 육지, 대기 중에 존재하며 독립된 분자나 화합물로 존재한다. 목적하는 물질은 촉매, 화학반응, 생물반응 등을 통하여 수득할 수 있다.The substances that humanity needs are in the form of solids, liquids, and gases in the ocean, land, and atmosphere, and exist as independent molecules or compounds. The desired material can be obtained through catalysts, chemical reactions, bioreactions and the like.
본 발명자는 상기 물질들을 얻기 위해, 정삼투압 및/또는 역삼투압막으로 구성된 피드챔버 및 π-이콜라이자 챔버를 포함하는 무삼투압 농축장치를 이용한 용질 함유 수용액의 농축방법을 개발한 바 있다(국제특허 제PCT/KR2014/000952호). 그러나 상기 기술에서는 농축액의 일부를 사용하여 유도용액으로 사용하고 있는데, 이럴 경우 적은 양의 π-이콜라이자용액을 사용할 경우에는 피드 챔버와 π-이콜라이자 챔버 사이에서 형성되는 높은 삼투압차를 극복하여야 하는 어려움이 있었으며, 이를 해결하기 위해 상기 특허에서는 π-이콜라이자 챔버에서 π-이콜라이자 용액의 체류시간을 줄이는 방법을 이용하였지만 π-이콜라이자 설계상의 문제점이 있음을 발견하여 새로운 공법의 필요하였다. The inventors have developed a method for concentrating a solute-containing aqueous solution using a non-osmotic pressure concentrator including a feed chamber composed of a forward osmosis and / or reverse osmosis membrane and a π-echolizer chamber to obtain the above materials (International Patent PCT / KR2014 / 000952). However, the above technique uses a part of the concentrate as an induction solution. In this case, when using a small amount of π-echolizer solution, the high osmotic pressure difference formed between the feed chamber and the π-echolizer chamber has to be overcome. In order to solve this problem, the above patent used a method of reducing the residence time of the π-echolizer solution in the π-echolizer chamber, but found a problem with the design of the π-echolizer and required a new method.
이에, 본 발명자들은 상기 문제점을 해결하기 위하여 예의 노력한 결과, 피드 용액의 절반을 피드 챔버로, 다른 절반은 π-이콜라이자 챔버로 운영할 경우, 상기 (1) 및 (2)식에서 △π를 제거/최소화하여 무삼투압차 상태(△π =0)혹은 저삼투압차에서 유압(△P) 만으로 피드 용액을 농축하고, 이로 인해 희석된 π-이콜라이저 용액은 낮은 삼투압을 가지게 되므로 역삼투압으로 순수한 물을 회수할 수 있으며, 최종적으로 피드용액의 농축을 최대화하면서 에너지 소모 및 운영비용을 최소화 할 수 있다는 사실을 확인하고, 본 발명을 완성하게 되었다.Therefore, the present inventors have made diligent efforts to solve the above problems, and when half of the feed solution is operated as the feed chamber and the other half as the π-echolizer chamber, Δπ is removed from the above formulas (1) and (2). By minimizing the osmotic pressure difference (△ π = 0) or low osmotic pressure, the feed solution is concentrated by hydraulic pressure (△ P) alone, which causes the diluted π-echolizer solution to have a low osmotic pressure. After recovering, finally confirming that the energy consumption and operating costs can be minimized while maximizing the concentration of the feed solution, the present invention has been completed.
발명의 요약Summary of the Invention
본 발명의 목적은 휘발성유기산 및 이와 비슷한 특성을 갖는 다양한 종류의 저분자물질을 함유하는 용질 함유 수용액의 농축을 최대화하면서, 에너지 및 운영비용을 최소화할 수 있는 수용액의 농축과 미량의 용질을 포함하는 순수한 물을 제공하는데 있다. It is an object of the present invention to maximize the concentration of solute-containing aqueous solutions containing volatile organic acids and various types of low molecular weight substances having similar properties, while the concentration of aqueous solutions and trace solutes can be minimized to minimize energy and operating costs. To provide water.
상기 목적을 달성하기 위하여, 본 발명은 (a) 용질 함유 수용액을 역삼투압 막으로 구획되어 있는 피드 챔버와 π-이콜라이저 챔버를 포함하는 무삼투압차 농축기를 이용하여 농축하되, 하기의 공정 중 하나 이상의 공정을 이용하여 피드챔버와 π-이콜라이저 챔버간의 △π저감 조건을 형성하는 단계: (i) (Feed) Input-split cascade 공정; (ii) (Feed) Output-split counter-current 공정; 및 (iii) Nano filtration membrane을 적용하는 공정 및 (b) 상기 농축된 용질 함유 수용액을 또다른 무삼투압차 농축기 또는 역삼투압을 이용하여 회수하는 단계를 포함하는 △π 저감 조건에서 용질 함유 수용액을 농축하는 방법을 제공한다.In order to achieve the above object, the present invention (a) concentrating the solute-containing aqueous solution using an osmotic pressure difference concentrator comprising a feed chamber and a π-echolizer chamber partitioned with a reverse osmosis membrane, one of the following processes Forming a Δπ reduction condition between the feed chamber and the π-echolizer chamber using the above process: (i) (Feed) Input-split cascade process; (ii) (Feed) Output-split counter-current process; And (iii) applying the nano filtration membrane and (b) recovering the concentrated solute-containing aqueous solution using another osmotic pressure difference concentrator or reverse osmosis. Provide a way to.
도 1은 본 발명의 무/저삼투압차 상태에서 3%(w/w)의 NaCl 용액에서 물 970g과 소금 30g을 분리하는 공정을 도식화한 도면이므로, ① 및 ②는 정상적인 역삼투압 공정을 나타내고 ③은 무삼투압차 농축기를 의미한다.1 is a diagram illustrating a process of separating 970 g of water and 30 g of salt in a NaCl solution of 3% (w / w) in a zero / low osmotic pressure difference of the present invention, ① and ② indicate a normal reverse osmosis process ③ Means an osmotic pressure difference concentrator.
도2는 본 발명의 무삼투압/저삼투압차를 달성하기 위해 사용하는 세가지 방법을 도식화한 도면으로, A는 Feed “Input-Split Cascade”방법을 의미하는데, 어떠한 농도의 고삼투압 용액도 유입용액을 두 갈래로 나누면 Feed chamber와 π(Pai)-Equalizer chamber로 분리하면 삼투압차는 zero가 된다 (즉 Δπ=0). 삼투압차는 zero에서 feed chamber에 압력을 가하면 Feed chamber는 농축이 되고 π-Equalizer chamber는 희석이 되어 삼투압 차가 생길 수 있다 B는 Feed Output-Split CC(counter-current) 방법으로, 농축된 용액의 일부 (예: 50%)를 π-Equalizer chamber로 보내면 또한 Feed output stream과 recycling (equalizer) stream간의 삼투압차는 Δπ=0가 된다. 이때는 두 stream은 counter-current로 흐른다. 예를 들면 feed chamber의 시작농도가 6%이면 π-Equalizer stream의 마지막 농도는 3%로 만들 수 있다. C는 용질의 reflection coefficient가 낮은 막을 사용하면 feed stream의 용질의 일부가 π-Equalizer stream으로 흘러 들어가 삼투압차를 낮출 수가 있음을 나타내며, 각각의 왼쪽 그래프는 이 세가지 방법을 활용하는 데 따라서 두 chamber간의 삼투압차를 나타낸 것이다.Figure 2 is a schematic diagram of three methods used to achieve the osmotic pressure / low osmotic pressure difference of the present invention, A means Feed "Input-Split Cascade" method, the high osmotic pressure of any concentration of the inlet solution Divided into two parts, the osmotic pressure difference becomes zero when the feed chamber and the π (Pai) -Equalizer chamber are separated (ie Δπ = 0). The osmotic pressure difference is zero and the feed chamber is concentrated, the feed chamber is concentrated and the π-Equalizer chamber is diluted so that osmotic pressure difference can occur. B is the Feed Output-Split CC (counter-current) method. Example: 50%) is sent to the π-Equalizer chamber and the osmotic pressure difference between the feed output stream and the recycling (equalizer) stream is Δπ = 0. At this time, both streams flow counter-current. For example, if the starting concentration of the feed chamber is 6%, the final concentration of the π-Equalizer stream can be made 3%. C indicates that the use of a membrane with a low solute reflection coefficient allows some of the solute in the feed stream to flow into the π-Equalizer stream to reduce the osmotic pressure difference.The graph on the left shows the difference between the two chambers according to these three methods. Osmotic pressure difference is shown.
도 3은 무삼투압차 상태에서 용액의 농도에 따른 ΔP와 투과량 (LMH=liters/(m2.h)을 측정하기 위한 장치를 나타낸 것으로, (가)는 고압피드챔버(A)와 상압 드로챔버(B)를 포함하는 무삽투압차 회분식 시스템의 설명도이며, (나)는 A와 B의 구체적인 설계도면이다.Figure 3 shows a device for measuring the ΔP and permeation (LMH = liters / (m2.h) according to the concentration of the solution in the osmotic pressure difference state, (a) is a high pressure feed chamber (A) and an atmospheric pressure draw chamber ( It is an explanatory drawing of a BPS system including B), and (B) is a specific design drawing of A and B.
도 4는 Input-split cascade system을 이용하여 3%의 NaCl용액을 물과 26.47% 포화용액(혹은 염)으로 분리할 수 있는 시스템 도면이다.FIG. 4 is a system diagram that can separate 3% NaCl solution into water and 26.47% saturated solution (or salt) using an input-split cascade system.
도 5는 본 발명의 일 실시예에 따른 flux 및 소모 에너지를 계산한 결과이다.5 is a result of calculating the flux and the energy consumption according to an embodiment of the present invention.
도 6은 본 발명의 일 실시예에 따른 무삼투압차 공정의 모사값을 계산한 결과이다.6 is a result of calculating the simulation value of the osmotic pressure difference process according to an embodiment of the present invention.
발명의 상세한 설명 및 바람직한 Detailed description of the invention and preferred 구현예Embodiment
다른 식으로 정의되지 않는 한, 본 명세서에서 사용된 모든 기술적 및 과학적 용어들은 본 발명이 속하는 기술분야에서 숙련된 전문가에 의해서 통상적으로 이해되는 것과 동일한 의미를 갖는다. 일반적으로 본 명세서에서 사용된 명명법은 본 기술분야에서 잘 알려져 있고 통상적으로 사용되는 것이다. Unless defined otherwise, all technical and scientific terms used herein have the same meaning as commonly understood by one of ordinary skill in the art to which this invention belongs. In general, the nomenclature used herein is well known and commonly used in the art.
역삼투압법에 소요되는 에너지가 재래식의 열을 이용하는 다단계 증발공법에 비해 훨씬 적다는 것은 이미 잘 알려져 있으며, 또한 최근에 연구되기 시작한 정삼투압법 역시 유도용액(draw solution)의 재생에 예상보다 많은 에너지가 소요되고, 유도용액이 손실되므로 정삼투압법이 역삼투압법에 비해 해수 담수화에 큰 실익이 없다는 점이 점점 부각되고 있다.It is well known that the energy required for reverse osmosis is much less than that of conventional multi-stage evaporation techniques. Also, recently studied forward osmosis is more energy than expected for regeneration of draw solutions. It is increasingly important that the forward osmosis method has no significant benefits in seawater desalination compared to the reverse osmosis method due to the loss of the induction solution.
해수 담수화와는 반대로 고부가가치의 저분자 물질을 포함하는 수용액을 역삼투압법으로 농축시킬 경우, 경제성이 있을 수도 있으나, 유도용액을 재생하는 데는 다단계증발법(MSF)과 같은 열적인 방법을 시용해야 하므로 시설비 및 에너지 비용이 부담으로 작용된다.Contrary to seawater desalination, it may be economical to concentrate an aqueous solution containing high value-added low molecular weight materials by reverse osmosis, but thermal methods such as multistage evaporation (MSF) must be used to recover the induction solution. Facility and energy costs are burdensome.
또한, 발효액의 농축에 있어서도 정삼투압법으로 3.5%를 4배로 농축하여 14%(용매기준)로 만드는 것은 어렵지 않았으나 실질적으로 발효산업에서 필요한 농도는 거의 발효산물의 20∼60무게%의 염포화수용액에서 에탄올의 순도 99.5% 액체까지 다양하나, 정삼투압법 만으로 소정의 목적을 달성하기는 어려운 실정이다.In addition, in the concentration of fermentation broth, it was not difficult to concentrate 3.5% four times by forward osmosis to 14% (solvent basis), but the concentration required in the fermentation industry was almost 20 to 60% by weight of saline solution. Although it varies from 99.5% purity of ethanol to the liquid, it is difficult to achieve a predetermined purpose only by forward osmosis.
또한 정삼투압에서는 고농도의 유도용액을 사용하므로 피드챔버로 확산되어 피드용액과 섞일 수가 있고 피드용액의 용질거절율 (rejection rate)이 100%이하인 경우 유도용액에 섞인 용질을 회수하는 것이 어렵다는 단점이 있다.In addition, in the case of forward osmosis, a high concentration of induction solution is used, which can be diffused into the feed chamber and mixed with the feed solution. If the rejection rate of the feed solution is less than 100%, it is difficult to recover the solute mixed in the induction solution. .
