Dairy Technology - Tapflo

Table of contentsMILKComposition of Danish Cow’s Milk 2002 .......... 3Density of Milk . ............................. 3Yields from Whole Milk etc. .................... 4Determination of Fat Content in Milk and Cream. ... 4Determination of Protein Content in Milk and Cream 6Detection of Preservatives and Antibiotics in Milk ... 7Acidity of Milk. .............................. 7The Phosphatase Test ........................ 10Standardisation of Whole Milk and Cream. ........ 10Standard Deviation. .......................... 13Calculating the Extent of Random Sampling ....... 14GENEREL MILK PROCESSINGPasteurisation. .............................. 17Homogenisation ............................. 18UHT/ESL TREATMENT OF MILKUHT/ESL. .................................. 21ESL - Extended Shelf Life ..................... 21UHT - Ultra High Temperature .................. 24High Heat Infusion Steriliser. ................... 31BUTTERComposition of Butter ........................ 33Yields ..................................... 33Buttermaking ............................... 33Calculating Butter Yield ....................... 36Churning Recovery. .......................... 36Adjusting Moisture Content in Butter. ............ 39Determination of Salt Content in Butter ........... 39lodine Value and Refractive Index ............... 40Fluctuations in lodine Value andTemperature Treatment of Cream. ............. 40CHEESECheese Varieties. ............................ 42Cheesemaking .............................. 43Standardisation of Cheesemilk and Calculation ofCheese Yield. ............................. 43Utilisation Value of Skimmilk in Cheesemaking ..... 47Strength, Acidity and Temperature of Brine for Salt ing 48

MILKComposition of Danish Cow’s Milk 2002Fat ............................. approx. 4.3%Protein .......................... - 3.4%Lactose. ......................... - 4.8%Ash ............................. - 0.7%Citric acid ........................ - 0.2%Water ........................... - 86.6%Density of MilkThe density of milk is equivalent to the weight in kilos of 1litre of milk at a temperature of 15°C.The easiest way to determine the density is to use a specialtype of hydrometer called a lactometer. The upper partof the lactometer is provided with a scale showing thelactometer degree, which, when added as the second andthird decimal to 1.000 kg, indicates the density of milk,ie, a lactometer degree of 30 corresponds to a density of1.030 kg/litre.The lactometer is lowered into the milk and when it hascome to rest, the lactometer degree can be read on thescale at the surface level of the milk.As milk contains fat and as the density depends on thephysical state of the fat, the milk should be healed to 40°Cand then cooled to 15°C before the density is determined.If the, determination of the density is not carried out atexactly 15°C, the reading must be converted by means ofa correction table.The density of milk depends upon its composition, andcan be calculated as follows:100% fat + % protein + % lactose+acid + % ash + % water0.93 1.45 1.53 2.80 1.0Density:1 litre whole milk ................. approx. 1.032 kg- skimmilk ................... - 1.035 kg- buttermilk .................. - 1.033 kg- skimmed whey 6.5% TS ...... - 1.025 kg- cream with 20% fat .......... - 1.013 kg- cream with 30% fat .......... - 1.002 kg- cream with 40% fat .......... - 0.993 kg3

Yields from Whole Milk etc.100 kg standardised whole milk yields:with 4.0 % fat approx. 4.75 kg butter- 4.0 % - - 13.0 - whole milk powder- 3.0 % - - 9.5 - 45% cheese *)- 2.5 % - - 9.1 - 40% - *)- 1.6 % - - 8.3 - 30% - *)- 1.0 % - - 8.0 - 20% - *)- 0.45 % - - 7.4 - 10% - *)100 kg skimmilk with 9.5% solids yields:approx. 9.8 kg skimmilk powder- 6.9 - skimmilk cheese *)- 7.5 - raw casein- 3.5 - dried casein100 kg buttermilk with 9.0% solids yields:approx. 9.3 kg buttermilk powder100 kg unskimmed whey with approx. 7.0% solids yields:approx. 0.4 kg whey butter- 7.2 - whey cheese100 kg skimmed whey with approx. 6.5% solids yields:approx. 6.7 kg whey powder- 3,5 - raw lactose- 3.0 - refined lactose- 8.0 - lactic acid- 2.2 - WPC 35- 1.2 - WPC 60- 0.9 - WPC 80*) ripened cheeseDetermination of Fat Content in Milk and CreamRöse-Gottlieb (RG)The fat globule membranes are destroyed by ammoniaand heat, and the phospholipids are dissolved with ethanol.After heat treatment, the fat is extracted with a mixtureof diethyl ether and light petroleum. Then the solvents areremoved by evaporation and the fat content is determinedby weighing the mass left after evaporation.Schmid-Bondzynski-Ratzloff (SBR)This method uses hydrochloric acid instead of ammoniato destroy the fat globule membranes and is used forcheese samples.4

The principal difference between RG and SBR is that thefree fatty acids are not extracted by the RG method sincethe analysis is made in alkaline media. The free fatty acidsare extracted by the SBR method since the analysis ismade in an acidic medium.Gerber’s methodWhole milk is analysed as follows:Measure into the butyrometer 10 ml sulphuric acid, 11 mlmilk (in some countries only 10.8 ml) and 1 ml amyl alcohol,in that order.Before measuring out the milk, heat to 40°C and mixcare- fully. Insert the stopper and shake the mixture whileholding the stopper upwards. Then turn the butyrometerupside down two or three times until the acid remainingin the narrow end of the butyrometer is mixed completelywith the other constituents.During the mixing process, the temperature rises to sucha degree that centrifugation can take place without furtherheating. The butyrometer is centrifuged for 5 minutes at1,200 rpm and the sample is placed in a water bath at 65-70°C before reading. The reading is made at the lowestpoint of the fat meniscus.Skimmilk and buttermilk are analysed as follows:The acid, milk and amyl alcohol are measured out as describedabove. Immediately after shaking, the sample iscooled to 10-20°C before the sulphuric acid remaining inthe narrow end of the butyrometer is mixed in by turningthe butyrometer up and down. Before centrifugation, thesample is heated to 65-70°C. The butyrometer is centrifugedfor 10-15 minutes at 1,200 rpm and the value readat 65-70°C.When skimmilk samples are read, the fat will be seen astwo small triangles. If these two triangles are just touchingeach other, the milk contains approx. 0.05 % fat. For buttermilksamples, the reading is taken at the lowest pointof the fat meniscus and the figure of 0.05 is then added togive the fat content.Cream is analysed as follows:Measure into the butyrometer 10 ml sulphuric acid, 5 mlcream, 5 ml water, and 1 ml alcohol. The water is usedfor removing the remainder of the cream from the creampipette into the butyrometer and must have a temperatureof 40°C. Insert the stopper and continue as described for5

whole milk. Before a reading is taken, the bottom of the fatcolumn must be set at zero on the butyrometer by turningthe rubber stopper to move it up or down.MilkoscanThe Danish company N. Foss Electric has developed aninstrument, the Milkoscan, for rapid and simultaneous,determination of fat, protein, lactose and water.In this instrument, the sample is diluted and homogenised.Then the mixture passes through a flow cuvette where thedifferent components are measured by their infrared absorption.Fat at 5.73 µmProtein at 6.40 µmLactose at 9.55 µmThe value for water is calculated on the basis of the sumof the values for fat, protein, and lactose plus a constantvalue for mineral content.The instrument requires exact calibration and must bethermostatically controlled.Determination of Protein Content in Milk and CreamKjeldahl’s methodKjeldahl’s method provides for accurate determination ofthe milk protein content. This method involves the combustionof the protein contained in a specific quantity ofmilk in sulphuric acid with an admixture of potassium sulphateand copper sulphate. This converts nitrogen fromorganic compounds into ammonium ions. The additionof sodium hydroxide liberates ammonia, which distilsover into a boric acid solution. The amount of ammoniais determined by hydrochloric acid titration. The proteincontent is found by multiplying the measured nitrogenquantity by 6.38.The amido black method (Pro-milk)When milk is mixed with an amido black solution at pH2.45, the positively charged protein molecules are linkedto the negatively-charged amido black molecules in a specificratio, and the protein is precipitated. When the precipitateof coloured protein pigment has been removed,the concentration of non-precipitated pigment, which ismeasured by means of the photometer, is inversely proportionalto the milk protein content.6

This method has been automated in an instrument, thePro- milk, from N. Foss Electric. The instrument filtersout the protein pigment by means of special syntheticfilters and a photometer displays the protein percentagedirectly.Detection of Preservatives and Antibiotics in MilkThe growth of lactic acid bacteria may be inhibited by thepresence in the milk of ordinary antiseptics (such as boricacid, borax, benzoic acid, salicylic acid, salicylates, formalin,hydrogen peroxide) or antibiotics (penicillin, aureomycin,etc). In order to find out which of the above mentionedsubstances is present, it is necessary to test foreach of them - which is both costly and time-consuming.However, tests for rapid determination ¯f antibiotics, especiallypenicillin, in milk have been developed. One of theseis the Dutch Delvotest P.A special substrate containing Bacillus colidolactis, whichis highly sensitive to penicillin and to some extent also toother antibiotics, is inoculated with the suspected milk.After 2 1/2 hours, the quantity of acid produced will besufficient to change the colour in the dissolved pH indicatorfrom red to yellow. This method gives a definite determinationof the penicillin concentration down to 0.06I.U./ml.Rapid detention of slow-ripening milk can be achieved bya comparison of the acidification process in the suspectedsample with that in a sample of mixed milk.Both samples are heat-treated at 90-95°C for approx.15 minutes, cooled to approx. 25°C, and mixed with 2%starter.After 6-8 hours there will be a distinct difference in thetitres (or pH) of the two samples if one of them containsantibiotics or other growth-inhibiting substances.Acidity of MilkNormally, fresh milk has a slightly acid reaction. The acidityis determined by measuring either the titrated acidity,i.e., the total content of free and bound acids, or by measuringthe pH value, which indicates the true acidity (thehydrogen ion concentration).The titrated acidity of fresh milk is 16-18, and pH is 6.6-6.8.7

TitrationNormally, the titrated acidity of milk is indicated by thenumber of ml of a 0.1 n sodium hydroxide solution requiredto neutralise 100 ml of milk, using phenolphthaleinas an indicator.By means of a pipette, 25 ml of milk is measured intoan Erlenmeyer flask. To this 13 drops of a 5% alcoholicphenolphthalein solution is added, and from a burette 0.1n sodium hydroxide solution is added, drop by drop, intothe flask until the colour of the liquid changes from whiteto a uniform pale red. Since for practical reasons only 25ml of milk is used in the analysis, the figure obtained mustbe multiplied by four.Consequently, supposing that the quantity of sodiumhydroxide solution used was 5 ml, the titratable aciditywould be:85 × 4 = 20The normal titratable acidity of fresh milk is 16-18. If thetitratable acidity increases to 30 or more, the casein contentwill be precipitated when the milk is heated.When cultured milk or buttermilk is titrated, part of themilk will stick to the inside of the pipette. This residue iswashed into the Erlenmeyer flask by milk taken from theflask after neutralisation takes place and the red colourstarts to appear. Titration then proceeds as explainedabove.The acidity of cream is determined by the same procedure.When the final result is calculated, the fat content of thecream must be taken into account. Supposing that the latteris 30% and that the quantity of sodium hydroxide solutionused was 2.8 ml, the titratable acidity of the creamwould be:2.8 × 4 ×100= 16100-30The acidity of milk is expressed in various ways in variouscountries.Soxhlet Henkel degrees (S.H.) give the number of ml of a0.25 n NaOH solution necessary to neutralise 100 ml ofmilk, using phenolphthalein as an indicator.Thörner degrees of acidity indicate the number of ml of a0.1 n NAOH solution required to neutralise 100 ml of milk

to which two parts of water have been added. Phenolphthaleinis used as an indicator.Dornic degrees of acidity give the number of ml of a 119n NAOH solution necessary to neutralise 100 ml of milk,using phenolphthalein as an indicator Divided by 100, thefigure gives the percentage of lactic acid.In the various methods of analysis, the milk is diluted todifferent degrees, and it is therefore only possible to makeapproximate comparisons of the various degrees of acidity.However, working only from the amount of NaOH usedand the normal acidity figure, the various degrees of aciditycan be compared as shown below:Degreesof acidity02.505.007.510.012.515.017.520.022.525.027.530.0Soxhlet-Henkel010203040506070809101112Thömer02.505.007.510.012.515.017.520.022.525.027.530.0Dornic02.2504.5006.7509.0011.2513.5015.7518.0020.2522.5024.7527.00Approx. %lactic acid0.02250.04500.06750.09000.11250.13500.15750.18000.20250.22500.24750.2700Measurement of pHThe true acidity of a liquid is determined by its content ofhydrogen ions.Acidity is measured in pH value, pH being the symbol usedto express the negative logarithm of hydrogen ion concentration.For example, a solution with a hydrogen ion concentrationof 1:1,000 or 10-3 has a pH of 3. The neutral point ispH 7.0. Values below 7.0 indicate acid reactions, and valuesabove 7.0 indicate alkaline reactions. A difference in pHvalue of 1 represents a tenfold difference in acidity, ie, pH 5.5shows a degree of acidity ten times higher than pH 6.5.In milk, it is the pH value and not the titratable aciditythat controls the processes of coagulation, enzyme activity,bacteria growth, reactions of colour indicators, taste,etc. The pH value is measured by a pH-meter with a combinedglass electrode, and the system must always becalibrated properly before use.9

The Phosphatase TestThe phosphatase test is used to control the effect of HTSTpasteurisation and batch pasteurisation of milk. Milk pasteurisedby one of these methods must be healed in sucha way that, when the phosphatase test is applied, a maximumof 0.010 mg free phenol is liberated per ml milk.However, the heat treatment must not be so effective thatthe reaction of the milk to Storch’s test (peroxidase test)is negative.The phosphatase test is performed as follows:Measure 1 ml milk into two test tubes, marked A and B.Transfer test tube B to a 80”C water bath for 5 minutesand then cool. To the milk in test tube A, add 5 ml distilledwater saturated with chloroform and 5 ml substratesolution (prepared by dissolving one small “Ewos” phosphatasetablet l in 25 ml of a solution consisting of 9.2 gpure an- hydrous sodium carbonate and 13.6 g sodiumbicarbonate in 1 litre distilled water saturated with chloroform).To test tube B, add 5 ml diluted phenol solution (0.010 mgphenol in 5 ml) and 5 ml substrate solution. Shake bothtest tubes and leave them in a water bath at 38-40°C forone hour. Then, to both tubes, add exactly six drops ofphenol reagent (three “Ewos” phosphatase tablets II in 10ml 93% alcohol), and shake the tubes vigorously. Leavethe two test tubes at room temperature for 15 minutes andcompare them. Only if the contents of test tube A appearpaler in colour than the contents of test tube B can themilk be considered sufficiently heated.If the milk fails this test, a sample for control testing shouldbe sent to an authorised research institute, which willcarry out the phosphatase test in such a way that colouris extracted after incubation. The colour extinction is ameasure of the content of phenol and can be measured ina Pullfricphotometer.Standardisation of Whole Milk and CreamIn many countries, milk and cream sold for consumptionmust contain a legally fixed fat percentage, although slightvariations are usually allowed.In Denmark, for example, the fat content of heat-treatedwhole milk must be 3.5%, in low-fat milk 1.5% and 0.5%,and in skimmilk 0.1%. The various types of cream musthave a fat content of 9, 13, 18, or 36%, respectively.In order to comply with these regulations, it is necessary10

to standardise the fat content. This can be done in variousways depending on the stage at which standardisation iscarried out.Standardisation before or during heat treatment is to bepreferred as the danger of subsequent contamination isthereby reduced. Standardisation will normally take placeautomatically during the separating and pasteurisingprocess. It may, however, be done manually as a batchprocess, in which case the table below may be used.Table for standardisation of Whole Milk% fat in% fat in standardised milkwholemilk 04.00 03.90 03.80 03.70 03.60 03.50 03.40 03.30 03.20 03.10 03.004.512.70 15.60 18.70 21.90 25.40 30.00 32.80 36.90 41.30 45.90 50.804.410.10 13.00 16.00 19.20 22.50 26.00 29.90 33.80 38.10 42.60 47.504.307.60 10.40 13.30 16.40 19.70 23.20 26.90 30.80 34.90 39.30 44.104.205.10 07.80 10.70 13.70 16.90 20.30 23.90 27.70 31.70 36.10 40.704.102.50 05.20 08.00 11.00 14.00 17.40 20.90 24.60 28.60 32.80 37.304.002.60 05.30 08.20 11.30 14.50 17.90 21.50 25.40 29.50 33.903.900.3802.70 05.50 08.50 11.60 14.90 18.50 22.20 26.20 30.503.800.7700.3802.70 05.60 08.70 11.90 15.40 19.00 23.00 27.103.701.1500.7700.3802.80 05.80 09.00 12.30 15.90 19.70 23.703.601.5401.1500.7600.3802.90 06.00 09.20 12.70 16.40 20.303.501.9201.5301.1500.7600.3803.00 06.10 09.50 13.10 16.903.402.3101.9201.5301.1400.7600.3803.10 06.30 09.80 13.603.302.6902.3001.9101.5201.1400.7500.3803.10 06.60 10.203.203.0802.6802.2901.9001.5201.1300.7500.3703.30 06.803.103.4603.0702.6702.2801.8901.5101.1300.7500.3703.403.003.8503.4503.0502.6602.2701.8901.5001.1200.7500.37The figures above the shaded lines indicate the amount inkg of skimmilk to be added per 100 kg whole milk whenthe fat content is too high.The figures below the shaded lines indicate the amount inkg of cream with 30% fat to be added per 100 kg wholemilk when the fat content is too low.Batch StandardisationFor batch standardisation the following equations may beused.Fat content to be reduced:To reduce the fat content in y kg whole milk, add x kgskimmilk.x kg skimmilk = y (% fat in whole milk - % fat required)% fat required - % fat in skimmilkTo obtain z kg standardised milk, mix y kg whole milk withx kg skimmilk.11

y kg whole milk = z (% fat required - % fat in skimmilk)% fat in whole milk - % fat in skimmilkx kg skimmilk = z - yFat content to be increased:To increase the fat content in y kg low-fat milk, add x kgcream (or high-fat milk).x kg cream =y (% fat required - % fat in low-fat milk)% fat in cream - % fat requiredTo obtain z kg standardised milk, mix y kg low-fat milkwith x kg cream (or high-fat milk).y kg low-fat milk =x kg cream = z - yz (% fat in cream - % fat required% fat in cream - % fat in low-fat milkln-line StandardisationFor in-line standardisation the following equations may beused.Fat content to be reduced:To obtain z kg standardised milk, use y kg whole milk.Surplus cream x kg.y kgz (% fat in surplus cream - % fat required)whole = % fat in surplus cream - % fat in whole milkmilkx kg surplus cream = y - zTo obtain x kg surplus cream, use y kg whole milk. Standardisedmilk z kg.y kgz (% fat in cream - % fat in standardised milk)whole = % fat in whole milk - % fat in standardised milkmilkz kg standardised milk = y - xy kg whole milk used will result in z kg standardised milkand x kg surplus cream.12

z kgy (% fat in surplus cream - % fat in shole milk)stand. = % fat in surplus cream - % fat in stand. milkmilkx kg surplus cream = y - zFat content to be increased:Standard in-line systems cannot be used for this purpose.The fat content of skimmilk is normally estimatedat 0.05%.Standard DeviationThe accuracy of an automatic butter fat standardising unitwill commonly be expressed in the term Standard Deviation(SD).By stating a SD figure, it is guarantied that a certain percentageof the fat standardised milk will be kept withinthe upper and lower limits, which are derived from thestandard deviation figure (cf. the below table).GuaranteedSigmaPercent within thespecificationDefects per1000Defects permillion1 68%0000000000.317.400-2 95%0000000000.045.600-3 99.73%00000000002.7002, 700.0000004 99. 99366%00000000.063, 0063.4000005 99. 9999426%000- , 0000.5740006 99. 9999998026%- , 0000.001974It is assumed that the data are distributed normally!99,9 9366%99,7 3%95%68%13

