Technological process.

COLLECTION

Consulting materials

For 5th year students

TO GAK

Certification of practical skills.

Production of medicines in the conditions of pharmaceutical production.


UDC 615.014 (076.1)

The collection of consulting materials is compiled by:

Reviewers:

Approved at the CMS meeting

CMS Chairman

This collection includes typical production tasks and their solutions for pharmaceutical technology of finished dosage forms (factory technology of drugs), intended for conducting an exam for attesting the practical skills of graduate students at the SAC stage, as well as for fourth-year students.

INTRODUCTION

Certification of practical skills of graduate students in the Department of Drug Technology is one of the stages of IGAK. In this regard, certification of practical skills in the technology of finished medicines includes:

– ability to work with normative documents regulating the composition, manufacture and dispensing of medicinal products, knowledge of the modern range of medicinal and excipients;

– knowledge of the principles of design and operation of various production equipment for the manufacture of tablets, dosage forms in ampoules, soft dosage forms, phytoextraction preparations, etc. and the ability to justify the choice of the optimal equipment for the implementation of a particular production process;

– knowledge of the production processes carried out in the manufacture of finished medicinal products and the ability to theoretically substantiate their effectiveness and expediency of use;

– the ability to carry out calculations for the preparation of working prescriptions, loading norms, consumption norms, technical and economic indicators of the efficiency of the production process and material balance;

– the ability to choose ways of recycling production waste, to offer the best options for the implementation of industrial sanitation and environmental protection;

– the ability to justify the organization of the production flow, the choice of containers and packaging materials, the standardization of the received preparations and the conditions for their storage.

This collection contains typical production tasks and their solutions, including the main listed practical skills.

The problems from this collection can be used by students of the IV year in preparation for the exam.

Task 1

How many liters of 96.45% alcohol and water will be required to make 120 liters of 40.04% alcohol?

Decision

1) The amount of 96.45% alcohol in l is calculated by the formula:

x u003d P b / a u003d 120 40.04 / 96.45 u003d 49.82 l

2) The amount of 94.45% alcohol by mass

P = V r = 49.82 0.8136 = 40.53 kg

3) Weight 120 l 40.04% alcohol (P)

P u003d V r u003d 120 0.9480 u003d 113.76 kg

4) Amount of water

113.76 – 40.53 = 73.23 kg (l)

Answer : 49.82 liters of 96.45% alcohol and 73.23 liters of water.

Task 2

How many liters of 95% alcohol must be mixed with water to make 400 liters of 40% alcohol? What is the contraction?

Decision

1) According to tab. 4, GF XI, no. 1, p. 319 to obtain 1 liter of 40% alcohol, mix 421 ml of 40% and 607 ml of water. Calculation for 400 l:

1 l – 421 ml 40% sp. 1 l – 607 ml of water

400 l – x 1 400 l – x 2

x 1 u003d 168.40 l x 2 u003d 242.80 l of water

2) Contraction: (168.4 + 242.8) – 400 = 11.2 liters

Answer : 168.4 liters of 95% alcohol and 242.8 liters of water.


Task 3

Determine how many liters of 70% alcohol will be obtained by mixing 1000 liters of 95% with 391 liters of water. Solve according to the alcoholometric table.

Decision

1) Mass of 1000 liters of 95% alcohol (Table 3, GF XI, issue 1. p. 318)

P u003d V r u003d 1000 0.8114 u003d 811.4 kg

2) Mass of 70% alcohol

811.4 + 391 = 1202.4 kg

3) Volume 70% alcohol

V u003d P / r u003d 1202.4 / 0.8856 u003d 1357.72 l

4) Contraction

(1000 + 391) – 1357.72 = 3.28 liters

Answer: 1357.72 liters

Task 4

Determine how many liters of 96% and 10% alcohol must be mixed to get 500 liters of 40% alcohol. What is the contraction?

Decision

For the convenience of carrying out calculations, it is necessary to enter the known data in the table.

