AU2008203793A1

AU2008203793A1 - Desalination of seawater in a vacuum tube

Info

Publication number: AU2008203793A1
Application number: AU2008203793A
Authority: AU
Inventors: Oktay Karacanta
Original assignee: Individual
Current assignee: Individual
Priority date: 2008-06-17
Filing date: 2008-08-07
Publication date: 2010-01-07
Anticipated expiration: 2028-08-07
Also published as: AU2008203793B2

Description

1 DESALINATION OF SEAWATER IN A VACUUM TUBE Evaporation is a very well known process that happens in nature. It is also a well-known fact that evaporation can be achieved at low temperatures under vacuum. From this principal, this patent proves when a vacuum is created over seawater in a closed space; the water will evaporate and leave the sea salt in the water behind. By cooling this steam in another space, such as a condenser, pure water can be obtained. From the following calculations and attached figures, the above claim will be proven. As can be seen on the attached figures, the device used for this process comprises a vacuum tube which is an inverted U shaped tube (2), one of its legs standing on the evaporation hood (1), which is located on the seawater pond (11), and the other leg standing on the pure water tank (7). It is possible to create vacuum, inside this tube and the evaporation hood, by the following two different procedures: Procedure 1: Vacuum will be created with a vacuum pump. See. Fig.1 1. Open the vacuum valve (17) and turn the vacuum pump (18) on. 2. Close the vacuum valve and the vacuum pump when the pressure inside the tube drops under the saturation pressure of the water inside the water pond (11) by observing the pressure reading at the vacuum gauge (16). At this stage, we would have two water columns inside each end of the U tube, the first one inside the evaporation hood (1), and the other one inside the condensed water pipe (6), each one having around 10,000 mm height, which indicates the atmospheric pressure. Evaporation will start in both of the water columns, and saturated steam will take place inside the tube. The pressure inside the tube will start to increase and it will reach up to the saturation pressure of the water, inside the ponds. As a numerical example, if the temperature inside ponds is 23*C, the pressure inside the tube will reach up to 0.0281 bar = 291 mm w.g., which is the saturation pressure for 23*C. This pressure in the tube will push the atmospheric pressure 291 mm down, and the new height of water columns will be 10,353-291= 10,062 mm w.g. If the cooling unit (5) of the condenser (4) is turned on, the saturated steam inside the condenser would then be cooled down, the steam inside the condenser will condense, and condensed water will run down into the pure water tank (7), and. If we fix the lower temperature of condensation at 3C (before freezing point), with the thermostat of the cooling 2 unit, a new kind of saturated steam will take place inside the condenser, which has a temperature of 3*C, and a saturation pressure of: 0.0076 bar = 79mm mm w.g. There will be a pressure difference inside the tube, between the evaporation hood, and the condenser, which is: 0.0281 - 0.0076 = 0.0205 bar = 212 mm w.g. This pressure difference will cause the saturated steam inside the evaporation hood, to flow from evaporation hood to the condenser, and the condensed water will run down through the condensed water pipe (6), into the pure water tank (7). The pure water, which is obtained in this tank, would then be sent down, via pure water pipe (8) to the hydroelectric generator (19), then to delivery pond (9). And delivery pipe (10) will deliver to the customers. Hydroelectric generator (19) is optional. By this generator we can meet some of the energy requirement, and make the cost of water cheaper. In case we need to built the vacuum tube away from sea shore, we can meet the energy requirement of seawater supply pump and brine pump by this generator. When the cooling unit (5) is shut down, condensation and evaporation would stop. And the vacuum inside the tube will be ready for the next operation. The evaporation hood (1), vacuum tube (2) and the condenser (4) will be externally heat insulated (3), to protect the process from atmospheric variations. Evaporation hood will so be designed that, sea salt inside the water, will not be dragged into the vacuum tube. In the evaporation hood, the salt inside seawater will be separated from the water and settle down into the salt settling bin (12). The salt in the settling bin, will be carried out of the sea pond (11), while it is in brine condition, with a brine pipe (13) and a brine pump (14), then the brine will be carried with another brine pipe (15), either to a salt drying and packing facility, or back to the sea. Procedure 2: Vacuum will be created with the aid of valves. See Fig.2 1. Close the evaporation hood seal valve (22), located underneath the evaporation hood and immersed into the sea water. 2. Close the pure water tank seal valve (23), located below the condensed water pipe and immersed into the pure water tank. 3. Close the pure water inlet valve (20) 4. Open the water overflow valve (21) At this stage the U tube is completely empty and open to atmosphere.

