ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 3 ppt
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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 4 pptx
... 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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 8 ppt
... 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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 13 ppt
... 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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 1 docx
... 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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 2 docx
...
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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 5 doc
... 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...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 6 potx
...
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....
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 7 docx
... 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 =...
ENVIRONMENTAL ENGINEER’S MATHEMATICS HANDBOOK - CHAPTER 9 docx
... $ 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...
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