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350hw7soln

# 350hw7soln - 278 Since the smallest volume truck is 9.0 m3...

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Unformatted text preview: 278 Since the smallest volume truck is 9.0 m3, Volume of truck = 9.0 m3 f. Number of trucks The number of trucks required is a function of the population that must be served. An assumption must be made about the number of work days, i.e. pickup days per week. I have assumed 5 days/wk. 44,000 people ......................... = 11,000 stops 4 people/residence 11,000 stops x 2 pickups/wk = 22,000 stops/wk 22,000 stops/wk ...................... = 4,400 stops/d 5 days/wk 4,400 stops/d ---------------------------------- = 11.57 or 12 trucks (190 stops/load)(2 loads/d) 8-5 Collection parameters for Darren, CA Given: Gobs of data, see problem. Solution: a. Volume per pickup V = (2.53 cans/stop)(0.1136 m /can) = 0.2874 m3 Solve Eqn. 8-1 for tp (mean time per collection stop) VP H 200 B [p = “““““ [ “'“ ‘ """“ ' 2 td ' [u ' ““"] VT (1') Nd S Nd th + tu = 1.0 h B = (.50 h for breaks) + (0.40 h for maintenance) B = 0.90 0 2874 8.0 0 90 = ----_---_.n..-_-_[ m-- - 1 00 - mm] (18.0)(3.97) 2 . 2 t9 = 0.00402 (2.55) . A - .‘ 8-6 tp = 0.010256 h or 0.6153 min or 36.9 s/stop b. Number of pickup locations per load 2.55 Np = ------------- = 248.63 or 249 locations/load 0.010256 0. Number of trucks Assume 4 people/collection stop 361,564 No. of stops = -------------- = 90,391 4 90,391 stops ------------------------------------------- = 36.30 trucks (5 d/wk)(249 stops/load)(2 loads/d) with 15% out of service 36.30 x 1.15 = 41.75 or 42 trucks required Truck size Given: Example 8-2 with no rear of yard pickup and one trip to disposal site Solution: a. Mean time per collection tp = 0.72 + 0.54 =‘1.26 min or 0.0210 b b. Volume of truck 0.1 l 8 2(6.4) 13 6 0.50 VT: -—-—-----«---~[----- - -----« - 2( ----- ) ------------ 1 (3.77)(0.0210) 1 27 60 / 60 1 V;- = (1.3894)(8 - 0.4741 - 0.4333 - 0.100 - 0.500) V;- = (1.3894)(6.4926) = 9.021 or 9 m3 truck c. Number of pickup locations 6.4926 Np = --------- = 309 pickups per load 0.0210 This exceeds the 250 stops per day required. 279 280 8—7 Rework collection data for Darren /‘ Given: Problem 8-5 and tbp = 28.20 s and a = 12.80 s/can; trips to disposal site = 1/day Solution: a. Calculate tp tp = 28.20 + 12.80 (2.53) = 60.58 s tp = 1.009667 min or 0.016828 h b. Number of pickup locations 0.016828 0.016828 N1) = 362.39 or 362 locations/load c. Number of trucks 90,391 ---------------------------- = 49.94 trucks (5 d/wk)(362 stops/d) With 15% out of service 49.94 x 1.15 = 57.43 or 57 trucks required a? 282 f. Cost per week per household \$115.20/Mg ................... x 32.76 kg/wk = \$3.77/wk 1,000 kg/Mg & Annual and weekly cost (3 person and 1 person crew) Given: Gobs of data, see problem statement. NOTE: density of compacted waste not given! 