If for instance the SD figures for a fat value range from1% to 5% are:SD of the automatic butter fat standardising unit: 0.015%*) SD of the controlling lab instrument: 0.01%Then the two SD figures shall be added as follows:(SD of the automatic standardising system) 2 +(SD on the measuring instrument) 2140.015 2 + 0.01 2 = 0.018%The summarised SD will thus be = 0.018%Conferring the above table, the accuracy to be obtainedwill be as follows:1s level: 68% of the production time the fat value will liewithin ± 0.018%2s level: 95% of the production time the fat value will liewithin ± 0.036%3s level: 99.7% of the production time the fat value will liewithin ± 0.054%4s level: 99.99366% of the production time the fat valuewill lie within ± 0.072%The above accuracy figures can now be used to calculatethe fat value set point of the automatic standardising unit.If a dairy for instance must guarantee minimum 3.4% fatin 99.7% (3s) of the milk delivered, then the fat value setpoint of the automatic standardising unit must be 3.4% +0.054% = 3.454%*) There is a degree of accuracy connected with the measuringequipment. The supplier of the measuring instrumentexpresses this by stating the standard deviation ofthe measurements to be xxx%.Calculating the Extent of Random SamplingHow many samples need to be taken in order to provethat the standardising unit will comply with the grantedguarantees?

Various methods are available for calculating the extent ofa random sampling – this is a simple method.From the below chart the relation between the Numberof Degree of Freedom Required (the number of samplestaken) to estimate the standard deviation within P% of ItsTrue Value with Confidence Coefficient g can be read.A Confidence Coefficient g = 95 would normally apply forthe dairy and food industry.Example (above example continued):Verification of the SD guarantee of 0.018%:- Number of samples 30 and- Confidence Coefficient (g = 95)Referring to the below chart, 25% (P%) deviation from ItsTrue Value (0.0018%) must be allowed for.Due to the analysis uncertainty, the calculated SD of the30 random samples must thus be better than 0.018% +25% = 0.023%.Logically, if the number of samples is increased the deviation(P%) from Its True Value to be allowed for will narrowin. The magnitude hereof is illustrated in the belowexamples:Number ofsamplesP%RequiredSDin sample set3025%0.023%8015%0.021%20010%0.020%N (Total) 0 % 0.018%15

Chart T *): Number of Degrees of Freedom Required toEstimate the Standard Deviation within P% of Its TrueValue with Confidence Coefficient g1,000800600500400300200g =.99g =.95Degrees of freedom g1008060504030g =.9020108655 6 8 10 20 30 40 50P %*) Adapted with permission from Greenwood, J. A. andSandomire, M. M. (1950). “Statistics Manual, Sample SizeRequired for Estimating the Standard Deviation as a Percentof Its True Value”. Journal of the American StatisticalAssociation, vol. 45, p. 258. The manner of graphing isadapted with permission from Crow, E. L. Davis, F. A. andMaxfield, M. W. (1955). NAVORD Report 3369. NOTS 948,U.S. Naval Ordnance Test Station, China Lake, CA. (Reprintedby Dover Publications, New York, 1960).16

GENEREL MILK PROCESSINGPasteurisationPasteurisation is a heat treatment applied to milk in order toavoid public health hazards arising from pathogenic microorganismsassociated with milk. The process also increasesthe sheIf life of the product. Pasteurisation is intendedto create only minimal chemical, physical and organolepticchanges in products to be kept in cold storage.Pasteurisation temperature and timeThe temperature/time combinations stated below are similarin effect and all have the minimum bactericidal effectrequired for pasteurisation.Pasteurised milk and skimmilk63°C/30 min.72°C/15 sec.Pasteurised cream (10% fat):75°C/15 sec.- - (35% fat): 80°C/15 sec.Pasteurised, concentrated milk,ice cream mix, sweetened products, etc.80°C/25 sec.In each case the product is subsequently cooled to 10°Cor less - preferably to 4°C.In some countries, local legislation specifies minimumtemperature/time combinations.In many countries, the phosphatase test is used to determinewhether the pasteurisation process has been carriedout correctly. A negative phosphatase test is consideredto be equivalent to less than 2.2 microgrammes of phenolliberated by 1 ml of sample or less than 10 microgrammespara-nitrophenol liberated by 1 ml of sample.In order to minimise the risk of failure in the pasteurisationprocess, the system should have an automatic controlsystem for:(1) Pasteurisation temperature. Temperature recorder andflow diversion valve at the outlet of the temperature holderfor diverting the flow back to the balance tank in case ofpasteurisation temperatures below the legal requirement.(2) Holding time at pasteurisation temperature. Capacitycontrol system which activates the flow diversion valvein case the capacity exceeds the maximum for which theholding tube is designed.17

(3) Pressure differential control. The system will activatethe flow diversion valve if the pressure on the raw-milkside of the regenerator exceeds a set minimum below thepressure on the pasteurised side, thus preventing possibleleakage of raw milk into the pasteurised milk.Calculation of residence time in holding tubeThe mean residence time (t) in the holding tube can becalculated as follows:18t = length of tube x volume per metrecapacity per secondValues for volume per metre can be found in the tableVolume in Stainless Steel Pipes.The individual particles spend different times in the holdingtube and this results in residence time variations. Toavoid bacteriological problems, it is necessary to heateven the fastest particles long enough.The holding tube must have an efficiency of at least 0.8(tmin/tmean) and this can best be achieved by avoiding alaminar flow, ie, ensuring a turbulent flow at a ReynoldsNumber >12,000 and choosing a ratio of length (m)/diameter>200 for the holding tube.HomogenisationMilk products are usually homogenised to prevent separationduring storage. Other dairy products are homogenisedto improve water binding, reduce free fat etc. Homogenisationtakes place in a high-pressure homogeniser, whichis basically a positive pump equipped with a narrow slitcalled the homogenising valve. The milk is forced throughthe homogenising valve at high pressure and this processcauses disruption of the fat globules. Advanced types ofhomogenising valves have been constructed for optimumhomogenising efficiency in various processes.In a pasteurisation plant the homogeniser is typicallyplaced upstream before the final heat treatment in a heatexchanger. Homogenisation of milk must take place at atemperature above the melting point of the milk fat. Thismeans that the homogeniser is often placed after the firstregenerative section. In indirect UHT milk plants (Fig. 3on page 21) the homogeniser is also generally placed upstream.

Fig. 1: The particle size distribution of fat globules in milkbefore and after homogenisation at 200 BAR total pressurewith 30 BAR on the 2nd stage (volume weighted distributions).However, in indirect UHT cream systems where the fatcontentis higher than approx. 10% (possibly as low as6%), and in milk products with higher protein content, thehomogeniser is preferably placed downstream. In directUHT systems the homogeniser is always placed downstreamon the aseptic side after UHT treatment (Fig. 4 onpage 22 and fig. 8 on page 31).Total homogenisation is most commonly applied for pasteurisedmilk and always used with UHT milk. In thesecases, the fat content is standardised prior to homogenisation.Two-stage homogenisation with a SEO or XFDhomogenising valve or single-stage homogenisation witha LW homogenising valve at a total pressure of 100 – 150BAR is often sufficient for the required stability of pasteurisedmilk. For UHT milk a total pressure of 200 – 250BAR is recommended (Fig. 1). For very high flow rates,two-stage homogenisation with a patented MicroGap homogenisingvalve is recommended. The MicroGap enablesreduction of the total pressure by approx. 20 – 30%(Fig. 2 on page 20).19

3.595%-Fractile (Milk) vs. Homogenising Pressure95% fractile diameter (µm)32.521.51MicroGapConventional Two-stage Homogenising Valvearrangement0.5060 80 100 120 140 160 180 200Pressure (BAR)Fig. 4: Micro-Gap valve compared with conventionaltwo-stage valve arrangement (95%-Fractile from volumeweightedparticle size distributions, analysed by HelosSympatec particle sizer).Another option is partial homogenisation in order to saveoperating costs. This can enable a reduction of totalpower consumption during homogenisation by approx.65% as only about one third of the milk volume is passedthrough the homogeniser. This type of homogenisation isonly applied for pasteurised milk (never for UHT milk). Inpartial homogenisation, 1/3 of the volume consists of homogenisedcream with up to max. 12% fat, while 2/3 ofthe volume consists of skimmed milk, which is bypassedand added to the homogenised cream.20

UHT/ESL TREATMENT OF MILKUHT/ESLAPV is focussed on being the leader within the UHT/ESL technology and has the largest product range withinUHT:Indirect:Direct:Plate UHT PlantTubular UHT Plant (Figure 3)Injection UHT PlantInfusion UHT PlantIn addition to the 4 main systems, APV has developed thefollowing variations:ESL - Extended Shelf LifePure Lac TMCombi UHT (2-4 systems in one)High Heat InfusionInstant Infusion8PRODUCT3 395ºC 140ºCFILLING975ºC21 145675ºC25ºCSTEAM10COOLINGWATER1. Tubular regenerativepreheaters2. Homogeniser3. Holding tubes4. Tubular final heater5. Tubular regenerativecooler6. Final cooler7. Sterile tank8. CIP unit9. Sterilising loop10. Water HeaterFig. 3: Flow diagram for Tubular SteriliserESL - Extended Shelf LifeIn many parts of the world the production of fresh milkpresents a problem in regard to keeping quality. This is dueto inadequate cold chains, poor raw material and/or insufficientprocess and filling technology. Until recently, the onlysolution has been to produce UHT milk with a shelf life of 3 -6 months at ambient temperature. In order to try to improvethe shelf life of ordinary pasteurised milk, various attempts21

have been made to increase the pasteurisation temperatureand this led to the extended shelf life concept.The term extended shelf life or ESL is being applied moreand more frequently. There is no single general definitionof ESL. Basically, what it means is the capability to extendthe shelf life of a product beyond its traditional wellknownand generally accepted shelf life without causingany significant degradation in product quality. A typicaltemperature/time combination for high-temperature pasteurisationof ESL milk is 125 - 130°C for 2 - 4 seconds.This is also known in the USA as ultrapasteurisation.APV has during the last years developed a pa tented processwhere the temperature may be raised to as high as140°C, but only for fractions of a second. This is the basisfor the Pure-Lac TM process.The APV infusion ESL is based on the theory that a hightemperature/ultra short holding time will provide an efficientkill rate as well as a very low chemical degradation.75ºCSTEAM29COOLINGWATERFILLINGPRODUCTCOOLINGWATER4735VACUUM5ºCSTEAM143ºC 75ºC 25ºC

Temperature135ºCPure-Lac TM 120ºCHigh pasteurisation72ºCLow pasteurisationFig. 5: Temperature profile for pasteurisation processes.TimeThe Pure Lac TM processIn co-operation with Elopak, APV has developed the PureLac TM concept which in a systematic way attacks thechallenge of improving milk quality for the consumer.Based on investigations of consumer requirements andthe present market conditions in a larger number of countries,the objective of Pure Lac TM was defined as follows:• A sensory quality equal to or better than pasteurisedproducts• A “real life” distribution temperature of neither 5°C, nor7°C but 10°C• A prolonged shelf life corresponding to 14 to 45 days at10°C depending on filling methods and raw milk quality• A method to accommodate changes in purchasing patternsof the consumer• An improved method for distribution of niche products• To cover the complete milk product range, i.e. milk,creams, desserts, ice cream mix, etc.• To provide tailored packaging concepts designed togive maximum protection using minimum but adequatepackaging solutions.After reviewing the range of “cold technologies” available,it became obvious that most of them were only suited forwhite milk. Furthermore, the actual microbiological reduc-23

tion rate for some of the processes was inadequate toprovide sufficient safety for shelf life of more than 14 daysat 10°C.Process Technology/Shelf LifeProcessLog. reductionaerobic, psycro-tropic sporesExtended shelflife max 4°CstorageExpected shelflifemax 10° CstoragePasteurisation0 10days1 - 2 daysCentrifugation1 14days4 - 5 daysMicrofiltration2-3 30days6 - 7 daysTMPureLacESL pasteurisationUHT processHigh Heat Process8 Over 45 days8 (*)40* Thermophilic spores** Depending on filling solution180 days at25°CUp to 45 days(**)180 days25°CatUHT - Ultra High TemperatureAll UHT processes are designed to achieve commercialsterility. This calls for application of heat to the productand a chemical sterilant or other treatment that render theequipment, final packaging containers and product free ofviable micro-organisms able to reproduce in food undernormal conditions of storage and distribution. In additionit is necessary to inactivate toxins and enzymes presentand to limit chemical and physical changes in the product.In very general terms it is useful to have in mind that anincrease in temperature of 10ºC increases the sterilisingeffect 10-fold whereas the chemical effect only increasesapproximately 3-fold. In this section we will define someof the more commonly used terms and how they can beused for process evaluation.24

ºC150Direct InfusionHigh Heat InfusionIndirect UHT100500TimeFig. 6: Temperature profiles for direct infusion, high heatinfusion and indirect UHT processesThe logarithmic reduction of spores and sterilisingefficiencyWhen micro-organisms and/or spores are exposed toheat treatment not all of them are killed at once.However, in a given period of time a certain number is killedwhile the remainder survives. If the surviving micro-organismsare once more exposed to the temperature treatmentfor the same period of time an equal proportion of them willbe killed. On this basis the lethal effect of sterilisation canbe expressed mathematically as a logarithmic function:K · t = log N/N twhere N = number of micro-organisms/spores originallypresentN t = number of micro-organisms/spores presentafter a given time of treatment (t)K = constantt = time of treatmentA logarithmic function can never reach zero, which meansthat sterility defined as the absence of living bacterialspores in an unlimited volume of product is impossible toachieve. Therefore the more workable concept of “sterilisingeffect” or “sterilising efficiency” is commonly used.The sterilising effect is expressed as the number of decimalreductions achieved in a process. A sterilising effect25

of 9 indicates that out of 10 9 bacterial spores fed into theprocess only 1 (10°) will survive.Spores of Bacillus subtilis or Bacillus stearothermophilusare normally used as test organisms to determine the efficiencyof UHT systems because they form fairly heatresistant spores.Terms and expressions to characterise heat treatmentprocessesQ 10 value. The sterilising effect of heat sterilisation increasesrapidly with the increase in temperature as describedabove. This also applies to chemical reactions which takeplace as a consequence of an increase in temperature.The Q 10 value has been introduced as an expression ofthis increase in speed of reactions and specifies howmany times the speed of a reaction increases when thetemperature is raised by 10ºC. Q 10 for flavour changes isin the order of 2 to 3 which means that a temperature increaseof 10ºC doubles or triples the speed of the chemicalreactions.A Q 10 value calculated for killing bacterial spores wouldrange from 8 to 30, depending on the sensitivity of a particularstrain to the heat treatment.D-Value. This is also called the decimal reduction timeand is defined as the time required to reduce the numberof micro-organisms to one-tenth of the original value, i.e.corresponding to a reduction of 90%.Z-Value. This is defined as the temperature change, whichgives a 10-fold change in the D-value.F 0 value. This is defined as the total integrated lethal effectand is expressed in terms of minutes at a selectedreference temperature of 121.1ºC. F 0 can be calculatedas follows:F 0 = 10 (T - 121.1) /z x t / 60, whereT = processing temperature (ºC)z = Z-value (ºC)t = processing time (seconds)26