Volume % Weight % Density
96% 93.86% 0.8079
40% 33.33% 0.9480
ten % 7.99% 0.9848

1) Mass of 40% alcohol

P = V r = 500 0.9480 = 474.00 kg

2) Weight 96% alcohol

x u003d P (bc) / (ac) x u003d 474 (33.33 – 7.99) / (93.86 – 7.99) u003d 139.85 kg

3) Mass of 10% alcohol

474.00 – 139.85 = 334.15 kg

4) Volume 96% alcohol

V u003d P / r u003d 139.85 / 0.8074 u003d 173.21 l

5) Volume 10% alcohol

V u003d P / r u003d 334.15 / 0.9848 u003d 339.31 l

6) Contraction

(173.21 + 339.31) – 500 = 12.52 liters

Answer: the contraction is 12.52 liters, the volume of 96% alcohol is 173.21 liters, the volume of 10% alcohol is 339.31 liters.

Task 5

It is necessary to prepare 300 liters of 70% alcohol from 90% alcohol and recuperate with an absolute alcohol content of 5%. How many liters of 90% alcohol and recuperate should be used up? What is the contraction?

Decision:

Volume % Weight % Density
90% 85.68 0.8292
70% 62.36 0.8856
ten % 4.02 0.9910

1) Mass of 70% alcohol

P = V r P = 300 0.8856 = 265.68 kg

2) Mass of 90% alcohol

x u003d P (bc) : (ac) x u003d 265.68 (62.36 – 4.02) : (85.68 – 4.02) u003d 189.81 kg

3) Mass of 5% alcohol (recuperate)

265.68 – 189.81 = 75.87 kg

4) Volume 90% alcohol

V u003d P : r V u003d 189.81 : 0.8292 u003d 228.91 l

5) The volume of 5% alcohol (recuperate)

V u003d P : r V u003d 75.87 : 0.9910 u003d 75.56 l

6) Contraction

(228.91 + 75.56) – 300 = 4.47 liters

Answer: 228.91 liters of 90% alcohol and 75.56 liters of recuperate, contraction 4.47 liters.

Task 6.

Determine the mass of 230 liters of an alcohol-water mixture if at 20 ° C the glass alcohol meter drops to 82%. How many liters of anhydrous alcohol are contained in the alcohol-water mixture?

Solution .

1. The mass of 82% alcohol is:

r 82% u003d 0.8536 m u003d V * r u003d 0.8536 * 230 u003d 196.238 kg

2. How many liters of anhydrous alcohol are contained in the alcohol-water mixture?

Answer : the alcohol-water mixture contains 188.6 liters of anhydrous alcohol

m=196.33 kg.

Task 7

Prepare 320 liters of 30% ethanol, necessary to obtain liquid thyme extract, based on rectified alcohol with a concentration of 70% and recuperate alcohol with a concentration of 10%.

Solution .

1. r 30% alcohol u003d 0.9622% m30% alcohol u003d 24.64

r 10% alcohol u003d 0.9848% m10% alcohol u003d 7.99

2. Weight of 320 l 30% ethanol:

m=V 30% alcohol *r 30% alcohol = 320*0.9622=307.9 kg

3. Quantity kg 70% alcohol:

70% alcohol

4. Amount of 10% alcohol:

m 10% =307.9-94.15=213.75 kg

5. The volume of 70% alcohol:

6. Volume of 10% alcohol:

Answer : 106.312 liters of 70% alcohol and 217.049 liters of 10% alcohol.

Task 8

Received 800 ml of a solution of basic aluminum acetate with a density of 1.017. How much should a solution with a density of 1.063 be added to get a standard solution with a density of 1.048?

Decision

Calculation of the amount of solution with a density of 1.063:

When strengthening solutions, the calculation is carried out by weight (m = V r)

Density of solutions Quantity Quantity

solutions, ml solutions, g

(x) 1.063 0.031 0.031 1.063 = 0.0330 g

1.048

(813.6 g) 1.017 0.015 0.015 1.017 = 0.0153 g

1) m solution with a density of 1.017 = 800 1.017 = 813.6 g

2) 0.33 g – 0.0153 g.

x — 813.6

x u003d (0.031 1.063 800 1.017) : (0.015 1.017) u003d 1757.49 g

Answer: 1757.49 g or 1653.33 ml of a solution with a density of 1.063.


Task 9

How much should a solution of basic lead acetate with a density of 1.5 be taken to obtain 850 g of a solution with a density of 1.230?