1W Solution: a. Annual truck cost crew of 3 Notes: VT = 18.0 m3 from problem 8-5 NT = (1 trip per d)(5 d/wk)(52 wk/y) = 260 xT = Average annual distance/NT 1,000(99,000) 0.0825(5 + 1) AT = -------------------------- [ 1 + ——————————————————— 1 (18.0)(400)(260)(5) 2 1 1,797 1,000( ----------- )(4.55) 260 + ............................. (18.0)(400) AT = (10.58)(1.25) + 28.67 AT = 13.22 + 28.67 = \$41.89/Mg for crew of 3 b. Volume of truck, crew of 1 Notes: from Problem 8-5: Vp = 0.2874 m3, r = 3.97 from Prob. 8-7: tp = 0.016828 0.2874 VT = ----------------------- (6.10) = 26.24 m3 (3.97)( 0.016828) Assume we choose a 27 m3 truck continued on next page 283 0. Annual truck cost crew of 1 1,000(125,000) 0.0825(5 + 1) AT = ----------------------------- [ 1 + ------------------ ] (26.24)(400)(260)(5) 2 1 1,797 1,000( ------------ )(4.60) 260 + ______________________________ (26.24)(400) AT = (9.16)(1.25) + 19.89 AT = \$11.45 + \$19.89 = \$31.34/Mg for crew of 1 d. Annual labor costs crew of 3 Weighted average of wages (1)(1 1.25) + (2)(11.00) 33.25 -------------------------------- = = \$11.08 3 3 (1,000)(3)(11.08)(8.0) AL = ------------------------------ [ 1 + 0.6764] (18.0)(400)(2) AL = (\$18.47)(1.6764) = \$30.97/Mg e. Annual labor costs crew of 1 (1,000)(1)(11.50)(8.0) AL = ------------------------------- [ 1 + 0.7504] (26.24)(400)(2) AL = (4.38)(l.7504) = \$7.67/Mg ; f. Total cost crew of 3 TC = \$41.89/Mg + \$30.97/Mg = \$72.86/Mg for crew of 3 g. Total cost crew of 1 TC = \$31.34/Mg + \$7.67/Mg = \$39.01/Mg for crew of l 284 g. Average weekly charge Solid waste generated per stop (100.76 kg/m3)(2.53 cans/wk)(0.1136 m /can) = 28.96 kg/wk For crew of 3 \$72.86/Mg ------------------- (28.96 kg/wk) = \$2.11/wk for crew of 3 1,000 kg/Mg For crew of 1 \$39.01/Mg ----------------- (28.96 kg/wk) = \$1.13/wk for crew of 1 1,000 kg/Mg B0 Volume of landﬁll Given: Table of mass fractions; assume cell height of 2.40 m and normal compaction; 20 y life Solution: a. Mass of solid waste/y (44 000 pop)(1.17 kg/cap d)(365 d/y) = 18 790 200 kg/y or 18 790 Mg/y b. Compute weighted compaction ratio (using Table 8-?) Component Food Waste Paper Plastic, rubber, wood Textiles Metals Glass Misc. Mass Normal Weighted Fraction Compaction ratio 0.0926 2.8 0.26 0.4954 5.0 2.48 0.0438 4.45* 0.19 0.0379 2.5 0.09 0.0741 4.25" 0.31 0.1668' 1.7 0.28 0.0894 1.2 0.11 Sum = 373— * Average of components; i.e. (5.6 + 3.3)l2 = 4.45 0. Compute density of compacted ﬁll 8 Dc = (144.7 k m3 )= 539.12 kg/m3 d. Volume per day (44,000 people)(l .17 kg/cap-d) .................................... = 95.49 m3/d 539.12 kg/m3 e. Area per day @Aﬁume spread in 95.49 m3/d 0.3 m layer ”l- = 318.30 m2/d area per day 0.3 m 296 - 2...»