F 0 = 1 after the product has been heated to 121.1 ºC forone minute. To obtain commercially sterile milk from goodquality raw milk, for example, an F 0 value of minimum 5to 6 is required.B* and C* Values. In the case of milk treatment, somecountries are using the following terms:• Bacteriological effect:B* (known as B star)• Chemical effectC* (known as C star)B* is based on the assumption that commercial sterility isachieved at 135ºC for 10.1 seconds with a correspondingZ-value of 10.5ºC; this reference process is giving aB* value of 1.0, representing a reduction of thermophilicspore count of 10 9 per unit (log 9 reduction). The B* valuefor a process is calculated similarly to the F 0 value:B* = 10 ( T - 135 ) / 10.5 x t / 10.1, whereT = processing temperature (ºC)t = processing time (seconds)The C* value is based on the conditions for a 3 percentdestruction of thiamine (vitamin B 1 ); this is equivalent to135ºC for 30.5 seconds with a Z-value of 31.4ºC. Consequentlythe C* value can be calculated as follows:C* = 10 ( T - 135 ) /31.4 x t / 30.5Fig. 6 shows that a UHT process is deemed to be satisfactorywith regard to keeping quality and organoleptic qualityof the product when B* is > 1 and C* is < 1.The B* and C* calculations may be used for designingUHT plants for milk and other heat sensitive products.The B* and C* values also include the bacteriological andchemical effects of the heating up and cooling down timesand are therefore important in designing a plant with minimumchemical change and maximum sterilising effect.The more severe the heat treatment is, the higher the C*value will be. For different UHT plants the C* value corre-27

sponding to a sterilising effect of B* = 1 will vary greatly. AC* value of below 1 is generally accepted for an averagedesign UHT plant. Improved designs will have C* valuessignificantly lower than 1.The APV Steam Infusion Steriliser has a C* value of 0.15.Residence timeParticular attention must be paid to the residence time ina holding cell or tube and the actual dimensioning will dependon several factors such as turbulent versus laminarflow, foaming, air content and steam bubbles. Since there isa tendency to ope-rate at reduced residence time in order tominimise the chemical degradation (C* value < 1) it becomesincreasingly important to know the exact residence time.In APV the infusion system has been designed with a specialpump mounted directly below the infusion chamber whichensures a sufficient over-pressure in the holding tube in orderto have a single phase flow free from air and steam bubbles.This principle enables APV to define and monitor theholding time and temperature precisely and makes it the onlydirect steam heating system, which allows true validation offlow and temperature at the point of heat transfer.Commercial sterilityThe expression of commercial sterility has been mentionedpreviously and it has been pointed out that completesterility in its strictest sense is not possible. In workingwith UHT products commercial sterility is used as amore practical term, and a commercially sterile product isdefined as one which is free from micro-organisms whichgrow under the prevailing conditions.Chemical and bacteriological changes at hightemperaturesThe heating of milk and other food products to high temperaturesresults in a range of complex chemical reactionscausing changes in colour (browning), development ofoff-flavours and formation of sediments. These unwantedreactions are largely avoided through heat treatment at ahigher temperature for a very short time. It is important toseek the optimum time/temperature combination, whichprovides sufficient kill effect on spores but, at the sametime, limits the heat damage, in order to comply with marketrequirements for the final product.28

Raw material qualityIt is important that all raw materials are of very high quality, asthe quality of the final product will be directly affected. Rawmaterials must be free from dirt and have a very low bacteriaspore count, and any powders must be easy to dissolve.All powder products must be dissolved prior to UHT treatmentbecause bacteria spores can survive in dry powderparticles even at UHT temperatures. Undissolved powderparticles will also damage homogenising valves causingsterility problems.Heat stability. The question of heat stability is an importantparameter in UHT processing.Different products have different heat stability and althoughthe UHT plant will be chosen on this basis, it is desirableto be able to measure the heat stability of the productsto be UHT treated. For most products this is possible byapplying the alcohol test. When samples of milk are mixedwith equal volumes of an ethyl alcohol solution, the proteinsbecome unstable and the milk flocculates. The higher theconcentration of ethyl alcohol is without flocculation, thebetter the heat stability of the milk. Production and shelflife problems are usually avoided provided the milk remainsstable at an alcohol concentration of 75%.High heat stability is important because of the need toproduce stable homogeneous products, but also to preventoperational problems as e.g. fouling in the UHT plant.This will decrease running hours between CIP cleaningsand thereby increase product waste, water, chemical andenergy consumption. Generally it will also disrupt smoothoperation and increase the risk of insterility.Shelf life. The shelf life of a product is generally definedas the time for which the product can be stored withoutthe quality falling below a certain minimum acceptablelevel. This is not a very sharp and exact definition andit depends to a large extent on the perception of “minimumacceptable quality”. Having defined this, it will beraw material quality, processing and packaging conditionsand conditions during distribution and storage which willdetermine the shelf life of the product.Milk is a good example of how wide a span the conceptof shelf life covers:29

60%400030002000region ofsterilisationloss of thiamine = 80%HMF 100 µmol/l1000900800700600500HMF 10 µmol/l400threshold range of discolouration300loss of thiamine = 3% / C*=1Heating time or equivalent heating time in seconds2001009080706050403020HMF 1 µmol/l40%20%10%lactulose 600 mg/llactulose 400 mg/lthermal death value = 9thermophilic spores / B*=1loss of lysine = 1%1098765UHTregion4321100 110120 130 140 150 160ºCProduct Shelf life StoragePasteurised milk 5 - 10 days refrigeratedESL/Pure-Lac TM 20 - 45 days refrigeratedUHT milk 3 - 6 months ambient temperatureThe usual organoleptic factors limiting shelf life are deterioratedtaste, smell and colour, while the physical and302.7 2.6 2.5 2.4 2.31T3 -1·10 in KFig. 7: Bacteriological and chemical changes of heatedmilk

chemical limiting factors are incipient gelling, increase inviscosity, sedimentation and cream lining.High Heat Infusion SteriliserThe growing incidents of heat resistant spores (HRS) ischallenging traditional UHT technologies and setting newtargets. The HRS are extremely heat resistant and requirea minimum of 145 - 150ºC for 3 - 10 seconds to achievecommercial sterility. If the temperature is increased to thislevel in a traditional indirect UHT plant it would have anadverse effect on the product quality and the overall runningtime of the plant. Furthermore, it would result in higherproduct losses during start and stop and more frequentCIP cycles would have to be applied. Using the traditionaldirect steam infusion system would result in higher energyconsumption and increased capital cost. On this basis,APV developed the new High Heat Infusion system.The flow diagram in fig. 8 illustrates the principle designincluding the most important processing parameters whilefig. 8 shows the temperature/time profile in comparison toconventional infusion and indirect systems.Note that the vacuum chamber has been installed priorto the infusion chamber. This design facilitates improvementin energy recovery and it is possible to achieve 75%regeneration compared to 40% with conventional infusionsystems and 80 - 85% with indirect tubular systems. Thekilling rate is F 0 = 40 - 70.PRODUCTVACUUM90ºC 125ºCSTEAMFILLING235COOLINGWATER95ºC 60ºC150ºC 75ºC 25ºC1 4127 6710 COOLING88WATERSTEAMSTEAM1. Tubular preheaters4. Non aseptic flavour dosing (option) 7. Tubular coolers2. Holding tube5. Steam infusion chamber8. Tubular Heaters3. Flash vessel (non aseptic) 6. Homogeniser (aseptic)9. Aseptic tank10. Non aseptic coolerFig. 8: Flow diagram for High Heat Infusion Steriliser31

UHT of products with HRS (comparative temperature profiles with Fo= 40)ºC150100500Direct UHT 150ºCHigh Heat Infusion 150ºCIndirect UHT 147ºCReference Indirect UHT 140ºCTimeFig. 9: Time/temperature profiles illustrating High Heat Infusionprocessing parameters32

BUTTERComposition of ButterButter must comply with certain regulations:Fat .................... Min. 80% (82%)Moisture ................ Max. 16%Milk solids non-fat (MSNF). . Max. 2%Salt (NaCl):Mildly salted .......... approx. 1%Strongly salted ........ - 2%Acidity:Sweet cream butter. .... pH 6.7Cultured butter ........ pH 4.6Mildly cultured butter ... pH 5.3Buttermilk normally contains:Sweet buttermilk ......... 0.5-0.7% fat. . . . . . . . . . . . . . . . . . . . . . . . approx. 8.5% MSNFCultured buttermilk. ....... 0.4-0.6% fat. . . . . . . . . . . . . . . . . . . . . . . . approx. 8.3% MSNFYields1 kg butter can be made from:approx. 20 kg milk with 4.2% fat- 2.2 kg cream with 38% fat- 2.0 kg cream with 42% fatButtermakingButtermaking may be carried out either as a batch processin a butter churn or as a continuous process in acontinuous buttermaking machine.In addition to cream treatment, buttermaking comprisesthe following stages:(1) churning of cream into butter grains and buttermilk;(2) separation of butter grains and buttermilk;(3) working of the butter grains into a cohesive mass;(4) addition and distribution of salt;(5) adjustment and distribution of moisture;(6) final working, under vacuum, to minimise the air content.A continuous buttermaking machine has existed for manyyears. It was invented by a German professor, Dr. Fritz.However, this machine was deficient in a number of respects.It could be used only for the treatment of sweet33

cream, and there were problems with the production ofsalted butter.APV manufactures continuous butter making machines withcapacities ranging from 500 kg to 12,000 kg butter/hour.The APV continuous buttermaking machine can produceall types of butter: cultured and sweet, salted and unsalted.Furthermore, the machine can produce butter accordingto the “NIZO” as well as to the “IBC” method. Blendedproducts (e.g. Bregott) in which some of the butter fat hasbeen replaced by vegetable fats can also be produced.The APV continuous buttermaking machine also guaranteesthat products are of the highest possible quality, andthat the operating economy is the best obtainable.The APV continuous buttermaking machine is designedaccording to the following principles:(1) The churning section is, in principle, designed in accordancewith the system of Dr. Fritz. The section consistsof a horizontal cylinder and a rotating beater. The beatervelocity is infinitely variable between 0 and 1,400 rpm.Since the churning process lasts only 1-2 seconds, it isimportant to adjust the beater velocity to obtain optimumbutter grain size. The moisture content of the butter andthe fat content of the buttermilk also depend on the beatervelocity.(2) The separating section consists of a horizontal rotatingcylinder. The velocity is infinitely variable.The first part of the cylinder is equipped with baffle platesfor further treatment of the mixture of butter grains andbuttermilk which is fed in from the churning section.The second part of the cylinder is designed as a sievefor buttermilk drainage. It is equipped with a very finelymeshed wire screen, which retains even small buttergrains. The buttermilk drainage from the butter grains isvery efficient and the rotation of the strainer drum preventsbutter clogging.(3) The working section consists of two inclined sections(I and II) with augers for transport of the butler, and workingelements in the form of perforated plates and mixingvanes. The velocity of each of the two sections is infinitelyvariable.In the production of salted butter, a salt slurry (40-60%)is pumped into working section I where it is worked intothe butter.34

ButterWaterButtermilk214335(1) Churning section(2) Separating section(3) Working section(4) Vacuum chamber(5) Butter pumpThe above is a diagram of APV’s continuous buttermakingmachine.Any adjustment of the moisture content also takes place inworking section I. Water dosing is carried out automatically.In order to reduce the air content in the butter from 5-6%or more to below 0.5%, a vacuum chamber has been insertedbetween working sections I and II. When the butterfrom working section l enters this chamber, it passesthrough a double perforated plate from which it emergesin very thin layers. This provides the best conditions forescape of air. The butter leaves the machine through anozzle fitted at the end of working section II. Mounted onthe nozzle is a butter pump, which conveys the butter tothe butter silo.Buttermaking according to the IBC method(Indirect Biological Culturing)This is a method for production of cultured butter fromsweet cream. After sweet cream churning and buttermilkdrainage, a so-called D starter, which has a high diacetyl(aroma) content, is worked into the butter. Also, lactic acidhas been added to this starter, producing a pH reductionin addition to the aroma, Furthermore, an ordinary B starteris worked into the butter to obtain the correct moisturecontent. When salted butter is produced, the salt is mixedinto the D starter.35

A similar production method is the well known “NIZO”method.The above methods provide for more flexible cream treatmentsince the incubation temperatures for the starters donot have to be taken into account. Besides, the productionof cultured buttermilk is avoided (sweet buttermilk ismuch more usable in other products than cultured buttermilk).Finally, butter produced according to this methodhas a longer shelf life.Calculating Butter YieldThe yield of butter from whole milk can be calculated usingthe following equations. (Loss and overweight are notconsidered.).kg cream = kg milk x (% fat in milk - % fat in skimmilk)% fat in cream - % fat in skimmilkkg butter = kg cream x (% fat in cream - % fat in buttermilk)% fat in butter - % fat in buttermilkIf the fat percentage in skimmilk, buttermilk and butter isnot known, the following estimated values rnay be used:Skimmilk = 00.05% fatButtermilk = 00.4% fatButter = 82.5% fatChurning RecoveryThe churning recovery value (CRV) is equal to the amountof fat remaining in the buttermilk expressed as a percentageof the total fat content of the cream before churning.It can be worked out from the following equation:CRV = (100-7/6 x % fat in cream) x % fat in buttermilk% fat in creamIn other words, the only data required are the cream andbuttermilk fat percentages.36

Churning Recovery Table% fatincream 0.10.2030.50.21.4231.00.21.4131.50.20.4032.00.20.3932.50.19.3833.30.19.3733.50.18.3634.00.18.3534.50.17.3535.00.17.3435.50.16.3336.00.16.3236.50.16.3137.00.15.3137.50.15.3038.00.14.2938.50.14.2939.00.14.2839.50.14.2740.00.13.2740.50.13.2641.00.13.2541.50.12.2542.00.12.2442.50.12.2443.00.12.2343.50.11.2344.00.11.2244.50.11.2245.00.11.21% fat in buttermilk0 0.300.400.500.600.700.800.900 0.630.851.061.271.481.691.900 0.620.821.031.241.441.651.850 0.600.801.001.211.411.611.810 0.590.780.981.181.371.571.760 0.570.760.961.151.341.531.720 0.560.750.931.121.311.491.680 0.550.730.911.091.271.461.640 0.530.710.891.071.241.421.600 0.520.690.871.041.211.391.560 0.510.680.851.011.181.351.520 0.500.660.830.991.161.321.490 0.480.640.810.971.131.291.450 0.470.630.790.941.101.261.420 0.460.610.770.921.081.231.380 0.450.600.750.901.051.201.350 0.440.590.730.881.031.171.320 0.430.570.720.861.001.141.290 0.420.560.700.840.981.121.260 0.410.550.680.820.961.091.230 0.400.530.670.800.931.071.200 0.390.520.650.780.911.041.170 0.380.510.640.760.891.021.150 0.370.500.620.750.871.001.120 0.360.490.610.730.850.971.090 0.360.470.590.710.830.951.070 0.350.460.580.700.810.931.040 0.340.450.560.680.790.911.020 0.330.440.550.660.770.881.000 0.320.430.540.650.760.860.970 0.320.420.530.630.740.840.95The result can also be taken from a table that has beenworked out on the basis of Report No. 38 from the DanishGovernment Dairy Research Institute. See below.37

Table for adjustment of Moisture Content in ButterAddition of water in kg per 100 kg butter when the% waterdesired % moisture is as follows:present16.015.9 15.8 15.7 15.6 15. 515.90.1215.80.240.1215.70.360.240.1215.60.470.360.240.1215.50.590.470.360.240.1215.40.710.590.470.360.240.1215.30.830.710.590.470.350.2415.20.940.830.710.590.470.3515.11.060.940.820.710.590.4715.01.181.060.940.820.710.5914.91.291.181.060.940.820.7114.81.411.291.171.060.940.8214.71.521.411.291.171.060.9414.61.641.521.411.291.171.0514.51.751.641.521.401.291.1714.41.871.751.641.521.401.2914.31.981.871.751.631.521.4014.22.101.981.871.751.631.5214.12.212.101.981.861.751.6314.02.332.212.091.981.861.7413.92.442.322.212.091.971.8613.82.552.442.322.202.091.9713.72.672.552.432.322.202.0913.62.782.662.552.432.322.2013.52.892.782.662.542.432.3113.43.002.892.772.662.542.4313.33.113.002.882.772.652.5413.23.223.113.002.882.772.6513.13.343.223.112.992.882.7613.03.453.333.223.102.992.8712.93.563.443.333.223.102.9912.83.673.563.443.333.213.1012.73.783.673.553.443.323.2112.63.893.783.663.553.433.3212.54.004.893.773.663.543.4312.44.114.003.883.773.653.5412.34.224.113.993.883.763.6512.24.334.214.103.993.873.7612.14.444.324.214.103.983.8712.04.554.434.324.214.093.9838