본 발명에서는 역삼투압 분리기를 이용하여 다양한 종류의 농축하고자 하는 용질 함유 수용액중 물은 외부로 배출시키고, 농축된 수용액을 역삼투압막 또는 정삼투압막으로 구획되어 있는 피드 챔버 및 드로우 챔버를 포함하는 농축장치의 피드 챔버에 투입시킨 다음, 피드 챔버에 투입된 수용액과 동일한 혹은 약간 삼투압을 가지는 용액을 드로우 챔버에 투입할 경우, 피드 챔버와 드로우 챔버간의 삼투압차가 0(△π =0)에 가깝게 되어 적은 압력만으로도 피드 용액을 농축할 수 있다는 것을 확인하고자 하였다. In the present invention, the water is discharged to the outside of the various types of solute-containing aqueous solution to be concentrated using a reverse osmosis separator, and the concentrated aqueous solution includes a feed chamber and a draw chamber partitioned by a reverse osmosis membrane or an forward osmosis membrane. When the solution is introduced into the feed chamber of the apparatus, and then a solution having the same or slightly osmotic pressure as the aqueous solution introduced into the feed chamber is introduced into the draw chamber, the osmotic pressure difference between the feed chamber and the draw chamber is close to 0 (Δπ = 0), and the pressure is low. It was intended to confirm that the feed solution can be concentrated alone.
본 발명자가 출원한 기존 특허(국제특허 제PCT/KR2014/000952호)에서는 열적인 방법으로 소금을 회수하는 방법을 채택하고 있고, 삼투압차를 줄이는 방식으로는 드로우용액의 체류시간을 조정하는 방법을 사용하였다. 그러나 위의 두가지 방법으로는 △π를 저삼투압/무삼투압 상태로 유지하는 것이 경제적/기술적인 어려움이 있다.Existing patent (international patent no. PCT / KR2014 / 000952) filed by the present inventors adopts a method of recovering salt by a thermal method, and a method of adjusting the residence time of a draw solution by reducing the osmotic pressure difference. Used. However, in the above two methods, it is economical and technically difficult to maintain Δπ at a low osmotic / no osmotic pressure.
본 발명의 일실시예에서는 (feed) input-split cascade, (feed) output-split count-current 또는 nano filteration membrane 적용 공정을 이용할 경우, 피드 챔버와 드로우 챔버사이의 삼투압차를 저삼투압/무삼투압 상태로 유지할 수 있다는 것을 확인하였다.In one embodiment of the present invention, when using the (feed) input-split cascade, (feed) output-split count-current or nano filteration membrane application process, the osmotic pressure difference between the feed chamber and the draw chamber is a low osmotic pressure / no osmotic pressure state It was confirmed that it can be maintained as.
본 발명에서는 꼭 무삼투압차 농축기에 feed를 역삼투압으로 물을 회수한 나머지 용질포함용액에 국한하여야 하는 데 대한 의문을 품게 되었다. 이는 용질포함 용액 (3%)에서 역삼투압으로 뽑는 물의 양을 줄이면서 전체에너지 소모가 줄어드는 것을 확인하고, 역삼투압-순수제조 공정을 거치지 않은 순수한 3% 소금물 (바닷물)을 무삼투압차에 농축기에 투입하여도, 용매 및 용질을 높은 효율로 분리할 수 있다는 것을 확인할 수 있었다.In the present invention, the question of whether the feed should be limited to the solute-containing solution after recovering the water by reverse osmosis in a non-osmotic pressure difference concentrator. This confirms that the total energy consumption is reduced while reducing the amount of water drawn from the solute-containing solution (3%) by reverse osmosis, and pure 3% brine (sea water), which has not undergone reverse osmosis-pure manufacturing, Even when added, it was confirmed that the solvent and the solute can be separated with high efficiency.
따라서, 본 발명은 일 관점에서, (a) 용질 함유 수용액을 역삼투압 막으로 구획되어 있는 피드 챔버와 π-이콜라이저 챔버를 포함하는 무삼투압차 농축기를 이용하여 농축하되, 하기의 공정 중 하나 이상의 공정을 이용하여 피드챔버와 π-이콜라이저 챔버간의 △π저감 조건을 형성하는 단계: (i) (Feed) Input-split cascade 공정; (ii) (Feed) Output-split counter-current 공정; 및 (iii) Nano filtration membrane을 적용하는 공정; 및 (b) 상기 농축된 용질 함유 수용액을 또다른 무삼투압차 농축기 또는 역삼투압을 이용하여 회수하는 단계를 포함하는 △π 저감 조건에서 용질 함유 수용액을 농축하는 방법에 관한 것이다.Accordingly, the present invention in one aspect, (a) the solute-containing aqueous solution is concentrated using an osmotic pressure concentrator comprising a feed chamber and a π-echolizer chamber partitioned with a reverse osmosis membrane, wherein at least one of Forming a Δπ reduction condition between the feed chamber and the π-echolizer chamber using the process: (i) (Feed) Input-split cascade process; (ii) (Feed) Output-split counter-current process; And (iii) applying a nano filtration membrane; And (b) recovering the concentrated solute-containing aqueous solution using another osmotic pressure difference concentrator or reverse osmosis pressure.
본 발명에 있어서, (c) 상기 (b) 단계의 또다른 무삽투압차 농축기는 (i) 내지 (iii) 공정 중 하나 이상을 이용하여 용질의 회수를 최대화하는 단계를 추가로 포함하는 것을 특징으로 할 수 있다.In the present invention, (c) another non-intrusive pressure difference concentrator of step (b) further comprises the step of maximizing the recovery of the solute using one or more of the processes (i) to (iii). can do.
본 발명에 있어서, 상기 (a) 단계에 이전에 역삼투압을 이용하여 농축 대상 용질 함유 수용액을 농축시키는 단계를 추가로 포함하는 것을 특징으로 할 수 있다.In the present invention, the step (a) may be characterized in that it further comprises the step of concentrating the solute-containing aqueous solution to be concentrated before using the reverse osmosis.
본 발명에 있어서, 또다른 무삼투압차 농축기를 이용하여 용질 함유 수용액을 농축시키는 (b) 단계는 (i) 무삼투압차 농축기의 피드 챔버로 상기 농축된 수용액을 이송시키고, 상기 피드 챔버로 이송된 농축된 수용액과 동일한 삼투압을 가지는 용액을 드로우 챔버로 투입하여, 피드 챔버와 드로우 챔버간의 무삼투압 상태를 형성하는 단계; 및 (ii) 무삼투압 상태에서 피드 챔버에 정삼투압(Forward Osmosis,FO)의 경우에는 0∼5기압, 역삼투압의 경우는 10∼200기압까지의 압력을 가하여, 농축된 수용액중의 물을 드로우 챔버로 이송시킴으로써 농축된 수용액을 추가로 농축시키는 단계를 포함할 수 있다.In the present invention, the step (b) of concentrating the solute-containing aqueous solution using another osmotic pressure difference concentrator (i) transfers the concentrated aqueous solution to the feed chamber of the osmotic pressure difference concentrator, and is transferred to the feed chamber. Injecting a solution having the same osmotic pressure as the concentrated aqueous solution into the draw chamber to form an osmotic pressure state between the feed chamber and the draw chamber; And (ii) a pressure of up to 0 to 5 atm for forward osmosis (FO) and 10 to 200 atm for reverse osmosis is applied to the feed chamber at zero osmotic pressure, and the water in the concentrated aqueous solution is drawn. And further concentrating the concentrated aqueous solution by transferring to the chamber.
본 발명에서는 피드 챔버로 공급되는 피드스트림과 π-이콜라이저 챔버로 공급되는 π-이콜라이저간의 삼투압차가 커질 때는 (i) Input-split cascade 공정; (ii) Output-split counter-current 공정; 또는 (iii) nano filtration membrane을 적용하는 공정을 사용하여 그 차이를 줄이는 것을 특징으로 할 수 있다.In the present invention, when the osmotic pressure difference between the feed stream supplied to the feed chamber and the π-echoizer supplied to the π-echolizer chamber is increased (i) an input-split cascade process; (ii) an output-split counter-current process; Or (iii) using the process of applying the nano filtration membrane to reduce the difference.
본 발명에서 용어 “(feed) Input-split cascade 공정”, “Input-split cascade 공정”은 무삼투압차 농축기로 공급되는 피드 용액의 절반은 피드 챔버로, 절반은 π-이콜라이저 챔버로 공급하여 무삼투압차를 형성한 다음, 저압으로 피드 챔버의 수용액을 농축하고, 농축된 용질 함유 수용액을 또 다른 무삼투압 농축기로 이송시키는 방법을 의미한다.In the present invention, the terms “(feed) input-split cascade process” and “input-split cascade process” refer to half of the feed solution fed to the osmotic pressure concentrator and half to the feed chamber, and half to the π-echoiser chamber. After forming the osmotic pressure difference, the aqueous solution of the feed chamber is concentrated at low pressure, and the concentrated solute-containing aqueous solution is transferred to another osmotic pressure concentrator.
본 발명에서 용어 “△π 저감 조건”은 피드 챔버와 π-이콜라이저 챔버 사이의 삼투압 차이가 없거나 매우 작은 값을 가지는 조건을 의미한다. 즉, 본 발명의 △π 저감조건은 △π가 0 이거나, 1 내지 100bar 이하인 경우를 의미한다.As used herein, the term “Δπ reduction condition” means a condition in which there is no osmotic pressure difference or a very small value between the feed chamber and the π-echolizer chamber. That is, the Δπ reduction condition of the present invention means a case where Δπ is 0 or 1 to 100 bar or less.
본 발명의 용어 “(feed) output-split counter-current 공정” 또는 “output-split counter-current 공정”은 무삼투압차 농축기에서 농축된 상태에서 π-이콜라이저 스트림의 일부를 피드스트림으로 보내는 것을 특징으로 할 수 있다.The term “(feed) output-split counter-current process” or “output-split counter-current process” of the present invention is characterized by sending a portion of the π-echolizer stream to the feed stream in a concentrated state in an osmotic pressure concentrator. You can do
본 발명의 용어 “π(Pai)-이콜라이저 챔버”는 드로우 챔버와 같은 의미로 사용되며, 피드스트림에서 필터액(물)을 받아 희석되는 점은 동일하지만 기동력이 정투압법에서는 드로우 챔버의 농축된 유도용액인데 비해 여기서는 수력학적 압력(ΔP)라는 점에서 차이가 있다. 참고로 본 발명자들은 정삼투압법에서도 ΔP를 시험했지만 3기압정도가 최고이었고 막도 달라 성공하지 못했다.The term “π (Pai) -echolizer chamber” of the present invention is used in the same sense as the draw chamber, and the dilution of the filter liquid (water) in the feed stream is the same, but the maneuverability in the orthostatic method is concentrated in the draw chamber. Compared to the induction solution, the difference here is the hydraulic pressure (ΔP). For reference, the present inventors tested ΔP in the forward osmosis method, but about 3 atmospheres were the best and the membranes did not succeed.
본 발명에서 상기 단계 (a)는 (i) 내지 (iii) 단계를 각각 적용할 수 있고, (i) 및 (ii), (i) 및 (iii), (ii) 및 (iii) 또는 (i) 내지 (iii) 단계를 모두 적용하여 △π 저감 조건을 형성할 수 있다.In the present invention, step (a) may apply steps (i) to (iii), respectively, and (i) and (ii), (i) and (iii), (ii) and (iii) or (i) )) To (iii) may be applied to form the Δπ reduction conditions.
본 발명의 다른 실시예에서는 3% 소금물을 농축하여, Input-split cascade 공정 및 Output-split counter-current 공정으로 소금 결정과 물을 얻을 수 있다는 것을 확인하였다.In another embodiment of the present invention it was confirmed that by concentrating 3% brine, salt crystals and water can be obtained by the input-split cascade process and the output-split counter-current process.
즉, 본 발명의 다른 실시 예에서는 3% 소금물이 Input-split cascade 공정 및 Output-split counter-current 공정으로 3%의 NaCl용액이 26.47%의 포화용액에서 역삼투압법으로 완전히 결정형태 (고형분)로 농축되는 공정을 모사하였고 결정화 전문가와 상의한 결과 가능하다는 조언을 얻었다. Input-split cascade 공정 (그림 2A)은 각 농도 구간마다 피드스트림과 π(삼투압)-이콜라이저 스트림으로 나뉘고 용질은 피드 챔버에 남아 있고 용매인 물만 π-이콜라이저로 움직여 피드챔버의 용액농도가 올라가고 π-이콜라이저는 감소한다. 그러나 두 챔버간의 삼투압차(Δπ)는 처음에는 0에서 어느 이상으로 올라가면 (예 60 bar)으로 올라가면 예를 들면 6% stream은 농축 stream은 9%와 희석 stream은 3%로 만들 수 있다. 이는 두 챔버 간의 크기의 비 (A)와 feed chamber에서 얼만 만큼의 물을 여과하느냐에 달려 있다 (여과량/feed chamber=T). 즉 어떠한 농도이든 이 두 가지 인자 (A, T)를 조정하면 원하는 농축농도, 희석농도를 만들 수 있으며 농축수는 위로, 희석수는 아래로 보내어 물과, 소금으로 분리할 수 있다 (실시예 3,4 참조).That is, in another embodiment of the present invention, 3% brine is completely converted into crystalline form (solid content) by reverse osmosis in 3% NaCl solution in 26.47% saturated solution by input-split cascade process and output-split counter-current process. The process of concentration was simulated and consulted with a crystallization expert to get advice. The input-split cascade process (Figure 2A) is divided into feed stream and π (osmotic pressure) -echolizer stream for each concentration section, solute remains in the feed chamber and only solvent, water, is moved to the π-echoiser to increase the solution concentration in the feed chamber. π-echolizer decreases. However, the osmotic pressure difference (Δπ) between the two chambers can initially be raised from zero to more than one (eg 60 bar), for example a 6% stream can be 9% concentrated and 3% dilution. This depends on the ratio of the size between the two chambers (A) and how much water is filtered out of the feed chamber (filtration / feed chamber = T). That is, by adjusting these two factors (A, T) at any concentration, the desired concentration and dilution can be made, and the concentrated water can be sent up and the diluted water can be separated into water and salt (Example 3 , 4).
