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Location

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Correct

xv

right column

National Resource
Conservation Service

Natural Resources
Conservation Service

16

9^th line
from top

methemoglobenemia

methemoglobinemia

40

Ex. 3.2, line 2

Keutucky

Kentucky

40,
41

Ex. 3.2, second table,
column headings

I/P    log_e(log_e[I/P])

1/P    log_e(log_e[1/P])
(In the first and last column headings, “I” is the capital letter “I” for rainfall
amount.)

43

Table 3-1 caption

Intensity Duration
Frequency Table

Depth Duration
Frequency Table

44

Ex. 3.3, line 1

Determine the rainfall
intensities

Determine the rainfall
depths

49

Par. 2, line 4

Equation 3.3

Equation 3.7

49

Ex. 3.7, line 2

3.3

3.7

54

Sec. 4.1, line above
Eq. 4.3

Rohwer (1931)
evaluated the constant C in
Equation 4.1

Rohwer (1931)
evaluated the constant C in
Equation 4.2

61

First line under Eq
4.10

Boltzman

Boltzmann

62

First line

where z
in

where z
is

64

Ex. 4.3, Step (2)

37.3

237.3

65

Ex. 4.3, Step (12)

0.003 cos

0.033 cos

65

Ex. 4.3, Step (13)

41.65 MJ m^-2 day^-1

41.63 MJ m^-2 day^-1

65

Ex. 4.3, Step (14)

41.65 =31.28 MJ m^-2 day^-1

41.63 =31.27 MJ m^-2 day^-1

66

Ex. 4.3, Step (15)

5.185
MJ m^-2 day^-1

5.186
MJ m^-2 day^-1

74

Line after Eq. 4.31

where ET_L
in

where ET_L
is

75

Ex. 4.6, Solution

<![if !vml]><![endif]>

Calculate ET_L
from Equation 4-31

<![if !vml]><![endif]>

The estimated volume of water to be delivered
per irrigation is

<![if !vml]><![endif]>

Thus the irrigation system should be set to
deliver about 1000 L of water every three days.

<![if !vml]><![endif]>

Calculate ET_L
from Equation 4-31

<![if !vml]><![endif]>

The estimated volume of water to be delivered
per irrigation is

<![if !vml]><![endif]>

Thus the irrigation system should be set to
deliver about 900 L of water every three days.

84

2^nd line
after Eq. 5.6

T = time (T)

t = time (T)

93

Example 5.4, Solution
(line 1)

Weighted CN = 3384/40

Weighted CN = 3544/40

101

Figure 5-7, table line
8

h 20 99

h 18 99

101

Line under Eq. 5.15

ham

ha-m

106

Problem 5.10

15-mm rainfall

10-mm rainfall

106

Problem 5.10

6-h, 25-y storm

6-h, 25-y storm during
the dormant season

109

Figure 6-1b

<![if !vml]><![endif]>

<![if !vml]><![endif]>

<![if !vml]><![endif]>

<![if !vml]><![endif]>

109

Figure 6-1c

P = t
+ 8d2
/ 3t

P = t
+ 8d² / 3t

110

Above Eq 6.6

It disregards
elevation potential, since elevation changes are negligible in short reaches
of mildly sloped conduits.

[Delete entire
sentence.]

119

Eq. 6.14 definitions

R = hydraulic radius (Equation 6.3)

R = hydraulic radius (Equation 6.2)

114

Figure 6-5

Insert horizontal rule
under

<![if !vml]><![endif]> (above 2g)

114

Below Eq. 6.12

where Q is the flow
rate (L/T³).

where Q is the flow
rate (L³/T).

115

Example 6.2, second
equation

= 0.48 m/s

= 0.48 m³/s

117

Ex. 6.3, Solution,
line 2

6-12 and 6-13

6.12 and 6.13

125

Example 6.5 label

Example 6.5

Example 6.6

133

Problem 6.7

slope of the channel
is 1 percent.

slope of the channel
is 0.08 percent.

133

Problem 6.7

(Insert sentence at
the end of the problem.)

Comment on the
practicality of the results.

