Errata for SWCE 6th Edition
Last updated: 12-Oct-2016
Page |
Location |
Existing |
Correct |
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22 | Line for Lindane | catttle | cattle | ||||||||||||||||
23 |
Table 2.3, footnote [c], last bullet, first line |
go above 5 nephelolometric turbidity |
go above 5 nephelometric turbidity |
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39 | Figure 3.5 caption | Theissen | Thiessen | ||||||||||||||||
46 |
Example 3.4, last line. |
106.4 mm. |
106.4 mm, and the intensity is (106.4 mm)/(6 h) = 17.7 mm/h. |
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52 |
Example 3.7, first equation |
0..76 |
0.76 |
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53 |
Internet Resources |
Precipitation Data Server |
Precipitation Frequency Data Server |
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55 | Paragraph 3, line 5 | guage | gauge | ||||||||||||||||
58 |
Last line of example |
E = (112.5 = 25.1 … |
E = (112.5 + 25.1 … |
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63 |
First line under Eq. 4.10 |
= 4.903 10-9 |
= 4.903x10-9 |
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86 |
Section 5.7, last line |
(Chapter16) |
(Chapter 16) |
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94 | Table 5.4, footnote [d] | (litter, grass, and brush overstory | (litter, grass, and brush overstory) | ||||||||||||||||
94 | Table 5.4, footnote [d] | Good > 70% ground cover | Good: > 70% ground cover | ||||||||||||||||
104 |
Figure 5.7 |
More precise |
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105 |
Paragraph 1, line 7 |
2620 |
2670 |
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105 |
Second paragraph |
2620 |
2670 |
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105 |
Equation 5.13 |
u = Q/2620 |
u = Q/2670 |
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105 |
Paragraph after Eq 5.14 |
4th line: runoff must equal 1/2620 |
4th line: runoff must equal 1/2670 |
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106 |
Example 5.7, Solution |
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107 |
5th line on page |
ungaugedwatershed |
ungauged watershed |
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113 | Problem 5.11, last sentence | point n | the point of inflection | ||||||||||||||||
116 | After Equation 6.6 | q = discharge (L3/T), | q = discharge (L3/T), | ||||||||||||||||
148 | After Equation 7.4, second line | (dimensionless) | (kg m-1 s-1 Pa-1.5) | ||||||||||||||||
149 | Second line | 0.1 kg Pa-1.5 | 0.1 kg m-1 s-1 Pa-1.5 | ||||||||||||||||
172 |
Figure 8.4(a) |
Cutslope, Sc |
Cutslope, zc |
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180 |
Example 8.3 |
Sf Sb |
zf zb |
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182 |
Equation 8.3, definition of A |
(L3) |
(L2) |
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195 |
Problem 1, 3rd sentence |
Determine, the |
Determine the |
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201 |
3rd sentence |
Values of C can be converted from to |
Values of C can be converted to |
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223 |
Equation 9.8 |
Vx |
Vs |
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230 |
Section 9.29, 3rd paragraph |
Insert after |
A spreadsheet |
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233 |
First paragraph, 7th line |
Table 9.3 or Figure 9.20 |
Table 9.2 or |
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255 | First paragraph, line 4 | than | then | ||||||||||||||||
300 |
Example 12.2, Solution, Line 2 |
Table 12.2 |
Table 12.3 |
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311 |
2nd paragraph, 3rd line |
(Figure 8.11) |
(Figure 8.10) |
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312 | Example 13.2, second to last line | Total Volume 144.45 | Total Volume 144.05 | ||||||||||||||||
329 |
Table 14.3, first two lines, right column |
3.5 |
10.0 |
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331 |
Example 14.1, (3) |
5.1 |
15.1 |
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331 |
Example 14.1, (3) |
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331 |
Example 14.1, (4) |
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331 |
Example 14.1, last paragraph, line 1 |
0.98 m |
1.11 m |
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331 |
Example 14.1, last paragraph, line 2 |
24.9 m |
26.1 m |
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331 |
Example 14.1, last paragraph, line 4 |
0.55 m and 17.8 m |
0.61 m and 18.3 m |
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333 |