Decision

Calculation of the amount of solution with a density of 1.5 :

The calculation is carried out by weight.

Density of solutions Quantity Quantity

solutions, ml solutions, g

1.5 0.230 0.23 1.5 = 0.345 g

1.230

1.0 (850 g) 0.270 0.27 1.0 = 0.270 g

(0.23 + 0.27) 0.500 0.5 1.23 = 0.615 g

1) 0.615 g – 0.345 g

850 g – x g

x u003d 850 0.345 : 0.615 u003d 476.83 g or 317.9 ml

Answer: 476.83 g or 317.9 ml of a solution with a density of 1.5.

Task 10.

Calculate how much a 70% calcium chloride solution must be mixed with a 45% solution to obtain 2 kg of a 50% solution.

Solution .

To solve, use the “asterisk” form, into which the initial data is substituted:

1. Calculation of the amount of 70% solution:

For 25 parts of a 50% solution – 5 parts of a 70% solution

for 2 kg of 50% solution – X kg of 70% solution

2. Calculation of the amount of 45% solution:

For 25 parts of a 50% solution – 20 parts of a 45% solution

for 2 kg of 50% solution – X kg of 45% solution

Answer : 0.4 kg of a 70% solution and 1.6 kg of a 45% solution.

Task 11.

Received 180 l of basic aluminum acetate with a density of 1.08. What amount of water must be added to obtain a preparation with a density of 1.048 g/cm3?

Decision.

1 way. Diluent calculation by volume:

V – x

2 way. Calculation according to the “asterisk” scheme:

0.048(1.08) – 0.032(water)

180s

Answer: 120 liters of water

Task 12.

Prepare 150 liters of camphor in oil for external use. The consumption coefficient is 1.05, the density of camphor oil is 0.93 g/cm 3. Give a description of the technological process.

Decision.

1. Calculation of the amount of camphor

for 1 liter – 100 g of camphor

for 150 l – X

2. Calculation of the mass of camphor oil

m=V camphor oil *r oil =150*0.93=139.5

3. Calculation of the amount of sunflower oil:

m camphor oil -m camphor u003d 139.5-15 u003d 124.5 kg

4. Calculation of the amount of initial ingredients, taking into account Crash

camphor – 15*1.05=15.75kg

oils – 124.5 * 1.05 u003d 130.725 kg.

Technological process:

In a steam boiler, sunflower oil is heated to 40-45°C, and with the mixer turned on, camphor is added, stirred until the camphor is dissolved. The solution is filtered under pressure. Standardization is carried out and the solution is packaged in vials.

Answer : camphor 15.75 kg; sunflower oil 130.725 kg.

Task 13.

Prepare 200 liters of camphor alcohol. The consumption coefficient is 1.02. Give a description of the technological process (the density of camphor alcohol is 0.886 g/cm 3 ).

Solution .

1. Calculation of the mass of camphor for the preparation of the solution:

for 1 liter of solution – 100 g of camphor

per 200 l of solution – X g of camphor

2. Calculation of the amount of 70% alcohol to prepare the solution:

mass of camphor alcohol

m=200*0.886=177.2kg

weight 70% alcohol

m camphor alcohol -m camphor u003d 177.2-20 u003d 157.2 kg

3. The composition of the working recipe:

camphor – 20 * 1.2 = 20.4 kg

alcohol 70% – 157.2*1.02=160.344kg

70% alcohol

Answer : 20 kg of camphor and 180.975 liters of 70% alcohol.

Technological process :

Camphor is weighed into a dry reactor and ethanol is added. The mixture is stirred until camphor is completely dissolved. The solution is filtered, standardized and packaged in bottles of 25; 50 ml.

Task 14

Calculate the amount of raw material and extractant to obtain 120 liters of lily of the valley tincture if the absorption coefficient is 2.0 cm 3 /g.

Decision

The raw material is lily of the valley grass, crushed to 8 mm (GF XI). The extractant is ethyl alcohol 70%. The ratio of raw materials and finished products is 1:10. Amount of raw materials: 120/10 = 12 kg.

