2018-01-21
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Protection of the environment from radioactive contamination provided by the following measures:
• application of perfect production technology that minimizes the amount of radioactive waste flowing and prevents their leakage (sealing of processes associated with their use and use of aerosols, application of the water cycle, etc.);
• Carrying out neutralization, centralized collection and storage of radioactive waste;
• organization of sanitary protection zones and planning activities.
At present, taking into account the requirements for sources of environmental pollution, radionuclides allow the discharge of waste with such activity, the level of which corresponds to the number of artificial radionuclides entering the body in excess of their annual receipts for individuals from among or from outside the air radionuclides Argon, krypton, Xenon and short-lived isotopes of carbon, nitrogen and oxygen.
Radioactive extracts include solutions, articles, materials, biological objects containing radionuclides in quantities exceeding the values, sanitary standards in force (SPORO-2002) and not subject to further use. The radioactive radiation also includes the spent resources of ionizing radiation.
Radioactive waste for aggregate equipment for liquid, solid and gaseous materials.
Liquid radioactive waste - organic and inorganic liquids, pulps and sludge, not subject to further use, including total volume activity of radionuclides more than 10 times greater than the values given in
Solid radioactive waste - spent radionuclide sources, not intended for further use materials, products, equipment, biological
Objects, as well as cured liquid radioactive waste, in which the specific activity of radionuclides is higher than the minimum significant specific activity, given in Appendix NRB-99/2009. In the case of an unknown radionuclide compound, solid wastes are considered radioactive if their specific activity is more
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• 100 kBq / kg for β-emitting radionuclides;
• 10 kBq / kg - for α-emitting radionuclides;
• 1 kBq / kg - for transuranic radionuclides.
Gaseous radioactive wastes are radioactive gases and aerosols that are not subject to use and are formed during production processes. For the purification of air from radioactive gases and aerosols, the following methods are most often recommended:
• filtration on fine-fibrous polymers in the form of fabrics (for aerosols);
• filtering on packed filters (for aerosols);
• absorption by solutions;
• absorption of gases on solid sorbents;
• exposure in time.
Methods for processing liquid radioactive waste At present, the choice of the scheme for processing liquid radioactive wastes is due, firstly, to the specific activity of the waste and its volume, and secondly to the qualitative composition of the liquid waste both in isotopes and in other components. The ultimate goal of these methods is to concentrate radionuclides for further curing.
To remove radionuclides from liquid wastes, distillation, sedimentation methods, coagulation and ion exchange, evaporation are most widely used.
Distillation is a simple and reliable method for treating liquid radioactive waste. When the solutions are evaporated, the radionuclides are concentrated in a small volume of the unpaired residue. The degree of purification of solutions (the ratio of the concentration of radioactive material in the initial solution to its concentration in the distillate) with this method reaches 10,000 or more. The appearance of radionuclides in the distillate can be due to the sublimation of certain isotopes (for example, 103Ru, 131I) and the drift of droplets and particles by steam during foaming
41.What is the amount of carbon dioxide taken in normal conditions, it is necessary to absorb the plant has grown to a tree with the following parameters: trunk diameter D = 0,8 m, height h = 7,53 m, wood density ρ = 0,6 g / cm 3. We assume that all the wood consists of carbon, and that the tree trunk has a regular cylindrical shape.
D=0.8 m r(D/2)=0.4 m
h=7.53m CO2=C+O2
P=0.6 g/cm3 m= 
r2hp
m =? m= 3.14*(0.4m)2*7.53m*0.6g/cm3=2.27 g
42.Find the mass of carbon monoxide emitted during the full combustion of coal, with the mass of fuel m = 21 kg; combustion factor is k = 0,85; coefficient, corresponding to the number of carbon, undergoing partial combustion (forming CO) ψ1 = 0.14; coefficient, corresponding to the number of carbon, forming CO in a secondary process, ψ 2 = 0.2.
m = 21 kg С + О2 = СО2 2С + О2 = СО
k = 0,85 СО2 + С = 2СО
ψ1 = 0.14 m1 = m*k = 21*0.85 = 17.85 kg
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ψ2 = 0.2. m2 = m1* ψ1 = 17.85*0.14 = 2.5 kg
m4 =? m3 = m1* ψ2 = 17.85*0.2 = 3.57 kg
m4 = m2 + m3 = 2.5 + 3.57 = 6.07 kg
43. Calculate the marginal drain (MD) and the extreme concentration of kerosene (Sst.pred.) in a reservoir at the following parameters (γ – rate of dilution of sewage is equal in a reservoir to 0,1): Maximum concentration limit = 0,7 mg/l, a waterway water discharge in the place of a water intake (Q) – 20 m3/s, a water discharge the enterprise (q) – 1 m3/s and at concentration of kerosene C = 1,5 mg/l.
γ = 0.1
Cmax.con = 0.7 mg/l K = 
= 
=3
Q = 20m3/s MD = 
= 
q = 1m3/s
C = 1,5 mg/l
MD =?