... “M 297 f. Time to complete cell U31 g 0.15 m/d cover then 0.3 m sw + 0.15 m soil = 0.45 m and it will take 2.4 m - 0.15 m .................... = 5 d to complete a cell 0.45 m g. Soil volumeljg ignoring soil separating cells for daily cover (3 cells/stack)(5 lifts/cell)(318.30 m2/d)(0.15 m) = 716.18 m3 plus 0.15 m for intermediate cover (3 cells/stack)(318.30)(0.15) = 143.24 m3 plus ﬁnal cover (318.30)(0.15) = 47.75 m3 plus soil separating stack (square @ (3183)“2 = 17.84 m) 0.03 m x 2.4 m high x 17.84 m long x 3 cells = 38.54 m3 for two sides 2(38.54 m3) = 77.08 m3 for total soil volume Vsoil = 716.18 + 143.24 + 47.75 + 77.08 V5011 = 984.25 m3 h. Volume of solid waste sz = (95.49 m3/d)(15 d/stack) = 1417.35 m3/stack i. Value for B 984.25 + 1,417.35 E = ......................... = 1.69 1,417.35 , 298 j. Volume of landﬁll (18,790 Mg/y)(1.69) VLF: ----------------------------- x20y= 1,178,034 m3 0.53912 Mg/m3 or VLF = 1,180,000 m3 k. Area of landﬁll 1,180,000 m3 Au: = - ------------------- = 157,333.33 or 157,000 m2 (3)(2.4) + 0.3 .. n....._...a.a..—...-—.. 8-21 Landﬁll for Binford, VT Given: 50 Mg/d, 5 d/wk, spread at 122 kg/m3 ; compacted from 0.5 m to 0.25 m; 3 lifts ; per day and daily cover = 0.3 m; ignore all other cover material ; Solution: a. Mass of solid waste per year 1; (50 Mg/d)(5 d/wk)(52 wk/y) = 13,000 Mg/y ‘ b. Compacted density 1 In compacting from 0.50 m to 0.25 m the density as compacted will be 0.50 m (122 kg/m3) ----------- = 244 kg/m3 0.25 m 0. Estimate of E (3 1ifts)(0.25 m) + 0.15 m 0.90 E = ---------------------------------- = -------- = 1.20 (3)(0.25) 0.75 d. Volume of landﬁll (13,000 Mg/y)(1.20) VLF = -------------------------- = 63,934.43 or 64,000 m3 0.244 Mg/m3 e. Daily area 50 Mg/d ALF = -------------------------------------- = 273.22 or 273 m2 (0.244 Mg/m3)(0.75 m deep) 299 8-22 Recycling Given: New equation for tp and Probs. 8-5 and 8-9 Solution: a. New pickup time tp = 22.6 + 3.80 (1.53) + 5.50 (3.00) tp = 44.91 s or 0.75 min or 0.0125 h b. New VT 0.2874 VT = -------------------- (2.55) = (5.80)(2.55) (3.97)(0.0125) VT = 14.80 m3 c. Recompute costs 1,000(99,000) 0.0825(5 + 1) AT = - --------------------------- [ l + ------------------- ] (14.8)(400)(260)(5) 2 1 1,797 1,000( ----------- )(4.55) 260 + .............................. (14.8)(400) AT = (12.864)(1.25) + 34.87 = 50.95 (1,000)(3)(11.08)(8.0) AL = ----------------------------- [ 1 + 0.6764] (14-8)(400)(2) AL = (\$22.46)(1.6764) = \$37.65/Mg (1. Total cost TC = \$50.95 + \$37.65 = \$88.60/Mg Without recycling the cost was \$72.86/Mg thus the savings would have to be \$88.60 - \$72.86 = \$15.74/Mg or greater 300 8-23 Reduction in landfill volume with recycling program Given: Example 8-5 and 50% paper and 80% glass and metal recycled Solution: a. Mass of solid waste generated Fraction removed Mass fract. removed Paper (0.4317)(0.5) 0.2159(780) = 168.40 Tin cans (0.0520)(0.8) 0.0416(780) = 32.45 Nonferrous (0.0150)(0.8) 0.0120(780) = 9.36 Ferrous (0.0430)(0.8) 0.0344(780) = 26.83 Glass (0.0749)(O.8) 0.0599(780) = 46.72 Total fraction removed = 0.3638 Total mass fraction removed = 283.76 Mg Alternate calc. 780 Mg/y x (1 — 0.3638) = 496.22 Mg/y b. New weighted compaction ratio Component New New Normal Weighted Mass Mass/ Compaction Compaction Fract.