Military Hygiene and Hygiene in Emergency Situations - K.O. Pashka 2005

Occupational hygiene of military personnel during the mitigation of emergency consequences and in wartime
Situational tasks for independent work

Section 2

Problem No. 1. Determine the volume of air that must be supplied to a shelter per 1 hour if it accommodates 50 people performing light physical work. The maximum allowable concentration of СО₂ must not exceed 2 %.

Sample Solution:

The required air volume is calculated using the formula:

Q= (CxN)/(Р₂ - Р₁) m³/h, where:

Q - volume of air supplied to the shelter per hour, m³;

С - amount of СО₂ emitted by one person - 22.6 dm³/h;

N - number of people in the shelter - 50 individuals;

Р₁ - amount of СО₂ in the shelter air in the absence of people - 0.4 dm³/m³;

Р₂ - maximum allowable amount of СО₂ in the shelter air - 20 dm³/m³;

Q = (22.6x50)/(20-0.4) = 1130/19.6 = 57.6 m³/h.

Problem No. 2. In a sealed shelter, there is 2 m³ of air per person. The concentration of СО₂ in atmospheric air is 0.04 %. The amount of СО₂ exhaled by a person in 1 h is 22.6 dm³/h (light physical work).

To what percentage will the СО₂ concentration increase after 3 hours?

Sample Solution:

The calculation is performed using the formula:

К= (р₂хр₁хt)/(qхр₁x10), where:

К - СО₂ concentration, %;

Р₁ - СО₂ concentration in atmospheric air - 0.04 %;

Р₂ - amount of СО₂ exhaled by one person - 22.6 dm³/h;

q - volume of air per 1 person - 2 m³;

t - duration of stay in the room - 3 h;

10 - conversion factor from volumetric units to percentages;

К = (22.6x0.04x3)/(2x0.04x10)= 2.71/0.8= 3.3 %.

Problem No. 3. Determine the dust concentration in the air if the mass of the filter before air sampling with an aspirator was 20.452 g, and after sampling it was 20.478 g. The sampling time is 10 minutes, and the volumetric air flow rate through the aspirator is 20 dm³/min.

Sample Solution:

The determination is carried out using the formula:

С = (а₂-а₁)х10⁶/Y, where:

С - dust concentration in the air, mg/m³;

а₁ - weight of the filter before air sampling, g;

а₂ - weight of the filter after air sampling, g;

10⁶ - conversion factor;

Y - volume of sampled air, dm³;

С = (20,478 - 20,452)х10⁶/200 = 130 mg/m³.

Section 3

Problem No. 1. What amount of bleaching powder containing 25% available chlorine is required to disinfect Water in a completely filled RDV-5000 container?

Sample answer:

Disinfection is carried out using elevated chlorine doses based on 10 mg of available chlorine per 1 dm³ of water. To disinfect 1 dm³ of Н₂О, 10 mg of available chlorine is needed, and for 5000 dm³ of Н₂О — X mg of available chlorine.

Class="center">Х = (5000х10)/1 = 50000 mg = 50 g of available chlorine.

Thus, to disinfect the 5000 dm³ of water contained in this vessel, 50 g of available chlorine is required.

We calculate the quantity of bleaching powder that contains the specified amount of available chlorine using the proportion:

in 100 g of lime — 25 g of available chlorine; in X g of lime — 50 g of available chlorine;

Х = (100х50)/25 = 200 g of bleaching powder.

50 g of available chlorine is contained in 200 g of bleaching powder.

Problem No. 2. It is necessary to determine the available chlorine content in water disinfection tablets ("Aquatabs", "Acuasept", "Neoaquasept") and justify the Conclusion regarding their suitability.

Sample answer:

One tablet from the test batch should be dissolved in a flask containing 50 ml of distilled water. After the tablet dissolves, 5 drops of an HCl solution (1:2) are added to the container, followed by 10 drops of a 5% starch solution.

Titration is performed dropwise with a 0,7% sodium thiosulfate solution (1 drop binds 0.04 mg of chlorine) until decolorization occurs. Tablets containing less than 1.5 mg of available chlorine are considered unsuitable for water disinfection.

Problem No. 3. Water from a tube well has the following parameters:

transparency greater than 30 cm;

color intensity - 4°;

odor - 1 PSU;

taste - 2 PSU;

coliform index - 2 CFU/dm³;

total bacterial count - 35 CFU/cm³.

Draw a conclusion regarding the compliance of the water with the requirements of State Sanitary Rules and Regulations of Ukraine No. 383 (DSanPiN) and justify its suitability for drinking without additional purification.