The total amount of extractant to obtain the tincture is calculated by the formula:

V = V 1 + P K, where

V is the total volume of the extractant, l (ml)

V 1 – volume of tincture (finished product), l (ml)

P is the amount of plant material, kg (g)

K is the absorption coefficient.

V u003d 120 + 12 2 u003d 144 l

Answer : lily of the valley herb – 12 kg, ethyl alcohol 70% – 144 liters.

Task 15

Prepare 120 liters of hawthorn tincture by percolation if Kn = 2 cm 3 /g. Make a material balance for absolute alcohol if the yield is 98%. Alcohol content in tincture 65%

Decision:

The ratio of raw materials and finished products is 1:10.

Raw materials – hawthorn fruits 120 : 10 = 12 kg.

Extractant – ethyl alcohol 70%

V u003d 120 + (12 2) u003d 144 l

The amount of tincture obtained (taking into account the yield of products):

120 l – 100% x u003d (120 98) : 100 u003d 117.6 l of tincture

x – 98%

Used up Received
Name Absolute alcohol content l Name Absolute alcohol content l
Ethyl alcohol 70% – 144 l X u003d 144 70%: 100% u003d 100.8 l Hawthorn tincture 117.6 l with 65% alcohol X u003d 117.6 65: 100 u003d 76.44 l
Losses: 100.8 – 76.44 u003d 24.36 liters
Total: 100.8 l Total: 100.8 l

1) Output h u003d 76.44 100 : 100.8 u003d 75.83%

2) Waste e = 24.36 100 : 100.8 = 24.17%

3) To flow. = 100.8 : 76.44 = 1.31

Answer: the material balance equation has the form: output 75.83%, expenditure 24.17%, consumption coefficient – 1.31.

Task 16.

Prepare 100 liters of eucalyptus tincture by fractional maceration according to VNIIF, Kp = 1.5 cm 3 /g. Calculate the amount of extractant based on 84% alcohol.

Decision.

Raw materials – eucalyptus leaf, extractant – 70% ethyl alcohol, eucalyptus tincture is prepared in a ratio of 1:5.

Quantity of raw materials: M raw materials=100:5=20kg

The amount of extractant: V extractant u003d 100 + 20 * 1.5 u003d 130l

Recalculation of 70% alcohol from 84%:

1. The mass of 70% alcohol is:

m 70% u003d 0.8856 (r 70% alcohol) * 130 u003d 115.1 kg

2. The mass of 84% alcohol is:

m 84% u003d (115.1 kg * 62.36): 78.16 u003d 91.94 kg

Note: % by mass 70%=62.36

% by mass 84%=78.16

3. Amount of water: m water =115.1kg-91.94kg=23.16kg

4. Volume 84% alcohol:

5. Contraction (108.3+23.16)-130=1.46l.

Answer: to obtain 70% alcohol, you need to mix 108.3 liters of 84% alcohol and 23.16 liters of water.

Technological process:

The essence of the process lies in the sequential extraction of plant materials with fractional parts of fresh extractant. Under production conditions, 4 plums are made.

Extractant calculation for 1 bay:

100(number of finished products):4(number of drains)

Task 17.

Calculate the amount of raw materials and extractant for the preparation of 280 liters of St. John’s wort tincture, K p u003d 1.5 cm 3 / g. How much water and available 96% alcohol will be required to prepare the calculated amount of extractant?

Decision

Raw materials – St. John’s wort, extractant – 40% ethyl alcohol, St. John’s wort tinctures in a ratio of 1:5.

Amount of raw material: M=280:5=56kg

Extractant quantity: V=280+5.6*1.5=364l

1. The mass of 40% alcohol is:

m 40% u003d 0.9480 (r 40% alcohol) * 364 u003d 345.07

2. The mass of 96% alcohol is:

33.33*345.07=93.86*х

m 96% u003d (345.07 kg * 33.33): 93.86 u003d 122.53 kg

3. Amount of water:

m water u003d m 40% alcohol -m 96% alcohol

345.07kg-122.53kg=22.54kg

4. Volume 96% alcohol:

5. Contraction:

V 96% alcohol + V water -V 40% alcohol u003d (151.7 + 222.54) -364 u003d 10.24 l

Answer : to get 40% alcohol, mix 151.7 liters of 96% alcohol and 222.54 liters of water and get 364 liters of 40% alcohol

Task 18.