133

Problem 6.10

Example 6.5

Example 6.6

133

Problem 6.10, line 2

stages of flow as
follows:

stages of flow,
yielding the following depth-discharge pairs:

133

Problem 6.11

head of 0.4 m

head of 0.3 m

145

Eq. 7.7b

where θ = field slope angle = tan^-1(S)
S = slope steepness (m/m)

where θ = field slope angle = tan^-1(s)
s = slope steepness (m/m)

147

Example 7.2, second
line

S = 5 percent

s = 5 percent

147

Example 7.2, step 6

From Equation 7.8c,
calculate the S-factor:
S
= 16.8 sin(2.86) – 0.6 = 0.34

From Equation 7.8b,
calculate the S-factor:
S = 10.8 sin(2.86) + 0.03 = 0.57

147

Example 7.2, step 7

<![if !vml]><![endif]>

<![if !vml]><![endif]>

157

Problem 7.4

S = 10 percent

s = 10 percent

158

Problem 7.7, second to
last line

density of water is
N/m3

specific weight of
water is N/m³

158

Problem 7.8, second
line

Flagstaff,Arizona

Flagstaff, Arizona

158

Problem 7.12

are collected in 20
seconds.

are collected in 1
minute.

161

Last two lines

… backs-
lope

… back-
slope

166

Example 8.1, line 1

K = 0.1, 1 = 120 m, S =
8

K = 0.1, l = 120 m, s = 8
(lowercase “L” instead of numeral “1” and lowercase “s”)

170

Example 8.3, last
equation

<![if !vml]><![endif]>

<![if !vml]><![endif]>

180

Table 8-4, Class D

Bermuda … 0.1

Bermuda … 0.06

180

Table 8-4, Class E

0.1

0.04

184

Last line of top
paragraph

revegeated

revegetated

185

Problem 8.4

freeboard is 0.8 m

freeboard is 0.08 m

186

Problem 8.15

[Change the last
sentence.]

Determine the top and
bottom widths and design depths for both the 2 and 7 percent sections.

192

Figure
9-5

Expression
for apron length (at top of figure)

<![if !vml]><![endif]>

<![if !vml]><![endif]>

199

Ex
9.2, first equation

<![if !vml]><![endif]>

<![if !vml]><![endif]>

200

Ex 9.4, paragraph 4,
line 1

58.80 m – 57.73 m =
1.07 m

58.80 m – 57.76 m =
1.04 m

200

Ex 9.4, par 4, line 5

1.005 m.

0.975 m.

200

Ex.9.4, par 4, last line

0.99 m

0.97 m

200

Ex 9.4, par 5, line 1

(2’) a = 1.27 m, R = 0.44 m, v = 1.22
m/s, Q = 1.54 m³/s

(2’) A = 1.22 m, R = 0.43 m, v = 1.21
m/s, Q = 1.47 m³/s

200

Ex 9.4, par 5, line 2

0.076 m, giving a flow
depth of 0.99 m.

0.074 m, giving a flow
depth of 0.97 m.

200

Ex 9.4, par 5, line 3

discharge is 1.54 m³/s

discharge is 1.47 m³/s

200

Ex 9.4, par 5, line 5

discharge is 1.54 m³/s

discharge is 1.47 m³/s

206

Ex. 9.5, Solution,
line2

= 61.8 m.

= 68.1 m.

215

Example 9.8, equation

11.7 m

11.5 m

222

Last line

absolute viscosity of
water (LM^-1T^-1)

absolute viscosity of
water (ML^-1T^-1)

228

Problem 9.14

weir coefficient of
3.0

K_c = 1.0

weir coefficient of
1.7

K_e = 1.0

238

Figure 10-6, for
Stream Type A

ER < 1.4

1.4 < ER < 2.2

238

Figure 10-6, for
Stream Type B

1.4 > ER < 2.2

1.4 < ER < 2.2

276

Figure 12-3(c)

(The fluctuations
should be approximately sinusoidal with approximately two peaks per day. The
illustration is conceptual rather than precise.)

300

Equation after 3^rd
paragraph

V_g = PE_o + QS_x + RS_y

V_g = PE_o + QS_x
+ RS_y                       13.4

(add
equation number)

301

Ex. 13.3, Solution
step 3

<![if !vml]><![endif]>

<![if !vml]><![endif]>

(delete equation
number)

303

Example 13.4, last two
lines

the volume of fills
(653 m³). The cut-fill ratio resulting from a lowering of the
plane is 1.16.

the volume of fills
(623 m³). The cut-fill ratio resulting from this lowering of the
plane is 1.22.