‘ Ratio Ratio Food waste 73.87 0.1487 2. 8 0.4164 Paper 168.40 0.3390 5.0 1.6950 Cardboard 50.70 0.1021 4.0 0.4082 Plastics 14.12 0.0284 6.7 0.1904 Textiles 1.56 0.0031 5.6 0.0176 Leather 11.70 0.0236 3.3 0.0777 Garden 111.70 0.2249 4.0 0.8994 Wood 27.30 0.0550 3.4 0.1869 Glass 11.68 0.0235 1.7 0.0400 Tin cans 8.11 0.0163 5.6 0.0914 Nonferrous 2.34 0.0047 5 . 6 0.0264 Ferrous 6.70 0.0135 2.9 0.0391 Dirt, etc. 8.58 0.0173 1.2 0.0207 Sum = 496.76 Sum = 4.1092 OR 4.1 1 * New Mass/496.76 301 c. The density of the compacted ﬁll Dc = (106)(4.11) = 435.66 kg/m3 or 0.436 Mg/m3 d. New volume 496.22 Mg/y 1 1 V = ---------------------- x ------ x ----- = 4.38 m3/d 0.43566 Mg/m3 52 5 e. In 0.3 m layers 4.38 m3/d --------------- = 14.60 m2/d 0.3 f. Soil volumes Soil separating stack (0.3)(2.4)(14.6)1/2(3 cells)(2 sides) = 16.5 For daily cover (3 cells/stack)(5 lifts/cell)(14.60 m2)(0.15 m) = 32.85 m3 Plus 0.15 m intermediate cover for week (3 cells/stack)(l4.60 m2)(0.15 m) = 6.57 m3 Plus ﬁnal 0.3 m (14.60 m2)(0.3 m) = 4.38 Total soil volume Vsou = 32.85 + 6.57 + 4.38 + 16.5 = 60.30 g. The solid waste volume sz = (4.38 m3/d)(15 d/stack) = 65.70 m3/stack h. The value for E 65.70 + 60.03 E = .................... = 1.9137 65.70 302 i. Volume of landﬁll (496.76)(1.9137) VLF = --------------------- X 20 y = 43,641.78 01' 44,000 m3 0.43566 j. Area of landﬁll 44,0000 ALF = -------------------- = 5,866.67 or 5,900 m2 (3)(2.4) + 0.3 i I :3 1:; ”l , i i 1 t 8-24 Lower heating value Given: Cellulose C6H1005 with HHV of 32,600 Solution: a. Compute GMW of cellulose 13 C=6(12) =72 ? H = 10(1) = 10 o = 506) = 80 Total = 162 ' b. Compute LHV using Eqn 8-13 ‘ 10 LHV = 32,600 — [(2,420)(9)( ............. ) 162 LHV = 32,600 - 1,344.44 = 31,255.56 or 31,300 kJ/kg 305 9-2 Estimating chronic daily intake Given: Child exposed for 5 years to 1,1,1—trichloroethane at drinking water limit. She swims, bathes. Average age is 8 years over exposure period. a. Routes of exposure are: 1. Ingestion of drinking water 2. Ingestion while swimming 3. Dermal contact while swimming 4. Dermal contact with water during bath 5. Inhalation during bath b. Drinking water standard from Table 3-6 on p. 162 is 0.2 mg/L c. Ingestion of drinking water (Eqn 9-10) (0.2 mg/L)(l.0 Ud)(365 d/y)(5 y) CD1: ———————————————————————————————————————————— (26 kg)(5 Y)(365 dIY) CDI = 7.69 x 10-3 mg/kg - d d. Ingestion while swimming (Eqn 9-11) NOTE: 30 min/wk ET = ------------ = 0.5 h/wk 60 min/h C (0.2 mg/L)(50 mL/h)(10-3 L/mL)(0.5 h/wk)(52 wk/y)(5 y) DI = ....................................................................... (26 ngS ”(365 d/Y) CDI = 2.74 x 105 mg/kg . d e. Dermal contact while swimming (Eqn 9-12). NOTE: Assume 100% of body is exposed during swimming ( a bit high but no other data given) PC = 6.0 x 10-3 m/h, and 30 min/wk ET = ------------ = 0.5 h/wk 60 min/h (0.2 mg/L)(0.925 m2)(6.0 x 10-3 m/h)(0.5 h/wk)(52 wk/y)(5 y)(1(}3 L/m3) AD = ............................................................................................ (26 kg)(5 Y)(365 d/Y) 306 f. Dermal contact while bathing (Eqn 9-12). NOTES: 80% submergence, PC = 6.0 x 10-3 m/h, and 10 min/d ET = ------------- = 0.1667 h/d 60 min/h (0. 2 mg/L)(0 925 m2)(6. 0 x 10 3 m/h)(0. 1667 h/d)(365 d/y)(5 y)(10- -3 Um3) AD: -------------------------------------------------------------------------------------------- (0.80) (26 kg)(5 y)(365 d/y) AD = 7.12 x10-9)(0.80)= 5.69 x 109 rug/kg - d g. Inhalation during bath ( Eqn 9-15) NOTE: IR = 5 m3/d = 0.2083 m3/h (2.0 ug/m3)(10-3 mg/ug)(0.2033 m3/h)(0.1667 h/d)(365 d/y)(5 y) CDI = --------------------------------------------------------------------------------- (26 kg)(5 Y)(365 (W) CDI = 2.67 x 10-6 mg/kg - d h. Total CD1 CD1 = 7.69 x 10-3 + 2.74 x 10-5 + 3.04 x 10—9 + 5.69 x 10-9 + 2.67 x 10—6 CD1: 7.72 x 10-3 or 7.7 x 10-3 mg/kg - d 9-3 Characterize risk Given: toluene, barium, and xylenes Solution: a. These are not carcinogens so calculate hazard index using Eqn 9-19 and Table 9—6. 0.03 0.06 0.3 HI = ---------- + ------------ + --------- 0.2 0.05 2.0 HI = 0.15 +1.2 + 0.15 = 1.5 9-5 307 Characterize risk Given: tetrachloroethylene, arsenic, dichloromethane Solution: a. These are all carcinogens so calculate risk using Eqn 9-17 and slope factors from Table 9-5 Risk = (1.43 x 10:4)(0.052) + (1.43 x 10-3)(1.5) + (1.43 x 10-4)(0.0075) Risk = 2.15 X 10-3 Identifying RCRA hazardous waste Given: Municipal wastewater containing 2.0 mg/L of selenium; spent pickle liquor from steel ﬁnishing operations Solution: a. Municipal wastewater containing 2.0 mg/L of selenium is not a RCRA hazardous waste because municipal wastewaters are excluded (Figure 9-6) b. Spent pickle liquor from steel ﬁnishing operations is a RCRA hazardous waste because it is K062 (See Table C-2 in Appendix C) Identifying RCRA hazardous waste Given: Sludgelfrom which TCLP extract yields 5.1 mg/L of silver; an empty pesticide container that a homeowner wishes to discard Solution: a. Sludge from which TCLP extract yields 5.1 mg/L of silver is a RCRA hazardous waste because concentration exceeds toxicity characteristic level of 5.0 mg/L (Table 9— 10). b. An empty pesticide container that a homeowner wishes to discard is not a RCRA hazardous waste because household waste is excluded (Figure 9-6) 308 9—7 Dry cleaner waste Given: Dry cleaner produces 10 kg/mo of carbon tetrachloride and wishes to store for 6 months Solution: a. Dry cleaner may store for 6 months (or more!) because quantities less than 100 kg/mo are excluded from regulation (page 731). 