Calculate the amount of raw materials and extractant to obtain 100 liters of valerian tincture. Kp is equal to 1.3 cm 3 / g. Describe the preparation of tincture by the method of fractional maceration according to VNIIF. Make a material balance for absolute alcohol. The tincture contains 64% alcohol.

Decision.

Raw materials : rhizomes with valerian roots, extractant : 70% ethanol.

1. Calculation of the amount of raw materials:

Valerian tincture is prepared in a ratio of 1:5; y=5

m raw material =100:5=20kg

2. Calculation of the amount of extractant:

V extractant u003d m raw materials * (Kp + y) u003d 20 * (1.3 + 5) u003d 126 l of 70% ethyl alcohol

20 kg of raw materials are loaded into the percolator and poured:

3. Calculation of the amount of extractant for one bay:

100:4=25l;

25l + (20 * 1.3) u003d 51l 70% alcohol,

where 20 kg is the mass of raw materials.

Leave for a day. The presentation of the technological process is placed in the solution to problem No. 16.

material balance.

Used up Received
Name Absolute alcohol content, l Name Absolute alcohol content, l
Ethyl alcohol 70% 126l Absolute alcohol in l: 1. Tincture 100l (alcohol content 64%) 2. Losses Absolute alcohol in l: 24.2
TOTAL 88.2 TOTAL 88.2

Losses: 88.2-64=24.2l

;

Answer : rhizomes and roots of valerian 20 kg, ethyl alcohol 70% 126l;

h=72.56%; Cr=1.37; e=27.44%

Task 19.

Calculate the amount of raw materials and extractant to obtain 150 liters of belladonna tincture (Kp = 2 cm 3 /g). Give a description of the technological process by the method of percolation. Calculate the amount of extractant based on 91.4% ethyl alcohol.

Decision.

1. Raw material – belladonna leaf, extractant – 40% ethyl alcohol, phase ratio 1:10.

2. Calculation of raw materials: 150L:10=15kg

3. Calculation of the extractant: 150+(15*2) = 180l 40% alcohol

4. Calculation of the mass of 40% alcohol:

r 40% alcohol = 0.9480

m 40% alcohol u003d 0.9480 * 180 u003d 170.64 kg.

5. Calculation of the mass of 91.4% alcohol:

6. Calculation of water: M=m 40% alcohol – m 91.4% alcohol=170.64-65.02=105.62kg

7. Calculation of the volume of 91.4% alcohol:

r 91.4% alcohol = 0.8246;

8. Contraction: V 91.4% alcohol + V water – V 40% alcohol =

u003d 78.85 + 105.82-180 u003d 4.47 l

Answer : 78.85 liters of 91.4% alcohol and 105.85 liters of water.

Technological process:

1st day. 15 kg of raw materials are loaded into the percolator and 30 + 30 = 60 liters of extractant are poured and left for 24 hours for spreading.

2nd day. The percolation rate is set equal to 1/24 – 1/48 of the occupied working volume of the percolator and pure extractant is added at the same rate. Percolate until the depletion of raw materials, while spending the calculated amount of extractant. The resulting tincture is kept at a temperature not exceeding 10°C for at least 2 days, filtered, standardized and packaged.


Task 20.

From 20 kg of raw belladonna leaves with an alkaloid content of 0.36%, 200 liters of a standard tincture were obtained, with an alkaloid content of 0.033%. Draw up a material balance for active substances. Give a description of the process.

Solution .

material balance.

Used up Received
Name The content of alacloids, kg Name Content of alkaloids, kg
Belladonna leaves (alkaloid content 0.36%) – 10kg 100 – 0.36 20 – х х=0.072 1. Tincture of belladonna (content of alkaloids 0.033%) – 200l 2. Losses 100 – 0.33 200 – x x=0.066 0.006
TOTAL 0.072 TOTAL 0.072

Losses: 0.072-0.066=0.006kg

;

Answer : h=91.67%; Kp=1.09; e=8.33%

Technological process: 20 kg of raw materials are loaded into the percolator, filled with 40% alcohol to the mirror and left for a day to infuse. A day later, the percolation rate is set equal to 1/24 – 1/48 of the working volume of the percolator, and pure extractant is fed into the percolator at the same rate. Percolate until the depletion of raw materials, while spending the calculated amount of extractant. The resulting tincture is kept at a temperature not exceeding 10°C for at least 2 days, filtered, standardized and packaged.