Adjusting Moisture Content in ButterConventional ChurnsThe churning of the cream should be carried out in sucha way that the moisture content of the butter is slightlybelow the maximum permitted amount. A test of the moisturecontent should be made as soon as the butter hasbeen worked sufficiently.When the amount of butler is known, the table above canbe used.If desired, the following equation may also be used:kg water to be added =kg butter x (% MD - % MP)100 - % MPwhere:MD = Moisture desiredMP = Moisture presentContinuous Buttermaking MachinesThe churning of the cream should be carried out in sucha way that the moisture content of the butter - withoutany addition of water - is below the maximum permittedamount.The moisture content of the butter and the regulation ofthe water dosing pump will normally be automatically controlled.When salted butter is manufactured, a salt slurry is continuouslydosed into the butter. This, however, will increasethe moisture content of the butter, reducing the amount ofwater to be added.Determination of Salt Content in ButterThere are several ways of determining the salt content ofbutter. The analysis can most conveniently be carried outwith a 10-gramme sample that has already been used fordetermination of the moisture content of the butter.The butter is melted and poured into a 150 ml beaker. Thebutter residue is washed into the beaker by means of 50-100 ml of water at 70°C. After addition of 10 drops of saturatedpotassium chromate solution, titration takes placewith the use of a 0.17 n silver nitrate solution (AgNO 3 ),added gradually until the colour changes from yellow tobrownish. The salt content is then determined in accordancewith the following equation:ml of silver nitrate solution used x 0.1 = percentage ofsalt.39

lodine Value and Refractive IndexThe iodine value is defined as the number of grammes ofiodine that can be absorbed in 100 g butterfat. The refractiveindex stales the angle of refraction measured in a socalledrefractometer, when a ray of light passes from the airthrough melted butterfat. Both the iodine value and the refractiveindex are an indication of the content of unsaturatedfatty acids (the most important being oleic acid), whichhave a lower melting point than saturated fatty acids.The relation between the iodine value and the refractiveindex is given in the table below.Hard fatSoft fatIodinevalue Refractive Index2640. 62740. 92841. 22941. 43041. 73142. 03242. 23342. 53442. 73543. 03643. 33743. 53843. 83944. 14044. 34144. 64244. 8Fluctuations in lodine Value and TemperatureTreatment of CreamMilk fat contains, on average, 35% oleic acid (iodine valueapprox. 35), but this percentage is subject to large seasonalfluctuations: the iodine value is high in the summerand low in the winter.The iodine value depends primarily on the fat content ofthe feed and on the composition and melting point of thisfat. It is therefore possible to influence the iodine valueand thereby the firmness of the butter through feeding. Itis usually difficult to regulate the various ingredients thatmake up coarse feed. Roots, for example, give hard andbrittle butter, while grass and hay give butter of a goodconsistency. On the other hand, concentrated feed shouldbe chosen only after taking into account the fat content40

and particularly the composition of the fat (iodine value).For example, feeding with soya beans, linseed and rapeseed cakes, etc, gives butterfat with a high iodine value,whereas the iodine value is lower when feeding with coconutand palm cakes.Other conditions being equal, Jersey cows yield butterfatwith a lower iodine value than, for example, Holsteins,but this difference can be adjusted by choosing the rightfeed. By means of temperature treatment of the cream, itis possible to change the structure of the butter in orderto improve its consistency. The temperatures used shouldbe determined partly on the basis of the iodine value ofthe butterfat and partly on the basis of the temperatureat which the butter will be consumed. It is therefore necessaryfor the creamery to know the iodine value of thebutterfat used, and this value should be determined oncea month.In periods with iodine values above 35, the 19-16-8 methodor a modification, for example, 23-12-8, should beused.In periods with iodine values below 32, the 8-19-16 methodor a modification, for example, 8-20-12, should beused.In transitional periods (iodine values between 32 and 35),a 12-19-12 treatment can be used in the autumn, whereasin the spring, the normal high iodine treatment should bestarted straightaway.41

CHEESECheese VarietiesIt would be an almost impossible task to list all cheesetypes. In general, we distinguish between two basiccheese classes: Yellow and white cheese, where yellowcheese is cheese produced from cow’s milk and whitecheese is cheese produced from ewe’s and goat’s milk, inwhich the fat does not contain carotene.Below are possible classifications of cheese types:Extra hard cheese:Hard cheese:Semi-hard cheese:Semi-soft cheese:Soft cheese:Pasta Filata:Mould cheese:Fresh cheese:Parmesan, Goya, GEmmental, Cheddar, etc.Gouda, Samsoe, Fontal, etc.Tilsit, Danbo, Butterkäse, Limburger,etc.Port Salut, Bel Paese, Feta, etc.Mozzarella, Pizza Cheese, Provolone, Kashkaval, etc.Blue veined cheese: Stilton, Roquefort, Danablu.White surface ripened cheese:Camembert, Brie.Unripened cheese: Queso Fresco,Quarg, Cottage Cheese etc.However, many cheeses are characterised solely by theirname. As an addition, the fat content of the cheese is oftenindicated, and very rarely the content of total solids(TS) in the cheese is also stated.The fat content of the cheese states the fat in the cheeseas a percentage of the TS content (50+, 45+, 30+, 20+).Furthermore, the designations “Full-Fat”, “Reduced Fat”and “Half Fat” are used, which means that the cheesescontain 50-53% fat in TS, 36-39% fat in TS and 26-29%fat in TS respectively.The TS content of the cheese normally varies between65% (Cheddar) and 40% (Feta), but it is constant for eachtype of cheese.42

CheesemakingThe feature common to all cheesemaking is that rennet isadded to the milk, rennet being an enzyme that makes themilk coagulate and the coagulum contract, which, in turn,causes whey exudation, so-called syneresis.Thus, the cheesemilk is separated into curd (cheese) andwhey.CHEESE: 10-15% of the milkFat: 89-94% of the milk fatProtein: 74-77% of the milk proteinsapprox. 100% of the milk caseinWHEY: 85-90% of the milkFat: 6-11% of the milk fatProtein: 23-26% of the milk proteins, incl. NPN*MSNF**: 6.5% of whey is MSNF* non-protein nitrogen** milk solids non-fatStandardisation of Cheesemilk and Calculation ofCheese YieldThe standardisation of cheesemilk has two separate objectives:(1) To obtain cheese with a composition that complieswith the agreed standards.(2) To obtain the most economic use of milk componentsconsistent with consumer demands.The two main elements in the standardisation of the fatpercentage of cheese milk are:(1) The protein percentage of the cheesemilk. The higherthe protein percentage, the higher the fat percentage.(2) The fat content required in the desired cheese type.The table below can be used as a guideline for fat standardisation.43

Wholemilk45% fatin TS40% fatin TS30% fatin TS20% fatin TS10% fatin TS% fat% protein% fat incheesemilk% whole milk% fat incheesemilk% whole milk% fat incheesemilk% whole milk% fat incheesemilk% whole milk% fat incheesemilk% whole milk4.33.553.20752.75641.71391.03230.5110. 84.23.503.20762.70641.69401.02230.5111. 04.13.453.15772.70651.67401.01240.5011. 14.03.403.10772.65661.65401.00240.5011. 23.93.353.05782.60671.65411.00240.4911. 33.83.303.05802.60681.60410.95240.4911. 63.73.253.00812.55691.60420.95240.4811. 63.83.202.95822.50701.55420.90240.4711. 73.53.152.95842.50711.55430.90250.4712. 0Example 1:The cheesemilk contains: 3.3% proteinThe cheese is to contain: 45% fat in TSIn the column “Whole milk” of the table, a value of 3.3%protein is found. From the column “45% fat in TS” it appearsthat the milk must be standardised to a fat contentof 3.05%.In case the protein content of the milk is not known, it ispossible to make an approximate calculation of the proteinpercentage of the milk by using the following equation:0.5 x fat% + 1.4 = protein%thus, for example,0.5 x 3.8% + 1.4 = 1.9 + 1.4 = 3.3% protein.The table is arranged in such a way that it can also beused in case only the fat content of the non-standardisedmilk is known.Example 2:The non-standardised milk contains: 04%fatThe cheese is to contain:40% fat in TSIn the column “Whole milk” of the table, a value of 4.0%fat is found. From the column “40% fat in TS” it appearsthat the milk must be standardised to 2.65% fat. Furthermore,it can be seen that this is obtained by mixing 66%44

non-standardised milk with a fat content of 4.0% with34% skimmilk.Cheese samples should be analysed regularly to makesure that the cheesemilk has contained the correct percentageof fat, and this should be adjusted on the basisof the chemical composition of the milk, which varies withthe seasons.It is important that care is taken when stirring the cheesemilkand when carrying out the fat analysis, as a readingerror of 0.1% means an error of 1.5% fat in TS in a 45%cheese, and more in cheeses of the low-fat type.If samples are taken for analysis of fresh, unsalted cheese,it must be taken into account that the salt increases theTS in the cheese by approximately 2%, reducing the fat inTS by approximately 1.5%.The final determination of fat in TS can only be carriedout after 4-6 weeks when the salt has spread throughoutthe cheese, but even then, variations of more than 1%fat in TS can be found in cheeses from the same vat. It istherefore advisable to operate with a safety margin of atleast 1% for ripened cheese and consequently 1.5% morefor the fresh cheese.Instead of using the table for adjusting the fat content inthe cheesemilk, the actual fat percentage can be calculated.Several equations can be used for this calculation,but the one used in the following gives a very high degreeof accuracy.(1) Cheese to be produced:Moisture ................ 41.5%Fat in TS ................ 51.0%Salt (NaCl). .............. 1.5%(2) Raw milk:Fat. .................... 4.0%Protein. ................. 3.4%(3) Retention figures:Fat. .................... 91.0%Protein. ................. 76.5%Protein in MSNF in cheese .. 87.6%45

(4) Calculations:(4.1) Cheese ................. 100.0% = 1,000.0 gMoisture ................ 41.5% = 415.0 gTS . . . . . . . . . . . . . . . . . . . . . . 58.5% = 585.0 gFat in TS ................ 51.0% = 298.4 gSolids non-fat ............ = 286.6 gSalt (NaCl). .............. 1.5% = 15.0 gMSNF .................. = 271.6 gProtein in MSNF .......... 87.6% = 237.9 g(4.2) Kg milk/kg cheese:FatProtein1,000 g cheese: 298.4 g = 91% 237.9 g = 76.5%Whey: 29.5 g = 9% 73.1 g = 23.5%Cheesemilk: 327.9 g = 100% 311.0 g = 100.0%Protein in fat-free milk = 3.4 x 100 = 3.54%(100 - 4)Per 1,000 g cheese:Fat-free = 311.0 x 100 = 8,785.3 gmilk 3.54Fat. ............. = 327.9 gCheesemilk ....... = 9,113.2 g(4.3) Fat percentage in cheesemilk:= 9.1132 kg milk/kg cheese327.9 x 1009.113= 3.60%(4.4) Cheese yield:1009.113= 10.97%46

Equations often used for the calculation of cheese yields are:Cheddar Y = (0.9 F + 0.78 P - 0.1) x 1.091 - MMozzarella: Y = (0.88 F + 0.78 P - 0.02) x 1.121 - MCheddar Y = (0.77 F + 0.78 P - 0.2) x 1.101 - Mwhere:Y = Yield in per centF = Fat percentage in milkP = Protein percentage in milkM = Moisture per kg cheese, 38% = 0.38 kgCheese yield is influenced by the loss of fat and curd finesin the whey. However, with modem production equipmentand correct processing technology, it is possible to reducethe fat loss to less than 7.0% and the loss of curdfines to approx. 100 mg/kg whey.Utilisation Value of Skimmilk in CheesemakingFor this calculation, the figures from the cheese yield calculationare used as an example:kg cheesemilk per kg cheese. ............... 9.1132kg fat in cheesemilk ....................... 0.3279kg skimmilk. ............................. 8.7853kg fat in whey ............................ 0.0295kg whey ................... 9.1132 -1.000 = 8.1132fat in whey . . . . . . . . . . . . . . . . . 0.0295 x 100 = 0.36%8.1132The fat in whey may be reduced to 0.05% by means ofseparation.In the following example, the values used are:Cheese = 23.00 krone/kg*Whey = 00.30 krone/kgButter fat = 27.00 krone/kg* 1 Danish krone = 100 øre47

Income per kg cheese:1 kg cheese ............. 2,300.0 øre8.11 kg whey at 30 øre/kg . . 243 ørefat from whey separation:8.11 x (0.36 -0.05) x 2.700100= 69.0 øre 2,612.0 øreCosts per kg cheese:butter value 0.3279 x 2,700 = 885.0 øreoperating costs. .......... 500.0 ørewhey separation 8.11 x 0.986 = 8.0 øre 1,393.0 øreValue of skimmilk per kg cheese . . . . . . . . . 976.2 øreUtilisation value of skimmilk . . .1,219.0 = 138.8 øre8.7853Strength, Acidity and Temperature of Brine forSalt ingThe saturated brine which is normally used for saltingcheese occasionally produces too hard a rind, but this canbe counteracted by using a weaker solution. The solutionshould, however, contain at least 20% salt, correspondingto 10°BÈ. The strength of the brine should be checkedevery day: otherwise there is a risk that the solution maybecome too weak. If this happens, the cheese protein exudedthrough the whey will quickly decompose, and theincrease in the growth of bacteria will cause defects notonly in the rind but also in the interior of the cheese.The strength of the brine should be measured with a hydrometerindicating degrees Baumè. When the brine hasbeen in use for a certain time, the hydrometer will show adeviation of 1-2°BÈ because of the substances dissolvedin the brine. In practice, this means that, when measuringthe strength of a 2-3 months old brine solution, degreesBaumè can be considered equal to the salt percentage.The acidity of the brine should be about the same as thatof the cheese, i.e. approx. pH 5.2, but in a freshly madesolution it will usually be somewhat higher depending uponthe acidity of the water supply. It will usually take a week forthe acidity to fail to the desired pH level, but to avoid anyrisk of damaging the cheese rinds during this time, the pHvalue should immediately be brought to the desired level bythe addition of hydrochloric acid to the solution. By means48

of a simple analysis of the creamery’s water supply, anylaboratory will be able to state the amount of hydrochloricacid required.The temperature of the brine, in particular, controls thespeed at which the salt is absorbed by the cheese, andshould be 10-12°C the whole year round. It is thereforeoften necessary to cool the brine in the summer and heatit in the winter.Strictly speaking, brine can be used for an indefinite timeprovided that the content of saltpetre (KNO 3 ) or bacteriaand moulds does not become too high.If the brine contains considerably more than 100,000 bacteriaor moulds per ml, it should be sterilised by boilingor by adding 1/2 litre sodium hypochlorite per 1,000 litresbrine. Sodium hypochlorite can also be added regularlyonce a month, and this will ensure that the content ofharmful bacteria in the brine is kept low. When used forthe manufacture of rindless cheese, the brine should besterilised regularly.49

MEMBRANE FILTRATIONDefinitionsMembrane filtration processes are pressure-driven molecularseparation processes to obtain either concentration,fractionation, clarification and/or even a sterilisationof a liquid. The separation is determined by the membranecharacteristics (molecular weight cut-off value – MWCO)and the molecular size of the individual componentspresent in the liquid.Membrane filtration changes the volume and/or the compositionof a liquid, as the feed is divided into two newliquids of altered chemical/microbiological composition:1) the retentate (what is rejected and concentrated by themembrane, e.g. proteins) and2) the permeate (i.e. filtrate, what is passing through themembrane, e.g. water and minerals).The volume of permeate produced by a certain membranesurface area per hour is called flux (measured in l/m 2 /h orsimply “lmh”). The volumetric concentration factor (VCF orCF) is the ratio between the incoming feed volume and theoutcoming retentate volume.Rejection is 100%, when the component is fully concentratedby the membrane (cannot pass the membrane), andthe rejection is 0%, when the component passes freelythrough the membrane, giving an identical concentrationon both sides of the membrane.The driving pressure is the transmembrane pressure(TMP), which is the pressure difference between the meanpressure on the retentate side (high) and the mean pressureon the permeate side (low or zero).All membrane filtration processes are cross-flow filtration(feed flow parallel to the membrane surface, alsocalled tangential flow), since a high velocity and shearrate across the membrane surface is essential to preventbuild-up of retained materials, which reduces run timesand flux and may alter the separation characteristics. Highcross-flow velocities are especially important in UF andMF systems.Membrane ProcessesConcentration: In true concentration all total solids are re-50

tained since only water can pass through the membrane(as in evaporation and drying processes). Example: ReverseOsmosis (RO).Fractionation: Changing the chemical composition byconcentrating some components, while others remain unchanged.Example: Nanofiltration (NF), Ultrafiltration (UF),Microfiltration fractionation (MFF).Clarification: Changing a turbid liquid into a clear solutionby removing all suspended and turbid particles. Example:Ultrafiltration (UF) and Microfiltration (MF)Sterilisation: Removing all microorganisms from a liquid.Example: Microfiltration (MF).Reverse OsmosisIn reverse osmosis practically all total solids componentsare rejected by the membrane allowing only waterto pass through the membrane. Since also practically allions (apart from H + and OH - ) are rejected by the membrane,the osmotic pressure in the retentate will increase,why high-pressure pumps are needed to overcome theosmotic pressure. The amount of permeate produced isoften referred to as “recovery”. 90% recovery means that90% of the feed is recovered as permeate (equal to 10xconcentration).Low molecular components like organic acids and NPNcomponentsare not fully rejected by the membrane, especiallywhen they appear uncharged (non-ionic), typicallyin acidic environments. This is the reason why COD levelsin the permeate are higher processing acid products (e.g.lactic acid whey) compared to sweet products (e.g. sweetwhey).Max. achievable solids by RO are in the range of 17-23%TS for whey and UF permeates.51

