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OPERATION AND MAINTENANCE OF DM PLANT OPERATION AND MAINTENANCE OF DM PLANT 1.0 OBJECTIVE : Objective of this procedure is to provide operation and maintenance instruction of Dematerialization Plant (DM Plant) as per safety and cGMP requirements. 2.0 SCOPE: This SOP covers the instructions for operation & maintenance of DM Plant. The SOP also covers preventive maintenance schedule and gives guideline for maintaining critical spares generally required for maintenance. The SOP specifies the responsibilities of staff for the various functions outlined here under. 3.0 SYSTEM REQUIREMENT: A number of minerals readily dissolve into raw water. DM Plant removes mineral salt by an ion-exchange process from the raw water. After demineralization water passes through the fine filtration and UV sterilizer so that the water meet all the specification of pharmaceutical for purified water 4.0 SYSTEM DESCRIPTION: The Indion Packaged Up flow DM Plant consists of water feed pumps (one working one stand by) with motor, activated carbon filter, strong acidic Cation unit, strong alkaline anion unit, Mixed bed unit, 5 micron filter, UV sterilizer, inter connected pipe work with valves & ejectors, pressure gauges, flow meter, conductivity meter etc., In Activated carbon filter (ACF), Activated carbon used is in granular form which is filled in mild steel pressure vessel. ACF eliminates suspended solids and absorbs most pollutants dissolve in water like detergents, phenols, obnoxious gases like chlorine & sulphur dioxide. It also helps to reduce the impurities like colour, taste etc. Cation Exchange column of MSRL material contains strongly acidic Cation Exchanger resin. The Cation Exchange material, which exchanges Cations such as Calcium, Magnesium and Sodium for hydrogen ions to form the corresponding acids. This process is reversed when the resin is regenerated by hydrochloric acid through column. Anion Exchange column of MSRL material contains a strongly basic Anion resin. The strongly basic Anion Exchanger will remove the acid formed by the cation exchange material and producing water, which is free from dissolved solids. This process is reversed when the resin is regenerated by Caustic Soda solution Mix bed Exchange column of MSRL material contains Cation and Anion resins and thoroughly mixed with compressed air. When the water passes through Mixed Bed Column the traces of dissolved salt leaked out will be retained in the Mixed Bed Column. Conductivity meter is used to show the purity of dematerialized water. The conductivity is proportional to be the residual ionized solids in the water. Low conductivity indicates water of high purity. 5 micron filters with SS housing removes the particular impurities from the water. UV as disinfections is used for removing microbial reduction. The unit is fitted with all necessary piping, valves & ejectors from the raw water inlet and to DM water outlet. 5.0 OPERATING PROCEDURE: Ensure prior to start up - First time after every maintenance:

Ensure that Activated carbon filter, Cation, Anion, Mix bed, vessel is filled with specified and required quantity resin. Ensure that the all valves are in operation. Check that all piping work is carried as per its piping drawing given in manufacturer’s manual. Ensure that there is no water leakage. Ensure that the DM plant feed water pump gives rated flow and pressure required for DM plant. Test the water inlet quality. It should be as per specification given in design data.

Ensure raw water pressure at the inlet is not less than 1.7 to 2 kg/cm2

Starting Procedure / Instruction of DM Plant

Open the GIDC raw water inlet valve leading to the suction line of the DM plant feed water pump. Switch ON the DM plant feed water pump. Slowly open the pump discharge valve leading to the inlet valve of the DM plant. Open the Cation inlet valve V1. Open the Anion inlet valve V5 and rinse outlet valve V7 Adjust V7 to get specified rinse flow to drain. Open valve V12 to allow some flow through the conductivity cell. Switch ON the conductivity meter to ‘CALIBRATE’ If the needle on meter indicates exactly 40, move to ‘READ’ position

Stopping Procedure / Instruction of DM Plant

Switch OFF the conductivity meter. Switch OFF the DM water feed pump. Close the Anion outlet valve V6. Close the Anion inlet valve V5. Close the Cation inlet valve V1. Close the GIDC inlet raw water valve.

