350hw3

350hw3 - 4-2 Theaetical oxygen demand first printing of...

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Unformatted text preview: 4-2 Theaetical oxygen demand first printing of the 3rd edition) :1. Compute gram molecular weighty» GMW of c,1-x,mo2 = 113 ‘ GMWofoxygen(502+2Og)=224 b.Calculatethe'IhOD 224 11.01) = (30 myL)--1-1-;-~ = 59.47 or 60 mg/L -0.2941 = 40-“ b. Divide through by -l and take log of both sides log (0.2941) = log (lo-Km) -0.5315 = -K(7) K a: 0.0759 d~1 Converting from 25 0C to 16 0C Given: K = 0.0759 (1-1 at 25 0C from Problem 4-4 Solution: tMusteonvenKntZSOCtoKatmonmt x’ or - 0.0759 s K20 (1.056)” - no l- v‘” \ / of bacterial cells and oxidntion reactions (NOTE: the concentration is not 7 4.12 Ultimate BOD's for diffcrent K's Given: BOD5 = 280.0; K = 0.0800 d-1 or K = 0.120 d-i Solution: 1. Fat K a 0.0800 d-l 280.0 a L (1 - 104mm) 280.0 L = - ------- -- (1 - 0.39817 L = 465.199 01' 465.2 mg/L b. For K = 0.120 d-1 280 = 1.. (1404012161) L = 373.93 or 373.9 lug/L 4-20 Tannery ultimate BOD after mixing Given: Tannery Q. = 0.011 m3ls. 300, a 590 tug/L. Creek Q, = 1.7 m3/s. BOD; Sol ‘ .6 mg/L. Tanner k = 0.115 M. Creek k = 3.7 d-1 “hon: “mus. L. (5911 4-25) = 1,349.2 mg/L b. Calculm the ultimate BOD of Cattanugus Creek 0.6 mg/L 0.6 1-0 == --~---~------~---~- = “mm-mm“ = 0.6 mg/L 1 - exp[(-3.7)(5)] l - 9.24 x 10-9 Thus. L (Eqn 4-25) = 0.6 [BI/L #1 Mixed ultimate BOD Given: Problem 4-20. data from glue factory and municipal WWTP Solution: a. The mixed ultimate BOD is the weighted avenge of the three wastewater streams and the excel: . L. (0.011)(1.349.2) + (1.7)(0.6) + (0.13)(255) + (0.02)(75) 0.011 + 1.7 +0.13 +0.02 14.84 + 1.02 + 33.15 + 1.50 50.51 L,=- ------------------------------ --= --------- --=27.l4or30mglL 1.86 1.86 4-22 moxygetiation and mention rate constants Given: Table of data Solution: a. Calculate the «oxygenation rate constant using Eqn 4-42 0.5 k4 = 0.20 + .- ------ -- (0.4) = 0.40 d-1 1.0 h. Calculate the teaeration rate constant using Bqn 4-43 3.9 (0.5)05 3.90.71) k. = = = 2.76 d-1 (1 .0)‘-5 1.0 4~24 Critical point and cn'tical DO Given: L. = 50 mg/L. D0 = saturation. river temp = 10 ’C, ltd = 0.30 M. k, = 0.30 d-1 Solution: _a. .Since D0 in the river is at saturation after the wastewater and river have mixed. the initial deficit (1).) is 0.0 mg/L. b.Note that this is a special case of ltd = k, so Eqn 4-45 applies. 1 0.0 _._ _______--.(1 . ----..-) = d 0.30 50 7 4-25 Critical point and critical DO Given: Problem 4—24 and new telnpmmles Solution: a. Calculate the new In using Eqn 4-5 with 0 = 1.135 Calculate k. at 20 'C 1:10 0.30 0.30 1:20 = ------------ - = ------------- -- = ------- -- = 1.0643 d~l (0)1040 (1.135)-10 0.2819 Calculate K: at 15 'C k, ; kls = 1.0643(1.l35)15 - 20 = (l.0643)(0.5309) = 0.5651 (14 1). Calculate new 1:, using Eqn 4-5 with 6 = 1.024 (See p. 311, discussion after Eqn 4-43). Calculate k. at 20 'C 1:10 0.30 0.30 - 1:20 = = = m»...- = 0.3803 d-l (0)1040 (1.02040 0.7889 Calculate k. at 15 'C k, = kls = 0.3803(l.024)l$ - 30 = (0.3803)(0.8882) = 0.3378 (‘1-1 0. Calculate the critical time using Eqn 4.44. Note: since the D0 is at saturation, D. = 0.0 Inle 1 0.3378 0.3378 - 0.5651 t.=--------------------lnl- -------- --(l -(0)-- ------------ --)1 0.3378 - 0.5651 0.5651 (0.5651)(50) 1.; = (-4.399)1n[0.5978] = 2.264 d. The clitical deficit is found using Eqn 4-40 with t = 2.264 (0.565 I)c = ............. ..