한편 Output-split CC공정(그림 2B)은 Input으로 들어간 용액은 압력을 가하면 용매만 피드챔버에서 π-이콜라이저 챔버로 흘러 들어가 input stream은 6%에서 농축농도 (26.47%)까지 농축이 된다. 피드 챔버의 마지막에 포화된 농축농도의 feed output stream의 50%는 π-이콜라이저 tank로 들어가 feed output과 π-이콜라이저 input이 무삼투압차를 형성한다. π-이콜라이저 스트림이 feed stream과 counter-current로 흐르면서 농도는 감소한다. Equalizer의 output은 피드스트림의 포화된 농축용액이 시스템 바깥으로 배출되는 경우는 970-83.34=886.66만 회수가 가능하다. RO로 회수한 물 500을 제외하면 428.33 만 회수가 가능하여 15/(15+428.33)=3.38%가 될 것이다. 물론 결정화 등으로 물을 470을 전부회수하게 되면 3.00%가 된다. In the output-split CC process (Figure 2B), when the input solution is pressurized, only solvent flows from the feed chamber into the π-echoizer chamber, and the input stream is concentrated from 6% to the concentration (26.47%). At the end of the feed chamber, 50% of the saturated feed output stream enters the π-echolizer tank and the osmotic pressure difference between the feed output and π-echolizer input. The concentration decreases as the π-echoizer stream flows into the feed stream and counter-current. The output of the equalizer can only recover 970-83.34 = 886.66 when the saturated concentrate of the feed stream exits the system. Excluding the 500 recovered by RO, only 428.33 would be recovered, resulting in 15 / (15 + 428.33) = 3.38%. Of course, the total recovery of water 470 due to crystallization, etc. is 3.00%.
3% input에서 RO를 통하며 50%의 물을 회수 할 수 있으며 나머지는 6% 470L는 Recycle unit으로 들어가 Δπ 가 0~60 bar를 유지하면서 π-이콜라이저 output으로 되돌아 RO-2 unit에서 물을 생산하고 나머지는 피드 스트림으로 되돌아가 재순화하게 되는 데 30g의 염을 제거하기 위해서는 최소한 두 번 재순환 하여야 한다 (실시예 5 참조).50% water can be recovered through RO at 3% input and the rest 6% 470L enters recycle unit and returns to π-equalizer output while maintaining Δπ 0 ~ 60 bar. The remainder of the product is returned to the feed stream and recycled, which must be recycled at least twice to remove 30 g of salt (see Example 5).
본 발명에서 염을 회수할 때는 포화용액에 당업자에게 잘 알려진 열에너지 또는 전기에너지를 이용하는 방법을 적용하거나, 피드 Input-split cascade 공정 및 피드 output-split counter-current 공정을 적용하는 것을 특징으로 할 수 있으나, 이에 한정되는 것은 아니다.When recovering the salt in the present invention may be characterized by applying a method using thermal energy or electrical energy well known to those skilled in the art in a saturated solution, or by applying a feed input-split cascade process and a feed output-split counter-current process It is not limited to this.
본 발명의 또다른 실시 예에서는 피드 챔버와 드로우 챔버 사이의 삼투압차를 줄이기 위해, 상기 방법으로 농축된 용질 함유 수용액의 일부를 상기 드로우 챔버로 재순환시키되, 나노필트레이션(nano filtration)을 이용할 경우, 삼투압차가 빠르게 줄어드는 것을 확인하였다.In another embodiment of the present invention, in order to reduce the osmotic pressure difference between the feed chamber and the draw chamber, a portion of the solute-containing aqueous solution concentrated by the above method is recycled to the draw chamber, but using nano filtration, It was confirmed that the osmotic pressure difference decreased rapidly.
따라서, 본 발명에서 삼투압을 줄이기 위해, 나노필트레이션 막을 이용하는 것을 특징으로 할 수 있다.Therefore, in order to reduce the osmotic pressure in the present invention, it may be characterized by using a nanofiltration membrane.
본 발명에 있어서, 상기 (a) 단계의 (iii) 공정은 (i) 공정에 적용할 때는 처음부터, (ii) 공정에 적용할 때는 농축수 회수정도와 삼투압차를 고려하여 초기에는 선택도가 높은 막을 후기로 가 삼투압차가 커질수록 선택도가 낮은 막을 선택하여 삼투압차를 줄일 수 있는 것을 특징으로 할 수 있으나, 이에 한정되는 것은 아니다.In the present invention, the step (iii) of the step (a) is selected from the beginning when applied to the process (i), the initial selectivity in consideration of the degree of concentrated water recovery and osmotic pressure difference when applied to the process (ii) The higher the late membrane, the greater the osmotic pressure difference may be characterized in that the osmotic pressure difference can be reduced by selecting a membrane with low selectivity, but is not limited thereto.
본 발명에서 순수한 물을 생산할 때 역삼투압법을 사용하는 것을 특징으로 할 수 있고, 물을 생산할 때 RO 공정을 시스템으로 사용하는 것을 특징으로 할 수 있다.In the present invention, the reverse osmosis method may be used when producing pure water, and the RO process may be used as a system when producing water.
본 발명의 방법으로 농축한 포화용액은 강물이나 해수를 이용하여 PRO(압력지연삼투압법, pressure retarded osmosis)공법을 사용하여 전력을 생산 할 수 있다.The saturated solution concentrated by the method of the present invention can produce power using PRO (pressure retarded osmosis) method using river water or seawater.
본 발명에 따른 무삼투압차 농축장치는 다수로 구성될 수 있다. 즉, 무삼투압차 농축장치를 구성하는 피드 챔버 및 π-이콜라이저 챔버는 다단계로 구성되어 있는 것을 특징으로 한다.The osmotic pressure difference concentrator according to the present invention may be composed of a plurality. That is, the feed chamber and the π-echolizer chamber constituting the osmotic pressure difference concentrator are characterized in that it is composed of a multi-stage.
본 발명에 있어서, 상기 무삼투압차 농축장치의 피드 챔버와 드로우 챔버를 구획하는 역삼투압막 또는 정삼투압막은 용질은 통과시키지 않고, 용매를 주로 통과시키는 막이라면 특별한 제한 없이 이용할 수 있다.In the present invention, the reverse osmosis membrane or the forward osmosis membrane partitioning the feed chamber and the draw chamber of the osmotic pressure difference concentrator may be used without particular limitation as long as it does not pass the solute and mainly passes the solvent.
본 발명에 있어서, 용질이라 함은 물을 용매로 하여 녹아 있는 액체 혹은 고체 상태의 물질을 의미한다. In the present invention, the solute means a liquid or solid substance dissolved in water as a solvent.
본 발명에 있어서, 상기 농축 대상 용질 함유 수용액은 해수(sea water), 기수(brackish water), 세포 대사산물, 반응액 등을 예시할 수 있으며, 세포 대사산물은 동물 세포, 식물 세포 또는 미생물의 배양액, 이들의 1차 산물, 2차 산물, 체외 분비 단백질, 생물 변환물(biotransformation) 등을 포함하는 개념이지만 분자량이 작아 삼투압이 높은 경우를 경우에 적합한 공정이라 할 수 있다. In the present invention, the solute-containing aqueous solution to be concentrated may include sea water, brackish water, cell metabolites, reaction solutions, and the like, and cell metabolites may be cultured cells of animal cells, plant cells, or microorganisms. The concept includes a primary product, a secondary product, an in vitro secreted protein, a biotransformation, and the like, but may be a suitable process when the molecular weight is small and the osmotic pressure is high.
상기 반응액으로는 화학반응을 통한 반응액 및 효소반응을 통한 반응액을 예시할 수 있다. Examples of the reaction solution include a reaction solution through a chemical reaction and a reaction solution through an enzyme reaction.
상기 미생물의 1차 산물로는 유기산(초산, 프로피온산, 부틸산, 유산, 구연산, 유산, 숙신산 등), 알코올(에탄올, 부탄올 등), 핵산, 아미노산(라이신, 트립토판 등), 비타민, 다당류 등을 예시할 수 있으나, 이에 한정되는 것은 아니다.The primary products of the microorganisms include organic acids (acetic acid, propionic acid, butyric acid, lactic acid, citric acid, lactic acid, succinic acid, etc.), alcohols (ethanol, butanol, etc.), nucleic acids, amino acids (lysine, tryptophan, etc.), vitamins, polysaccharides, and the like. It may be illustrated, but is not limited thereto.
상기 미생물의 2차 산물로는 항생제(폐니실린 등), 효소저해제, 생리활성물질(탁솔 등)을, 상기 미생물의 체외 분비 단백질로는 아밀라제, 셀룰라제 등의 효소, 인슐린, 인터페론, 단일군 항체 등을, 상기 미생물의 생물변환물(biotransformation)은 미생물 또는 효소를 활용하여 생산된 물질로서, 스테로이드 등을 예시할 수 있으나, 이에 한정되는 것은 아니다.The secondary products of the microorganisms include antibiotics (such as lung nicillin), enzyme inhibitors, physiologically active substances (taxols, etc.), and the in vitro secreted proteins of the microorganisms include enzymes such as amylase and cellulase, insulin, interferon, and single group antibodies. For example, the biotransformation of the microorganism is a substance produced by using a microorganism or an enzyme, and examples thereof may include steroids, but are not limited thereto.
예를 들어, 본 발명의 방법을 이용하여, 에탄올을 농축시킬 경우, 역삼투압 농축기(RO-1)에서 에탄올을 농축시킬 수 있는 농도는 약 20% 정도이며, 이론적으로 무삼투압차 방법으로 20~100%까지 농축할 수 있는 것으로 알려졌다. 그러나 에탄올을 휘발성 유기산(VFA)이나 소금처럼 91~100%까지 농축시킬 수 있는 막은 아직 개발되지 않고 있다.For example, when ethanol is concentrated using the method of the present invention, the concentration of ethanol can be concentrated in a reverse osmosis concentrator (RO-1) is about 20%, theoretically 20 ~ by the osmotic pressure difference method It is known to be able to concentrate up to 100%. However, membranes that can concentrate ethanol up to 91-100%, such as volatile organic acids (VFA) or salt, have not yet been developed.
발성 유기산염 (VFA)의 경우는 포화도가 50∼60wt% 정도 이므로 이론적으로 100% 농축이 가능하며, 또 용질 거절율 (rejection rate)도 높아 무삼투압차 상태에서 100% 농축이 가능하다.In the case of VFA, the saturation degree is about 50 to 60 wt%, so it is theoretically possible to concentrate 100%, and the high solute rejection rate is also 100% concentrated in the absence of osmotic pressure difference.
본 발명에서 용질 저 선택성 막이라 함은 선택도가 1보다 낮고 0 보다 막으로 통상적으로는 나노필터레이션 막에 해당하지만 이에 국한되는 것은 아니다. In the present invention, the solute low selectivity membrane is a membrane having a selectivity lower than 1 and higher than 0, which generally corresponds to a nanofiltration membrane, but is not limited thereto.
본 발명에서 대상물질은 고체인 경우는 온도, pH에 따라 결정화가 용이하고 또 고농도에 점도가 높지 않은 물질에 적합하다. 액체의 경우 연료용 알콜이 좋은 응용 예 일 수 있지만 여기에 국한 되는 것은 아니다.In the present invention, when the target material is a solid, it is easy to crystallize according to temperature and pH, and is suitable for a material having high viscosity at high concentration. For liquids, alcohol for fuel may be a good application but is not limited to this.
본 발명에 있어서, 상기 수용액의 pH는 2-13, 온도는 물이 액체를 유지하는 온도 (통상 0∼100℃, 바람직하게는 15∼50℃, 더욱 바람직하게는 20∼40℃) 혹은 그 이상이나 이하가 될 수 있다. 예를 들면 다른 용질/용매와 혼합물인 경우는 위의 온도를 벗어날 수 있다.In the present invention, the pH of the aqueous solution is 2-13, the temperature is the temperature at which water maintains the liquid (usually 0 to 100 ℃, preferably 15 to 50 ℃, more preferably 20 to 40 ℃) or more Or can be For example, mixtures with other solutes / solvents may deviate from the above temperatures.
본 발명에 있어서, 피드 챔버와 드로우 챔버간의 무삼투압 상태를 형성하기 위하여 π-이콜라이저 챔버로 투입하는 용액으로는 피드 챔버로 이송된 농축된 수용액, 사용 후 분리가 용이한 용액 등을 이용할 수 있으나, 상기 피드챔버로 이송된 농축된 수용액과 동일한 조성의 수용액을 이용하는 것이 바람직하다.In the present invention, as a solution introduced into the π-echolizer chamber to form an osmotic pressure between the feed chamber and the draw chamber, a concentrated aqueous solution transferred to the feed chamber, a solution that can be easily separated after use, and the like may be used. It is preferable to use an aqueous solution of the same composition as the concentrated aqueous solution transferred to the feed chamber.
본 발명에 있어서, 상기 무삼투압차 농축기를 이용한 농축은 그 효과를 극대화하기 위하여 회분식 또는 연속식으로 수행될 수 있다.In the present invention, the concentration using the osmotic pressure difference concentrator may be performed in a batch or continuous manner to maximize the effect.
상기 회분식은 양챔버와 외부시스템과 흐름이 없는 경우, 상기 연속식은 외부시스템과 흐름이 있는 경우 수행될 수 있다.The batch may be performed when there is no flow with both chambers and the external system, and the continuous may be performed when there is a flow with the external system.
본 발명에 있어서, 상기 피드 챔버 및 π-이콜라이저 챔버는 다단계로 구성되어 있는 것을 특징으로 한다.In the present invention, the feed chamber and the π-equalizer chamber is characterized in that it is composed of a multi-stage.