Acid Injection: Prepare the stipulated quantity of acid + dematerialized water solution by stirring in the measuring tank. Open the main raw water inlet valve V1, acid ejector power water valve V3. Open the acid outlet valve V10. Open the acid ejector suction valve V4. Adjust V3 to maintain the specified flow to drain, such that the acid measuring tank is emptied in the specified time, leaving only a dead level of 25 mm in the tank. Now close valve V4. Acid Rinse: Adjust valve V3 to maintain specified flow for the specified time. At the end of this specified period, it is advisable to take a sample from valve V10 and test it for pH with the help of pH strip. If the result of the test is satisfactory, close valves V10 and V3. Soda Injection: Prepare the stipulated quantity of caustic soda + dematerialized water solution by stirring in the measuring tank. Open the main raw water inlet valve V1, caustic ejector power water valve V8. Open the caustic outlet valve V11. Open the caustic ejector suction valve V4. Adjust V8 to maintain the specified flow to drain, such that the acid measuring tank is emptied in the specified time, leaving only a dead level of 25 mm in the tank. Now close valve V9. Soda Rinse: Adjust valve V3 to maintain specified flow for the specified time of 30 min.. At the end of this specified period, close valves V8 and V11. Final Rinse:

Open valve Anion inlet valve V5 and final rinse outlet valve V7. After @ 20 minutes, open conductivity cell isolation valve V12 Switch ON the conductivity tester. Reading must be below 30 uS/cm. Close final rinse outlet valve V7. Open Anion outlet valve V6 to send DM water to service. After 30 to 40 minutes we can get an acceptable level of conductivity

Ensure prior to start up of Mixed Bed Unit – Only first time:

Ensure that mixed bed vessel is filled with specified and required quantity resin. Ensure that the all valves are in operation. Check that all piping work is carried as per its piping drawing given in manufacturer’s manual. Ensure that there is no water leakage. Ensure the specified flow rate. Test the water inlet quality. It should be as per specification given in design data.

Preparation of Regenerants: Prepare the 6L Hydrochloric Acid + 30L dematerialized water solution (10%) by stirring in the measuring tank. Prepare the 2.6 kgs Caustic + dematerialized water solution (10%) by stirring in the measuring tank.

Description This section is from the book "Alcohol, Its Production, Properties, Chemistry, And Industrial Applications", by Charles Simmonds. Also available from Amazon:Alcohol: Its Production, Properties, Chemistry, And Industrial Applications.

Alcohol Calculations (1). To convert percentage ofalcoholby volume into percentage by weight. Multiply the volume percent age by thespecific gravityofabsolute alcohol(0 7936), and divide the product by the specific gravity of the liquid in question. For let S be the specific gravity of the liquid (at 15.6°/l5.6o). Then 100 c.c. weigh 100 Sρ grams, where ρ is the sp. gr. ofwaterat 15.6°/4°. If V be the percentage of alcohol by volume, then 100 c.c. of the liquid contain V c.c. of absolute alcohol. That is, 100 Sρ grams contain V c.c.; .'. 100 grams contain V/Sρ c.c. abs. ale.

But 1 c.c. of abs. ale. weighs 0 7936 ρ gram, .-. V/Sρ c.c. weigh V/Sρ X 0.7936 ρ gram; = V/S X 0.7936 gram. That is, 100 grams of the liquid contain V X0.7936/S gram of alcohol, i.e., percentage by weight = Percent. by volume x 0.7936 /.Sp. gr. of the liquid (2). To convert percentage by volume into grams per 100 c.c. Multiply by 0.79284. For if 100 c.c. of the liquid contain V c.c. of absolute alcohol, the weight of the V c.c. is V X 0.7936 x ρ grams; = V X 0 7936 x 0.999037 = V X 0.79284. (3). To convert percentage of alcohol by volume into percentage ofproof spiritby volume, Multiply by 1 7535. For absolute alcohol is 75.35 "over proof " - that is, 100 volumes of absolute alcohol contain the same quantity of alcohol as do 175.35 volumes of proof spirit. Therefore 1 vol. of alcohol = 1.7535 vols. of proof spirit. Or for summary calculations, multiply by 7/4 (= 1.75). From the foregoing threeexamplesit will be seen how any one denomination can be expressed in terms of any other. Thus from (3) a quantity of alcohol in terms of proof spirit can be expressed in terms of absolute alcohol by volume on dividing by 1.7535, and then either as percentage of absolute alcohol by weight, or as grams of absolute alcohol per 100 c.c, as shown in (1) and (2) respectively. The various relations can be summarised for reference in the following "conversion equations." Let S denote the sp. gr. of a specimen of alcohol, P the percentage of proof spirit by volume, V the percentage of alcohol by volume, W „ „ „ weight, and G the grams per 100 c.c. Then P = 1.7535 V, = 2.2095 WS. V = 0.5703 P, = 1.2601 WS: W= 1/2.2095 x P/S = 0.7936 X V/S and G = 0.7928 F, = 0.4521 P. An operation frequently required is the reduction of alcohol from a higher to a lower strength. In the laboratory, with convenient measuring vessels at hand, the operation is simple. The volumes are inversely as the strengths. We therefore take a convenient quantity of the alcohol, measured at the standard temperature, and make it up with water to such a volume, at the same temperature, that the ratio of this volume to the first shall be the ratio of the given strength to the required strength. Example: Given strength 90 per cent., required strength 20 per cen. Dilute 20 c.c. to 90 c.c, or 222 to 100, or 1111 to 500, as may be convenient. Since the temperature of the liquid rises during the mixing, for precise work it must be adjusted before completing the volume.