312

Between Equations 14.5
and 14.6

drain flow rate

drainage rate per unit
area

312

Equation 14.6

<![if !vml]><![endif]>

<![if !vml]><![endif]>

312

Equation 14.7

<![if !vml]><![endif]>

<![if !vml]><![endif]>

314

Ex. 14.1, step (2)

x = … = 0.36 m

x = … = 0.36

315

Step (3)

… = 1.07

… = 1.07 m

316

Line 1

(b_o – b) [non-italic right parenthesis]

(b_o – b)

320

Figure 14-7

(10.5 m contour
missing)

(Insert 10.5 contour
approx. midway between 10.4 m and 10.6 m contours.)

323

Paragraph 2, line 2

spacing DC

spacing, DC

324

Figure 14-10

Spacing at 15.6 m
Spacing at 12.4 m

Spacing at 14.9 m
Spacing at 11.8 m

336

Example 15.1

<![if !vml]><![endif]>

<![if !vml]><![endif]>

336

Example 15.1, Columns
for Calculated P_e & Actual P_e

Calculated P_e:
57, 67, 68, 40, 31, 263

Actual P_e:
40, 67, 68, 40, 20, 235

Calculated P_e:
48, 55,58,41,13,215

Actual P_e:
40, 55, 58, 41, 13, 207

336

Example 15.1, last
line

235 mm

207 mm

337

Table 15-2, last line

(1.19-1.32)

(1.19-1.44)

350

Second line of text

Irrigation should
begin July 20

Irrigation should
begin June 20

366

Ex. 16.1, 3^rd
line

20(mm/h^0.45)
t(h)^0.45 + 5 (mm/h) t (h)

20(mm/h^0.45)
τ(h)^0.45 +
5(mm/h) τ(h)

379

Figure 16-9

[Insert label for the
x-axis.]

Discharge (L/s)

379

Figure 16-9

[Insert “L” in gap
between arrows above the siphon.]

379

Figure 16-9

[Insert “d” before “=
Outside”.]

379

Figure 16-9

[Insert “d” before “=
… mm” on each curve.]

379

Figure 16-9

[Insert “L” before “=
2 m” or “= 3 m” on each curve.]

382

Prob. 16.5, 3^rd
line

the elevation of the
water surface elevation

the elevation of the
water surface

397

Table 17-5, bottom of
left column

>130

≥130

401

First sentence

H_o in m =

H_d in m =

415

Problem 17.7

[Insert before the
last sentence.]

The static water level
is 30 m
below the surface.

428

Step (3), line 3

Table 15.4

Table 15-4

428

Step (3), line 4

the available water is
12 percent

the available water is
8 percent

428

Step (3), line 7

<![if !vml]><![endif]>

<![if !vml]><![endif]>

428

Step (3), line 8

<![if !vml]><![endif]>

<![if !vml]><![endif]>

429

Step (12), line 7

3.35 L/s

3.35 L/h

429

Step (12), last line

the lateral discharge
is 3.35 L/h.

the average emitter
discharge is 3.35 L/h.

430

Step (16), line 5

3.41

3.41 L/h

446

Example 19.2, last two
lines

<![if !vml]><![endif]>

<![if !vml]><![endif]>

449

Exp 19.3 Solution

Perforamnce =

Performance =

456

Figure 19-12, Brake
Power scale

0   20
40    60

[Delete the 0.
Change 20 to 30 and change 60 to 50.]

465

Equation 20.12

<![if !vml]><![endif]>

<![if !vml]><![endif]>

467

Equation 20.14

<![if !vml]><![endif]>

<![if !vml]><![endif]>

467

Step 1 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>

467

Step 5 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>

468

Step 7 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>

468

Step 8 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>

468

Step 11 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>[*]

468

Step 12 Eq

<![if !vml]><![endif]>

<![if !vml]><![endif]>

468

After Step 12

Thus, the estimated
soil loss is = 57.9 Mg ha^-1y^-1.

Thus, the estimated
soil loss is = 80 Mg ha^-1y^-1.