본 발명에 있어서, 상기 무삼투압차 농축기에서 추가로 농축된 수용액으로부터 용질 및 물을 회수하는 방법은 각각 독립적으로 통상적으로 알려져 있는 다단 증발법, 투석증발, 열분해법, 황산법, 칼슘법 및 Input-split cascade 등을 이용할 수 있으나, 이에 한정되는 것은 아니다.In the present invention, the method of recovering the solute and the water from the aqueous solution further concentrated in the osmotic pressure difference concentrator is independently a commonly known multistage evaporation method, dialysis evaporation, pyrolysis method, sulfuric acid method, calcium method and input-split. cascade may be used, but is not limited thereto.
본 발명에 있어서, 상기 농축방법은 (d) 상기 (c) 단계의 (i) 내지 (iii) 공정 중 어느 하나를 이용하여 용질 또는 물 중 어느 하나의 회수를 최대화하는 단계를 추가로 포함하는 것을 특징으로 할 수 있다.In the present invention, the concentration method further comprises the step of (d) maximizing the recovery of either solute or water using any one of the steps (i) to (iii) of step (c). It can be characterized.
본 발명에 있어서, 압력지연 정삼투압(PRO) 발전, 자원이용 및 희토류 회수 공정 등을 추가로 포함하여 상기 공정의 부가가치를 높이는 것을 특징으로 할 수 있다.In the present invention, the pressure-added forward osmosis (PRO) power generation, resource utilization and rare earth recovery process may be further included to increase the added value of the process.
본 발명에 있어서, 물질수지 및 에너지수지를 최적화하는 공정을 추가로 포함하는 것을 특징으로 할 수 있다.In the present invention, it may further comprise a step of optimizing the material balance and the energy balance.
본 발명은 또한, 상기의 농축 방법을 이용하여 용질 함유 용액에서 용매와 용질을 분리하는 방법에 관한 것이다.The present invention also relates to a method for separating a solvent and a solute from a solute-containing solution using the above concentration method.
본 발명에 있어서, 상기 용질은 염이고, 상기 용매는 물인 것을 특징으로 할 수 있다.In the present invention, the solute is a salt, the solvent may be characterized in that the water.
이하, 실시예를 통하여 본 발명을 더욱 상세히 설명하고자 한다. 이들 실시예는 오로지 본 발명을 예시하기 위한 것으로, 본 발명의 범위가 이들 실시예에 의해 제한되는 것으로 해석되지 않는 것은 당업계에서 통상의 지식을 가진 자에게 있어서 자명할 것이다.Hereinafter, the present invention will be described in more detail with reference to Examples. These examples are only for illustrating the present invention, and it will be apparent to those skilled in the art that the scope of the present invention is not to be construed as being limited by these examples.
실시예 1: 물 및 용질의 회수 방법Example 1 Recovery of Water and Solutes
1-1. 물의 회수 방법1-1. Recovery method of water
3%소금물에서 압력을 가하면 물의 flux는 Jw=Lp (ΔP - Δπ)로 쉽게 구할 수 있다. 표1/표4에 개시된 바와 같이 3% 소금물의 삼투압은 23.743bar이고 이때 최소에너지는 0.659kwh/m3이다. 물을 500을 RO로 회수하면 용질 30kg (solute rejection 100%가정)과 물 470kg이 남게 되며 농도는 6%가 된다. 6%의 삼투압은 50.467 bar이고 역삼투압에 소요되는 에너지는 When pressure is applied at 3% salt water, the flux of water can be easily obtained as Jw = Lp (ΔP-Δπ). As disclosed in Table 1 / Table 4, the osmotic pressure of 3% brine is 23.743 bar with a minimum energy of 0.659 kwh / m3. Recovery of water at 500 RO leaves 30 kg (solute rejection 100% home) and 470 kg water, leaving a concentration of 6%. The osmotic pressure of 6% is 50.467 bar and the energy required for reverse osmosis
3%와 6%의 삼투압의 산술평균을 이용하여 계산하면 (23.743+50.467)/2*0.5/36=0.515 kwh 이다.Using the arithmetic mean of osmotic pressure of 3% and 6%, it is (23.743 + 50.467) /2*0.5/36=0.515 kwh.
1-2. Recycle stream에서 물을 회수 하는 방법1-2. How to recover water from recycle stream
Recycle stream에서는 recycle energy 와 물 회수용 RO에 에너지가 소모된다. Recycle stream에 관한 에너지는 방법에 따라 다르므로 물을 100% 회수하는 경우는 470L가 회수되며 0.515*0.470/0.500=0.484kwh가 추가된다. 여기에 recycle energy를 추가된다.In the recycle stream, energy is consumed for recycle energy and RO for water recovery. The energy associated with the recycle stream depends on the method, so for 100% recovery of water, 470 L is recovered and 0.515 * 0.470 / 0.500 = 0.484 kwh is added. Recycle energy is added here.
염을 회수하지 않고 포화용액으로 배출할 경우는 한번 회수할 때 물 41.67kg, 소금 15kg만 배출하므로 두 번 recycle하여야 한다. 즉 470-41.67=428.33Kg이 회수된다. 그런데 6%용액으로 두 번 째 recycle하는 데 물이 235kg이 공급되어야 하므로 회수할 수 있는 양은 193.33이 된다. 그래서 두 번 recycle해서 회수되는 물의 양은 386.66kg이 된다. 이 양은 대표도 1의 ② RO unit에서 회수되는 양이다. 즉 970kg-83.34kg=886.66kg (=500kg+386.66kg)으로 일치한다. 중요한 원칙은 두번째 stream에서 물을 회수할 수 있는 양은 product stream이 3%로 돌아오면 235×2=470kg이 회수되지만 그렇지 않으면 회수될 수 있는 양이 줄어든다는 점에 특색이 있다.In case of discharging to saturated solution without recovering salt, 41.67kg of water and 15kg of salt are discharged twice. That is, 470-41.67 = 428.33 Kg is recovered. By the way, the second recycle with 6% solution has to be supplied with 235kg of water, so the recoverable amount is 193.33. Thus, the amount of water recovered by recycling twice is 386.66 kg. This amount is the amount recovered from ② RO unit of representative figure 1. 970kg-83.34kg = 886.66kg (= 500kg + 386.66kg). An important principle is that the amount of water recoverable in the second stream is 235 × 2 = 470 kg when the product stream returns to 3%, but otherwise the amount that can be recovered is reduced.
Figure PCTKR2016008104-appb-T000004
Figure PCTKR2016008104-appb-T000004
상기 표 4는 3% NaCl의 삼투압을 계산한 것이며, 표 각 열의 설명은 하기와 같다.Table 4 calculates the osmotic pressure of 3% NaCl, the description of each column of the table is as follows.
#1: w/w%는 물과 소금의 무게 비율이다. 3%의 소금물은 물 970kg과 소금 30kg으로 구성되어 있다# 1: w / w% is the weight ratio of water and salt. 3% brine consists of 970kg water and 30kg salt
#2: 삼투압계산에 쓰이는 몰분율 (2*NaCl/(총몰수) 이다.# 2: Mole fraction (2 * NaCl / (total moles)) used for osmotic pressure calculation.
#3: 각 w/w%에서 OLI_Analyzer software를 사용하여 계산한 삼투압이다. (https://en.wikipedia.org/wiki/OLI_Aanlyzer)# 3: Osmotic pressure calculated using OLI_Analyzer software at each w / w%. (https://en.wikipedia.org/wiki/OLI_Aanlyzer)
#4: Differential pressure로 각 % 사이의 % 삼투압을 계산할 때 사용한다.# 4: Differential pressure is used to calculate the% osmotic pressure between each%.
#5: 각%의 농도에서 원수가 무한 공급되어 %가 변하지 않는 상태에서의 물 1톤을 생산할때 필요한 에너지 양이다.# 5: The amount of energy required to produce one ton of water with an unchanged percentage of raw water at each concentration.
#6: Volume1은 소금 30kg을 기준으로 각%용액에 포함되어 있는 물의 양이다.# 6: Volume 1 is the amount of water contained in each% solution based on 30 kg of salt.
#7: Differential volume으로 각 구간간의 Volume의 차이이다.# 7: Differential volume is the difference in volume between each section.
#8: 각 구간별 differential energy 량이다.# 8: Differential energy amount for each section.
#9: #8을 누적한 값이다. 6%에서 26.47%까지 필요한 에너지는 1.7512-0.4750=1.2402이다.# 9: Accumulated # 8 The energy required from 6% to 26.47% is 1.7512-0.4750 = 1.2402.
검토 결과, 3%는 농도는 낮으나 구간에 많은 물이 포함되어 있어 differential energy가 높은 반면 고농도 예 25%이후는 압력이 높으나 구간에 포함된 물이 워낙 적어 큰 영향을 못 미친는 것을 확인할 수 있었다.As a result, it was found that 3% had a low concentration but a lot of water was included in the section, so that the differential energy was high, whereas after 25% of the high concentration, the pressure was high, but the water contained in the section had little effect.
실시예 2: Δπ=0 상태에서 농도에 따른 플럭스( LMH, liters/ (m2.h) Example 2 Flux with Concentration at Δπ = 0 (LMH, liters / (m2.h)
도 3 (a)와 같은 장치에 시료(NaCl 수용액)를 넣고 압력을 가하여 주어진 시간동안에 고압 A 챔버에서 RO막을 통해서 투과한 양이 B챔버 (상압)에 무게로 나타난 양을 측정한다.A sample (NaCl aqueous solution) is put in the apparatus as shown in FIG.
본실시예에 사용된 막은 RE2521-TL (Woongjin Chemical Co, Seoul, Korea;http://www.csmfilter.com) Ltd로서 thin-film composite형이며 막은 negatively charged, polyamide, spiral-wound module에 사용되고 있는 RO사용막이다. 투과능력은 1.1m3/day이고 유효면적은 1.1m2이다. 1,500mg/L 소금용액, 1.0MPa에서 99% rejection 하며, 최대압력은 4.14MPa이고 최대 유량은 1.36m3/hr, 최저 유량(농축액)은 0.23m3/hr이다. 최대 온도는 45℃이고 pH는 3.0∼10.0이며 또한 2.0∼11.0까지 견딜 수 있다.The membrane used in this example is RE2521-TL (Woongjin Chemical Co, Seoul, Korea; http: //www.csmfilter.com) Ltd, which is a thin-film composite type and the membrane is used in a negatively charged, polyamide, spiral-wound module. RO membrane. The permeability is 1.1m 3 / day and the effective area is 1.1m 2 . 99% rejection at 1,500mg / L salt solution, 1.0MPa, maximum pressure is 4.14MPa, maximum flow rate is 1.36m 3 / hr, minimum flow rate (concentrate) is 0.23m 3 / hr. The maximum temperature is 45 ° C. and the pH is 3.0-10.0 and can withstand 2.0-11.0.
하기 표5의 조성을 가진 NaCl용액을 A, B 챔버에 500ml씩 채운 실험을 실하면 시간에 따른 투과량이 B챔버의 무게로 나타나 이를 컴퓨터로 모니터 한다. NaCl의 경우는 conductivity meter (CM-31p, TOADKK, Japan)를 사용했고 VFA-Na의 경우는 HPLC(YoungLin, Korea)를 사용하였다(표 5).When the NaCl solution having the composition shown in Table 5 below was filled with 500 ml in the A and B chambers, the amount of permeation with time appeared as the weight of the B chamber and monitored by a computer. For NaCl, a conductivity meter (CM-31p, TOADKK, Japan) was used, and for VFA-Na, HPLC (YoungLin, Korea) was used (Table 5).
Figure PCTKR2016008104-appb-T000005
Figure PCTKR2016008104-appb-T000005
표 5에 개시된 바와 같이, 1.5~3% 부분은 플러스가 임계 압력을 넘어 서면 직선 관계가 성립되나 그 이후의 농도에서는 (ΔP)α(C)β의 비례 했는데 여기서 α는 0~1, β는 -1~0을 표시했다. 즉 농도가 높을 수록 플럭스가 감소했으며 압력이 증가할 수록 증가하였으며, 최고 압력은 40 bar 였다. As shown in Table 5, the 1.5 to 3% portion has a linear relationship where the plus is above the critical pressure, but at subsequent concentrations it is proportional to (ΔP) α (C) β where α is 0 to 1 and β is -1 to 0 are displayed. The higher the concentration, the lower the flux and the higher the pressure. The maximum pressure was 40 bar.
실시예 3: Feed Input-Split Cascade 공정 기반 용질/용매 100% 분리Example 3: 100% Separation of Solute / Solvent Based on Feed Input-Split Cascade Process
도 4에 개시된 바와 같이 본 발명의 공정은 3%의 input stream, 하부의 물 생산시스템, 최 상부의 포화농축용액 혹은 이로 염의 생산 가능하다.As shown in FIG. 4, the process of the present invention is capable of producing an input stream of 3%, a water production system at the bottom, a saturated concentrated solution at the top or a salt thereof.
3% 용액은 500L의 순수를 생산하고 6%, 470L물(합계 500kg 용액)으로 6% chamber로 진입한다. 여기서 Input-Split 방법으로 9%의 농축액과 3%의 용액으로 분리되어 6%는 상부의 9% chamber로 3%는 RO-2장치에서 235kg의 원수를 생산하고, 나머지 235kg은 Recycle로 chamber로 들어간다. 9→12→15→18→21→24→26.47%(sat.)로 농축되고, 희석된 스트림은 26.47%→21%→18→15→12→9→6→3→0 (순수)로 회수된다.The 3% solution produces 500 L of pure water and enters the 6% chamber with 6%, 470 L water (total 500 kg solution). Here, the input-split method is used to separate 9% concentrate and 3% solution, 6% of which is a 9% chamber at the top, 3% of which produces 235 kg of raw water in the RO-2 unit, and the remaining 235 kg enters the chamber as Recycle. . 9 → 12 → 15 → 18 → 21 → 24 → 26.47% (sat.) And the diluted stream was recovered to 26.47% → 21% → 18 → 15 → 12 → 9 → 6 → 3 → 0 (pure) do.