RONFUFMFFMFPore size(nm)0.1- 1 0.5- 2 5 - 10050- 200800 - 1400MWCO< 100100- 5005,000 - 20,000Typicalpressure(bar)Typical temp.(°C)30- 40 20- 303 - 8 0.1- 0. 8 0.1 - 0. 810- 30 10- 3010or 505050ApplicationsConcentrationDemineralisation/concentrationProteinconcentration(WPC/MPC)ProteinfractionationWhey fatremoval (WPI)BacteriaremovalCheese milkESL milkNanofiltrationNanofiltration is very similar to the RO process, but the NFmembranes are slightly more open than in conventionalreverse osmosis. Nanofiltration allows passage of monovalentions like Na + , K + and Cl - , whereas divalent ions likeMg ++ and Ca ++ are rejected by the membrane. In this waythe nanofiltration process demineralises the feed by typical30-40%. The degree of demineralisation is the %removalof minerals (or ash) from the feed to the permeate.Since some of the monovalent ions are removed from theretentate, the osmotic pressure will be lower than for conventionalRO. For this reason it is possible to obtain higher%TS in the retentate compared to the RO process.Max. achievable solids by NF are in the range of 21-25%TS for whey and UF permeates.Example of NF mass balance of UF permeate from cheesewhey (indicative):NanofiltrationUFpermeateRetentatePermeateTrueprotein% 00,000.010,000.040,000.00NPN%00,000.20 0,000.40 0,000.10Lactose%00,004.60 0,016.00 0,000.20Acids%00,000.20 0,000.60 0,000.02Totalash% 00,000.50 0,001.00 0,000.30Totalsolids% 00,005.50 0,018.00 0,000.60Capacitykg/h 10,000. 002,820. 007,180.0052

UltrafiltrationUltrafiltration has many applications, but basically it is aprocess for concentration of protein (and milk fat).In the dairy ingredients industry UF is used for concentrationof whey proteins from whey into WPC products or for concentratingmilk proteins from skim milk into MPC pro ducts.The protein content may be concentrated up to 23-27%protein, and in many cases the retentate can be spray drieddirectly without an evaporation step. Dia fil tra tion is necessaryfor higher purity products like WPC 80 (80% protein inthe powder or in the solids). In diafiltration, water is addedto the retentate to increase “washing out” of dissolved substanceslike lactose and minerals to the permeate.UF of whey for the production of WPC retentates (a fatremoval step is essential for producing WPI):CompositionWheyWPC 35WPC 55WPC 70WPC 80WPI 90Protein%0.80 3.3 08.3 17.9 23.3 23. 3Lactose%4.60 4.9 04.7 04.0 01.7 01.3Ash%0.50 0.5 00.7 01.0 00.9 00.5Fat%0.060.3 00.8 01.7 02.3 00.2TS%6.00 9.0 14.5 24.7 28.1 25. 4VCFratio1x5x13x29x38x38xDiafiltration- - - - + +Ultrafiltration of cheese milkProtein standardisation: The protein content in the cheesemilk is increased (e.g. from 3.2% up to 4.0-4.5%). Whenthis method is used, traditional cheesemaking equipmentmay be used after UF and the cheesemaking technologyinvolved is largely the same as that used in the traditionalcheesemaking. The advantages of this method are savingsin cheese rennet, and higher and more standardisedcheese yields (throughput capacity) in existing cheeseequipment.Total concentration: Total concentration is a process inwhich the TS content in the retentate and in the freshcheese is the same, i.e. a cheese process without wheydrainage. This method is used for fresh cheeses likeQuarg, Cream Cheese, Queso Fresco and Cast Feta.Ymer, Yoghurt and Pate Fraiche may also be produced bytotal UF concentration.53

MicroparticulationMicroparticulation is a thermal and mechanical treatmentprocess that is used to denature whey protein concentrate(WPC) and form ideal protein particle sizes similarto fat globules in milk. Due to the increasing demand forreduced-fat products, microparticulated whey protein isan attractive option in the dairy and food industries to enhancetaste and texture in reduced-fat products, and alsoas a multi-functional protein source.APV has developed a unique microparticulation process,the APV LeanCreme process that comprises anultrafiltration system for the production of WPC and amicroparticulation system. The APV LeanCreme TM processis designed for optimum denaturation and results in aproduct called LeanCreme TM . In more detail, the LeanCremeTM process comprises a plate heat exchanger (PHE) forpreheating the WPC and a number of ASA’s (APV ShearAgglomerators – purpose-built, scraped surface heat exchangers),a holding tube, an ASA for the first cooling, anda PHE for the second cooling in the regeneration section.During the APV LeanCreme TM process the particle size iscontrolled very accurately by the ASA’s.WheyMembraneloopsWPC60PermeateUF PlantASAASAPHEpreheatingHoldingcellCoolingLeanCremeCoolingHeatingMP PlantApplicationsFig. 10: The APV LeanCreme processParticle size distributionLeanCremeTM particle quality is mainly a question of particlesize distribution, which is determined and controlledin the process. The curves in the graph below show howthe ASA speed has a direct influence on the characteristicsof two different LeanCreme qualities.54

2.001.751.501.251.000.750.500.25Helos SympatecParticle size distribution1)LeanCreme 60 - low speed (35%)2)LeanCreme 60 - medium speed (65%)Volume Weighted Density Distribution00.10.5 1 5 10Particle size / µmFig. 11: Particle size distributionDegree of denaturationThe quantity of LeanCreme TM particles is measured by thedegree of denaturation.This is defined as:(Total protein (TOP) - Non casein nitrogen (NCN))Degree of denaturation = × 100(Total protein (TOP) - Non protein nitrogen (NPN))In other words, the degree of denaturation is the percentageof aggregated proteins divided by the true proteins.Types of LeanCreme TMThe below table shows the different feed sources (WPC’s)resulting in the different types of LeanCreme TM :LeanCremeNeutralLeanCremeLacticLeanCremeAcidLeanCremeIdealLeanCremePlusLeanCremeMixSweetcheesewheyWPCXLactic acidwheyWPCX X*XFeed Source – WPC28 to WPC80AcidcaseinwheyWPCXIdeal wheyWPCXMilk fat/vegetableoilWPCXCasein/wheyWPC* Lactic acid whey WPC originating from thermo quargwhey is not recommended. The reason is the small quantityof whey proteins left after the cheese heating process. Theresulting lactic acid whey contains a high amount of NPN,which cannot be transformed into LeanCreme TM particles.X55

and between the elements. SW elements can be mountedin series inside a housing (also called pressure vessel ormodule). Spacer height, flux curves, pump performancesand pressure drops determine the configuration of a SWplant.Plate & frame (P&F), module 37 (M37) is the only P&Fmod ule still in use and only for high viscosity productslike cream cheese (Philadelphia type). This module cango high in protein% (more than 29%), when operated witha positive pump up to 12 bar. The crossflow rate shouldbe 25 l/plate/min.When assembling new membranes, the module shouldbe compressed applying 240kN (or 24 tons) of pressure(or until the module stops leaking!). The M37 module isincreasingly challenged by newer module types, like speciallydesigned SW elements and tubular ceramic membranes.Inorganic Membranes (Ceramics)Unlike the polymeric membranes (especially RO/NF), theceramic material is very resistant to heat and chemicals,and ceramic membranes will last for typically 5-10 yearsor more. However, they are much more expensive, andgenerally require more pumping energy. Due to the ceramicnature, they are sensitive to mechanical vibrations(should always be installed vertically) and thermal shock.Tubular membranesAPV’s experience is largely based on the French “Exekia”membrane (formerly SCT). The membranes are tubular,with the feed circulating inside tubular channels. The diameterof these channels is 3, 4 and 6 mm, which is selectedaccording to the viscosity of the product. The mainapplication for ceramics is MF, since the ceramic elementcan be operated with permeate back-pressure, so as toachieve a low TMP, which is crucial for successful results.Two products are available: The standard element, whereUTP operation is required (permeate recirculation to createpermeate back-pressure) and the newer GP element,where the permeate back pressure/resistance is integratedinside the membrane structure (GP = Gradient Pressure).Available MF pore sizes are: 0.1 – 0.2 – 0.5 – 0.8 – 1.4 – 2– 3 – 5 microns, which are alumina membranes on aluminastructure. UF pore sizes available are: 20 – 50 – 100 nm,60

which are zirconia material on alumina structure. For UFprocesses it is not necessary to control a low TMP.Exekia Membralox membranes and their membrane areas:Channel size Ø3 mm(P37-30 GL)4 mm(P19-40 GL)6 mm(P19-60 GL)1P housing (m 2 ) 0.350.240.363P housing (m 2 ) 1.050.721.087P housing (m 2 ) 2.451.68Not available12P housing (m 2 ) NotavailableNotavailable4.3219P housing (m2 ) 6.654 .567.92 (22P)CIPCleaning of membranes is nothing like cleaning of standarddairy equipment made of stainless steel. Membraneelements are often organic polymeric membranes madeof materials, which only tolerate certain cleaning limits interms of pH and temperature (and desinfectants/oxidisers).Therefore it is almost always necessary to use formulatedcleaning products including enzymatic productsfrom approved suppliers like Henkel, Ecolab, Diversey-Lever, Novadan and others. In the table below some limitsare listed for different membrane materials.Membrane materialSupport/backingPolyamide(RO/NF)PolyesterPolysulphone(UF)PolyesterPolysulphone(UF pHt)PolypropyleneCeramic(MF/UF)AluminaMaxtemp (°C)50507085 (not critical)Cooling rateNotcriticalNotcriticalNotcriticalMax 10°/minPH range1.5-11.5 1.5-11.5 1-131-14FreechlorinePhosphoricacidSurfactantsSanitationNoYesOnlyanionicMax200 ppmYesOnlyanionicMax200 ppmYesOnlyanionicNot criticalNoNot critical0.2% bisulfite0.2% bisulfite0.2% bisulfite0.5% nitric acidWater flux: After installation and cleaning of new membranes,the water flux should be registered to be used for futurereference. Organic membranes always stabilise within thefirst few weeks. Cleaning of membranes should always befollowed by a water flux reading, which must be recordedat the same pressure, temperature, time and cleaning step,so the cleaning efficiency can be monitored.CIP Water Quality GuidelinesFor optimal cleaning and flushing of membranes, the waterused should be within the following specifications61

ParameterIron(Fe)Manganese(Mn)Aluminium (Al)Silica(SiO 2 )Chlorine(Cl 2 /HOCl)GermanHardnessFoulingIndexTurbidityTotal plate count 22°CTotal plate count 37°CColiformsUnitsRO/NForganicmembraneUF/MForganicmembraneUF/MFceramicmembranemg/l < 0.05< 0.05

Notes on parametersmg/l: In practice equal to ppm (parts per million)Silica: Total = colloidal + soluble silica. Silica is practicallyinsoluble in water at any temperature and is very hard toremove from the membrane, especially once precipitated.Colloidal silica should be absent, or as low as possible.Chlorine: Must be analysed on site as the chlorine quicklydisappears from the sampleHardness: Is determined from the content of calcium andmagnesium (see formula for German hardness °dH).˚dH = 5.61 x ( ppmCa2+ + ppmMg2+ )40.1 24.3Total hardness = temporary + permanent hardnessSoft water < 8°dH medium water < 16°dH hard water.1°dH equals 10 ppm CaOor 07.14 ppm MgOor 17.9 ppm CaCO 3or 24.3 ppm CaSO 4or 15.0 ppm MgCO 3Equivalent units are listed below:UnitGerman°dHDanish°dHEnglish°HAmerican°HFrench°THF1°dH German 1.001.001.2517.851.791°dH Danish 1.001.001.2517.851.791°H English 0.801.001.0014.301.431°H American 0.0560.0560.071.000.101°THFFrench 0.560.560.7010.001.00Conductivity: If water is demineralised one should expectless than 30 µS/cm. In comparison, drinking water is in therange of 300-800 µS/cm.Turbidity: Method: Particles scatter light (expressed inNTU, equal to JTU or FTU). Turbidity may also be expressedin SiO 2 (mg/l), where 10 mg/l equals 4 JTU.Silt Density: Equal to Fouling Index, Colloid Index or ColmatationIndex. This index is related to “Suspended Solids”and replaces this analysis.Method: Pass the water through a 0.45 micron CA filterØ 47 mm (ref. Milli-pore HAW PO 47000) at a constantpressure of 2.1 bar (30 psi). The time to pass 500 ml water63

is measured at test start (t0) and 15 minutes (t15). SDI0-3: Non-fouling, SDI 3-6: Some fouling, SDI 6-20: Highfouling.SDI = 100 x ( 1-(t0/t15) )15CIP and hardnessThe hardness of the water is an important factor, as it governsthe dosage concentration of the cleaning chemicalsand the flushing time. Soft water is the most gentle forthe membranes, with a low risk of mineral precipitation onthe membrane surface. However, soft water has a muchreduced buffering effect when dosing cleaning chemicals,which means that pH limits are reached at lower concentrations.As a rule of thumb, if 2% may be tolerated in20°dH before the pH limit is reached, only 1% may betolerated in 10°dH (when applying Divos 124). However,these figures are not true for all caustic products, but theprinciple is the same. Lower concentrations reduce thecleaning efficiency even at the same pH, as there are lesscleaning agents (surfactants, carriers, complexing agents)to bind or “carry” the soil and to keep it in solution untilflushing. Severe foaming may also be a result of using softwater. The flushing time is prolonged with higher waterconsumption as a result (ever washed hands using softwater?). Some enzymatic products need certain minerals(e.g. calcium) in order to work. When using soft water,these minerals will have to be added. When using hardwater extra complexing agents such as EDTA or NTA mustbe added in order to prevent mineral precipitation. Thesolubility of calcium salts is much reduced at higher temperaturesresulting in heavy fouling of the membrane.Pre-treatment methodsIf some of the parameters do not meet the requirements,the following pre-treatments may be applied:Cartridge filter: Reduces SDI and remove particles byraw water filtration (5-10 micron pore size).Sand filter: Removes Fe and Mn.Sand filter: Special filling material removes fouling particles(SDI/turbidity).Active carbon: Removes organic matter and neutraliseschlorine.Bisulfite: Neutralises chlorine.64

Ion exchange: Removes SiO 2, Al, Fe, Mn, softens hardwater.Chlorination: Kills bacteria (e.g. from surface water). Onehour chlorination followed by dechlorination is recommended.Milk and Whey CompositionRaw milk quality (Denmark, 2001):Extra1stclass2nd class3rd classTotalcounts/ml 0< 30.000030.000-100.000100.000-300.000>300.000Somaticcells/ml < 300.000300.000-400.000400.000-650.000>650.000Anaerobicspores/l < 400< 400400-1100>1100Freezingpoint °C-0.543 to -0.516AntibioticsNegativeComposition of milk in Northern Europe (average values):Raw milk(DK/NL 1999)Skim milk(Germany 2002)F AT4 .3%0.06%T OP (total protein)3 .4%3.63%N PN (NPNx6.38)0 .19%0.19%T RP (true protein)3 .21%3.44%T WP (true whey proteins) 0 .55%0.60%C AS (casein)2 .66%2.84%A CD (citric acid)0 .18%0.20%L AC (lactose)4 .65%4.84%T OA (total ash)0 .73%0.77%T S (total solids)13.3%9.50%C AS/TRP ratio83-84%82.6%C AS/TOP ratio77-79%78.2%T WP/TOP ratio16.5-15.5%N PN/TOP ratio5 .0-6.5%5.2%65

Components in milk and whey and their approximate size:Large particlesDiameter size inmicron (my)Somaticcells (leukocytes)10-20Yeastcells5-30Bacteriacells0.5-5Bacteriaspores (Bacillus/Clostridium)0.8 x 1. 5F at globules in raw milk0.1-10 (2-6)Fatglobules in skim milk/homogenised milk < 1Protein particles (colloidal)Lipoprotein particles(protein + P-lipids)Casein micelle (app. 500 subunits)(casein micelle = 70% water + 30% casein)Subunit of casein micelle(10 casein molecules)Diameter size innanometer (nm)1010-30010-12Individual proteinsMolecular Weight(MW = Daltons)Caseinmolecule20-25.000Paracasein12.200Wheyproteins (= serum proteins)3-6 nmImmunoglobulins(IgG)150.000I mmunoglobulins (IgM)900.000 (= 30 nm)ß-lactoglobulin(ß-LG)2 x 18.000Alpha-lactalbumin14.000BovinSerum Albumin (BSA)66.000Lactoferrin/Transferrin(LF)77.000Caseinomacropeptide(CMP/GMP)6.800EnzymesLactoperoxidase(LP)77.500Cheeserennet (chymosin/rennin)31.000X anthin Oxidase (XO) ( in fat globules)283.000M ilk Lipase (mLPL) ( in casein micelle)50.000P hosphatase ( in fat globule membrane)2 x 85.000M ilk Plasmin ( in casein micelles)89.000Molecular WeightNon-Protein Nitrogen (NPN)66(MW = daltons)Cholin(vitamin)121

M ilk Lipase (mLPL) ( in casein micelle)50.000P hosphatase ( in fat globule membrane)2 x 85.000Components M ilk Plasmin ( in in casein milk and micelles whey ) and their approximate 89.000size(continued):Non-Protein Nitrogen (NPN)Molecular Weight(MW = daltons)Cholin(vitamin)121Aminoacids75-200Peptides200-1500Urea-N60Creatin/creatinin131Carbohydrates/AcidsLactose342Glucose180Galactose180Lactulose342Lacticacid90Citricacid192Aceticacid60Minerals – positively chargedSodium (Na+)23Magnesium (Mg++)24Potassium (K+)39Calcium (Ca++) soluble40Minerals – negatively chargedChloride(Cl-)35Phosphate(PO4—) soluble95Sulphate(SO4—)96Carbonate(HCO3-)6167