44.What volume will occupy the carbon monoxide which is emitted at the complete combustion of wood, coal or other fuel with the following parameters: m l=3,0 – room length; m n=6,0 – room width; m h=2,7 – room height. VCO = 4,670 m3 ‒ the volume which will occupy this amount of carbon monoxide under normal conditions. T1=44 °C = 317K; P1=789 mm Hg. To define with what height of the room the zone filled with carbon monoxide will begin.
l = 3.0V = P0*V0*T1/P1*T0 =
n = 6.0 = 760*4.670*317/789*273 = 5.2 m3
h = 2.7 S = l*n = 3*6 = 18 m2
VCO = 4.670 m3h = V/S = 5.2/18= 0.29 m
T1 = 317 K h-hx = 2.7 – 0.29 = 2.41 m
P1 = 789mm Hg
T0 = 273 K
P0 = 760mm Hg
h-hx =?
Answer: the area filled with carbon monoxide is above the level of 2.41 m.
45. Calculate the weight of the total emission of particulate matter in the flue gas boiler, if the ash content of the fuel qT = 31%, and the burnt fuel mass m = 150 t / year. (dimensional coefficient f = 0.002, and the efficiency of dust collector ε = 85%).
q = 31% M1 = q ·m· f· (1 - e/100) =
f = 0.002 =0.31*150*0.002*(1-0.85/100)=0.092t=92.2kg
ε = 85%
m = 150 t / year
M1=?
46. Calculate the diameter of the cyclone (CN-11) D and to determine the gas velocity in the cyclone ω, while purified gas volume Q = 12 m3 / s. The optimum value ωop speed = 3.5 m / s.
Q = 12 m3 / s D = 
= 
=
ωop = 3.5 m / s. = 2.1 m
D =?ω = 4*Q/ 
= 4*12/3.14*1*2.12 =
ω =? = 3.47 m/s
47.It is required to determine the volume of V and height of h of a deposit which, with a mass of the deposit which is subject to placement on a site of equal 756 t are admissible to use as fertilizer for agricultural object on the area of S = 1,1 hectares. The fertile layer of a site is presented by gray forest soils with loamy mechanical structure with density ρp = 1,56 t/m3.
m -756 t V=m/ρp=756/1.56=484.6m3
S =1,1 hectare=11000m2 h=V/S=484.6/11000=
ρp = 1,56 t/m3 =0.044m=4.4cm
V,h-?
48. Calculate a payment for superlimit emission PS (polluting substance), if at combustion of 1t of coal carbon oxide is emitted. At gross emission of the polluting substance Mi = 3 tons/year, and emission of PS within the set MLi = 2,1 tons/year. At the same time the standard of a payment for emission of 1 ton of PS within the set limit of НbL i = 3 tg/t. (coefficients of an ecological situation and the importance are equal to КE = 1,9 and КG= 1,2).
Mi = 3 tons/yearSLi=НbL i *КE*КG=3*1.9*1.2=6.84
MLi = 2,1 tons/year PSL = 5SLi*(Mi - MLi)=5*6.84*(3-2.1)=
НbL i = 3 tg/t = 30.78
КE = 1,9
КG= 1,2
PSL =?
49.Define the coefficient of hydraulic resistance of the exact cyclone (CN-24) ζ, if the correction coefficient considering diameter of cyclone k1= 0,9; the correction coefficient considering dust content of gas k2 = 0,95; coefficient of hydraulic resistance of a cyclone with a diameter 500 mm ζ500 = 75.
CN-24 ζ=k1*k2* ζ500
k1=0.9 ζ=0.9*095*75=64.125
k2=0.95
d=500mm
ζ500 = 75.
ζ =?
50.Calculate the demographic capacity of the territories (D1, people), suitable for industrial and civil construction, if the area of the region Tr = 216 650 ha; the coefficient showing the share of the territory that received the highest average suitability K1 = 0,06; and the estimated demand in an area of 1000 inhabitants, depending on the nature of the production bases H1 = 25 ha.
Tr = 216 650 ha D1 = Tr*K1*1000/H1 =
K1 = 0,06 = 216650 *0.06*1000/25 = 519960
H1 = 25 ha.
D1 =?
51.Set the maximum allowable limit runoff and the concentration of mercury in the reservoir at MAC = 0.01 mg/l, at a flow rate of water (q) is 1 m3/s and when the lead concentration C= 0.04 mg/l dilution ratio K = 0,8.
MAC=0.01mg/l 
=0.8*0.01=0.008mg/l
C= 0.04 mg/l MD = =0.008*1/0.04 = 0.2 mg/l
K = 0,8
q =1 m3/s
MD =?
52. Calculate the capacity of the territory by surface waters (D2, people), if the amount of expenditure in the watercourses at the entrance to the area E = 3 600 000 m3/day; coefficient taking into account the need for dilution of wastewater K2 = 0,25; normative availability 1000 inhabitants R = 1500 m3/day.