이를 조정할 수 있는 방법은 α(alpha: A), θ(theta, T)로 조정이 가능하다. This can be adjusted by using α (alpha: A) and θ (theta, T).
6%가 9%와 3%로 되는 가정을 계산하면, A=2, T=1/3: A=feed chamber(FC)/equalizer chamber(EC), T=fraction of water moved from FC to EC. Feed input-split will allocate 2/3 of input to FC and the other 1/3 EC. T=0.5 will move 1/3 (a half of FC) to EC. Then Water volume of FC will be 1/3 and that of EC will be 2/3. The concentration of EC will be 6%/(1-1/3)=9% while 6%/(1+2*0.5)=3%. 이와 같은 방식으로 두 챔버간의 삼투압차를 최소화 하면서 분배를 할 수 있으며. 궁극적으로 염을 위로 물은 아래로 이동되는 것을 확인할 수 있다.Calculating the assumption that 6% is 9% and 3%, A = 2, T = 1/3: A = feed chamber (FC) / equalizer chamber (EC), T = fraction of water moved from FC to EC. Feed input-split will allocate 2/3 of input to FC and the other 1/3 EC. T = 0.5 will move 1/3 (a half of FC) to EC. Then Water volume of FC will be 1/3 and that of EC will be 2/3. The concentration of EC will be 6% / (1-1 / 3) = 9% while 6% / (1 + 2 * 0.5) = 3%. In this way, dispensing can be done with minimal osmotic pressure difference between the two chambers. Ultimately the salt moves up the water down.
실시예 4: Feed “Input-Split Cascade”시스템에 소모되는 에너지 계산Example 4 Calculation of Energy Consumed in a Feed “Input-Split Cascade” System
3% input 1000kg은 실시 예 1의 순수 회수공법에 의해 500kg의 순수와 500kg(소금 30kg+470kg물)의 6% 혼합액으로 분리된다. 이것이 Feed-Split Cascade 시스템의 원료가 되며 이는 다시 Feed-Split에 의해 농축된 feed 용액 9%와 3% (혹은 3.38%)의 product stream을 분리된다. 1000 kg of 3% input is separated into a 6% mixed solution of 500 kg of pure water and 500 kg (30 kg of salt +470 kg of water) by the pure water recovery method of Example 1. This is the raw material of the Feed-Split Cascade system, which in turn separates the 9% and 3% (or 3.38%) product stream of the feed solution concentrated by the Feed-Split.
input 6% 소금물 용액 500kg은 Feed 용액 총량에서 얼마만큼 물을 Product Stream으로 여과하느냐의 분률을 θ(편의상 “T”)로 표시하고 Feed chamber/Product Chamber의 크기 비를 α(편의상 “A”)라고 하면 product stream을 9%로 하고 싶으면 6%/(1-T)=9%, T=1/3이 된다. 다음 두 stream간의 압력 차를 최소화하기 위해 A는 클수록 좋지만 편의상 “2.32”로 정하면 low 3% 값은 3.38%로 되며 6% Feed input stream에서 3.38% π-이콜라지자 stream으로 이동한 물의 양은 176.79kg이 된다. 평균 압력차를 30bar로 계산하면 필요한 최소 에너지는 30*0.17679/36=0.1473kwh이다. 참고로 π-이콜라지자 stream의 농도는 6%/(1+2.3204*1/3)=3.38%이다. The input 6% brine solution 500kg represents the fraction of how much water is filtered through the product stream in the total feed solution in θ (“T” for convenience) and α (for convenience “A”) for the size ratio of the feed chamber / Product Chamber. If you want to make the product stream 9%, then 6% / (1-T) = 9% and T = 1/3. In order to minimize the pressure difference between the next two streams, the larger A is better, but for convenience, a low value of 3% is 3.38%, and the amount of water transferred from the 6% feed input stream to the 3.38% π-Eco-rigid stream is 176.79 kg. do. Calculating the average pressure difference as 30bar, the minimum energy required is 30 * 0.17679 / 36 = 0.1473kwh. For reference, the concentration of π-Eco-Laziza stream is 6% / (1 + 2.3204 * 1/3) = 3.38%.
용질의 농도별로 flux movement와 소모 에너지는 Feed input-split cascade에서 막을 통해 이동한 총 물의 양은 371kg이고 평균 압력 차를 30bar로 했을 때 총 0.309Kwh가 필요하고 30kg의 소금을 제거 하는 데는 2회의 recycle이 필요하여 총 에너지 필요한 0.618kwh이다. 총 최소 에너지는 RO-1에너지 + RO-2 energy + recycle (2회)= 0.515+0.618+0.390=1.523kwh인 것을 확인하였다. 그러나 실제는 3%에서 26.47%로 가는 도중 무려 9단계회의 mixing 및 demixing이 일어나기 때문에 energy loss가 많아 의미가 없을 수 있다. 그러나 1~2 단계에서도 상당히 많은 물을 회수 할 수 있어 물을 전부 회수 하지 않는 다면 고려 대상이 될 수 있다. Flux movement and energy consumption by solute concentration are 371 kg of total water moved through the membrane in feed input-split cascade. A total pressure of 0.309 Kwh is required at an average pressure difference of 30 bar, and two recycles are required to remove 30 kg of salt. The total energy required is 0.618kwh. The total minimum energy was found to be RO-1 energy + RO-2 energy + recycle (2 times) = 0.515 + 0.618 + 0.390 = 1.523 kwh. In practice, however, 9 steps of mixing and demixing occur from 3% to 26.47%. However, a considerable amount of water can be recovered in steps 1 and 2, which can be considered if not all of the water is recovered.
실시예 5: Feed output-Split Counter Current (“Output-split CC”) 무삼투압차 공정 모사Example 5 Simulation of Feed Output-Split Counter Current (“Output-split CC”) Osmotic Pressure Difference Process
실시예 4의 Input-split cascade와는 달리 feed-input부터 feed-output까지는 pressure release가 없어 에너지 손실이 Input-split cascade에 비해 적다. Unlike the input-split cascade of Example 4, there is no pressure release from feed-input to feed-output, resulting in less energy loss than input-split cascade.
Output-Split CC에서 무삼투압차 농축공정에서 feed input이 6%, 470L, 30g으로 Feed chamber에서 압력이 가해지면서 물은 π-이콜러아져 chamber로 움직인다. 오른 쪽의 R은 소금총량이 얼마나 Feed chamber end stream에서 Product (Draw) chamber로 움직이는 비율을 보여준 것으로 50% 가 Product chamber로 들어갔을 때의 Feed chamber end에서는 농도가 26.47%로 삼투압 차가 “0”가 되지만 π-이콜러아져 챔버의 끝에서는 3.38%가 되며 회수되는 물의 양은 428.33kg이다. 이는 product를 saturated solution으로 했을 때의 양이다.In the output-split CC, the feed input is 6%, 470L, 30g in the osmotic pressure difference condensation process. As pressure is applied from the feed chamber, the water moves to the chamber by π-evaporator. The R on the right shows how much salt total flows from the feed chamber end stream to the product (draw) chamber. When 50% enters the product chamber, the concentration is 26.47% at the feed chamber end. However, at the end of the chamber, the π-echoarizer is 3.38% and the amount of water recovered is 428.33 kg. This is the amount when the product is a saturated solution.
이때 1번 recycle했을 때의 삼투압을 이기기 위한 최소에너지는 =0.38772*2=0.775 kwh 인 것으로 계산되었다.At this time, the minimum energy to overcome the osmotic pressure when recycled once was calculated as = 0.38772 * 2 = 0.775 kwh.
총에너지는 =0.515+0.775+0.390=1.680kwh이다.The total energy is = 0.515 + 0.775 + 0.390 = 1.680 kwh.
실시예 6: Output-Split CC에서 물과 농축소금물의 분리 Example 6 Separation of Water and Concentrated Salts in Output-Split CC
6-1. 탈염공정 물질수지 및 최소에너지 6-1. Desalination Process Balance and Minimum Energy
표 6에서 개시된 바와 같이, 탈염조(desalting tank)의 상층과 하층에 각각 3.0S, 3.0Ws(26.47% 포화상태)의 농축수가 Δπ=0있다고 가정한다. 시작은 “0”지만 마지막에는 0.5Ws의 물이 상층수에서 하층수로 움직이고 그 결과로 0.5S (15g)의 소금이 침전이 되어 시스템에서 제거되며 한편 3.0S, 3.0Ws의 형태가 되어 하층수로 재순환 된다. 상층수에서 필터 된 0.5Ws가 하층수로 와서 3.0S의 소금물은 휘석 되어 23.57%의 농도로 41.67L가 하층수로 상층수의 삼투압차를 줄이는 역할 (Δπ=0)을 수행한다.As disclosed in Table 6, it is assumed that the concentrated water of 3.0S and 3.0Ws (26.47% saturation) is Δπ = 0 in the upper and lower layers of the desalting tank, respectively. The beginning is “0” but at the end 0.5Ws of water moves from the upper water to the lower water, and as a result 0.5S (15g) of salt precipitates and is removed from the system, while it is in the form of 3.0S, 3.0Ws. Recycled. 0.5Ws filtered from the supernatant comes into the lower layer, and 3.0S of brine is crushed, and the concentration of 23.57% decreases the osmotic pressure difference of the upper layer into the lower layer (Δπ = 0).
한번 재순환에 15g의 소금과 물 41.67만 제거되므로 정상상태를 위해서는 2번 recycle이 필요하다. 이때 소요되는 에너지는 평균 농도로 계산하여 (343.707-288.559)=51.148; 51.148*0.04167/36=0.0638wkh/15kg이고, 30kg을 제거하는 데는 0.127kwh가 필요하며, 26.47%, 343.707 bar에서 RO로 물을 제거한다면 소요되는 에너지는 343.707*0.08334/36= 0.795 Kwh인 것으로 계산 되었다.Only 15g of salt and 41.67 of water are removed in one recycle, so two recycles are required for normal operation. At this time, the energy required is calculated as the average concentration (343.707-288.559) = 51.148; 51.148 * 0.04167 / 36 = 0.0638wkh / 15kg, 0.127kwh is required to remove 30kg, and the energy required to remove water with RO at 26.47%, 343.707 bar is calculated to be 343.707 * 0.08334 / 36 = 0.795 Kwh It became.
Figure PCTKR2016008104-appb-T000006
Figure PCTKR2016008104-appb-T000006
6-2. Feed의 크기와 에너지 소모량 계산6-2. Calculate Feed Size and Energy Consumption
위의 탈염공정에서 보았듯이 대상 RO 장치에 들어가는 시료의 크기에 따라 에너지 소모가 많이 차이가 남을 알 수 있다. 예를 들어 100톤의 원료 수에 1톤 (물 970L, 소금 30kg을 제거한다면 최소에너지는 아주 작을 수 밖에 없다. 물론 많은 물을 펌핑 하여야 하므로 pumping 에너지를 무시할 수 없겠지만 분리에 들어가는 작을 수밖에 없다.As seen in the desalination process above, it can be seen that the energy consumption varies greatly depending on the size of the sample entering the target RO apparatus. For example, if you remove 1 ton (100 liters of water and 30 kg of salt) from 100 tons of raw water, the minimum energy will be very small.
가령, 7S, 7W시스템에서 물 1W를 만든다면 3%에서 7S, 6W가 되고 농도는 3.48%가 된다. 삼투압차가 (27.849-23.743)=4.106bar이다. 이때 필요한 에너지는 4.106*0.970/36=0.110kwh이다. 이는 물 1m3을 얻는 데 들어가는 이론적 에너지가 1.14kwh임을 감안할 때 거의 10% 수준이다. 위의 탈염조에서도 RO에서 보다 3배 정도 큰 분리조를 쓰긴 했지만 큰 분리조는 0.127Kwh, RO에 의한 탈염조는 0.795kwh의 최소에너지를 필요로 하였다. For example, if you make 1W of water in a 7S, 7W system, you get 3S to 7S, 6W and the concentration is 3.48%. The osmotic pressure difference is (27.849-23.743) = 4.106 bar. The required energy is 4.106 * 0.970 / 36 = 0.110kwh. This is almost 10%, considering that the theoretical energy required to obtain 1 m 3 of water is 1.14 kwh. In the above desalination tank, although the separation tank was used about three times larger than that of RO, the desalination tank by the large separation tank required a minimum energy of 0.795 kwh.
6-3. 1톤의 3% 소금물에서 물과 소금을 소금을 완전히 분리했을 때의 최소 에너지 계산6-3. Calculation of minimum energy when salt is completely separated from water and salt in 1 ton of 3% brine
Feed input에서 큰 분리조를 사용했을 때 에너지 소모가 거의 10%에 불과했지만 사실은 에너지소모는 회수율과 많은 관련이 있다. 다음 예에서는 실시예 6-1의 1톤 3%용액에서 물을 1차 RO로 0.250m3을 회수하고 나머지는 물질 수지를 만족하고자 재순환(0.970-0.250=0.720)도 하고 또 탈염도 수행한다. 실시예 6-2의 경우는 물의 반인 970kg*0.5=485kg을 일차 RO로 회수하고 나머지 재순환수 (0.970-0.485=0.485)에서 추가 회수한다. F500의 경우는 1차 RO에서 0.500톤을 회수하고 나머지는 재순환 수(0.970-0.500=0.470)에서 회수한다.The energy consumption was only about 10% when a large separator was used at the feed input, but in fact energy consumption is highly related to the recovery rate. In the following example, 0.250m3 of water is recovered from the 1 ton 3% solution of Example 6-1 as the primary RO and the rest is recycled (0.970-0.250 = 0.720) and desalting is performed to satisfy the material balance. For Example 6-2, half of the water, 970 kg * 0.5 = 485 kg, is recovered to the primary RO and further recovered from the remaining recycle water (0.970-0.485 = 0.485). The F500 recovers 0.500 tonnes from the primary RO and the remainder from the recycle water (0.970-0.500 = 0.470).