Task 21.

Prepare 200 liters of motherwort tincture by maceration, Kp = 1.8 cm 3 /g. Make a material balance for absolute alcohol, provided that the yield of the tincture was 95%. The alcohol content in the tincture is 65%.

Solution .

1. Raw material – motherwort herb,

phase ratio 1:5, extractant – 70% ethyl alcohol

2. Calculation of the amount of raw materials:

motherwort tincture is prepared in a ratio of 1:5

200:5=40 kg

3. Calculation of the extractant:

200+(40*1.8)=272 l 70% alcohol

4. The output of the tincture was 95%

200 l – 100%

Xl – 95% X=190l

material balance.

Used up Received
Name Absolute alcohol content, l Name Absolute alcohol content, l
Ethyl alcohol 70% – 272 l 1. Motherwort tincture (alcohol content 65%) – 190l 2. Losses 66.9
TOTAL 190.4 l TOTAL 190.4l

Losses: 190.4-123.5=66.9

;

Answer : h=64.86%; Cr=1.54; e=35.14%

Task 22.

Prepare 150 l of valerian tincture by percolation, Kp = 1.2 cm 3 /g. Make a material balance for absolute alcohol, if the yield of tincture was 98%. The tincture contains 66% alcohol.

Solution .

1. Valerian tincture is prepared in a ratio of 1:5, so you need to take:

raw materials :

150:5=30 kg,

extractant (70% alcohol):

150+(30*1.2)=186l

2. The yield of the tincture was 98%, so the tinctures received:

material balance.

Used up Received
Name Absolute alcohol content, l Name Absolute alcohol content, l
Ethyl alcohol 70% – 186 l 1. Tincture of belladonna (alcohol content 66%) – 147l 2. Losses 33.18
TOTAL 130.2 l TOTAL 130.2 l

Losses: 130.2-97=33.18

;

Answer : h=74.52%; Cr=1.34; e=25.48%

Problem 23

Prepare 150 ml of liquid extract of nettle by the Chulkov repercolation method in five cycles, in a battery of 5 percolators. Calculate the amount of raw material and extractant, the load of one percolator, the duration of the process. Give a statement of the technological process within 6 days. Kn=3 cm 3 /g.

Decision

1. Calculation of the amount of raw materials.

Liquid extracts are prepared in a ratio of 1:1, which means that to prepare 150 liters of extract, you need to take raw materials: Q u003d V / y u003d 150/1 u003d 150 kg

V is the volume of finished products

y – coefficient of removal of finished products.

2. Calculation of the extractant for the entire process.

W u003d Q (Kn + y) u003d 150 (3 + 1) u003d 600 l

W is the volume of the extractant

Q – the amount of raw materials

y – coefficient of removal of finished products

Kn – absorption coefficient of raw materials cm 3 /g

3. Calculation of raw materials for one diffuser

G u003d Q / (n s) u003d 150 / (5 5) u003d 6 kg

G – the amount of raw materials

n is the number of diffusers

s – number of cycles

4. Calculation of the extractant for injection into one diffuser:

V 1 u003d G (Kn + y) u003d 6 (3 + 1) u003d 24 l

5. Calculation of raw materials for one cycle

G 1 u003d Q : c u003d 150 : 5 u003d 30 kg

6. Calculation of the extractant for one cycle:

W 1 u003d n G (Kn + y) u003d 5 6 (3 + 1) u003d 120

7. Calculation of the dose volume of the extract taken as a finished product:

a = G y = 6 1 = 6 l

8. How many days will the process take:

t = c n+n = 5 5+5 = 30 days

t – time

s – number of cycles

n is the number of percolators

1 day.

We load 6 kg of cut nettle leaves into the first percolator, compact it and fill it with 50% alcohol with the tap open to force out the air. Then we close the tap, pour the drained liquid into the percolator and add the remaining alcohol (24 liters in total). We leave for a day.