It is to be carefully noted that on account of thecontractionwhich occurs the required strength would not be given by adding 70 c.c. of water to 20 c.c. of the alcohol. More than 70 would be required. For this reason, the calculation is less simple when, as in large operations, no suitable vessels may be available for accurately making up the diluted spirit to a required volume at a particular temperature. It is then necessary to calculate the actual quantity of water which must be added. This may be done as follows. To find the volume of water which must be added to a given volume V1 of alcohol, in order to reduce it from a given strength S1 to a lower strength S2, the densities D1 and D2 respectively corresponding with these strengths, being known. Let x denote the weight in grams of the water required, and V2 the resulting volume, in c.c, of the diluted spirit. Then the weight of the given volume of alcohol is V1D1 and thai of the resulting volume is V2D2. But the latter weight = the former weight + x, .-. V1D1 + x = V2D2,and x = V2D2 - V1D1 . (1) Also, since the strengths are inversely as the volumes, V2 / V1 = S1 / S2; OR V2 = V1 S1 / S2 . . . . (2) Substituting from (2) in (1) we get: x = V1/S2 (D2S1 - D1S2) ..... (3) which gives the weight of water required, in terms of the known quantities, and expressed in grams. For ordinary work this may be taken as the required volume of water, in c.c. The precise volume will of course depend upon the temperature of the water. At 4°, x grams = x c.c.; at 156°, x grams = x x 1.0009 c.c. Example (I). How much water at 156° must be added to 100 c.c. of 90 per cent. alcohol in order to reduce its strength to 60 per cent. ? (Strengths by volume.) Here V1 = 100, S1 = 90, and S2 = 60. With sufficient accuracy the values of D1 and D2 may be taken from the ordinaryalcohol tables: D1 = 0.8337, and D2 = 0.9134. .-. x = 100/60 (0.9134 x 90 - 0.8337 X 60) x 1.0009 = 53.69 c.c, or practically, 53.7 c.c. Strictly, however, the values of D1 and D2 as taken from the ordinary alcohol tables are not the true densities (mass of unit volume), but the specific gravities at 15.6°, referred to water at that temperature as unity. Since the density of water at that temperature is 0.999037 (Despretz), and not 1, the values of D1 and D2 should be corrected accordingly. If we therefore multiply these values by 0.999037, we find the true densities D1 and D2 to be 0 8329 and 0 9125 respectively, and the corrected result is x = 53.64 c.c. But it is to be noted that we get the same result by simply taking the values of D1 and D2 from the ordinary alcohol tables and using them in equation (3), omitting the factor 1.0009: x = 100/60 (09134 x 90 - 08337 x 60), = 53 64. The fact is that the reciprocal of 0.999037 is 1/1.0009; and this cancels out the factor 1.0009 used in the first calculation. So that, finally, although x in equation (3) denotes the weight of water required, if we take the values of D1 and D2 as specific gravities from the ordinary alcohol tables the result expresses the required volume of water, in c.c.