9-8 Mass flow condensate tank Given: Flows and concentrations ll Solution: a. Mass ﬂow into stripper (Sample location #1) (5,858 mg/L)(40.0 L/min)(1,440 min/d)(10-6 kg/mg) = 341.63 kg/d b. Mass flow in wastewater (Sample location #2) (0.037 mg/L)(44.8 Umin)(l,440 min/d)(10-6 kg/mg) = 0.002386 kg/d c. Mass flow from vent GMW of methylene chloride (from Appendix A, Table A-8, note that methylene chloride = dichloromethane) = 84.93 g/mole Volumetric ﬂow rate of gas = (57 L/min)(0.4413) = 25.154 L/min Assuming ideal gas law applies (Eqn 6-2) (P)(V) (101.325 kPa)(25.154 L/min) n -. ----------------------------------------------- = 1.046 moles/Mm (R)(T) (8.3143 J/K-mole)(293 K) Mass ﬂow (1.046 moles)(84.93 g/mole)(l,440 min/d)(10-3 kg/g) = 128.03 kg/d d. Mass balance Mass in = Mass out W + Mass out vent + Mass to condensate Mass to condensate = Mass in - Mass out WW - Mass out vent Mass to condensate = 341.63 - 0.002386 - 128.03 = 213.59 kg/d 9-24 9-25 9—26 Mass ﬂow of methylene chloride in aqueous feed Given: Methylene chloride concentration = 5,858 mg/L, ﬂow rate of aqueous stream = 40.5 L/min Solution: (5,858 mg/L)(40.5 Umin) = 237,249 mg/min or 237.25 g/min Mass ﬂow of methylene chloride in ﬂue gas Given: Methylene chloride concentration = 211.86 ug/L, ﬂow rate of ﬂue gas = 597.55 m3/min Solution: (211.86 ug/L)(597.55 m3/min) = 126,596.9 ttg/m3 or 0.1266 g/min DRE for Problems 9—23 and 9-24 Given: Problems 9-23 and 9-24 Solution: 21. Using Eqn. 9—31 237.25 - 0.1266 DRE = ------------------------ (100) = 99.947% 237.25 DRE for xylene Given: mass flow into incinerator = 481 kg/h, mass ﬂow in stack = 72.2 g/h Solution: a. Using Eqn. 9-31 481 - 0.0722 DRE = ................... (100) = 99.985% 481 Incinerator does not comply with 99.99% DRE 327 335 9—32 Estimate of leachate volume Given: 100 hectare landﬁll, permeability of 10-7 cm/s Solution: a. Convert cm/s to mld k = (10-7 cm/s)( 10-2 m/cm)(86,400 s/d) = 8.64 x 10-5 m/d b. Calculate flow Q = (8.64 x 10-5 m/d)(100 hectares)(104 m2/hectare) = 86.4 m3/d \$1 Time for leachate to migrate 47/37} Given: Three soils with differing permeability and depth y Solution: SOIL A a. Hydraulic gradient H + T 0.3 m + 3 m i = --—--«---— = ---------------- = 1.10 T 3 m b. Darcy velocity v = (1.8 x 10-7 cm/s)(1.10) = 1.98 x 10-7 cm/s c. Seepage velocity 1.98 x 10-7 cm/s v’ = ------------------------ = 3.60 x 10-7 cm/s 0.55 (1. Travel time (3.0 m)(100 crn/m) ' t = ------------------------------------- = 9,645 d (3.60 x 107 cm/s)(86,400 s/d) SOIL B a. Hydraulic gradient H+T 0.3m+3m+10m i = ----------- = -------------------------- = 1.33 T 10 m 336 b. Darcy velocity v = (2.2 x 10-5 m/s)(1.33) = 2.93 x 10-5 m/s c. Seepage velocity 2.93 x 10-5 m/s V’ = ------------------------ = 1.17 x10-4 m/s 0.25 d. Travel time (10.0 m)(100 cm/m) t= --------------------------------------- = 0.99 d (1.17 x 104 m/s)(86,400 