CLEANING AND DISINFECTINGThe design of modern dairy equipment allows cleaningand disinfecting to take place without the equipment havingto be taken apart, i.e, cleaning-in-place (CIP). Thismeans that the processing equipment must be madeof materials (eg, stainless steel) that are resistant to thecorroding effects of the cleaning agents. The processingequipment must also be designed in such way that all surfacesin contact with the product can be cleaned.CIP Cleaning in GeneralMilk components are excellent substrates for microorganismsand a careful cleaning is thus very important. Thisdoes not alone apply to the parts in contact with the product,but also to the external parts and rooms etc.The effectiveness of the cleaning is determined by the followingfour factors:1. A chemical factor2. A mechanical factor3. A thermal factor4. A time factor1. The chemical factor is determined by the cleaningagent and the concentration in which it is used.The cleaning agent is chosen according to the type ofpollution to be removed, in this way:PollutionBasicAcidF at+ -P rotein+ +Ash(milk residues)- +Waterresidues- +In the central CIP plant the majority of the cleaning solutionsis led back to the CIP tanks and reused. Therefore,the concentration may be fixed at a suitable highlevel without too much waste.68

The functions of the cleaning agents are:- To loosen the pollution- To keep the impurities dissolved in the cleaningsolutions to prevent them from precipitation on thecleaned surfaces- To prevent sedimentation of lactic salts.Guiding concentrations: Acid (HNO 3 ) 0.8-1.2%, andlye (NaOH) 0.8-1.5%.2. The mechanical factor is determined by the speedof the liquid over the surfaces. The faster the liquidmoves, the more efficient the cleaning will be. It is importantthat the movement of the liquid is turbulent, i.e.that the liquid parts continuously change place mutually.Consequently, the pump speeds are considerablyhigher during CIP than during production.The cleaning turbines in the tanks make up an effectivemechanical factory, but partial blockings of theturbines may appear. In consequence, the turbinesshould be inspected regularly.3. The thermal factor (the temperature) is very important.Within chemistry it is said that the reaction speed isdoubled if the temperature is increased by 10 o C. However,a too high temperature also presents disadvantages,as residues of proteins and lactic salts are precipitatedat too high temperatures, and the solubility ofthe salts in the water is reduced.Guiding temperatures: Lye solution 70 - 75 o C and acidsolution 60 - 65 o C.4. The time factor is important to the softening and solutionpart of the pollution.In the program survey, approximate periods for thesingle steps in the programs are indicated. The indicatedperiods should only be regarded as a broadguidance, as there may be considerable differences69

70between the single routes, both as regards equipmentto be cleaned and the fouling degree.DisinfectionThe purpose of a disinfection is to kill the largest possiblenumber of bacteria to avoid an infection of the products.Heat in the form of steam or especially hot water is themost used form of disinfection. The central CIP plant includesprograms for sterilisation with hot water, and thereturn temperature is set to 85 - 90 o C.Cleaning of dairy equipment is carried out as follows:A. Pre-rinseThe processing equipment is rinsed with cold or warmwater. The object is to remove any possible productresidue before cleaning. The rinsing water containing theproduct residue should be led to suitable reception facilitiesin order to minimise pollution.B. Cleaning with sodium hydroxideThe process equipment is cleaned by means of circulationof a hot sodium hydroxide cleaning solution. Today,special cleaning agents are commonly used instead ofsodium hydroxide. After cleaning, the cleaning solutionis collected and re-used. Re-use should not take placebefore the concentration of the returning solution (%) hasbeen checked and adjusted accordingly.C. Intermediate rinseAny remaining cleaning solution is flushed out with eithercollected rinse water or fresh water.D. Cleaning with nitric acidThe process equipment is cleaned by means of circulationof a hot nitric acid cleaning solution. Today, special cleaningagents are commonly used instead of nitric acid.After cleaning, the cleaning solution is collected and reused.Re-use should not take place before the concentrationof the returning solution (%) has been checked andadjusted accordingly.E. Final rinseAny remaining cleaning solution is flushed out with eithercold or hot water. Chemical free water is collected andused for pre-rinse.

F. DisinfectionThis is carried out immediately before the product plant isput into operation. Disinfection can be carried out thermallyor chemically. The CIP plant is normally designed to allow fordisinfection by circulation of either hot water at 90-95°C or asolution of e.g. hydrogen peroxide. Today special agents fordisinfection is widely used in place of hydrogen peroxide.Disinfection must always be followed by a rinse with cleanand drinkable water.Cleaning MethodsCleaning agents:The following cleaning agents can be used for CIP-cleaning.Lye, NaOH, Sodium hydroxide:- 30% concentrated solution.Acid, HNO3,Nitric acid:- 30% concentrated solution.- 62% concentrated solution.Hydrochloric acid, (HCl), and/or chlorine-containing cleaningagents, (Cl ), must never be used.Normally used cleaning solutions:Lye: NaOH - Solution for cleaning oftanks and pipes 0.8-1.2%Above corresponds to a titter of 20.0-30.0Lye: NaOH - Solution for cleaning ofpasteuriser 1.2-1.5%Above corresponds to a titter of 30.0-37.5Acid: HNO3 - Solution for cleaning oftanks and pipes. 0.8-1.0%Above corresponds to a titter of 12.7-15.9Acid: HNO3 - Solution for cleaning ofpasteuriser 0.8-1.2%Above corresponds to a titter of 12.7-19.0Note:Titter corresponds to ml 0.1 N (NaOH or HCL),per 10 ml against phenolphthalein (8.4).71

Reagents: 0.1 N Sodium hydroxide, (NaOH), solution.0.1 N Hydrochloric acid, (HCl), solution.5% Alcoholic phenolphthalein solution.General maintenance of CIP plant:Daily check: Control of lye and acid cleaning concentrations.Weekly check:Control of stone deposits in lye tank/tanks and water tank/tanks.Drawing off of bottom sludge from lyeand acid tanks.Monthly check: Control of various gaskets and replacementof these, if necessary.Quarterly check: Change of cleaning solution in the lyeand acid tanks.CIP Cleaning Programs for Pipes and TanksPipesPicking up of residual productsPre-rinse, cold water/recyclable waterLye cleaning 1% solution at 70°C(The time stated is only started when returnconcentration and return temperatureare identical with the above)Intermediate rinse, cold water/recyclablewater - Special software solutionAcid cleaning 0.8% solution at 60°C(The time stated is only started when returnconcentration and return temperatureare identical with the above)Final rinse, cold water(The time stated is only started when returnconcentration indicates clean water)Total cleaning timeCleaning Time* minutes1-3 minutes6-10 minutes1-3 minutes4-6 minutes1-3 minutes** minutes72

Hot water sterilisation at 85°C(The time stated is only started whenreturn temperature is identical with theabove)3-5 minutesCold water disinfection with hydrogen peroxide, H 2 O 2 ,solution 200 ppm.*)Time is dependent on the physical conditions in andaround various pipes/pipelines to be cleaned.**)Time is dependent on the physical conditions in andaround various pipes/pipelines to be cleaned as well asthe software to control cleaning of pipes/pipelines.Above times are stated as efficient cleaning times andshould be seen as recommendable values. These valuesmay change dependent on the physical conditions in andaround various pipes/pipelines as well as the complexityof various products with regard to the physical/chemicalconditions, as well as the complexity of various physical/chemical as well as microbiological deposits.TanksPicking up of residual productsPre-rinse, cold water/recyclable waterCleaning Time* minutes1-3 minutesLye cleaning 1% solution at 70°C 10-15 minutes(The time stated is only started when returnconcentration and return temperatureare identical with the above)Intermediate rinse, cold water/recyclablewater - special software solutionAcid cleaning 0.8% solution at 50-60°C(The time stated is only started when returnconcentration and return temperatureare identical with the above)1-3 minutes4-6 minutes73

Final rinse, cold water 0.5-1 minute(The time stated is only started when returnconcentration indicates clean water)Total cleaning timeHot water sterilisation at 85°C(The time stated is only started whenreturn temperature is identical with theabove)** minutes3-5 minutesCold water disinfection with hydrogen peroxide, H 2 O 2 ,solution 200 ppm*)Time is dependent on the physical conditions in andaround various tanks to be cleaned (tank dimension).**)Time is dependent on the physical conditions in andaround various tanks to be cleaned (tank dimension), aswell as the software to control cleaning of tank/tanks.Above times are stated as efficient cleaning times andshould be seen as recommendable values. These valuesmay change dependent on the physical conditions in andaround various tanks (tank dimensions) as well as thecomplexity of various products with regard to the physical/chemical conditions, as well as the complexity of variousphysical/chemical as well as microbiological deposits.CIP Cleaning Programs for Plate PasteurisersPasteurisersPicking up of residual productsPre-rinse, cold water/recyclable waterCleaning Time* minutes5-10 minutesLye cleaning 1.5% solution at 70°C 45-60 minutes(The time stated is only started when returnconcentration and return temperatureare identical with the above)Intermediate rinse, cold water/recyclablewater - special software solution5-10 minutes74

Acid cleaning 0.8% solution at 50-60°C 20-30 minutes(The time stated is only started when returnconcentration and return temperatureare identical with the above)Final rinse, cold water(The time stated is only started when returnconcentration indicates clean water)Total cleaning time2-5 minutes** minutesHot water sterilisation at 85°C15-20 minutes(The time stated is only started whenreturn temperature is identical with theabove)Cold water disinfection with hydrogen peroxide, H 2 O 2 ,solution 200 ppm.*)Time is dependent on the physical conditions in and aroundvarious pasteuriser/pasteuriser plants to be cleaned.**)Time is dependent on the physical conditions in andaround various pasteuriser/pasteuriser plants to becleaned as well as the software to control cleaning of pasteuriser/pasteuriserplants.Above times are stated as efficient cleaning times andshould be seen as recommendable values. These valuesmay change dependent on the physical conditions in andaround various pasteuriser/pasteuriser plants as well asthe complexity of various products with regard to thephysical/chemical conditions, as well as the complexityof various physical/chemical as well as microbiologicaldeposits.PasteurisersContinuous buttermaking machinesUltrafiltration plants (UF)EvaporatorsCIP*CIP** specialCIP*** specialCIP75

*) As a consequence of both a higher detergent concentrationand a longer cleaning period compared withthe cleaning of pipes and tanks, it may be appropriateto clean the pasteurisation plant independently of theCIP plant for pipes and milk tanks.At the end of the production run, the pasteurisers, includingpumps, valves and pipes, are flushed out withcold water until the water is clear and free of milk atthe outlet.A closed circulating flow is then established by leadingthe water from the outlet back to the balance tankand slowly adding approx. 3.5-4.0 l 30% sodiumhydroxide (NaOH) per 100 kg water in the system. Ifthe sodium hydroxide is in dry form, it should be dissolvedin approx. 10 l cold water per kg NaOH beforeit is added to the balance tank.Warning: NaOH should always be mixed slowly intocold water - never water into NaOH as it will boil upwith explosive force. Always use facial protectionwhen working with concentrated detergents. If thevolume of the plant is unknown, the concentrationmust be checked as described below.If the water is very hard, 300-500 g trisodium phosphateshould also be added.The temperature is raised to 70-75°C and circulationis continued for at least 45-60 minutes.The NaOH solution is flushed out with water and thecirculating flow is re-established. Then, approx. 2.5 lnitric acid (30%) is added slowly and circulated for 20-30 minutes at 60-65°C after which the acid is flushedout with water.Before start-up of the next production run, the pasteurisationsystem is disinfected by circulation of hotwater at 90°C for 15-20 minutes. Cooling and pasteurisingtemperatures are adjusted to normal productionbefore the water is forced out with milk.**) CIP of buttermaking machines is always carried outwithout the use of the ordinary CIP plant, because relativelylarge amounts of fat residue must be removedby the detergent and because the cleaning of buttermakingequipment must give the machine surfaces aprotective coating, which serves to prevent the butterfrom adhering to the surfaces. For cleaning, an internalcirculating flow is established.***) CIP of a UF plant is always carried out by means of aninternal circulating flow as special detergents are used76

in order to prevent any damage to the membranes,which would reduce the permeate flow.General Comments to Defects/Faults in CIPCleaningIn case of unsatisfactory cleaning, the following defects/faults may be the cause:1. CIP flow speed too low2. Cleaning time too short3. Cleaning concentration (lye or/and acid) too low4. Cleaning temperature too high/low5. Time of production without cleaning too long6. Etc.Manual CleaningCIP is automatic cleaning, but firstly the external surfacesare not cleaned by CIP, secondly there will always be a fewmachine parts that have to be cleaned every day. Futhermore,requirements for disassembling of large machineparts, a.o. plate heat exchangers and pipe connections,will arise at intervals.Dirty surfaces, e.g. due to leakage, must be cleaned everyday with hot soapy water and rinsed with clean water.Cleaning also includes the rooms, and plans for regularmanual cleaning of both rooms and equipment should beworked out.A visual control of the effectiveness of the cleaning maybe difficult. Although a surface seems clean, there may bea large number of bacteria per cm 2 .Check of the Cleaning EffectHygienic controlApart from the daily visual control with the hygienic conditionof the production equipment and the productionrooms, microbiological examinations should be made fordetermination of the state of cleaning effect, for instanceby means of the swabbing method.Equipment:1. Swabs made of cotton wool coiled around the end ofa small stick.77

2. Test tubes with 10 ml Ringer’s liquid.3. Ordinary equipment for bacteriological examinations.Procedure:1. The swab is sterilised in the test tube with Ringer’s liquid.2. Approx. 100 cm 2 (10 x 10 cm) of the surface to be examinedare rubbed with the swab.3. The swab is transferred to the test tube (1) again, andthe upper part of the stick, which has been touched, isbroken off.4. Dependent on the degree of pollution, 1 ml or 0.1 ml,maybe 0.01 ml is transferred to a sterile Petri dish, andsubstrate is poured on according to the type of bacteriato be examined.After incubation, the state of the cleaning effect is judgedafter the following scale:Number of total bacteria2per 100 cm surfaceState of cleaning effect0-10Very good10-100Goodover100BadControl of the cleaning liquids and temperatureNaturally, it is important to keep the right strength in thecleaning agents and the right temperature.The mentioned guiding figures may be summarised here:ConcentrationTemperature85- 9o CRoom temperaturRoom temperatur60– 6o C70– 7o CHotwater0C oncentrated acid 0 or 60 - 62%C oncentrated lye0-33%A cid cleaning solution .8 - 1.2%L ye cleaning solution .8 - 1.5%3 e3 e0 50 5Control of the strength of the cleaning agents should bemade twice a day.78

Emptying of the tanks will be necessary at intervals dependingon fouling and may take place by opening thebottom valves manually.Control of Cleaning SolutionsDetermination of the strength of lye by titrationIn order to obtain a satisfactory cleaning effect it is importantthat during the whole course of cleaning the lyesolution keeps the right strength according to the directionsfor use.Equipment:1. Titration burette (25 ml)2. 10 ml pipette or measuring glass3. Drop bottle4. Phenolphthalein solution (2%)5. Titration flask 100 ml6. 0.1 N hydrochloric acid.Method:1. Hot cleaning solution is removed from the lye tank witha ladle, and the solution is cooled to approximately20 o C.2. 10 ml lye solution is measured with a measuring glassor a pipette, and this solution is transferred to a flask.3. Five drops of phenolphtalein solution are added, bywhich the lye solution is coloured red.4. Under careful shaking this is titrated with 0.1 ml normalhydrochloric acid until the colour changes. The colourchanges from red to colourless.5. Number of ml consumed of 0.1 normal acid is read onthe burette and corresponds to the titer of the lye solution.The titer of the lye solution corresponds to the concentrationof the cleaning solution.79

The concentration in the cleaning solution can be calculatedas follows:Concentration in %: a x b x c = xx.x %100Where:a = ml titration fluid until colour change/10 ml solutionb = normality of titration fluid (0.1)c = molecular weight (NaOH = 40.0)Example:Concentration in % 25.0 x 0.1 x 40.0 = 1.00 %100Determination of the strength of the acid by titrationAcid cleaning solutions containing nitric acid (technicallyclean, approximately 62%) are used at the dairies withmechanical cleaning of pipes and tanks of completelystainless material. Acid solutions dissolve calcium oxidecoatings, and lye solutions dissolve protein coatings. Thisis why combined cleaning is used, e.g. lye solution at first,then acid solution, or in reverse order, depending on whichcleaning technique gives the best result on the spot.Equipment:1. Titration equipment (see under lye solution).2. 0.1 N sodium hydroxide.Method:1. The acid solution is removed from the acid container,and this solution is cooled to approximately 20 o C.2. 10 ml acid solution is measured with a measuring glassor a pipette, and this solution is transferred to a titrationflask.3. Five drops of phenolphtalein solution are added.4. Under careful shaking this is titrated with 0.1 normalsodium hydroxide until the colour changes. The colourchanges from colourless to red.80

5. Number of ml consumed of 0.1 normal lye is read onthe burette and corresponds to the titer of the acid solution.The titer of the acid solution corresponds to the concentrationof the cleaning solution.The concentration in the cleaning solution can be calculatedas follows:Concentration in %: a x b x c = xx.x %100Where:a = ml titration fluid until colour change/10 ml solutionb = normality of titration fluid (0.1)c = molecular weight (HNO 3 = 63.02)Example:Concentration in % 15.9 x 0.1 x 63.02 = 1.00 %100In order to make the calculation easier it is possible towork out tables for the lye or acid strength and titer, e.g.from 0.1%-2% so that it is possible to read the lye or acidstrength directly.(see Table: Concentration of Cleaning Solution)To compare the strength of the cleaning solution and theconductivity measured in milli-siemens mS please look inthe manual of Henkel P3-LMIT 08.81

Concentration of Cleaning SolutionLyeNaOHSodium HydroxideTitration0.1 n 30%HCL NaOHml/10 ml l/100 l02.5 .2505.0 .5007.5 .7510.0.0012.5.2515.0.5017.5.7520.0.0022.5.2525.0.5027.5.7530.0.0032.5.2535.0.5037.5.7540.0.0042.5.2545.0.5047.5.7550.0.0082ConcentrationAcidHNO3Nitric acidTitration%30% 62% 0.1HNO3 HNO3 nNaOHl/100 l l/100 l ml/10 ml0 0.1 0.300.1001.600 0.2 0.550.2503.200 0.3 0.850.3504.801 0.4 1.150.4506.301 0.5 1.400.6007.901 0.6 1.700.7009.501 0.7 2.000.8011.102 0.8 2.250.9512.702 0.9 2.551.0514.302 1.0 2.801.1515.902 1.1 3.101.3017.503 1.2 3.401.4019.003 1.3 3.651.5020.603 1.4 3.951.6522.203 1.5 4.251.7523.804 1.6 4.501.8525.404 1.7 4.802.0027.004 1.8 5.102.1028.604 1.9 5.352.2030.105 2.0 5.652.3531.70Dairy EffluentIncreasing discharge costs make it important to haveknowledge of both the quantity of effluent and the contentof pollutants. The pollutants in dairy effluent are primarilythe organic substances fat, protein, and lactose, but nitrateand phosphate are also important substances.Two methods are used to determine the content of organicmaterial in effluent: BOD and COD. The result isexpressed in mg oxygen per litre.BOD (Biological Oxygen Demand) is determined by the demandof dissolved oxygen for oxydising the organic materialin an aqueous sample of the effluent in 5 days at 20°C.COD (Chemical Oxygen Demand) is determined by treatinga sample with a potassium dichromate solution andneutralising excess dichromate by titration with ferrousammonium sulphate.