This example has been elaborated a little, because the point in question is sometimes found puzzling by persons unfamiliar with alcohol calculations. Example (2). How much water is required in order to reduce 100 gallons of spirit at 60 over proof to a strength of 20 over proof ? From what precedes, it will be seen that equation (3) will give the answer, x, in gallons, if the specific gravities corresponding with the strengths are taken from the ordinary alcohol tables. Here V1 = 100, S1 = 160, S2 = 120; D1 = 0.8295, and D2 = 0.8936. Hence x = 100/120 (0.8936 x 160 0.8295 x 120) = 36.2 gallons. Problems of a slightly different character are set in the next two questions. (1). What weight of water must be added to 100 grams of an alcohol (A) of given strength (percentage by weight) in order to produce an alcohol (B) of given lower strength ? Let a and b be the respective given strengths (percentages by weight), and let x be the weight in grams of the water required. Then the total water present is 100 - a + x, and the weight of B produced is 100 + x. .'. in 100 grams of B there are - 100/100+x (100 - a + x) grams of water. But the weight of water in 100 grams of B is also 100 - b. Hence, equating, 100/100+x (100 - a + x) = 100 - b. Solving this equation, we get x = 100/b (a - b) . . . . . . (l). Thus if A is alcohol of 90 per cent. strength by weight, and we require to dilute it to 70 per cent., the weight of water to be added to 100 grams of A is: 100/70 (90 - 70) = 28 4/74 grams. (2). If in the foregoing example we have 100 c.c. of A instead of 100 grams, what is the quantity of water required ? Let D be the density of the alcohol A. Then 100 c.c. will weigh 100 D grams. Hence the quantity of water to be added is 100 D/100 x100/b (a - b) grams == 100 (a - b) D/b - grams . . . (ii). Here, as explained above, the value of D, if taken from the ordinary alcohol tables, is not the true density (mass of unit volume), but the specific gravity referred to water at 15.6°. Hence 100 c.c. do not weigh exactly 100 D grams if these tables are used. In this case, as in that explained above, the quantity of water given by the expression (ii) must be taken as the volume in c.c, not the weight in grams. The correction for reducing the value of D to true density, and that for converting grams of water into c.c, cancel each other out.

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Ensure that Activated carbon filter, Cation, Anion, Mix bed, vessel is filled with specified and required quantity resin. Ensure that the all valves are in operation. Check that all piping work is carried as per its piping drawing given in manufacturer’s manual. Ensure that there is no water leakage. Ensure that the DM plant feed water pump gives rated flow and pressure required for DM plant. Test the water inlet quality. It should be as per specification given in design data.

Ensure raw water pressure at the inlet is not less than 1.7 to 2 kg/cm2

Starting Procedure / Instruction of DM Plant

Open the GIDC raw water inlet valve leading to the suction line of the DM plant feed water pump. Switch ON the DM plant feed water pump. Slowly open the pump discharge valve leading to the inlet valve of the DM plant. Open the Cation inlet valve V1. Open the Anion inlet valve V5 and rinse outlet valve V7 Adjust V7 to get specified rinse flow to drain. Open valve V12 to allow some flow through the conductivity cell. Switch ON the conductivity meter to ‘CALIBRATE’ If the needle on meter indicates exactly 40, move to ‘READ’ position

Stopping Procedure / Instruction of DM Plant

Switch OFF the conductivity meter. Switch OFF the DM water feed pump. Close the Anion outlet valve V6. Close the Anion inlet valve V5. Close the Cation inlet valve V1. Close the GIDC inlet raw water valve.

Acid Injection: Prepare the stipulated quantity of acid + dematerialized water solution by stirring in the measuring tank. Open the main raw water inlet valve V1, acid ejector power water valve V3. Open the acid outlet valve V10. Open the acid ejector suction valve V4. Adjust V3 to maintain the specified flow to drain, such that the acid measuring tank is emptied in the specified time, leaving only a dead level of 25 mm in the tank. Now close valve V4. Acid Rinse: Adjust valve V3 to maintain specified flow for the specified time. At the end of this specified period, it is advisable to take a sample from valve V10 and test it for pH with the help of pH strip. If the result of the test is satisfactory, close valves V10 and V3. Soda Injection: Prepare the stipulated quantity of caustic soda + dematerialized water solution by stirring in the measuring tank. Open the main raw water inlet valve V1, caustic ejector power water valve V8. Open the caustic outlet valve V11. Open the caustic ejector suction valve V4. Adjust V8 to maintain the specified flow to drain, such that the acid measuring tank is emptied in the specified time, leaving only a dead level of 25 mm in the tank. Now close valve V9. Soda Rinse: Adjust valve V3 to maintain specified flow for the specified time of 30 min.. At the end of this specified period, close valves V8 and V11. Final Rinse:

Open valve Anion inlet valve V5 and final rinse outlet valve V7. After @ 20 minutes, open conductivity cell isolation valve V12 Switch ON the conductivity tester. Reading must be below 30 uS/cm. Close final rinse outlet valve V7. Open Anion outlet valve V6 to send DM water to service. After 30 to 40 minutes we can get an acceptable level of conductivity

Ensure prior to start up of Mixed Bed Unit – Only first time:

Ensure that mixed bed vessel is filled with specified and required quantity resin. Ensure that the all valves are in operation. Check that all piping work is carried as per its piping drawing given in manufacturer’s manual. Ensure that there is no water leakage. Ensure the specified flow rate. Test the water inlet quality. It should be as per specification given in design data.