Day 2

We load the second percolator with raw materials. Open the tap of the I percolator, drain 6 liters of the extract and transfer it to the II percolator. Pour 24 liters of 50% ethanol into the first percolator, drain 18 liters and transfer to the second percolator. We leave for a day.

Day 3

We load 6 kg of raw materials into percolator No. 3. From percolators No. 1 and No. 2 we drain 6 liters of extracts. We make the movement of extractions. We transfer the extraction from percolator II to III, from I to II. Pour 24 liters of 50% alcohol into the I percolator, drain 18 liters and transfer to II, drain 18 liters and transfer to III. We leave for a day.

Day 4

We load the IV percolator with raw materials. We open the taps of three percolators and drain 6 liters of extracts. We carry out the movement of extractions. From I to II, from II to III, from III to IV. Pour 24 liters of 50% alcohol into the first percolator and drain 18 liters, transfer to II. From II we drain 18 liters and transfer to III, from III we drain 18 liters and transfer to IV. We leave for a day.

Day 5

We load the V percolator with raw materials. From percolators I, II, III and IV we drain 6 liters of extracts. We move extracts from IV to V, from III to IV, from II to III, from I to II. In the first, pour 24 liters of 50% alcohol, drain 18 liters and transfer to II, from II to III, from III to IV, from IV to V. Leave for a day.

Day 6

We open the taps of all five percolators and get 6 liters of extracts. Extraction from V percolator – finished product. Since the pure extract entered the first percolator, the raw material in it was depleted. We carry out the distillation of alcohol in the first percolator and load it with fresh raw materials. We move the extracts: from I to II, from II to III, from III to IV, from IV to V. We transfer the extract from V to I percolator. Pour fresh extractant into percolator II 24 l, drain 24 l and transfer to III, drain 24 l and transfer to IV, drain 24 l and transfer to V, drain 18 l from V and transfer to percolator I. We leave for a day.

Answer: raw material – 150 kg, extractant – 600 l, loading of one percolator – 6 kg, process duration – 30 days.

Task 24.

Prepare 120 liters of liquid extract – adonis concentrate 1:2. Give a description of the process. Calculate the raw material and the extractant if Kp=1.5 cm 3 /g. Calculate the amount of extractant based on 86.6% ethanol.

Solution .

1. Raw material – crushed herb adonis.

m raw material =120/2=60kg

2. Extractant – ethyl alcohol 25%

V extractant u003d 120 + 60 * 1.5 u003d 210 l

r25 % =0.9682; r86.6 % =0.8402

m 25% alcohol u003d 210 * 0.9682 u003d 203.32 kg

3. The volume of 86.6% ethanol:

4. Mass of 86.6% ethanol:

m 86.6% u003d V 86.6% alcohol * r 86.6% alcohol u003d 60.62 l * 0.8402 u003d 50.93 kg

5. Water volume:

V water u003d m water u003d m 25% alcohol – m 86.6% alcohol u003d 203.32-50.93 u003d 152.39 l

Answer : To get 210 liters of 25% ethanol, you need to mix 60.62 liters of 86.6% ethanol and 152.39 liters of water.

Technological process :

Extraction is carried out in a battery of 3 percolators, each is loaded with 60:3=20 kg of adonis grass. In the 1st percolator, 70 liters of 25% alcohol are poured (120: 3 + m of raw materials * K p u003d 40 + 20 * 1.5)

After 2 hours, 40 liters of extract are drained and the extract is transferred to the 2nd percolator. 70 l of fresh extractant is poured into the 1st percolator and 30 l of the extract is drained, which is transferred to the 2nd percolator. The percolators are left to infuse.

After 2 hours, 40 liters of extract are drained from the 1st and 2nd percolators. The extract from the 2nd percolator is transferred to the 3rd percolator, from the 1st to the 2nd. AT
The 1st percolator is filled with 70 liters of fresh extractant, 30 liters of extract are drained and transferred to the 2nd percolator, 30 liters of extract are drained from the 2nd percolator and transferred to the 3rd.

A day later, the taps of all 3 percolators are opened and 40 liters of extract are drained, the extract from the 3rd percolator is a finished product. The extract from the 2nd percolator is transferred to the 3rd, from the 1st to the second. The percolators are left to infuse for 2 hours.