It is not possible to convert BOD directly to COD as thevalues for the two methods are dependent on the varyingcomposition of the organic matter. For dairy effluent thefollowing conversion can be used as a guideline:1 mg BOD = 1.3-1.5 mg COD1 mg COD = 0.75-0.67 mg BODThe table below lists COD values and thus the “pollutiondegree” of whole milk, skimmilk, and whey:SubstanceWholemilkContentmg/lSkimmilkmg ContentCOD/kg mg/lWheymg ContentCOD/kg mg/lmgCOD/kgFat40,000120,00000,40001 ,20000,40001,200Protein34,000046,00034,00046,24010,00013,600Lactose46,000052,00047,00053,11047,00053,110Total,approx.220,000100,00070,000A term often used to describe the “pollution degree” is“person equivalent” (p.e.). One p.e. corresponds to 250 lof water polluted to a COD value of 600. In other words, 1p.e. corresponds to 250 x 600 = 150,000 mg COD.Example:A dairy receives a daily quantity of 300,000 litres of milk.The loss is estimated to be 1%, ie, 3,000 l/day.COD: 3,000 x 218 = 4,360 p.e.150,000Or, in other words, effluent pollution equal to the pollutionfrom 4,360 people.83

TECHNICAL INFORMATIONStainless Steel PipesCapacity, friction loss and velocity of flow1¼" 1½" 2" 2½" 3" 4" 5" 6"6 7 8,000 100,000 1,000,000Capacity l/h84

Example:10,000 l/h in a 2” stainless steel pipe.Velocity: 1.5 m/sec.Friction loss: 5.5 m H 2 O per 100 m pipe.When pipe dimensions are determined, the water velocitymust not exceed 3 m/sec in small pipeline dimensionsup to about 3”. However, in bigger pipeline dimensions. avelocity of up to 3.5 m/sec. might be accepted.In milk lines, especially for unpasteurised milk, with pipedimensions below 3”, the velocity should not exceed 1.5m/sec. in the suction line and 2 m/sec. in the pressurelines. As concerns pipe dimensions of 3” and 4”, a velocityof up to 2 and 2.5 m/sec. is acceptable, and for pipedimensions 5” and 6” or bigger even higher velocities canbe acceptedIn pipelines for cream (40% fat) and other viscous dairyproducts, the velocity should be kept at a lower level. Forspecial products like fermented milk products, the velocityshould be kept at only 25-40% of the levels for milk.Friction Loss Equivalent in m Straight StainlessSteel Pipe for One FittingFittingNominaldiam.25mm38mm51mm63.5mm76mm101.6mmValve(two-way) 6 8 8 9 1010Valve(three-way) 7 9 9 101212Elbow0.8 1 1 1 1.5 1. 5Tee2 3 3 4 5 5The figures for pressure loss taken from the diagram arefairly good approximations for liquids having viscositiesbelow 5 cPs, such as water, whole milk and skimmilk.Velocity in Stainless Steel PipesThe velocity in stainless steel pipes should not exceedthe values (in m/sec.) stated below:ProductSuctionlines25mm ø 01.6 mm øMilk1.5. 0Cream1.5. 5Water3.0. 01 ø2 01 03 0Pressure lines25mm 101.6 mm ø2.2. 52.2. 03.3. 585

For CIP cleaning, the velocity should not be less than 1.5m/sec.Volume in Stainless Steel PipesOutsidediameter InsidediameterLitre/metre025.0mm022.6mm00.4011038.0mm035.6mm00.9954051.0mm048.6mm01.8551063.5mm060.3mm02.8558076.0mm072.9mm04.1739101.6mm097.6mm07.4815129.0mm125.0mm12.2718154.0mm150.0mm17.671586

Friction Loss in m H 2 O per 100 m Straight Pipewith Different Pipe Dimensions and Capacities(Non-stainless steel)Small figures: Velocity in metres per second.Large figures: Loss of head in m H 2 O per 100 m pipe.A: Friction loss in 90°C elbow or sluice valve indicated inmetres of straight pipe.B: Friction loss in Tee or non-return valve indicated in metresof straight pipe. (For foot, valves, multiply by 2).Friction loss: pipe length in metres x figures from table100 (metre head)Q uantity of waterNominal diameter in inches and inside diameter in m mm ³/h l /min . l/sec .0 .6 1 0 0.1 60 .9 1 5 0.2 51 .2 2 0 0.3 31 .5 2 5 0.4 21 .8 3 0 0.5 02 .1 3 5 0.5 82 .4 4 0 0.6 73 .0 5 0 0.8 33 .6 6 0 1.0 04 .2 7 0 1.1 24 .8 8 0 1.3 35 .4 9 0 1.5 06 .0 1 0 0 1.6 77 .5 1 2 5 2.0 89 .0 1 5 0 2.5 01 0.5 1 7 5 2.9 21 2 2 0 0 3.3 3½”15.750.8559.9101.28220.111.71033.532.13849.932.56569.342.99391.54¾”21.250.4702.4070.7054.8620.9408.0351.17411.911.40916.501.64421.751.87927.662.34941.402.81957.743.28876.491"27.00.2920.7840.4381.5700.5842.5880.7303.8340.8765.2771.0226.9491.1688.8201.46013.141.75118.282.04324.182.33530.872.62738.302.91946.493.64970.411¼”35.750.2490.4160.3310.6770.4151.0040.4981.3790.5811.8110.6642.2900.8303.4030.9964.7181.1626.2311.3287.9401.4949.8281.66011.902.07517.932.49025.112.90433.323.31942.751½”41.250.2490.3460.3120.5100.3470.7000.4360.9140.4491.1600.6231.7190.7482.3750.8733.1320.9973.9881.1224.9271.2475.9721.5588.9671.87012.532.18216.662.49321.362"52.500.2310.2230.2690.2910.3080.3680.3850.5440.4620.7510.5390.9880.6161.2540.6931.5510.7701.8750.9622.8021.1543.9031.3475.1791.5396.6242½”68.000.2290.1590.2750.2180.3210.2870.3670.3630.4130.4490.4590.5420.5740.8090.6881.1240.8031.4880.9181.9013"80.250.2310.1310.2630.1640.2960.2030.3290.2440.4120.3650.4940.5060.5760.6700.6590.8553½”92.500.2480.1240.3100.1850.3720.2560.4340.3380.4960.4314"105.00.2410.1010.2890.1400.3370.1840.3850.2345"130.00.2510.0841 5 2 5 0 4.1 74.14964.863.11732.321.92410.031.1472.8600.8231.2820.6200.6460.4810.3500.3140.1266"155.587

1 0.5 1 7 5 2.9 21 2 2 0 0 3.3 32.90433.323.31942.752.18216.662.49321.361.3475.1791.5396.6240.8031.4880.9181.9010.5760.6700.6590.8550.4340.3380.4960.4310.3370.1840.3850.2340.2510.0841 5 2 5 0 4.1 71 8 3 0 0 5.0 02 4 4 0 0 6.6 73 0 5 0 0 8.8 33 6 6 0 0 10. 04 2 7 0 0 11. 74 8 8 0 0 13. 35 4 9 0 0 15. 06 0 1 00 0 16. 77 5 1 25 0 20. 89 0 1 50 0 25. 01 05 1 75 0 29. 21 20 2 00 0 33. 31 50 2 50 0 41. 71 80 3 00 0 50. 02 40 4 00 0 66. 73 00 5 00 0 83. 34.14964.863.11732.323.74045.524.98778.17A 1 . 0 1 . 0 1 . 1 1 . 2 1 . 3 1 . 4 1 . 5 1 . 6 1 . 6 1 . 7 2 . 0 2. 5B 4 . 0 4 . 0 4 . 0 5 . 0 5 . 0 5 . 0 6 . 0 6 . 0 6 . 0 7 . 0 8 . 0 9. 01.92410.032.30914.043.07824.043.84836.714.61851.841.1472.8601.3774.0091.8366.8282.29510.402.75314.623.21219.523.67125.204.13031.514.58938.430.8231.2820.9681.7921.3173.0531.6474.6221.9766.5052.3068.6932.63511.182.96513.973.29417.064.11726.104.94136.970.6200.6460.7440.9030.9921.5301.2402.3151.4883.2611.7364.3561.9845.5822.2326.9832.4808.5213.10013.003.72018.424.34024.764.96031.940.4810.3500.5770.4880.7700.8290.9621.2541.1551.7571.3472.3451.5403.0091.7323.7621.9254.5952.4067.0102.8879.8923.36813.303.85017.164.81226.260.3140.1260.3770.1750.5020.2940.6280.4450.7530.6230.8790.8311.0051.0661.1301.3281.2561.6161.5702.4581.8833.4682.1974.6652.5116.9953.1399.2163.76713.055.02322.720.2630.0740.3510.1240.4390.1870.5260.2600.6140.3470.7020.4450.7900.5550.8770.6741.0971.0271.3161.4441.5351.9341.7542.4962.1933.8072.6325.4173.5098.9264.38614.4288

Units of MeasureThe MKSA SystemThe unit of weight is one kilogramme (kg).The unit of force is one kilogramme-force (kgf).In certain countries the designation kilopond (kp) is used.1 kp = 1 kgf.The unit of length is one metre (m).The unit of time is one second (s).The unit of temperature is one degree Celsius (IC).The terminal unit is one kilocalorie (kcal).One kilocalorie (kcal) is equal to the amount of heat requiredto heat or cool 1 kg water one degree Celsius.The specific gravity (density) is equal to the weight ingrammes (g) of one cubic centimetre (cm 3 ) of a substance.The unit of work, one kilogramme-force metre (kgfm) isequal to the energy required to raise one kilogramme to aheight of one metre.The unit of effect, one horse power (hp), is equal to a workperformance of 75 kilogramme-force metres per second(kgfm/s).One horse power hour (hph) is equal to the work that canbe carried out by one horse power (hp) in one hour.Specific heat is equal to the number of kilocalories requiredto heat 1 kg of a substance 1°C.Example: water 1iron 0.114copper 0.09air 0.24The latent heat of fusion is equal to the number of kilocaloriesrequired to change I kg of solid substance to liquidwhen it has previously been heated to melting point.Example: ice 8089

The thermal conductivity coefficient is equal to the numberof kilocalories that are transmitted in one hour through a1 m² cross section of a 1 m thick plate when the temperaturedifference is 1°C.The latent heat of evaporation is equal to the number ofkilocalories necessary to change 1 kg of liquid to vapourof the same temperature.Example: water at 100°C: 607water at 100°C: 536The degree of humidity, relative humidity, is equal to therelation between the actual water vapour content of theair, and the amount of water vapour the air can hold at thetemperature in question.The absolute humidity is equal to the weight in grammesof the water vapour contained in 1 cubic metre of air.The dew point is equal to the temperature reached whenair is cooled to saturation point.A technical atmosphere, 1 at, is equal to a pressure of:(1) 1 kgf per cm²(2) a 10 m column of water (H 2 O) at 0°C, or(3) 73.6 em mercury (Hg).1 ata is absolute pressure,1 ato is the pressure above atmospheric pressure (i.e. 1ato = 2 bar).A normal atmosphere, 1 atm, is equal to a pressure of:(1) 1.033 kgf/cm²(2) 1013 millibars of 76.0 cm mercury (Hg).The unit current intensity, one ampere (A), is equal to acurrent which, when passed through a solution of nitrateof silver, is capable of depositing silver at the rate of 1.118milligrammes per second.The unit of resistance, one ohm (Ω), is equal to the resistancein a column of mercury, 106.3 cm long and with across section of 1 mm², at a temperature of 0°C.The unit of potential, one volt (V), is equal to the differencein electrical potential between two separate points90

on a conductor with a resistance of 1 ohm, and where theelectric current is one ampere.The unit of power, one watt (W), is equal to the energy producedwhen the strength of the electric current is I ampereand the potential difference 1 volt.The unit of electric energy, one kilowatt hour (kWh) is equalto the energy that is (produced or used) by 1 kilowatt (kW)working for 1 hour (h).Conversion TablePower, heat flow ratehpkgfm/sIW3.8kcal/ h638.40.86hp*)1 757 62-2kgfm/s1.33x101 9 13-3W 1.36x100.1021 0-3kcal/h1.58x100.1191.161Energy, work, quantity of heathphkgfmkWh-52.70x1.736-1 .75x100.367x16-427.16x10kcal63.34x186hph1 0 0 2kgfm63.75x1026kWh1.36-0 1 0kcal31.58x101-31* metric- -32 0The SI Unit SystemSI (Système International d’Unités) is a metric systemof international units which lends itself to simplificationand systemisation. The SI system is gaining popularitythroughout the world and forms the basis of the first trulyinternational system of measurement. Such units as metre,kilogramme, litre, etc, will eventually be used worldwide.There is a definite advantage in applying the sameunits for all sizes, irrespective of the area measured. Forexample, the unit of power (Watt) can be used for electricmotors and combustion engines. Horsepower will graduallydisappear from the language. Thanks to uniformityand systemisation, no conversion factors will be requiredunder the SI unit system.SI includes a range of basic units, derivatives, multiplesand sub-multiples. There are also supplementary units,primarily associated with subdivision of the 24-hour day.91

Basic SI units:Length ........................... (m) metreMass. ............................ (k) kilogramTime ............................. (s) secondElectric current ..................... (A) ampereThermodynamic temperature .......... (K) kelvinLuminous intensity .................. (cd) candelaAmount of substance ................ (mol) moleSupplementary units:Plane angle. ....................... (rad) radianSolid angle ........................ (sr) steradianThe table below can be used to convert MKSA units usedin this booklet and other common units to SI units.Force newton N kg x m/s²WorkEnergy joule J kg x m²/s²= N x m = W x sQuantity of heatPower watt W kg x m²/s³ = J/sPressurepascal Pa N/m²bar bar 10 5 Pa92

Tables showing conversion Factors between SIUnits and other Common Unit Systems.Example showing use of pressure/stress table:1450 p.s.i. converted to bar?Find factor for bar, line p.s.i. = 16.9 x 10 -2 x 1450 ~ 100 barLengthSIunitmin(inch)ft(foot)Otherunitsyd(yard)mile1 3 9. 43 .2 81 .0 9.621 x 1 022 .54 x 10- 1 .33 x 1 00328 - 2 .77 x 1 0 - 21 5.8 x 1 0 60 .3051 21 0 .33 3.189 x 1 0030 .9143 63 1 .568 x 1 0033 3 3 31 .161 x 106 3.4 x 1 05 .28 x 1 01 .76 x 1 01AreaSIOther unit sunitm 2 i n2f t2(square inch)(square foot )y d2(square yard )31 1 .55 x 1 01 0. 81.2 030 .645 x 10- 1 .94 x 1 036 - 0 .772 x 1 0 3- 29 .29 x 101 4 41 0.11 130 .8361 .30 x 1 09 1-----93

VolumeSIOther unit sunitm 3 i n3f t3 y d3(cubic inch) (cubic foot ) (cubic yard )gallon(UK)gallon(US)1 61.0 x 1 061 6.4 x 10- 1 .579 x 1 022 .83 x 10- .73 x 1 033 5.31 .3 12 2 026 430 - 0 .214 x 1 0 - 63 .60 x 1 0 - 34 .33 x 1 0 331 1 .70 x 1 032-6 .2 37.4 830 .7654 6.7 x 1 02 71 1 6 820 2- 34 .55 x 102 7 70 .16 1.95 x 1 053- 33 .79 x 102 3 10 .13 4.95 x 1 043-1 1.2 0-0 .83 31VelocitySI unitm/sOtherunitsk m/hf t/ smile/ h1 3 . 63 .2 82.2 40 .2781 0 .91 10.62 10 .3051 .1 01 0.68 20 .4471 .6 11 .4 71-94