Preparation of Regenerants: Prepare the 6L Hydrochloric Acid + 30L dematerialized water solution (10%) by stirring in the measuring tank. Prepare the 2.6 kgs Caustic + dematerialized water solution (10%) by stirring in the measuring tank.

Description This section is from the book "Alcohol, Its Production, Properties, Chemistry, And Industrial Applications", by Charles Simmonds. Also available from Amazon:Alcohol: Its Production, Properties, Chemistry, And Industrial Applications.

Alcohol Calculations (1). To convert percentage ofalcoholby volume into percentage by weight. Multiply the volume percent age by thespecific gravityofabsolute alcohol(0 7936), and divide the product by the specific gravity of the liquid in question. For let S be the specific gravity of the liquid (at 15.6°/l5.6o). Then 100 c.c. weigh 100 Sρ grams, where ρ is the sp. gr. ofwaterat 15.6°/4°. If V be the percentage of alcohol by volume, then 100 c.c. of the liquid contain V c.c. of absolute alcohol. That is, 100 Sρ grams contain V c.c.; .'. 100 grams contain V/Sρ c.c. abs. ale.

But 1 c.c. of abs. ale. weighs 0 7936 ρ gram, .-. V/Sρ c.c. weigh V/Sρ X 0.7936 ρ gram; = V/S X 0.7936 gram. That is, 100 grams of the liquid contain V X0.7936/S gram of alcohol, i.e., percentage by weight = Percent. by volume x 0.7936 /.Sp. gr. of the liquid (2). To convert percentage by volume into grams per 100 c.c. Multiply by 0.79284. For if 100 c.c. of the liquid contain V c.c. of absolute alcohol, the weight of the V c.c. is V X 0.7936 x ρ grams; = V X 0 7936 x 0.999037 = V X 0.79284. (3). To convert percentage of alcohol by volume into percentage ofproof spiritby volume, Multiply by 1 7535. For absolute alcohol is 75.35 "over proof " - that is, 100 volumes of absolute alcohol contain the same quantity of alcohol as do 175.35 volumes of proof spirit. Therefore 1 vol. of alcohol = 1.7535 vols. of proof spirit. Or for summary calculations, multiply by 7/4 (= 1.75). From the foregoing threeexamplesit will be seen how any one denomination can be expressed in terms of any other. Thus from (3) a quantity of alcohol in terms of proof spirit can be expressed in terms of absolute alcohol by volume on dividing by 1.7535, and then either as percentage of absolute alcohol by weight, or as grams of absolute alcohol per 100 c.c, as shown in (1) and (2) respectively. The various relations can be summarised for reference in the following "conversion equations." Let S denote the sp. gr. of a specimen of alcohol, P the percentage of proof spirit by volume, V the percentage of alcohol by volume, W „ „ „ weight, and G the grams per 100 c.c. Then P = 1.7535 V, = 2.2095 WS. V = 0.5703 P, = 1.2601 WS: W= 1/2.2095 x P/S = 0.7936 X V/S and G = 0.7928 F, = 0.4521 P. An operation frequently required is the reduction of alcohol from a higher to a lower strength. In the laboratory, with convenient measuring vessels at hand, the operation is simple. The volumes are inversely as the strengths. We therefore take a convenient quantity of the alcohol, measured at the standard temperature, and make it up with water to such a volume, at the same temperature, that the ratio of this volume to the first shall be the ratio of the given strength to the required strength. Example: Given strength 90 per cent., required strength 20 per cen. Dilute 20 c.c. to 90 c.c, or 222 to 100, or 1111 to 500, as may be convenient. Since the temperature of the liquid rises during the mixing, for precise work it must be adjusted before completing the volume.