After 2 hours, the 2nd portion of the finished product is drained from the 3rd percolator. We transfer the extract from the 2nd percolator to the 3rd. The percolators are left to infuse for 2 hours.

After 2 hours, the third, last portion of the finished product is drained from the 3rd percolator. All three extracts are combined, settled at a temperature not exceeding 10°C for 2 days, filtered. Percolators are unloaded, alcohol is regenerated.

Task 25.

Prepare 160 liters of liquid water pepper extract by the Chulkov method in five cycles, in a battery of 4 percolators. Calculate the amount of raw material and extractant, the load of one percolator, the duration of the process. Give a description of the process. within 5 days. K p u003d 2.2 cm 3 / g.

Decision.

1. Liquid extract of water pepper is prepared in a ratio of 1: 1, so the raw materials must be taken:

m raw materials u003d 160: 1 u003d 160 kg

2. The volume of the extractant (70% ethanol) is

V extractant u003d m raw materials (Kp + y) u003d 160 + (160 * 2.2) u003d 512l

3. The mass of raw materials loaded into one percolator is:

m raw material : number of cycles : number of percolators=(160:5):4=8kg

4. The duration of the process is:

number of percolators * number of cycles + number of percolators=4*5+4=24 days

Technological process.

1st day. 8 kg of water pepper grass is loaded into the 1st percolator and 8+(8*2.2)=25.6 l of 70% alcohol is poured. The percolator is left for a day to infuse.

2nd day. From the 1st percolator, 8 liters of extract are drained, transferred to the 2nd percolator, where 8 kg of raw materials are previously placed.

25.6 liters of fresh extractant are poured into the 1st percolator and 17.6 liters of the extract are drained, which is transferred to the 2nd percolator. Percolators leave for a day to infuse.

3rd day. From the 1st and 2nd percolators, 8 liters of extract are drained, the extract from the 1st percolator is transferred to the 2nd, from the 2nd to the third, where 8 kg of raw materials are preliminarily loaded.

25.6 liters of pure extractant are poured into the 1st percolator, 17.6 liters of the extract are drained, which is transferred to the 2nd percolator. From the 2nd percolator, 17.6 liters of extract are drained, transferred to the 3rd percolator. Percolators leave for a day to infuse.

4th day. Drained from 3 percolators, 8 liters of extract. Load 8 kg of raw materials into the 4th percolator. The extract from the 1st percolator is transferred to the 2nd, from the 2nd to the 3rd, from the 3rd to the 4th. 25.6 liters of pure extractant are poured into the 1st percolator and 17.6 liters of extract are poured, which are transferred to the 2nd percolator, 17.6 liters of extract are drained from the 2nd percolator and transferred to the 3rd, 17 are drained from the 3rd, 6l extract and transfer to the 4th. Percolators leave for a day to infuse.

5th day. Drain 8 liters of extract from each percolator. The extract from the first percolator is a finished product. The extract from the 1st percolator is transferred to the 2nd, from the 2nd to the 3rd, from the 3rd to the 4th.

The first percolator is unloaded and loaded with fresh raw materials (8 kg of water pepper grass). 25.6 l of fresh extractant is poured into the 2nd percolator, 17.6 l of the extract is drained, which is transferred to the 3rd percolator, from the 3rd percolator 17.6 l of the extract is transferred to the 4th percolator, from the 4th percolator 17.6 l extracts are transferred to the 1st percolator. Percolators leave for a day to infuse.

Task 26.

Prepare 120 liters of hawthorn liquid extract using the Chulkov method. Use a battery of 3 percolators and 4 cycles. Kp is equal to 2 cm 3 / g. Determine the load of percolators, the amount of raw materials and extractant, give a description of the technological process from the 1st to the 6th day.

Solution .

Liquid extract of hawthorn is prepared in a ratio of 1:1 in 70% ethyl alcohol.

1. Quantity of hawthorn fruits (raw materials): 120:1=120 kg.

2. Amount of raw material to load one single percolator: 120(3*4)=10kg

3. Extractant (70% alcohol): 120+(120*2)=360l.

4. Extractant in one percolator 10+(10*2)=30l

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