s/d) SOIL C a. Hydraulic gradient H+T 0.3m+3m+10m+l2m i: ----------- = ------------------------------------ = 2.11 T 12 m b. Darcy velocity v = (5.3 x 10-5 mm/s)(2.ll) =1.12 x 10-4 mm/s c. Seepage velocity 1.12 x10-4 mm/s v’ = ------------------------ = 3.19 x 104 mm/s 0.35 d. Travel time (12.0 m)(1000 mm/m) t = -------------------------------------- = 435 d (3.19 x 104 mm/s)(86,400 s/d) TOTALTIME 9,645 (:1 + 0.99 d + 435 d 10,081 (1 Time = -------——---------------------—-- = --------------- = 27.6 or 28 y 365 d/y 365 d/y 341 10-4 What particles emitted in decay chain? Given: Lead-214 decay to lead-206 Solution: a :3sz --> 33Bi Beta particle b. 33Bi --> 34Po Beta particle c. 34Po --> 32Pb Alpha particle d. gng --> 83Bi Beta particle e. 33Bi --> 34Po Beta particle f. 34Po --> 32Pb Alpha particle 10—5 Show annihilation yields 1.02 MeV Given: Positron and electron annihilation Solution: a. From text discussion ( p. 811, 813, 814) ﬁnd Energy equivalent of 1 u = 931.634 MeV Electron has mass of 0.0005486 11 Positron has mass = electron b. Assuming annihilation converts all mass to energy Total mass = 0.0005486 u + 0.0005486 u = 0.001097 u Energy = (931.634 MeV/u)(0.001097 u) = 1.022 or 1.02 MeV 10-6 Holding time for 32P Given: 0.5 uCi/L of 32F is to be discharged 0p _ 5' _ Solution: qX/O/A 0%“, TM 1° 7/ a. Allowable discharge of 32 'w i/mL from Table 10-2 b. Convert original concentration to same units (0.5 uCi/L)(103 L/mL) = 0.5 x 103 uCi/mL 0. Calculate decay constant From Table 10-1; Tm = 14.3 d 342 0.693 0.693 Decay constant = ---—----- = --------- = 0.048462 d-l T1/2 14.3 d d. Using log form of Eqn. 10-6 .xse» ,;.u.m...,a.‘. .1” ln( -------------- ) = (-0.048462)(t) 0.5 x 10-3 -3.2189 0Q M ( «,r h t = -------------- =.6:642’0r 7 days or .- 34 “(h 0. “I “ 0048462 4’ 0mm? 2, . «we. \ “.54.; 10—7 Radium content in year 2,000 Given: 20.00 mg of RaC12 prepared in 1911 Solution: a. Calculate fraction of RaC12 that is radium GMW of Ra = 226.0254 GMW of 2C1 = 2(35.453) = 70.906 GMW of RaC12 = 226.0254 + 70.906 = 296.931 226.0254 Mass of Ra = --------------- (20.00 mg) = 15.22 mg 296.931 b. Calculate decay constant From Table 10-1, Tm = 1,622 y 0.693 Decay constant = ------------ = 4.27 x 10-4 y-1 1,622 y 0. Calculate number of years 2000 -1911= 89 years d. Calculate mg of Ra remaining using Eqn. 10-6 N = (15.22)exp(—(4.27 x 104)(89)) = 14.66 mg 349 10— 14 Thickness of lead 7". Given: 60Co source, transmission reduction of 99.6 % required using lead shield Solution: .. ._ ,..__ _..__.,. ——-—-—-———~—-—-—-—— .‘A a. Compute transmission allowed 1.000 - 0.996 = 0.004 b. From Figure 10-5 using 60Co line Lead thickness required = 6 cm ; (Allow some tolerance for graph reading) 10—15 Equivalent thickness of concrete é—i Given: Problem 10-14 Solution: a. Using a transmission of 0.004 from Problem 10-14...
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