Answer key:

All the listed parameters comply with the requirements of DSanPiN of Ukraine No. 383 for drinking water; therefore, this water can be considered suitable for drinking without additional purification.

Section 4

Problem No. 1. It is necessary to determine the amount of Vitamin C in 300 g of mashed potatoes prepared in a KP-130 field kitchen.

Answer key:

To do this, take a sample of mashed potatoes weighing 20 g and dilute it in 60 ml of extracted Hydrochloric acid; take 10 ml of the filtrate for testing. Titration required 2 ml of Tillmans' reagent. The titer correction factor is 0.98. The calculation is performed using the formula:

X= (AxKxVx0.088xС)/(рха), where:

X - amount of ascorbic acid, mg;

A - volume of Tillmans' reagent used for titration, ml;

K - titer correction factor for Tillmans' reagent;

B - mass of the dish, g;

0.088 - amount of ascorbic acid corresponding to 1 ml of Tillmans' reagent, mg;

C - volume of extracted hydrochloric acid, ml;

p - mass of the food sample, g;

a - volume of the filtrate taken for titration, ml;

X= (2x0.98x300x0.088x60)/(20x 10) = 14.3 mg. 300 g of mashed potatoes contain 14.3 mg of vitamin C.

Problem No. 2. Determine the moisture content of bread if the bread sample weighs 5 g, the mass of the weighing bottle with the bread sample before drying was 15 g, and after drying it was 13 g. Does the moisture content of the bread comply with hygienic standards?

Answer key:

The moisture content of bread is determined by the formula: X= [(a-b)/(c)]x100, where:

X - moisture content of bread, %;

a is the mass of the weighing bottle with the lid and the bread sample before drying, g;

b is the mass of the weighing bottle with the lid and the bread sample after drying, g; c is the mass of the bread sample, g;

100 is the conversion factor to percentages;

Х= [(15-13)/5]х100 = 40 %.

The moisture content of the bread is 40 %, which complies with hygienic standards.

Problem No. 3. During an inspection of a food warehouse in an earthquake-affected area, 5 tons of herring in wooden barrels were discovered. Due to power line damage, electricity was not supplied to the warehouse for 10 days, meaning the refrigerators were inoperable. Samples taken from several barrels had an unpleasant odor, a stale/overheated appearance (sustainability defect), rust, and sliminess; some barrels lacked brine. The reaction of the herring meat to ammonia using Eber's reagent was positive. Formulate a well-reasoned expert conclusion and provide recommendations regarding the further use of the herring.

Answer key:

Considering the unsatisfactory organoleptic qualities of the herring—unpleasant odor, staleness, rusting, sliminess—as well as a positive ammonia test indicating spoilage, the batch of herring is unfit for human consumption. Therefore, it is subject to seizure, with selected samples subsequently sent to the Sanitary and epidemiological station (SES) to determine whether it can be used for animal feed or technical utilization.

Section 5

Problem No. 1. Calculate the energy flux density (EFD) near a building located 100 m away from a radar station; Р_сР = 250 W, Д = 500.

Answer key:

The energy flux density is calculated using the formula:

ГПЕ= (Р_ср хДх10⁶)/4 HR² = μW/cm², where:

P is the average power of the radar station, W;

Д is the antenna gain;

10⁶ is the conversion factor from W to μW;

R is the distance from the radar station to the inspected object, cm;

ГПЕ = (250х500х10⁶)/(12,6х10⁸)= (125х10⁹)/(12,6х10⁸)=100 μW/cm².

Problem No. 2. To what height must the radar antenna be raised so that the "exclusion zone" is at least 50 m? The negative operating angle of the antenna is 2°, and the radiation pattern angle is 3°.

Answer key:

To calculate the required installation height of the radar antenna, the formula h = (L)/(tga)+ 2 m is used, where:

h is the installation height of the radar, m;

L is the required size of the exclusion zone, m;

2 m is the human presence zone; a = 90°-(2°+1.5°)= 86.5°;

h - 50/16.35+2 = 5m;

The angle a is calculated using the formula a = 90°-(ß+1/2y) where:

a is the angle to be determined;

ß is the antenna working angle;

Y is the antenna beamwidth angle, in degrees.

To obtain a restriction zone of at least 50 m in size, the radar antenna must be elevated to a height of 5 m.

Problem No. 3. During military exercises, the Temperature of a tank's armor reached 45 °C, the relative humidity inside the tank was 15 %, and the temperature was 38 °C. Name the modes of Heat transfer and indicate which one will be the primary one. Answer key:

Out of 85 % of the heat released by the body through the Skin, 30 % is lost via conduction, 45 % via radiation, and 10 % via evaporation. In this case, the primary mode of heat transfer for the crew members will be evaporation.