순수와 소금을 얻는 데 필요한 에너지는 쉽게 계산이 되지만 재순환 에너지는 상당히 복잡하고 방법에 갑도 많이 차이가 나기 때문에 다음과 같은 방법으로 비교한다. “Output-Split CC” 공정의 시작 (26.47%)와 끝의 산술평균이라 정확한 값은 아니겠지만 간편하고 방법간의 차이는 표 7에 개시된 바와 같다. 아래 공정은 모두 소금 30kg 물 970kg을 생산한다. The energy needed to obtain pure water and salt can be easily calculated, but the recycling energy is quite complicated and the methods vary greatly. The arithmetic mean at the beginning (26.47%) and the end of the “Output-Split CC” process is not exact but simple and the differences between the methods are shown in Table 7. The following processes all produce 30 kg of salt and 970 kg of water.
Figure PCTKR2016008104-appb-T000007
Figure PCTKR2016008104-appb-T000007
주: RO-2 (0.432~recycle energy, 0.483~water recovery), WR= water per recycleNote: RO-2 (0.432 ~ recycle energy, 0.483 ~ water recovery), WR = water per recycle
참고: F00공정의 총 에너지 1.020 (RO-1공정 생략, 삼투압차 농축기 직 투입)Note: Total energy 1.020 of F00 process (RO-1 process omitted, osmotic pressure concentrator directly)
상기 표에서 F250, F485, F500 RO-1의 input은 3% 소금물이고 Recycle stream으로 들어가는 즉 RO-1의 회수 정도에 따른 output의 농도이다. 회수를 적게 하면 농도가 낮고 F500 에서와 같이 RO-1 에서 500L을 회수하면 6%가 되는 것을 확인하였다. In the table, the inputs of F250, F485, and F500 RO-1 are 3% brine and the concentration of the output according to the recovery degree of RO-1 entering the recycle stream. When the recovery was small, the concentration was low, and as in the F500, the recovery of 500L in RO-1 was confirmed to be 6%.
첫 번째 Row 가 RO-1 unit에서 3% 소금물에서 순수를 회수하는 데 필요한 에너지이고, RO-2는 recycle unit에서 소모한 에너지 마지막 것이 실시 예 2의 F250에서 나온 탈염공정 (3.5S, 3.5Ws)에서 필요한 에너지이다.The first row is the energy required to recover pure water from 3% brine in the RO-1 unit, and the RO-2 is the energy consumed in the recycle unit. The last one is the desalination process from F250 of Example 2 (3.5S, 3.5Ws). Is the energy required.
RO-2의 계산은 아래와 같이 수행했다. RO-2 input (상층수) 하층수의 output의 차이 압력에다가 recycle 되는 물의 양을 곱하여 36을 나누면 바로 kwh를 얻을 수 있다. 한편 feed-output (26.47%)과 하층수-input (23.57%)의 압력차에 41.67을 곱하여 에너지를 구했다. 즉 Recycle System에서 처음과 마지막의 에너지 차를 구해 산술 평균을 내고 여기에 총 recycle 되는 물의 양을 곱하여 간단한 recycle 에너지를 구했다. [Δπ at end (4%/3%)*Q (0.470/time) + Δπ at begin (26.47%/23.57%)]/36/2*2로 구한다.The calculation of RO-2 was performed as follows. RO-2 input (top water) The difference in the output of the bottom water is multiplied by the amount of recycled water divided by 36 to get kwh. On the other hand, the energy difference was obtained by multiplying the pressure difference between feed-output (26.47%) and sub-water input (23.57%) by 41.67. In other words, the energy difference between the first and last energy was calculated from the recycling system, the arithmetic mean was calculated, and the total recycled water was multiplied to obtain a simple recycle energy. Obtain [Δπ at end (4% / 3%) * Q (0.470 / time) + Δπ at begin (26.47% / 23.57%)] / 36/2 * 2.
RO-1 (4%)의 계산 예를 보면 평균을 사용한다. 평균은 (23.743+32.298)/2=28.020 RO-1(F250)은 28.020*0.25/36=0.194 (여기서 28.02는 RO-1 tank input (3%)과 output 4%의 삼투압차 물회수량 0.25m3을 곱한 것이다. RO-2(F250)은 recycle 누적에너지 량 [0.171 + 0.063]/2*2=0.234가 나온다. 여기서 보면 RO-1 뿐만 아니라 RO-2에서도 아주 큰 영향을 미치고 있음을 볼 수 있다. RO-2의 물생산 비용은 Δπ공정비용과 떠 3%로 원수 탱크로 부터 진입한 용액으로 부터 물 생산비용을 추가한 값이다. 세가지 공정의 누적 총계를 보면 F250은 1.236로 낮지만 F485나 F500은 총 에너지 필요량이 1.528과 1.540으로 상당히 높아 “Input-split Cascade”나 “Output-split CC”와 크게 차이가 없음을 알 수 있다. 이 값들은 1 톤의 용질포함 용액을 사용한 값이므로 현재의 원수의 50%만 회수하는 상업적 공정에 비해 최소 에너지가 높지만 Δπ=0 RO 기술을 이용한 π-equalizing 공정으로 낮은 압력에서 salt도 회수하고 물도 100% 회수하는 공정으로 크게 의미를 둘 수 있다. In the example calculation of RO-1 (4%), we use the mean. The average is (23.743 + 32.298) /2=28.020 RO-1 (F250) is 28.020 * 0.25 / 36 = 0.194 (where 28.02 is the RO-1 tank input (3%) and the output osmotic pressure difference of 4% 0.25m 3 RO-2 (F250) yields a cumulative amount of recycled energy [0.171 + 0.063] /2*2=0.234, which shows a significant effect not only on RO-1 but also on RO-2. The cost of water production of RO-2 is the Δπ process cost, plus 3% of the water production cost from the solution entering the raw water tank.The cumulative total of the three processes shows a low F250 of 1.236 but F485 or The F500 has significantly higher total energy requirements of 1.528 and 1.540, which is not significantly different from “Input-split Cascade” or “Output-split CC.” These values are based on the use of one ton of solute solution. Although the minimum energy is higher than that of a commercial process that recovers only 50% of raw water, the π-equalizing process using Δπ = 0 RO technology The recovery of salt at low pressure and 100% recovery of water can be significant.
실시예 7: Input-Split Cascade와 Output-Split CC에서 low solute selectivity σ 막의 응용 비교. Example 7 Application Comparison of Low Solute Selectivity σ Membranes in Input-Split Cascade and Output-Split CC.
7-1. Low solute selectivity membrane을 이용한 순수 및 소금 회수 공정의 중요성7-1. Importance of pure and salt recovery using low solute selectivity membrane
실시예 2의 실험결과에서 보았듯이 용액이 고농도로 갈수록 flux가 많이 낮아지며 그렇다고 압력을 무한정 높일 수도 없다. 따라서 solute selectivity (o)는 떨어지더라도 flux가 큰 막을 사용하여야 경제성을 확보 할 수 있다. 현재 연구는 역삼투압이 아닌 나노-여과 (Nano filtration)영역에서 수행하는 것이 바람직하다. As can be seen from the experimental results of Example 2, the higher the solution, the lower the flux, and the pressure cannot be increased indefinitely. Therefore, even if the solute selectivity (o) falls, it is necessary to use a membrane with a large flux to secure economic feasibility. The current study is preferably performed in the nano-filtration zone, not reverse osmosis.
Input-split cascade와 Output-split CC에 low selectivity membrane을 적용하여 membrane flux을 대폭 향상시키고 recycle공정에 소요되는 에너지를 절약함을 목적으로 한다. 참고로 현재의 사용 RO 공정은 삼투압을 극복하는 데는 물 톤당 생산에 최소 삼투압 차를 극복하는데 1kwh/m3가 필요하나 실제로는 RO 공정의 삼투압 공정에 2kwh, 또 기타 공정에 2kwh가 소모되어 총 4kwh가 소모되는 것으로 알려져 있다. 이 전기 에너지 비용은 물 1톤 당 생산원가가 1$인 점을 감안하면 0.1$/kwh로 가정하면 물 원가의 40%가 전기에너지비용이 되므로 조금이라고 줄이는 것은 아주 중요하다.Low selectivity membranes are applied to the input-split cascade and output-split CC to significantly improve the membrane flux and to save energy for the recycling process. Note that the current use of the RO process is one There needs 1kwh / m 3 to overcome the minimum osmotic pressure difference in the production of water per ton to overcome the osmotic pressure actually is 2kwh is consumed 2kwh, yet other processes in the osmotic process, the RO process, the total 4kwh Is known to be consumed. Considering that the cost of electric energy is $ 1 per ton of water, it is very important to reduce it to a little since 40% of the cost of water is assumed to be $ 0.1 / kwh.
따라서 상기 무삼투압차 RO 공정에서 농 축수 재순환공정에 추가 전력이 필요한 것이 사실이지만 원료 수를 적게 사용하며 (현재는 원수 2톤에서 물 1톤 생산, 50% 활용) 또 농축수도 아주 고농도 (포화 농도, 혹은 결정화 형태)로 FO를 이용한 전력생산에도 유리할 뿐 아니라 바닷물 속에 미량 포함되어 있는 Lithium, Magnesium, 금, uranium 등의 생산에도 종전의 공법에 비해 경제성이 있는 공법으로 응용 될 수 있다. 즉 농축 배출 수의 활용에 있어서도 유리해 질 수 있어 종합적인 경제성 (RO물 판매 + 농축수를 이용한 FO (forward Osmosis)발전 + 미량 금속 회수)에서 유리할 수 있다. Therefore, it is true that the above-mentioned osmotic pressure difference RO process requires additional electric power for the recycling of the brine, but it uses less raw material (now produces 1 ton of water from 2 tons of water and utilizes 50%) and also concentrates very high concentration (saturated concentration). , Or crystallization form), which is not only advantageous for power generation using FO, but also can be applied to the production of Lithium, Magnesium, Gold, uranium, etc. contained in a small amount of seawater as a more economical method than the conventional method. In other words, it can be advantageous in terms of utilization of concentrated effluents, which can be advantageous in terms of overall economic feasibility (RO water sales + forward osmosis (FO) generation using concentrated water + trace metal recovery).
7-2 Feed Input-Split Cascade와 Feed Output-Split CC에서 low solute selectivity (o) 막의 응용 비교. 7-2 Comparison of Applications of Low Solute Selectivity (o) Membranes in Feed Input-Split Cascade and Feed Output-Split CC.
표 8-1의 Feed Input-Split cascade 공법에서 A=2, T=50% (상층수의 50%, 전체의 1/3)가 움직인다고 가정한다. Feed-Stream을 편의상 “상층수“ Product (Draw) stream을 “하층수라고 하자. σ=1인 경우 시작점 농도가 6%이고 끝나는 농도가 11.32%가 된다. 용매의 50%가 상층수에서 하층수로 옮아갔는 데도 상층수의 농도가 12%가 되지 않는 것은 용액(소금+물)의 반이 아닌 용매(물)의 반만 움직였기 때문이다. 상층수 평균이 8.66%, 하층수 평균이 4.55%로 농도차는 4.12%가 된다. 반면 σ=0.5인 경우는 상층수 6%, 하층수 8.74%로 평균이 5.29%로 평균의 차이가 2.08%이다. In the Feed Input-Split cascade method of Table 8-1, assume that A = 2, T = 50% (50% of the top water, 1/3 of the total) is moving. For convenience, let's call the "top water" Product (Draw) stream a "bottom water." For σ = 1, the starting point concentration is 6% and the ending concentration is 11.32%. Even if 50% of the solvent is transferred from the upper water to the lower water, the concentration of the upper water is not 12% because only half of the solvent (water) moves, not half of the solution (salt + water). The upper water average is 8.66%, the lower water average is 4.55%, and the concentration difference is 4.12%. On the other hand, in case of σ = 0.5, 6% of the upper floors and 8.74% of the lower floors had an average of 5.29% and the difference was 2.08%.
표 8-2의 경우는 Product-Split CC의 경우인데 A=1, T=0.5의 시스템 결과이다. 상층수에서 최종적으로 15g의 소금이 배출되어 2번 recycle을 하면 시스템에서 30g이 제거되도록 하였다. 여기서는 σ=1가 인 경우는 농도차가 1%인데 비해 σ=0.5로 하면 오히려 평균 농도차가 4.06%로 늘어나 오히려 σ=1의 경우가 유리하게 나타난다. Table 8-2 shows the case of Product-Split CC, which is the system result of A = 1 and T = 0.5. Finally, 15g of salt was discharged from the supernatant and recycled twice so that 30g was removed from the system. Here, when σ = 1 is 1%, the difference in concentration is 1%, whereas when σ = 0.5, the mean concentration difference is increased to 4.06%.
이상의 예를 볼 떼 Input-Split Cascade의 경우는 무삼투압차 효과가 Feed-stream의 시작 3% 근처에서 잘 나타나고 있고 Output-Split CC의 경우는 무삼투압차 효과가 뒤쪽 (Feed-stream의 끝)에서 나타나기 때문이다. In the case of the input-split cascade, the osmotic pressure difference effect is shown near 3% of the start of the feed-stream in the case of the input-split cascade. Because it appears.