3 5. Keep the valves (22) and (23) closed, and open the pure water inlet valve (20), to fill evaporation hood and the U tube with pure water. 6. Observe water overflowing from the over flow valve (21), then close the pure water inlet valve (20) and also the overflow valve (21). Now both the U tube and evaporation hood are completely full of pure water. 7. When valves (22) and (23) are opened, the water inside the U tube and the evaporation hood, will run down into the seawater pond (11) and pure water tank (7), respectively. At this stage, we should have two water columns inside each end of the U tube, the first one inside the evaporation hood (1), and the other one inside the condensed water pipe (6), each one having 10,353 mm height, which shows the atmospheric pressure. There will be an absolute zero vacuum inside the tube. From this point the process will be the same as procedure 1. Some numeric examples: How much pure water can be obtained? The calculations below are performed with the aid of "Piping Calculations Manual, written by E.S. Menon" Since sonic velocity exists in all examples below, modified Darcy formula will be applied. W = 1891YD 2 - K= fL/D Kv Where W - Mass flow rate, lb/h Y - Expansion factor for pipe D - Pipe inside diameter, in AP - Pressure drop in psig K - Resistance coefficient L - Pipe length, in f - Darcy friction factor v - Specific volume of steam at inlet pressure ft 3 / lb

P

1 Inlet pressure = 0.0281 bar = 0.413 psi v =788 m 3 /lb AP/P and Y values are taken from table 4.6 of mentioned book. Equivalent length of fittings are the same for all examples below. Two 900 elbows: K = 0.02 x 2 x 30= 1.2 One entrance: K = 0.5 One increaser: K = 1.0 4 Total K for fittings K = 2.7 D-in L-in K AP/P, Y A P-psi W-m 3 / day H-m 1 340 mm 10,000 mm 3.3 0.653 0.662 0.653x0.413 25 30 13.39 in 393.7 in =0.270 2 1,000 mm 50,000 mm 3.7 0.667 0.666 0.667x0.413 207 75 39.37in 1,968.5 in =0.275 3 10,000 mm 150,000 mm 3 0.648 0.658 0.648x0.413 22,326 260 393.7 in 5,905.5 in =0.268 4 10,000 mm 900,000 mm 4.5 0.689 0.674 0.689x0.413 19,255 1,020 393.7 in 35,433 in =0.285 5 20,000 mm 900,000 mm 3.6 0.664 0.665 0.664x0.413 83,306 1,110 787.4 in 35,433 in =0.274 Cost of energy for cooling unit Enthalpy of saturated vapor at 23*C = 2,543.6 KJ/Kg Enthalpy of saturated vapor at 3*C = 2,506.9 KJ/Kg Cooling load of 1 Kg condensed vapor: 2,543.6 - 2,506.9 = 36.7 KJ/Kg = 8.77 Kcal/Kg Cooling load of 1 m 3 condensed water: 8.77 Kcal/Kg x 1.000 Kg = 8,770 Kcal Assuming that the energy that needs to run the cooling unit is, half the cooling load of the condenser. Energy consumption in cooling unit for 1 mn condensed water: 8,770 / 2 = 4,385 Kcal = 5.1 Kwh Cost of 1 m 3 water = 5.1 Kwh x $AUO.17128149 / Kwh = $AUO.87 / m 3 How much salt can be obtained in the sea pond Sea water has %3.5 salt in it. For each m 3 of pure water we can obtain: 1,000 x 35 / 965 = 36.23 Kg salt