- [exP((-0.5651)(2.264)) - cxP((-0.3378)(2.264))] + 0 = 23.27 lug/L 0.3378 - 0.5651 2. Elle (lusitiéal D0 is found by solving Eqn 4—30 for DO with DOs = 10.15 mg/L Appendix - at ' mglnl<_fl1fl‘l_ 16|n_.nn _ . __ 7 4-35 Allowable mass discharge Given: Assume D.I = 0.0; L, = 0.0; D0 must be kept > 4.00 m3] at 8.05 km; K, = 1.80; K, = 2.20; T == 12 0C; Q, = 5.95 3113/3; v = 0.300 m/s; Q,v = 0.0130 10% Solution: 21. From Table A-2 at T = 12 0C find D0, = 10.83 mg/L b. Allowable deficit D = D00 ' Doonowable D = 10.83 - 4.00 z 6.83 mg/L 0. Travel time to 8.05 km downstream (s 05 km)(l,000 m/km) ’ 1 = ------------------------------- -- = 0.3106 d (0.300 mls)(86.400 s/d) d. Solve D0 Sag Eqn. for 1... (NOTE: in base 10) K: - K0. 1 L. = D (--------.--)[---------.-------—-------1 + 0 K. 101K410) - 100cm) ' 2.20 - 1.80 l L. = 1.80 10.“.sox03106) . 10-(2.20)(0.3106) 1 L. = (6.83)(0.222) [ .................. ...1 0.2760 -'0.2073 L. a -(6.83)(0.222)(l4.5560) = 22.0707 mg/L 0. Solve Eqn. 4-25 for QVL. (NOTE: mglL = g/m3) QwLw = L.(Qw + Q) - QrLr 0*“ = [22.0707 z/m3(0.0130 + 5.95)] - [(5.95)(o.00)1 .-. 131.6076 315 QwLw = (131.6076 nlflllfld Int-var. Ann .14\ . .-_. _ _ ‘ £27 Nitrogcnous BOD Given: Problem 4-35 and 3.00 mg/L ammonia-N. Solution: 3. Calculated values ofD0., D and t same as in 4-35 h. Solve Eqn. 4.46 for L. K: - K4 1 L.=( -------- --)[ ------------------- --1x Kd 10mm - 100cm) Kill-n [D - "—..-"muomxo - 10mm) - D.(10-(KIX0)] K. - K. c. Note from Problem 4-35 D. = 0.0 d. Compute L. L. = NBOD a (4.57 gig of N)(3.00 tag/L) 2 13.71 tug/L 6. Calculate L. 2.20 - 1.80 l L. = «mu-m».- ----»-~-------—-—-—--.—---—--. 7: 1.80 10-(l.80)(0.3106) - 1mz.mxo.31w) (0.900)(l3.7l) [5,33 . -....-..-------------(10-(0.900)(0.3106) - 10-(2.20)(0.3106))] 2.20 - 0.900 ‘ l L. = (0.22)( ------------- --) [6.83 - [9.49(0.53 -.21)|] 0.28 - 0.21 L.‘ = (3.14)(3.81) a 11.98 tug/L f. Find l.w (NOTEfrom 4-35 L. = 0) La(Qw + Q) - erq 1,w = ............. -.. Q. [(11.98)(0.0130 + 59‘“ - l(( cam nu 7 DISCUSSION QUESTIONS 4-4 .— Location of critical point Treated wastewater will have a lower rate constant (k) (See p. 293). This in turn will reduce the value of Ra (See 5:11: 4-42). The critical point will be displaced downstream (See No. 4 on p. 317). Although the effect of decreasing L... is to make LI smaller and hence to make tc smaller. the decrease in k4 is controlling. This would be classified as a eutrophic lake. High turbidity eliminates oligotrophic productivity. Large mats of floatin algae point to a eutrophic lake. Assuming summer stratification. the hypolimnion DO 0 l. mglL is close to anaerobic which would eliminate mesotrophic productivity. 5-2 Time of log growth and number of generations Given: Time and count data Solution: a. From plot of time and count data (see next page) find that: log growth starts at 10 h log growth stops at 35 h b. Numberof generations by solving Eqn. 5-2 as in Prob. 5-1 1.05 x 106 n = 3.3 log -—-----«--- = 3.30.85) = 9.39 or 9 generations 1.5 x 103 - Pflouga'M 5'3. - CONT'D J 0 ~ - ......w. _ .3..—.'..-...-f—. if...- a. . ... ' in"... .7 {.4-+£LE.J..{..§.__:=_‘..,- !. f-w—I.f. ,J. _. I. . .. _ k: ?- '.,.,....!._..:..'....'_1. -. ...-.._...... . .-... .. .. ..-. 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350hw3 - 4-2 Theaetical oxygen demand first printing of...

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