DensitySI unitkg/m(mass/volume)33g /cm ,g/mlOtherunits3 3lb/inlb/ft27.71 1 03- 3 6.1 x 1 0 - 66 .24 x 1 0 21 03 1 .61 x 1 0x1031 6.0 1.60 x 1 032-62. 42 7.71 1.73 x 1 0-25.79x10-313Force, weightOtherSI unitNkpunitslbf(pound force)1 0 .10 20.22 59 .811 2.2 14 .450 .45 41MassOtherunitsSIunitkgmetrictech.unit of masslb(pound)1 0 .10 22.2 19 .811 21. 70 .4544.63 x 1 0-21MomentofSI unitNmforceOtherunitsk pmlbf x f t1 0 .10 20.73 89 .811 7.2 31 .360 .13 81-95

Pressure,PaSI unitN /m2(pascal)stressbar2kp/cm at(tech. atmosph.)1 1 0 - 51 0.2 x 1 0 - 630 .10 27 .50 x 1 0 - 0 .145 x 1 0 31 05 31 1 .0 210.2 x 1 07 5014. 53398.1 x 100 .9811 10 x 1 07 3614. 269 .819 8.1 x 1 0 - 0 .1 x 1 0 - 31 7 .36 x 1 0 - 21 .42 x 1 0 3-3-3-21 331.33 x 1 01.36 x 101 3.61 1.93 x 1 03-2-26.90 x 106.90 x 107.03 x 107 035 1. 712S tandard atmosphere (atm), 1 atm = 101325 N/mOtherunitsm mH 2OmmHgtorrlbf/lnp.s.i.2Energy,J,work,SI unitNm,WsquantityofheatOtherk Whk p mkca lunits(Brit.Btuthermalunit)(footft x lbfpound-force)- 631 0 .278 x 1 00 .10 20 .239 x 1 0 - 0 .948 x 1 0 - 30.73 86633.6 x 101 0.367 x 1 08 603.41 x 1 02.66 x 1069 .812 .72 x 1 0 - 1 2 .34 x 1 0 - 39 29 x 1 0 - 37.2 33 - 334 .19 x 101 .16 x 1 04 2 71 3 .9 73 .09 x 1 03 - 31 .06 x 100 .293 x 1 01 0 80 .25 21 77 9-6-3-31 .360.377 x 1 00 .1380.324 x 1 01.29 x 1016,--96

Power,W,heatSI unitNm/s,flow rateJ/sOtherk pm/sk cal/ hBtu/ hunits(Brit.hphorsepower)(metr.hKhorsepower)1 0 .10 20 .86 03 .4 1.34 x 1 09 .811 8 .4 33 3. 5.32 x 1 01 .160 .11 91 3 .9 7.56 x 1 00 .293.99 x 1 022-0 .25 21 .393 x 1 031 - 1 .36 x 1 0 321 - 1 .33 x 1 0 231 - 1 .58 x 1 0 330 - 0 .399 x 1 0 337 467 6. 06 4 12 .55 x 1 01 1.0 17 .367 56 3 2.51 x 1 02360 .981----97

Input and Output of Electric MotorsAlternating current1 phase 3 phasesCurrent input (kW) =Mechanical output (hp)U x I x cos3 x U x I x cos10001000U x I x cos3 x U x I x cos736736U = Voltage; for thre-phase networks,U represents tension between two phasesI = Amperagecos ϕ: See table belown: See table below3 =1.73kW, hp and Full-load Current for 3x380 Volt, 50 CycleElectric Motors, and Approximate Values of cos j and n(at 1500 rpm)kWhpFull-loadcurrent cosϕ namp.0.370.5 1.0 0.7370. 50.550.751.450.7571. 00.751.0 1.850.7872. 01.11.5 2.6 0.8277. 01.52.0 3.4 0.8378. 02.23.0 4.9 0.8378. 03.04.0 6.3 0.8479. 03.75.0 7.8 0.8480. 04.05.5 9.0 0.8482. 05.57.5 11.5 0.8484. 07.510.0 15.0 0.8586. 011.015.0 22.0 0.8687. 015.020.0 29.0 0.8688. 018.525.0 36.0 0.8789. 022.030.0 42.0 0.8890. 030.040.0 56.0 0.9091. 037.050.0 69.0 0.8692. 045.060.0 83.0 0.8792. 055.075.0 104.0 0.8792. 075.0100.0 136.0 0.8792. 098

Fuel TableFuelCalorific valuekcal. kgPrice per tonDKKThermal efficiencyin boiler %Effective kcal.Price per 1000 effectivekcal. Ørekg steam per kg fuel(7 atm. abs.)Price per kg steam ØreLightfuel oil 9850338075739014.8911.209.82Heavy fuel oil(1500 sec.)*Heavy fuel oil(3500 sec.)977526357270409.5910.666.33975025137068259.5210.346.29Steam coal 7000167562434012.106.257.99Singles,Stoker 6800147569469010.347.116.82Screenedcoal 6500114055357510.775.427.10*) The viscosity measured in Redwood seconds at 100°F.1 kg steam at a pressure of 7 atm. abs. = 659.4 ~ 660kcal.In the part of the table dealing with oil-firing, the expensesof atomising the oil have not been considered.99

Saturated Steam Table(according to Mollier)AbsolutepressureAtmos.Temperature°CEnthalpykg°AbsolutepressureAtmos.Temperature°CEnthalpykg°0.1045.45617.0 02.5 126.79648. 30.2059.67623.1 03.0 132.88650. 30.3068.68626.8 03.5 138.19651. 90.4075.42629.5 04.0 142.92653. 40.5080.86631.6 04.5 147.20654. 70.6085.45633.4 05.0 151.11655. 80.7089.45634.9 05.5 154.72656. 50.8092.99636.2 06.0 158.08657. 80.9096.18637.4 06.5 161.21658. 71.0099.09638.5 07.0 164.17659. 41.1101.76639.4 07.5 166.97660. 11.2104.25640.3 08.0 169.61660. 81.3106.56641.2 08.5 172.13661. 41.4108.74642.0 09.0 174.53662. 01.5110.79642.8 09.5 176.83662. 51.6112.73643.5 10.0 179.04663. 01.7114.57644.1 12.5 188.92665. 11.8116.33644.7 15.0 197.36666. 61.9118.01645.3 17.5 204.76667. 72.0119.62645.8 20.0 211.38668. 5100

Atomic Weights, Melting and Boling Points of the ElementsNameSymbolnumber weight notes (°C)(°C)Atomic Atomic Foot-Melting point Boiling pointActiniumAc89227.028L 10503200±300AluminiumAl1326.9815660.372467AmericiumAm95( 243)994±4 2607Antimony(Stibium) Sb51121.75630.741750ArgonAr1839.948g,r - 189. 2 - 185. 7ArsenicAs3374.9216817(28 alm)613 (sub)AstatineAt85( 210)302337BariumBa56137.33g 7251640BerkeliumBk97(247)BerylliumBe4 9.012181278±5 2970 (5 mm)BismuthBi83208.980271.3 1560± 5BoronB 5 10.81m,r 20792550 (sub)BromineBr3579.904- 7. 2 58.78CadmiumCd48112.41g 320.9 765Caesium (Cesium) Cs55132.9052840±0.01669. 3CalciumCa2040.08g 839±2 1484CalifomiumCf98(251)CarbonC 6 12.011r , t 3652(sub)1CeriumCe58140.12g 7983443Cesium (Caesium) Cs55132.90542840±0.01669. 3ChlorineCl1735.453- 100.98- 34. 6ChromiumCr2451.9961857±202572CobaltCo2758.933214952870Copper(Cuprum) Cu2963.546r 1083.4±0.2 2567CuriumCm96( 247)1340±40DysprosiumDy66162.5014122567EinsteniumEs99(252)ErbiumEr68167.2615292868EuropiumEu63151.96g 8221527FermiumFm100(257)FluorineF 9 18.9984- 219.62- 188.14FranciumFr87( 223)( 27)(677)GadoliniumGd64157.25g 13133273GalliumGa3169.7229.782403GermaniumGe3272.59937.4 2830Gold(Aurum)Au79196.9671064.4342808± 2HafniumHf72178.492227±204602HeliumHe2 4.00260g - 272.226 atm- 268.934HolmiumHo67164.93014742700HydrogenH 1 1.00794g,m, r - 259.34- 252.87IndiumIn49114.82g 156.612080IodineI 53126.905113.5 184.35IridiumIr77192.2224104130Iron(Ferrum)Fe2655.84715352750KryptonKr368380g,m - 156. 6 - 152.30±0.10LanthanumLa57136.906g 9183464LawrenciumLr103(260)Lead(Plumbum) Pb82207.2 g,r 327.5021740LithiumLi3 6.941g,m, r 180.541342LutetiumLu71174.96716633402MagnesiumMg1224.305g 648.8±0.5 1090ManganeseMn2554.93801244±3 1962MendeleviumMd101(258)Mercury(Hydrargyrum) Hg80200.59- 38.87356.58MolybdenumMo4295.54g 26174612NeodymiumNd60144.24g 10213074NeonNe1020.1179g,m - 248.67- 246.048NeptuniumNp93237.048L 640±1 3902NickelNi2858.6914532732Niobium (Columbium) Nb4192.90642468±104742NitrogenN 7 14.0067- 209.86- 195. 8NobeliumNo102(259)OsmiumOs76190.2 g 3045±305027±100101

Atomic Weights,(continued)Melting and Boling Points of the ElementsNameSymbolnumber weight notes (°C)(°C)Atomic Atomic Foot-Melting point Boiling pointOxygenO 8 15..9994g,r - 218. 4 - 182.962PalladiumPd46106.42g 15543140PhosphorusP 1530.973844.1(white)280 (white)PlatinumPt78195.0817723827±100PlutoniumPu94( 244)6413232PoloniumPo84( 209)254962Potassium (Kalium) K 1939.098363.25759. 9PraseodymiumPr59140.9089313520PromethiumPm61( 145)10423000 (est.)ProtoactiniumPa91231.0359L 1600RadiumRa88226.025g,L 7001140RadonRn86( 222)- 71- 61. 8RheniumRe75186.20731805627 (est.)RhodiumRh45102.9061965±3 3727±100RubidiumRb3785.4678g 38.89686RutheniumRu44101.07g 23103900SamariumSm62150.36g 10741794ScandiumSc2144.955915412836SeleniumSe3478.96217684.9±1. 0SiliconSi1428.085514102355Silver(Argentum) Ag47107.868g 961.932212Sodium (Natrium) Na1122.989897.81±0.03882. 9StrontiumSr3887.62g 7691384SulfurS 1632.06r 112.8 444.674TantalumTa73180.947929965425±100TechnetiumTc43( 98)21724877TelluriumTe52127.60g 449.5± 0. 3 989.8±3. 8TerbiumTb65158.92513563230ThalliumTl81204.383303.5 1457±10ThoriumTh90232.038g,L 17503800 (approx.)ThuliumTm69168.93415451950Tin(Stannum)Sn50118.71231.96812270TitaniumTi2247.881660± 103287Tungsten(Wolfram) W 74183.853410± 205660U nnihexium( Unh)106(263)U nnilpentium( Unp)105(262)U nnilquadium( Unq)104(261)U nnilseptium( Uns)107(262)UraniumU 92238.029g,m 1132± 0. 8 3818VanadiumV 2350.94151890± 103380Wolfram (see Tungsten)XenonXe54131.29g,m - 111. 9 - 107.1 ± 3YtterbiumYb70173.048191196YttriumY 3988,905915525338ZincZa3065.39419.58907ZirconiumZr4091.224g 1852± 2 4377g geological exceptional specimens are known in which the element has an isotopic compositionoutside the limits for normal material. The difference between the atomic weightof the element in such specimens and that given in the Table may exceed the implieduncertainty considerably.m modified isotopic compositions may be found in commercially available material becauseif has been subjected to an undisclosed or inadvertent isotopic separation. Substantialdeviations in atomic weight of the element from that given in the Table may occur.rtrange in isotopic composition of normal terrestrial material prevents a more preciseatomic weight being given; the tabulated Ar (E) value should be applicable to any normalmaterial.triple point; (graphite-liquid-gas), 3627 ± 50°C at a pressure of 10.1 Mpa and (graphitediamond-liquid),3830 to 3930°C at a pressure of 12 to 13 Gpa.L Longest half-life isotop mass is chosen for the tabulated Ar (E) value.The atomic weights presented in the above Table are the 1981 atomic weights as presentedin Pure and Applied Chemistry, Vol. 55, No. 7, pp. 1101-1136, 1983.102

Prefixes with Symbols used in Forming DecimalMultiples and SubmultiplesNameSymbolFactor by which theunit is multipliedexaE 10petaP 10teraT 10gigaG 109megaM 106kilok 103hectoh 102decada10decid 10centic 10millim 10m icroµ 10nanon 10picop-1210femtof-1510attoa-1810The symbol representing the prefix is fixed to the unit symboland raises the latter to the stated power:Example: 12000 N = 12 x 10 3 N = 12 kN0.00394 m = 3.94 x 10 -3 m = 3.94 mm140000 N/m 2 = 140 x 10 3 N/m 2 = 140 kN/m 2or 1.4 x 10 5 N/m 2 = 1.4 bar0.0003 s = 0.3 x 10 -3 s = 0.3 ms103

Thermometric ScalesCelsius and Fahrenheit Degrees *)° C =/9 ( °F - 32° )° F = (°C x/5 + 32°° C ° F ° C ° F ° C ° F ° C ° F- 17.8 00.0 35095.0 074165.2 113235. 4- 15. 0 05.0 36096.9 075167.0 114237. 2- 10. 0 14.0 37098.6 076168.8 115239. 00- 5. 0 23.0 38100.4 077170.6 116240. 8- 10. 0 32.0 39102.2 078172.4 117242. 6- 11. 0 33.8 40104.0 079174.2 118244. 4- 12. 0 35.6 41105.8 080176.0 119246. 2- 13. 0 37.4 42107.6 081177.8 120248. 0- 14. 0 39.2 43109.4 082179.6 121249. 8- 15. 0 41.0 44111.2 083181.4 122251. 6- 16. 0 42.8 45113.0 084183.2 123253. 4- 17. 0 44.6 46114.8 085185.0 124255. 2- 18. 0 46.4 47116.6 086186.8 125257. 0- 19. 0 48.2 48118.4 087188.6 126258. 8- 10. 0 50.0 49120.2 088190.4 127260. 6- 11. 0 51.8 50122.0 089192.2 128262. 4- 12. 0 53.6 51123.8 090194.0 129264. 2- 13. 0 55.4 52125.6 091195.8 130266. 0- 14. 0 57.2 53127.4 092197.6 131267. 8- 15. 0 59.0 54129.2 093199.4 132269. 6- 16. 0 60.8 55131.0 094201.2 133271. 4- 17. 0 62.6 56132.8 095203.0 134273. 2- 18. 0 64.4 57134.6 096204.8 135275. 0- 19. 0 66.2 58136.4 097206.6 136276. 8- 20. 0 68.0 59138.2 098208.4 137278. 6- 21. 0 69.8 60140.0 099210.2 138280. 4- 22. 0 71.6 61141.8 100212.0 139282. 2- 23. 0 73.4 62143.6 101213.8 140284. 0- 24. 0 75.2 63145.4 102215.6 141285. 8- 25. 0 77.0 64147.2 103217.4 142287. 6- 26. 0 78.8 65149.0 104219.2 143289. 4- 27. 0 80.6 66150.8 105221.0 144291. 2- 28. 0 82.4 67152.6 106222.8 145293. 0- 29. 0 84.2 68154.4 107224.6 146294. 8- 30. 0 86.0 69156.2 108226.4 147296. 6- 31. 0 87.8 70158.0 109228.2 148298. 4- 32. 0 89.6 71159.8 110230.0 149300. 2- 33. 0 91.4 72161.6 111231.8 150302. 0- 34. 0 93.2 73163.4 112233. 6*) All temperatures in this booklet are in °C104

Conversion Table1 inchx 0002.5400= cm1 footx 0000.3048= m1 yardx 0000.9144= m1 milex 1609. 0000= m1 square inch x 0006.4520 = cm21 square footx 0000.09292= cm1 square yard x 0000.8360= cm21 acrex 4086.80002= cm1 cubic inchx 0016.3900= cm21 cubic footx 0028.3200= litre1 pint (liquid UK) x 0000.5680 = litre1 pint (liquid US) x 0000.4730 = litre1 UK quartx 0001.1360 = litre1 US quartx 0000.9460 = litre1 US gallonx 0003.7850 = litre1 UK gallonx 0004.5500= litre1 ouncex 0028.3500= g1 lbx 0000.4540 = kg1 short tonx 0907.1800= kg1 long tonx 1016.0600= kg1 pound per sq. inch x 0000.0700kg/cm=21 cmx 0000.3940 = inch1 mx 0003.2810 = foot1 mx 0001.0936= yard1 kmx 0000.6213= mile1 cm2x 0000.1550 = square inch1 m2x 0010.7640 = square foot1 m2x 0001.1970 = square yard1 hectarex 0002.4711= acre1 cm3x 0000.0610 = cubic inch1 m3x 0035.3200= cubic foot1 litrex 0001.7600= pint (liquid UK)1 litrex 0002.1100= pint (liquid US)1 litrex 0000.2640 = US gallon1 litrex 0000.2200= UK gallon1 gx 0015.4320 = grains1 kgx 0002.2046= lb1 tonnex 0001.1023= short ton1 tonnex 0000.9842= long ton1 kg/cm2x 0014.2200= pound per sq. inch° C =/9 ( °F - 32° )° F =/5 (°C + 32° )105

Notes

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