ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 7 docx

... 162 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK (7. 12) Equation 7. 12 assumes that ambient air is used as the source of O 2 and that ... Table 7. 2). Table 7. 2 Approximate Composition of Dry Air (by Volume) Component Symbol Concentration (%) Concentration (ppm) Nitrogen N 2 78 .084 78 0,840 Oxygen O 2 20.9 476 209, 476 Argon ... converting milligrams per cubi...

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... Direct Solution 5 87 17. 7 .7. 1 Storage Volume 588 17. 7 .7. 2 Rainfall Intensity 589 17. 7 .7. 3 Maximum Storage Volume 591 17. 7 .7. 4 Information Needed 591 17. 7.8 Stage–Storage Curve 595 17. 7.8.1 Storage ... Calculations 578 17. 7.2 Allowable Release Rates 578 17. 7.3 Storage Volume Requirements Estimates 579 17. 7.4 Graphical Hydrograph Analysis — SCS Methods 579 17. 7.4...

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... Problem : Round off the following to one decimal: 34 .73 ; 34 .77 ; 34 .75 ; 34.45; 34.35. Solution : 34 .73 = 34 .7 34 .77 = 34.8 34 .75 = 34.8 34.45 = 34.4 34.35 = 34.4 Grade 140 mm 22 m 100(%)=× = × ×= 140 ... foot: Example 2. 67 Problem: The pressure in a pipeline is 6 476 psf. What is the head on the pipe? Solution: 2,000,000 gal 7. 48 2 67, 380 cu ft= Volume .78 5 D depth 2...

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... lb/ft-sec (USEPA-84/09, p. 1 67) µm (1 µm = 10 –6 ) = 3.048 × 10 5 ft (USEPA-84/09, p. 183) v = 0.153g (d ) (p ) /(µ p 0 .71 p 1.14 p 0 .71 0.43 0.29 )) 0.153(32.2) [(400)/(25,400)(12)] ( 0 .71 1.14 = ... 2.0/4 0.5 ft ppp pp − == =2 .75 (62.4) 171 .6 lb/ft 3 L1681_book.fm Page 2 27 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 232 ENVIRONMENTAL ENGINEER’S MATHEMATICS HA...

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... 201 274 ++ × +++ ××+++ × ++ 201) (201 140 201 140) (140 1110 140 110) (110 75 110 75 )+ ×+++ × + (75 30 75 30) (30 1 30 0)]++ ×+++ × = = 1/3[6823] 2 274 acre-ft D L A) L = 2 π D 8.60 miles 0 .79 69 ... mi 3.16 = 2 .72 D V A = 2 274 acre-ft 510 acres = 4.46 ft L1681_book.fm Page 309 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 306 ENVIRONMENTAL ENGINEER’S MATHEMATIC...

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... 2.6 2.3 Q peak (cfs) 65 52 47 44 38 34 Area (in. 2 ) 7. 07 7.33 7. 31 7. 53 7. 45 7. 48 Storage volume (ft 3 ) 63,630 65, 970 65 ,79 0 67, 770 * 67, 050 67, 320 t c 60 50 40 30 10 20 0 Discharge ... (including right-of-way) 83 89 92 93 Gravel (including right-of-way) 76 85 89 91 Dirt (including right-of-way) 72 82 87 89 Urban districts: Commercial and business 85% 89...

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... 10 –2 9 .7 × 10 –1 Chlorobenzene 3 .7 × 10 –3 1.65 × 10 –1 Chloroform 4.8 × 10 –3 2.0 × 10 –1 Cyclohexane 0.18 7. 3 1,1-Dichloroethane 6 × 10 –3 2.4 × 10 –1 1,2-Dichloroethane 10 –3 4.1 × 10 –2 cis-1,2-Dichloroethene ... 10 –3 0.25 trans-1,2-Dichlorethene 6 .7 × 10 –3 0.23 Ethane 4.9 × 10 –1 20 Ethanol 6.3 × 10 –6 Ethylbenzene 8 .7 × 10 –3 3 .7 × 10 –1 Lindane 4.8 × 10 7 2.2 × 10 –5 M...

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... LLC 87 C HAPTER 4 Algorithms and Environmental Engineering 4.1 INTRODUCTION In Chapter 3, we pointed out that environmental modeling has become an important tool within the well-equipped ... uncomplicated terrain. • CALPUFF — a multilayer, multispecies non-steady-state puff dispersion model that simulates the effects of time- and space-varying meteorological conditions on...

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... Water Specific Gravity 150 65 ==23. L1681_book.fm Page 97 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 124 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK Solution: Step 1. Calculate the pounds ... (gpm) 21,000 gal 30 min = = 70 0 gpm pumping rate L1681_book.fm Page 132 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 120 ENVIRONMENTAL ENGINEER’S MATHEMATI...

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... 2640 4 275 5 5 2860 6 2650 7 2340 8 2350 9 2888 10 2330 7- Day Moving Ave. for Day 7 2330 3360 = ++22640 275 5 2860 2650 2340 7 ++++ = 270 5 mg/L 7- Day Moving Ave. for Day 8 3360 2640 = ++ 2 275 5 2860 ... 2340 2350 7 ++++ = 270 8 mg/L 7- Day Moving Ave. for Day 9 2640 275 5 = ++ 22860 2650 2340 2350 2888 7 ++++ = 2640 mg/L 7- Day Moving Ave. for Day 10 275 5 2860 = ++2650...

Ngày tải lên: 11/08/2014, 17:20