이상의 두 가지를 비교해보면 초기에는 Input-Split을 하다가 뒷부분에서 Output-Split으로 전환하는 것도 하나의 방법이 될 수 있다. Low solute selectivity (예: σ=0.5) 막을 사용하면 high selectivity 막 (예: σ=1.0) 보다 막 면적이나 전반적인 에너지 소모가 줄어 들 수 있음을 볼 수 있다. Comparing the two above, one method may be to input-split at first and then switch to output-split later. Using low solute selectivity membranes (eg σ = 0.5) can be seen to reduce membrane area or overall energy consumption than high selectivity membranes (eg σ = 1.0).
따라서, Input-Split Cascade 공정에서는 초기에 low membrane을 사용하고, 만약에 output-Split CC을 사용하면 초기에는 σ=0.8~0.9정도의 막을 사용하다가 농축이 상당히 진행되면 예를 9~12%정도에서 σ=0.5로 농축하여 20% 이상에서는 σ=0.2~0.1정도의 막을 사용하는 것이 바람직하며, 두 공정을 결합하여 사용하는 것도 좋은 효과를 나타내는 것을 확인하였다.Therefore, in the input-split cascade process, the low membrane is used initially, and if the output-split CC is used, the membrane is initially used at σ = 0.8 ~ 0.9 and the concentration is considerably increased. When sigma was concentrated to sigma = 0.5 and 20% or more, it was preferable to use a film of sigma = 0.2 to 0.1, and it was confirmed that the combination of the two processes had a good effect.
Figure PCTKR2016008104-appb-T000008
Figure PCTKR2016008104-appb-T000008
Figure PCTKR2016008104-appb-T000009
Figure PCTKR2016008104-appb-T000009
실시예 8: R1 없는 경우 (모두 재순환 하는 경우)Example 8: Without R1 (all recycled)
실시예 6에서 F500, F485, F250은 3% 원수의 970L물에서 RO를 통하여 먼저 500L, 485L, 250L 회수를 검토하였다, 그러면 극한 상항인 RO-1이 “zero”이고 모두 재순환을 검토해보자.In Example 6, F500, F485, and F250 first examined 500L, 485L, and 250L recovery through RO in 970L water of 3% raw water. Then, let's consider recirculation of RO-1, which is the extreme condition, all zero.
Feed Input 3%, 970L+30kg소금이고 π-이콜라이저 챔버의 recycle stream은 물 970L+15kg소금으로 1.52% 일 것이다. 3%의 삼투압은 23.743bar, 1.52%의 삼투압은 11.746bar이다. 두 삼투압의 차이는 11.997bar, 평균은 17.744bar이다. Feed input (3%, 970L) Equalizer (1.52%, 970L)의 에너지는 11.99*0.970/36=0.323, 끝(26.47%, 23.57%) 0.107이므로 평균(1/2)*2=0.430이다. RO에 필요한 에너지는 17.744*0.485/36*2=0.463kwh인 것으로 계산되었다. Feed input 3%, 970 L + 30 kg salt and the recycle stream of the π-echolizer chamber will be 1.52% with 970 L + 15 kg salt water. The osmotic pressure of 3% is 23.743 bar and the osmotic pressure of 1.52% is 11.746 bar. The difference between the two osmotic pressure is 11.997 bar, the average is 17.744 bar. The energy of the feed input (3%, 970L) equalizer (1.52%, 970L) is 11.99 * 0.970 / 36 = 0.323 and the end (26.47%, 23.57%) 0.107, so the average (1/2) * 2 = 0.430. The energy required for RO was calculated to be 17.744 * 0.485 / 36 * 2 = 0.463 kwh.
총 에너지는 0.430+0.463+0.127=1.020 kwh이다. Input의 압력 차는 11.997 bar이고 input 출구 (26.47%)/π-Equalizer 시작점의 압력차는 55.148bar정도이다. 중간과정에 높은 압력은 low σ membrane 나 Input-split cascade등 해결방법이 많이 있어 크게 문제가 되지 않는다.The total energy is 0.430 + 0.463 + 0.127 = 1.020 kwh. The pressure difference at the input is 11.997 bar and the pressure difference at the input outlet (26.47%) / pi-Equalizer start point is 55.148 bar. High pressure in the middle process is not a problem because there are many solutions such as low σ membrane or input-split cascade.
결과적으로 3.0% input에 1.5%output을 얻으면 20~30 bar정도의 압력만 있어도 바닷물에서 물을 얻을 수 있다는 점이다. π-이콜라이저 (1.5%)를 다시 무삼투압차 농축기의 피드로 넣어면 π-이콜라이저-output은 0.75% recycle 에너지~0.157+0.107 (0.264), RO-2~0.311+0.129=0.704이다. 압력차는 5.85 bar 물 생산시 압력은 11.56 bar이다. 재순환에 따라 총 최소에너지는 1.020에서 0.704로 줄었다. 20~30 bar에서 움직일 수 있는 역 삼투압법도 가능할 것이다.As a result, if you get 1.5% output at 3.0% input, you can get water from the sea with only 20 ~ 30 bar pressure. When the π-echolizer (1.5%) is fed back into the feed of the osmotic pressure concentrator, the π-echolizer-output is 0.75% recycle energy ~ 0.157 + 0.107 (0.264) and RO-2 ~ 0.311 + 0.129 = 0.704. The pressure difference is 5.85 bar and the water pressure is 11.56 bar. Recirculation reduced the total minimum energy from 1.020 to 0.704. Reverse osmosis methods that can move from 20 to 30 bar will also be possible.
요약하면 현재 RO공정 (70/50bar.46,50%,0.985), F250 (50/32.29bar, bar 수율 100%, 1.236) F00-3%/1.5%(35/23.74bar, 수율 100%, 1.020), F00-1.5%/0.75% (20/11.56 bar, 100%,0.704+1.020=1.724)에 현재 RO공정에 최소에너지 0.985에 추가로 들어가는 약 3kwh가 전 공정의 운영압력이 70=>50=>35>20로 변함에 따라 얼마나 줄어드는 지를 검토하면 적절한 최적화 운영 방법이 도출 될 것이다. In summary, the current RO process (70 / 50bar.46,50%, 0.985), F250 (50 / 32.29bar, bar yield 100%, 1.236) F00-3% / 1.5% (35 / 23.74bar, yield 100%, 1.020 ), F00-1.5% / 0.75% (20 / 11.56 bar, 100%, 0.704 + 1.020 = 1.724), about 3 kwh, which adds 0.985 minimum energy to the current RO process, has an operating pressure of 70 => 50 = Examining how shrinking as it changes to> 35> 20 will yield appropriate optimization operations.
계산에 착오가 있기 쉬우므로 구체적인 방법은 하기와 같다.Since the calculation is easy to be mistaken, the specific method is as follows.
일반적인 계산 방법 (F250): (1) 물을 먼저 계산한다. RO-1 Feed input은 항상 3%이다. RO-1 장치 Output은 얼마나 뽑느냐에 달려 있다. 전연 안 뽑으면 “0”이고 RO-1 output은 3%이다. RO-1 input(1 톤, 1000kg, 30S, 970W)로 구성되어 있어 250W (250kg물)을 뽑으면 RO-1 output은 750kg용액에 되며 30S, (970-250)=720W가 되며 30/(30+720)= 4%이며 이것이 Δπ-recirculation 장치의 input이 된다. Energy 계산: 소금물의 농도가 3%에서 4%로 변화므로 시작의 끝의 평균삼투압을 구해 계산하면, General calculation method (F250): (1) Calculate the water first. The RO-1 Feed input is always 3%. The RO-1 device output depends on how far you pull. If not pulled out, it is “0” and RO-1 output is 3%. It consists of RO-1 input (1 ton, 1000kg, 30S, 970W), so if you pull out 250W (250kg water), the RO-1 output will be 750kg solution, and 30S, (970-250) = 720W and 30 / (30+ 720) = 4%, which is the input to the Δπ-recirculation device. Energy calculation: Since the brine concentration changes from 3% to 4%, calculate the average osmotic pressure at the end of the start,
28.020*0.250/36=0.194(Kwh)이다. 두 농도의 압력평균과 압력차가 두 번 사용이 되는 데 혼돈을 하면 안된다. RO-1은 3%와 4%의 평균만 이용한다, 후에 재순환 에너지에서는 4%와 2.04%의 평균과 압력차를 둘 다 이용한다 28.020 * 0.250 / 36 = 0.194 (Kwh). Do not confuse the pressure average and the pressure difference between the two concentrations. RO-1 uses only an average of 3% and 4%, later using both an average of 4% and 2.04% and pressure difference in recycle energy.
(2)재순환 에너지를 계산 한다(2) Calculate recycle energy
상기 Δπ-recirculation에서는 feed stream π-equalizer chamber 전부(소금회수시, 970W) 혹은 일부 (농축수 제외 970-83.34=886.66W)가 E-chamber 되돌아온다. 즉 970W/886.66W와 소금 30S가 있다. 이것이 농축 후에는 반으로 나뉘어 외부로 배출되거나, 소금은 결정으로 회수 된다.In the Δπ-recirculation, all of the feed stream π-equalizer chambers (970 W during salt recovery) or a part (970-83.34 = 886.66 W excluding concentrated water) are returned to the E-chamber. That is 970W / 886.66W and salt 30S. After concentration, it is split in half and discharged to the outside, or the salt is recovered as crystals.
π-equalizer chamber의 input은 15S, 41.67W(농축수 성분의 1/2)이고 23.57%(실시예 6)로 정하자. Output은 모든 물이 다 회수되므로 15/(15+720)=2.04%가 된다. 4%와 2.04%의 압력차는 16.279이며 에너지 계산 16.279*0.720/36=0.3251kwh가 나온다.Let the input of the π-equalizer chamber be 15S, 41.67W (1/2 of the concentrated water component) and 23.57% (Example 6). The output is 15 / (15 + 720) = 2.04% since all the water is recovered. The pressure difference between 4% and 2.04% is 16.279 and the energy calculation is 16.279 * 0.720 / 36 = 0.3251kwh.
π-equalizer-chamber input에서 에너지가 0.107이므로 평균을 내고 두 번을 회전해야 30S가 제거되므로 (0.325+0.107)/2*2=0.432Kwh가 된다. (3) RO-2를 계산한다. 이때는 한번에 360W가 생산되므로 4%와 2.04%의 평균(24.150)이 필요하다. 두 번 recycle에 생산되는 물생산에너지는 (32.298+16.019)/2*0.360/36*2=0.483이며, 총합계는 0.194+0.433+0.483+0.127=1.236로, 종합하면 RO-1의 계산에는 평균 (3%와 4%)만 필요하고 재순환 공정은 압력차와 평균 둘 다 필요하며, 압력차는 재순환 에너지 계산에 평균은 RO-2의 물생산에너지 계산에 이용할 수 있다. Since the energy at the π-equalizer-chamber input is 0.107, it must be averaged and rotated twice to remove 30S, resulting in (0.325 + 0.107) /2*2=0.432Kwh. (3) Calculate RO-2. In this case, 360W is produced at a time, so an average of 4% and 2.04% (24.150) is required. The water production energy produced in the second recycle is (32.298 + 16.019) /2*0.360/36*2=0.483, and the total is 0.194 + 0.433 + 0.483 + 0.127 = 1.236. Only 3% and 4%) are required, and the recirculation process requires both a pressure difference and an average, and the pressure difference can be used to calculate the recycle energy and the average can be used to calculate the water production energy of RO-2.
실시 예 9: 무삼투압차 기술의 연료용 알코올에의 응용 Example 9 Application of Fuel Free Alcohol to Osmotic Pressure Differential Technology
연료용 에탄올은 현재 99.5% 혹은 99.6%의 순도로 사용되고 있다. 열 적 방법 (가스는 스팀을 발생, 전기 쓰임) 있는 데 최근의 에너지 사용량 중 최저치는 아래와 같다. 미국의 Renewable Fuels Association (March 06, 2016년)에 의하면 최소의 에너지는 Iowa WDG의 자료로 23,424BTU/gal이다. 이를 kwh/kg-fuel ethanol로 바꾸면 2.27kwh/kg-ethanol이 된다. 여기에 무삼투압차 RO기술을 적용하여 99.5%까지 농축하는 것을 검토해 보기로 한다. Ethanol for fuel is currently used in 99.5% or 99.6% purity. There is a thermal method (gas generates steam, electricity is used). According to the Renewable Fuels Association (March 06, 2016) in the United States, the minimum energy is 23,424 BTU / gal from Iowa WDG. If you change this to kwh / kg-fuel ethanol, you get 2.27kwh / kg-ethanol. Here, the concentration of up to 99.5% by applying an osmotic pressure difference RO technology will be considered.
표 10은 알코올의 삼투압을 Lewis식(eq.-4)으로 계산하였다. 처음에는 5%에서는 28bar이나 99.50%에서는 6010bar까지 올라간다. 높은 압력과는 달리 높은 압력에는 부피가 아주 극소하여 에너지 함량은 아주 낮은 편이다. 3%-NaCl과 마찬가지로 대부분의 에너지 함량은 낮은 농도에 집중되어 있다.Table 10 calculated the osmotic pressure of the alcohol by Lewis equation (eq.-4). Initially it goes up to 28 bar at 5% or 6010 bar at 99.50%. Unlike high pressures, high pressures have very low volume, which results in very low energy content. Like 3% -NaCl, most of the energy content is concentrated at low concentrations.
