ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 3 ppt

... chloride 3 × 10 3 1 .3 × 10 –1 n-Octane 2.95 121 Pentachlorophenol 3. 4 × 10 –6 1.5 × 10 –4 n-Pentane 1. 23 50 .3 Perchloroethane 8 .3 × 10 3 3.4 × 10 –1 Phenanthrene 3. 5 × 10 –5 1.5 × 10 3 Toluene ... 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...

Ngày tải lên: 11/08/2014, 17:20

... 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 environmental engineer’s ... a refined modeling analysis and are conservative. • CAL3QHC/CAL3QHCR (CALINE3 with queuing and hot spot calculations) — a CALINE 3- based CO model with a traffic model to...

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... USEPA-81/12, p. 4-2 0.) Q = 84.9 m 3 /min Y 1 = 0. 03 Y 2 = 0.0 03 X 1 = ? L = ? X 2 = 0 Y3%byvolume 0. 03 1 == Y 90% reduction from Y or only 10% of Y 21 = 112 ;therefore Y (0.10)(0. 03) 0.0 03= = YHX 11 = ... Institute (APTI), EPA45 0-2 -8 1-0 05. USEPA-81/05 (1981). Control of gaseous emissions, Course 415, USEPA Air Pollution Training Institute (APTI), EPA45 0-2 -8...

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... 22 y += ′ =q–Kbdh/dl L1681_book.fm Page 32 1 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 31 8 ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK ( 13. 1) Two basic types of porosity are ... = V V void total TKb= KT/b (30 ,000 gpm/ft)/60 ft== = 500 gpm/ft 2 L1681_book.fm Page 31 8 Tuesday, October 5, 2004 10:51 AM © 2005 by CRC Press LLC 32 0 ENVIRONMENTAL ENGINE...

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... Friction 131 5 .3. 2 Basic Pumping Calculations 131 5 .3. 2.1 Pumping Rates 132 5 .3. 3 Calculating Head Loss 133 5 .3. 4 Calculating Head 134 5 .3. 5 Calculating Horsepower and Efficiency 134 5 .3. 5.1 Hydraulic ... Chemical 184 8 .3. 3 Adsorption Equilibrium Relationships 185 8 .3. 3.1 Isotherm 185 8 .3. 3.2 Isostere 186 8 .3. 3 .3 Isobar 186 8 .3. 4 Factors Affecting Adsorption 187 8...

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... Example 2. 13 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 ... other: Monday 31 0 mg/L SS Tuesday 32 2 mg/L SS Wednesday 30 5 mg/L SS Thursday 32 6 mg/L SS Friday 31 3 mg/L SS Saturday 31 0 mg/L SS Sunday 32 0 mg/L SS...

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... 5. 13: (5.12) (5. 13) CO H O H CO 22 23 + ↔ HCO CaCO Ca(HCO ) 23 3 32 + ↔ Ca(HCO ) Na CO CaCO 2NaHCO 32 2 3 3 3 + ↔ + CaCO H SO CaSO 2H CO 32 4 4 23 + ↔ + Ca(HCO ) H SO CaSO 2H CO 32 2 4 4 2 3 + ... produce 5.0 psi is: 52 ft 0. 43 psi/ft 22 .36 psi×= 40 psi 2 .31 ft/psi 92.4×= 37 psi 2 .31 ft/psi 85.5 ft (rounded)×= 14.7 psi 2 .31 ft/psi 33 .957 or 34 ft×= 5.0 psi 2 .31 f...

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... 1 233 0 2 33 60 3 2640 4 2755 5 2860 6 2650 7 234 0 8 235 0 9 2888 10 233 0 7-Day Moving Ave. for Day 7 233 0 33 60 = ++22640 2755 2860 2650 234 0 7 ++++ = 2705 mg/L 7-Day Moving Ave. for Day 8 33 60 ... 1 20 Test 2 0 Test 3 180 Test 4 2 133 Test 5 69 Test 6 96 Test 7 19 Test 8 44 Log Test 1 20 1 .30 1 03 Test 2 1 0.00000 Test 3 180 2.25527 Test 4 2 133 3. 32899 Test 5 69 1....

Ngày tải lên: 11/08/2014, 17:20

... L. Sometimes converting milligrams per cubic meter (mg/m 3 ) — a weight-per-volume-ratio — into a volume-per-unit weight ratio is necessary. If 1 g-mol of an ideal gas at 25°C occupies 24.45 L is an ... Kelvins (by adding 2 73) because Celsius cannot be used in this equation. To do this, we get that the initial temperature is 40 + 2 73 = 31 3 K and the final temperature is 80 + 2 73 =...

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... $ 26.00 38 .00 10.00 20.00 Standard size 8 in. × 16 ft 10 in. × 16 ft 1 ft × 16 ft 1 ft × 20 ft A b = 3. 14(0.2 03) (3. 66) = 2 .33 m 2 Number of bags 118/2 .33 = = 51 bags L1681_book.fm Page 238 Tuesday, ... lb/ –5 °= × fft-sec P p 2.65(62.4) 165.4 lb/ft 3 == QQ(T/T) asas = =++70.6(446 460)/ (32 460) = 130 acfs η [gp (d ) /36 µ)(BL/Q) pp 2 = (32 .2)(165.4)(10.8)(15)(d ) /[ (36 )(1.75 p...

Ngày tải lên: 11/08/2014, 17:20