It is to be carefully noted that on account of thecontractionwhich occurs the required strength would not be given by adding 70 c.c. of water to 20 c.c. of the alcohol. More than 70 would be required. For this reason, the calculation is less simple when, as in large operations, no suitable vessels may be available for accurately making up the diluted spirit to a required volume at a particular temperature. It is then necessary to calculate the actual quantity of water which must be added. This may be done as follows. To find the volume of water which must be added to a given volume V1 of alcohol, in order to reduce it from a given strength S1 to a lower strength S2, the densities D1 and D2 respectively corresponding with these strengths, being known. Let x denote the weight in grams of the water required, and V2 the resulting volume, in c.c, of the diluted spirit. Then the weight of the given volume of alcohol is V1D1 and thai of the resulting volume is V2D2. But the latter weight = the former weight + x, .-. V1D1 + x = V2D2,and x = V2D2 - V1D1 . (1) Also, since the strengths are inversely as the volumes, V2 / V1 = S1 / S2; OR V2 = V1 S1 / S2 . . . . (2) Substituting from (2) in (1) we get: x = V1/S2 (D2S1 - D1S2) ..... (3) which gives the weight of water required, in terms of the known quantities, and expressed in grams. For ordinary work this may be taken as the required volume of water, in c.c. The precise volume will of course depend upon the temperature of the water. At 4°, x grams = x c.c.; at 156°, x grams = x x 1.0009 c.c. Example (I). How much water at 156° must be added to 100 c.c. of 90 per cent. alcohol in order to reduce its strength to 60 per cent. ? (Strengths by volume.) Here V1 = 100, S1 = 90, and S2 = 60. With sufficient accuracy the values of D1 and D2 may be taken from the ordinaryalcohol tables: D1 = 0.8337, and D2 = 0.9134. .-. x = 100/60 (0.9134 x 90 - 0.8337 X 60) x 1.0009 = 53.69 c.c, or practically, 53.7 c.c. Strictly, however, the values of D1 and D2 as taken from the ordinary alcohol tables are not the true densities (mass of unit volume), but the specific gravities at 15.6°, referred to water at that temperature as unity. Since the density of water at that temperature is 0.999037 (Despretz), and not 1, the values of D1 and D2 should be corrected accordingly. If we therefore multiply these values by 0.999037, we find the true densities D1 and D2 to be 0 8329 and 0 9125 respectively, and the corrected result is x = 53.64 c.c. But it is to be noted that we get the same result by simply taking the values of D1 and D2 from the ordinary alcohol tables and using them in equation (3), omitting the factor 1.0009: x = 100/60 (09134 x 90 - 08337 x 60), = 53 64. The fact is that the reciprocal of 0.999037 is 1/1.0009; and this cancels out the factor 1.0009 used in the first calculation. So that, finally, although x in equation (3) denotes the weight of water required, if we take the values of D1 and D2 as specific gravities from the ordinary alcohol tables the result expresses the required volume of water, in c.c.

This example has been elaborated a little, because the point in question is sometimes found puzzling by persons unfamiliar with alcohol calculations. Example (2). How much water is required in order to reduce 100 gallons of spirit at 60 over proof to a strength of 20 over proof ? From what precedes, it will be seen that equation (3) will give the answer, x, in gallons, if the specific gravities corresponding with the strengths are taken from the ordinary alcohol tables. Here V1 = 100, S1 = 160, S2 = 120; D1 = 0.8295, and D2 = 0.8936. Hence x = 100/120 (0.8936 x 160 0.8295 x 120) = 36.2 gallons. Problems of a slightly different character are set in the next two questions. (1). What weight of water must be added to 100 grams of an alcohol (A) of given strength (percentage by weight) in order to produce an alcohol (B) of given lower strength ? Let a and b be the respective given strengths (percentages by weight), and let x be the weight in grams of the water required. Then the total water present is 100 - a + x, and the weight of B produced is 100 + x. .'. in 100 grams of B there are - 100/100+x (100 - a + x) grams of water. But the weight of water in 100 grams of B is also 100 - b. Hence, equating, 100/100+x (100 - a + x) = 100 - b. Solving this equation, we get x = 100/b (a - b) . . . . . . (l). Thus if A is alcohol of 90 per cent. strength by weight, and we require to dilute it to 70 per cent., the weight of water to be added to 100 grams of A is: 100/70 (90 - 70) = 28 4/74 grams. (2). If in the foregoing example we have 100 c.c. of A instead of 100 grams, what is the quantity of water required ? Let D be the density of the alcohol A. Then 100 c.c. will weigh 100 D grams. Hence the quantity of water to be added is 100 D/100 x100/b (a - b) grams == 100 (a - b) D/b - grams . . . (ii). Here, as explained above, the value of D, if taken from the ordinary alcohol tables, is not the true density (mass of unit volume), but the specific gravity referred to water at 15.6°. Hence 100 c.c. do not weigh exactly 100 D grams if these tables are used. In this case, as in that explained above, the quantity of water given by the expression (ii) must be taken as the volume in c.c, not the weight in grams. The correction for reducing the value of D to true density, and that for converting grams of water into c.c, cancel each other out.

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