Figure PCTKR2016008104-appb-T000010
Figure PCTKR2016008104-appb-T000010
3% NaCl과 마찬가지로 농축하는 데 필요한 에너지를 계산 할 수 있다. 5%99.5%까지 농축에 필요한 에너지는 0.08819kwh/kg, 10%→99.5%는 0.0373kwh/kg, 7%→99.5%까지는 0.04787kwh/kg이다. 5%에서 99.5%까지 농축하는 데 필요한 최소에너지는 4.4093kwh인데 이를 알코올 함량으로 나누면 위와 같은 값을 얻을 수 있다. 5%는 4.4094/50=0.08819kwh/kg이다.As with 3% NaCl, the energy required for concentration can be calculated. The energy required for concentration up to 5% 99.5% is 0.08819 kwh / kg, 10% → 99.5% is 0.0373kwh / kg and 7% → 99.5% is 0.04787kwh / kg. The minimum energy required to concentrate from 5% to 99.5% is 4.4093 kwh, which can be obtained by dividing this by the alcohol content. 5% is 4.4094 / 50 = 0.08819 kwh / kg.
현재 가능 효과적이라는 열 적 농축공정과 비교하고자 한다. 아래의 website는 2.27kwh/kg이다.We would like to compare it with a thermal enrichment process that is currently available. The website below is 2.27kwh / kg.
http://www.ethanolproducer.coicles/13134/rfa-analysis-finds-improvement-of-corn-ethanol-net-energy-balance Accessed on July 17. 2016 Iowa WDG(lowest energy consumption)=23,428 BTU/gal =23,428/3,412BTU/(3.785*0.8) (kwh/kg) http: //www.ethanolproducer.coicles/13134/rfa-analysis-finds-improvement-of-corn-ethanol-net-energy-balance Accessed on July 17. 2016 Iowa lower energy consumption (WDG) = 23,428 BTU / gal = 23,428 / 3,412 BTU / (3.785 * 0.8) (kwh / kg)
한편 에탄올 예를 들면 7(wt%)의 에탄올을 RO로 99.5%까지 농축하는 데는 최소 에너지가 3.389kwh가 되며 여기에 추가로 3kwh를 추가하여 6.389kwh/70kg=0.091kwh/kg이 된다On the other hand, ethanol, for example, to concentrate 7 (wt%) ethanol to 99.5% with RO, the minimum energy is 3.389 kwh, and additionally 3 kwh is added to 6.389 kwh / 70 kg = 0.091 kwh / kg.
총열에너지가 삼투압공정의 2.27/0.091=24.94배가 된다. 이상의 예에서 99.6%의 삼투압은 무려 6000bar가 됨을 감안할 때 무삼투압차 기술을 쓰면 우리가 감당할 수 있는 압력수준 (예; 100 bar이하)으로 만들 수 있으며 더구나 막을 쓰게 되면 물과 에탄올과의 azeotrope 현상도 해결할 수 있어 유리하다고 할 수 있다.The total heat energy is 2.27 / 0.091 = 24.94 times that of the osmotic process. In the above example, since the osmotic pressure of 99.6% is 6000 bar, the osmotic pressure difference technology can make the pressure level that we can handle (e.g. below 100 bar). Moreover, when the membrane is used, the azeotrope phenomenon between water and ethanol is also achieved. It can be said that it can be solved.
실시 예 10: 무삼투압차 RO 농축 수에서 Li, Mg, 금, Uranium의 생산Example 10 Production of Li, Mg, Gold, Uranium in Osmotic RO Concentrated Water
세계적으로 총 담수화 시장은 2016년 (google image:해수담수화시장 accessed on 07-24-2016)에 의하면 60million/d에 이르고 있다. 현재 대한민국 부산 직할시 기장군에 있는 65000톤/d의 해수 담수화 시설에 관해서 경제성 평가를 수행하여 보았다. 경제성은 물의 판매가를 톤당 1$로 하고 계산한 값을 기준으로 각 원소들의 현재 국제가를 기준으로 비교하여 보았다. The global desalination market is 60 million / d, according to 2016 (google image: desalination market accessed on 07-24-2016). An economic evaluation was conducted on the 65,000 ton / d seawater desalination plant in Gijang-gun, Busan, Korea. Economics compared the price of water at $ 1 per tonne based on the current international price of each element.
하기 표 11에 개시된 바와 같이, 대부분이 너무 미량이라 물 판매가에 훨씬 못 미치나 Mg, 소금 등은 검토해 볼 가치가 있어 보인다. 본 기술은 바닷물 보다 농도를 최소한 10배 혹은 현재 RO 배출수의 5배 정도 농축된 상태이므로 다른 공정에 비해 유리할 것이다. As disclosed in Table 11 below, most of them are so small that much less than the water sales price, Mg, salt and the like seems worth examining. The technology is at least 10 times more concentrated than seawater or 5 times more concentrated than current RO effluents, which may be advantageous over other processes.
Figure PCTKR2016008104-appb-T000011
Figure PCTKR2016008104-appb-T000011
이상으로 본 발명 내용의 특정한 부분을 상세히 기술하였는바, 당업계의 통상의 지식을 가진 자에게 있어서 이러한 구체적 기술은 단지 바람직한 실시 태상일 뿐이며, 이에 의해 본 발명의 범위가 제한되는 것이 아닌 점은 명백할 것이다. 따라서, 본 발명의 실질적인 범위는 첨부된 청구항들과 그것들의 등가물에 의하여 정의된다고 할 것이다. The specific parts of the present invention have been described in detail above, and it is apparent to those skilled in the art that such specific descriptions are merely preferred embodiments, and thus the scope of the present invention is not limited thereto. something to do. Thus, the substantial scope of the present invention will be defined by the appended claims and their equivalents.
본 발명에 따른 무삼투압차 상태에서 유압으로 수용액을 농축하는 방법은 에너지의 소비가 적고, 추출용매를 사용하지 않아도 용질의 최대 포화 수용액 농도 또는 용질의 농도가 100% 될 때까지 농축할 수 있으며, 별도의 삼투압 유도 용액을 사용하지 않아도 되는 장점이 있다.The method of concentrating an aqueous solution by hydraulic pressure in a non-osmotic pressure difference state according to the present invention consumes less energy and can be concentrated until the maximum saturated aqueous solution concentration or the concentration of the solute is 100% without using an extraction solvent. There is an advantage that does not need to use a separate osmotic induction solution.

Claims (12)

  1. 다음 단계를 포함하는 △π 저감 조건에서 용질 함유 수용액을 농축하는 방법:A method of concentrating a solute-containing aqueous solution under a Δπ reduction condition comprising the following steps:
    (a) 용질 함유 수용액을 역삼투압 막으로 구획되어 있는 피드 챔버와 π-이콜라이저 챔버를 포함하는 무삼투압차 농축기를 이용하여 농축하되, 하기의 공정 중 하나 이상의 공정을 이용하여 피드챔버와 π-이콜라이저 챔버간의 △π저감 조건을 형성하는 단계:(a) Concentrating the solute-containing aqueous solution using a non-osmotic pressure concentrator comprising a feed chamber and a π-echolizer chamber partitioned with a reverse osmosis membrane, using one or more of the following steps to feed the feed chamber and π- Forming a Δπ reduction condition between the equalizer chambers:
    (i) (Feed) Input-split cascade 공정;(i) (Feed) Input-split cascade process;
    (ii) (Feed) Output-split counter-current 공정; 및 (ii) (Feed) Output-split counter-current process; And
    (iii) Nano filtration membrane을 적용하는 공정; (iii) applying a nano filtration membrane;
    (b) 상기 농축된 용질 함유 수용액을 또다른 무삼투압차 농축기 또는 역삼투압을 이용하여 회수하는 단계.(b) recovering the concentrated solute-containing aqueous solution using another osmotic pressure concentrator or reverse osmosis.
  2. 제1항에 있어서, The method of claim 1,
    (c) 상기 (b) 단계의 또다른 무삽투압차 농축기는 (i) 내지 (iii) 공정 중 하나 이상을 이용하여 용질의 회수를 최대화하는 단계를 추가로 포함하는 것을 특징으로 하는 방법. (c) another stepless pressure difference concentrator of step (b) further comprises maximizing recovery of the solute using one or more of processes (i) to (iii).
  3. 제1항에 있어서, 상기 (a) 단계에 이전에 역삼투압을 이용하여 농축 대상 용질 함유 수용액을 농축시키는 단계를 추가로 포함하는 것을 특징으로 하는 수용액을 농축하는 방법.The method of claim 1, further comprising concentrating the aqueous solution containing the solute to be concentrated using reverse osmosis pressure before the step (a).
  4. 제1항에 있어서, 상기 (b) 단계에서 또다른 무삽투압차 농축기는,According to claim 1, Another stepless pressure difference concentrator in the step (b),
    (i) 무삼투압차 농축기의 피드 챔버로 상기 용질 포함 상태 혹은 1차 농축된 수용액을 이송시키고, 상기 피드 챔버로 이송된 수용액과 동일한 삼투압을 가지는 용액을 π-이콜라이저 챔버로 투입하여, 피드 챔버와 π-이콜라이저 챔버간의 무삼투압 상태를 형성하는 단계; 및(i) transferring the solute-containing state or the first concentrated aqueous solution to the feed chamber of the osmotic pressure difference concentrator, and introducing a solution having the same osmotic pressure as the aqueous solution transferred to the feed chamber into the π-echoiser chamber, Forming an osmotic state between the π-echolizer chamber; And
    (ii) 무삼투압 상태에서 피드 챔버에, 역삼투압막은 10∼100기압의 압력을 가하여 수용액의 물을 π-이콜라이저 챔버로 이송시킴으로써 수용액을 농축시키는 단계;를 포함하는 것을 특징으로 하는 수용액을 농축하는 방법. (ii) concentrating the aqueous solution by transferring the water of the aqueous solution to the π-echolizer chamber by applying a pressure of 10 to 100 atm to the feed chamber in an osmotic pressure-free state. How to.
  5. 제1항에 있어서, 상기 (a) 단계의 Input-split cascade 공정은 상기 2차 농축된 용질 함유 수용액을 다른 무삼투압차 농축기로 이송시키되, 절반은 피드 챔버로 절반은 π-이콜라이저 챔버로 이송한 다음, 저압으로 피드 챔버의 수용액을 농축하고, 농축된 용질 함유 수용액을 또 다른 무삼투압 농축기로 이송시키는 것을 특징으로 하는 수용액을 농축하는 방법.The method of claim 1, wherein the input-split cascade process of step (a) transfers the secondary concentrated solute-containing aqueous solution to another osmotic pressure difference concentrator, half to a feed chamber and half to a π-echoiser chamber. Then concentrating the aqueous solution of the feed chamber at low pressure, and transferring the concentrated solute-containing aqueous solution to another osmotic pressure concentrator.
  6. 제1항에 있어서, 상기 (a) 단계의 Output-split counter-current 공정은 π-이콜라이저 스트림의 일부를 피드스트림으로 보내는 것을 특징으로 하는 수용액을 농축하는 방법.The method of claim 1, wherein the output-split counter-current process of step (a) sends a portion of the π-echolizer stream to the feed stream.
  7. 제1항에 있어서, 상기 (a) 단계의 (iii) 공정은 (i) 공정에 적용할 때는 처음부터, (ii) 공정에 적용할 때는 피드 챔버와 π-이콜라이져 챔버의 삼투압차와 물 회수량을 고려하여 무삼투압차 농축기 앞부분에는 용질 선택도가 높은 막을 뒷부분에는 낮은 막을 적용하는 것을 특징으로 하는 수용액을 농축하는 방법. The process according to claim 1, wherein the step (iii) of step (a) is performed from the beginning when applied to the process (i), and the osmotic pressure difference and water ash of the feed chamber and the π-echolizer chamber when applied to the process (ii). A method for concentrating an aqueous solution, characterized in that the membrane is applied to the high solute selectivity in the front portion and the low membrane in the rear portion in consideration of the quantity.
  8. 제2항에 있어서, The method of claim 2,
    (d) 상기 농축된 용질 함유 수용액을 열에너지, 전기에너지 또는 압력을 이용하여, 고형화된 용질을 회수하는 공정을 추가로 포함하는 것을 특징으로 하는 수용액을 농축하는 방법. and (d) recovering the solidified solute using thermal energy, electrical energy, or pressure in the concentrated solute-containing aqueous solution.
  9. 제1항에 있어서, 압력지연 정삼투압(PRO) 발전, 자원이용 및 희토류 회수 공정 등을 추가로 포함하여 상기 공정의 부가가치를 높이는 것을 특징으로 하는 수용액을 농축하는 방법.The method of claim 1, further comprising pressure delayed forward osmosis (PRO) power generation, resource utilization, and rare earth recovery processes to increase the added value of the process.
  10. 제1항에 있어서, 물질수지 및 에너지수지를 최적화하는 공정을 추가로 포함하는 것을 특징으로 하는 수용액을 농축하는 방법.The method of claim 1, further comprising the step of optimizing the mass balance and the energy balance.
  11. 제1항 내지 제10항 중 어느 한 항의 방법을 이용하여 용질 함유 용액에서 용매와 용질을 분리하는 방법.A method of separating a solvent and a solute from a solute-containing solution using the method of any one of claims 1 to 10.
  12. 제11항에 있어서, 상기 용질은 염 혹은 액체이고, 상기 용매는 물인 것을 특징으로 하는 방법.The method of claim 11, wherein the solute is a salt or a liquid and the solvent is water.
PCT/KR2016/008104 2015-07-24 2016-07-25 Method for concentrating solute-containing aqueous solution at high concentration by reverse osmosis method in non-osmotic pressure difference state WO2017018764A1 (en)

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EP16830797.3A EP3326977A4 (en) 2015-07-24 2016-07-25 Method for concentrating solute-containing aqueous solution at high concentration by reverse osmosis method in non-osmotic pressure difference state
US15/744,945 US10953367B2 (en) 2015-07-24 2016-07-25 Method of osmotic pressure free reverse osmosis for enriching solute-containing solution to high concentration
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