Example 14.2, (3), line 2 |
5.1 mm |
15.1 mm |
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333 |
Example 14.2, (3) |
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333 |
Example 14.2, (4) |
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333 |
Example 14.2, (4) |
Additional iterations give de = 1.05 m and a drain spacing of 30.5 m. … Soil 2 are 0.59 m |
Additional iterations give de = 1.18 |
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335 | Example 14.3, Solution step (3) | Equation 8b | Equation 14.8b | ||||||||||||||||
336 |
Example 14.3, (4) |
5.1 mm |
15.1 mm |
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336 |
Example 14.3, (4) |
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336 |
Example 14.3, (4) |
Additional iteration gives de = 0.73 m and a drain spacing of 14.9 m. … Soil 2 are 0.48 m |
Additional iteration gives de = 0.87 |
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336 |
Example 14.3, summary table |
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337 | Figure 14.8 | Spacing = 12.4 m | Spacing = 12.1 m | ||||||||||||||||
339 |
Figure 14.9 |
An 6e-Figure_14.9 includes |
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354 |
Example 15.1, second equation |
Pe = 0.92 [1.25(100)0.824 |
Pe = 0.92 [1.25(90)0.824 … |
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360 |
Second line after Eq. 15.5 |
D = yield decline from Table 15.4 |
D = yield decline from Table 15.3 |
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360 |
Fourth line after Eq. 15.5 |
salt tolerance |
salt tolerance |
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365 | Example 15.7, Solution, line 3 | Average depth = (0.95 + 0.98 … + 84 +0.88)/16 = 0.84 | Average depth = (0.95 + 0.98 … + 0.84 +0.88)/16 = 0.84 | ||||||||||||||||
365 |
Example 15.8, Solution, first line |
Table 15.4 |
Table 15.3 |
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365 |
Example 15.8, Solution, IR equation |
235 |
207 |
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750 mm |
796 mm |
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410 |
1st paragraph, 2nd sentence |
Average root depths of crops are given in Table 15.4 and … MAD are given in Table 15.5. |
Average root depths of crops are given in Table 15.3 and … |
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410 | Example 17.1, Solution, 2nd line | Table 15.4 | Table 15.3 | ||||||||||||||||
410 | Example 17.1, Solution, 4th line | (0.9 m × 120 mm/m) 108 mm | 0.9 m × 120 mm/m = 108 mm | ||||||||||||||||
410 | Example 17.1, Solution, 4th line | Table 15.5 | Table 15.4 | ||||||||||||||||
411 |
First line |
Figure 17.7 be should be |
Figure 17.7 should be |
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420 |
Step (5), first line |
From Step 3 |
From Step 2 |
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434 |
Problem 1 |
Two 186-m sprinkler laterals |
Two laterals |
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434 |
Problem 1 |
Insert at the end: |
The first sprinkler is one-half spacing from the main. |
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434 |
Problem 2 |
moving each 186-m sprinkler lateral |
moving each lateral |
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445 |
Section 18.5, 2nd paragraph, 1st |
numbers required depends |
numbers required depend |
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446 |
Example 18.2, Solution, 1st line |
From Table 15.2 |
From Table 15.3 |
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450 | Step (9) | Ho = 13.83 + … | Ho = 13.88 + … | ||||||||||||||||
451 | Step (16) | EU = … = 95.6% | EU = … = 95.3% | ||||||||||||||||
447 |
Last paragraph, 2nd line |
with multiple outlets in Table 17.7 |
with multiple outlets in Table 17.5 |
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454 | Step (10) | … flow in the manifold is 0.211 L/s. | .. flow in the manifold is 0.106 L/s. | ||||||||||||||||
504 |
Table B.1, line 44 |
CPT, Corrugated plastic tubing, 76 to 203 mm. dia. |
CPT, Corrugated plastic tubing, 76 to 152 mm dia. |
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504 |
Table B.1, line 45 |
45 CPT, 254 to 305 mm dia. 0.017 |
45a CPT, 203 mm dia. 0.016 |
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504 |
Table B.1, line 45 |
Insert new line after line 45a: |
45b CPT, 254 mm dia. 0.017 |
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504 |
Table B.1, line 45 |
Insert new line after line 45b: |
45c CPT, 305 mm dia. 0.018 |
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506 |
Table C.2, 1st column, 5th line |
10 (4) |
102 (4) |