E = 3 600 000 m3/dayD2 = Е×К2×1000/R =
K2 = 0,253600000*0.25*1000/1500=600000
R = 1500 m3/day
D2 =?
53.Calculate capacity by underground waters (to, the people) if the territory of the area 261 184 hectares; operational module of an underground drain E = 0,09; and special standard of water supply of 1000 inhabitants of = 40m3/d.
Tr = 261184 ha D3 = E*Tr*1000/RS=0.09*216184*1000/40=
E = 0,09 = 486414
RS = 40 m3/d
D3 =?
54. Calculate territory capacity under the terms of the organization of rest in the wood (, the people) if woodiness of the region of L = 82%; indicative standard of need of 1000 inhabitants for recreational territories of = 200 hectares; the coefficient considering distribution of vacationers in the wood and at water = 0,3.
L = 82%D4 = Tr*L*0.5*10/(H2*Mi) =
H2 = 200 ha=?*0.82*0.5*10/(200*.0.3) =
Mi= 0.3
Tr=?
D4=?
55.Calculate capacity under the terms of the organization of rest at water (D5, the people) if length of the water currents suitable for bathing V = 24 km; the coefficient considering a possibility of the organization of beaches C = 0,3; and the coefficient considering distribution of vacationers in the wood and at water = 0,13.
V = 24 kmD5 = 2V×С×1000 / (0,5 × М2)=
C = 0,3 = 2*24*0.3*1000/(0.5*0.13) = 221538.46
M2 = 0.13
D5=?
56.Calculate the capacity of the territory under the terms of the organization of suburban agricultural base (D6 pers.), If the area of land Tr = 187 434 ha; coefficient taking into account the share of the district K3 = 0.25; coefficient taking into account the use of agricultural land to suburban base K4 =0.30; coefficient, the approximate indicator reflecting needs of 1000 residents of the area for lands of suburban agricultural base P = 1200 hectares.
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Tr = 187 434 haD6 = Тr ×К3 ×К4 ×1000 / P =
K3 = 0.25 = 187434*0.25*0.30*1000/1200 = 11714.63
K4 =0.30
P = 1200 ha
D6=?
57.The fertile layer of the site is represented by gray forest soils of loamy texture density ρs = 1.62 t/m 3. It is required to determine the volume V and the height h of sludge, which is acceptable to use as fertilizer for a/c object on the area S = 0,9 hectares, with a mass of sediment to be placed on a plot of 800 t.
ρs = 1.62 t/m3 V=m/ρs=800/1.62=493.8m3
S=0,9hectares=9000m2 h=V/S=493.8/9000=0.055m=5.5cm
m=800t
V,h=?
58. Determine maximum allowable flow and to limit the concentration of lead in the water with the following parameters (γ - the dilution of waste water in the reservoir is 0.5): MPC = 0.01 mg / L, the water usage of the watercourse at the point of water intake (Q) - 60 m3/ s, water usage in facility (q) - 1 m3 / sec, and C = concentration of lead of 0.04 mg / l.
MPC = 0.01 mg/L K= (γ*Q+q)/q=(0.5*60+1)/1=31
Q=60 m3/s 
=31*0.01=0.31 mg/l
q=1 m3/secMD = 
=0.31*1/0.04=7.75 mg/l
C =0.04 mg/l
γ - 0.5
MD =?
59.Set the mass of carbon monoxide emitted during the full combustion of coal, with the mass of fuel m = 26 kg; combustion factor k = 0,78; coefficient corresponding to the number of carbon undergoing partial combustion (forming CO) ψ1 = 0,21; coefficient corresponding to the number of carbon forming CO in a secondary process, ψ2 = 0,13.
m = 26 kg C+O2→CO2↑
k = 0,78 2C+O2→2CO↑
ψ1 = 0,21 CO2+C→2CO↑
ψ2 = 0,13 m1=m*K=26kg*0.78=20.28kg
m(CO)-? m2=m1* ψ1=20.28kg*0.21=4.26 kg
m3=m1*ψ2=20.28kg*0.13=2.64 kg
m4=m2+m3= 4.26+2.64=6.9 kg
m(CO)=m(C)*M(CO)/M(C)= 6.9kg*28/12= 16.1kg
m(CO)=16.1kg
60.Evaluate the impact on the environment and calculate the demographic capacity of territory by the presence of (people) that are suitable for industrial and civil construction, if the territory of the district is Tr = 213 549 ha; coefficient showing the share of the territory, which received the highest score on the suitability of K1 = 0.07; and the estimated need for 1,000 inhabitants in the territory, depending on the nature of the production base of H1 = 35 m.
Tr =213 549haD1 = Tr*K1*1000/H1=
K1 = 0.07 = 213549*0.07*1000/0.0035 =
H1 = 35 m2=0.0035ha = 4.2*1010
D1=?
