1. No category

Practice Exam 1
AM (A)
Solutions
Practice Exam 1
AM (A)
Practice Exam 1
AM (A)
Solutions
Q. 1
A storm drain inlet for a 2.7 acre parking area is to be designed. The rainfall intensity is 8.6 in/hr
and the runoff coefficient is 0.96. The peak discharge, in cubic feet per second, is most nearly:
(A)
(B)
(C)
(D)
18
19
20
22
Solution
Given:
C = 0.96
I = 8.6 in/hr
A = 2.7 acre
Qp = C*I*A = 0.96*8.6*2.7 = 22.3 ft3/sec
THE CORRECT ANSWER IS (D)
Practice Exam 1
AM (A)
Solutions
Q. 2
The cantilevered retaining wall shown is under design. What is most nearly the active resultant
per unit length of wall?
118.4lbf / ft 3
31o
(A)
(B)
(C)
(D)
280 lbf/ft
1900 lbf/ft
4260 lbf /ft
8530 lbf/ft
Hint: Calculate the coefficient of active earth pressure and use it to find the active resultant.
Solution
The Rankine method may be used to find the active earth pressure acting at the base of the wall
and granular backfill. Determine the active earth pressure acting at the base of the wall and then
determine the active resultant. The equation for active pressure at the base of the wall is
pa kav kaH
The equation for the coefficient of active earth pressure is
1 sin  1 sin 31
ka 

1 sin  1 sin 31
= 0.32
Calculate the resultant of the triangular pressure distribution.
Ra 12 pa H 12 kaH 2
lbf 
1 

 (0.32) 
118.4 3 
(15ft) 2
ft 
2 

= 4262 lbf/ft of wall length (4260 lbf/ft)
Note that the resultant acts horizontally at the centroid of the triangular distribution at one third
the height of the wall measured upward from the base.
THE CORRECT ANSWER IS (C)
Why Other Options Are Wrong
Practice Exam 1
AM (A)
Solutions
(A) This solution fails to square the height of the wall in the resultant calculation. Note that
the units are incorrect as well.
(B) This incorrect solution assumes the active pressure acts at the centroid of the pressure
distribution diagram; actually, it is the resultant active force that acts. This answer is
obtained by multiplying by a centroid depth of 10 ft rather than 15 ft.
(D) This solution fails to divide the result by two, as required by a triangular distribution.
Q. 3
The maximum specific energy in a rectangular channel is 9 feet. The critical velocity, in ft/sec,
is most nearly:
(A) 8.7
(B) 12.3
(C) 14.1
(D) 15.4
Solution
Solve for the critical depth, which will allow you to solve for the critical velocity.
dc = 2/3 Ec = 2/3 * 9 = 6 feet
vc = (gdc)1/2 = (32.2 * 6)1/2 = 13.9 ft /sec
THE CORRECT ANSWER IS (C)
Practice Exam 1
AM (A)
Solutions
Q. 4
A 10ft by 10ft square, reinforced concrete footing is installed so that the footing bearing surface
is 5 ft below the soil level, at a point where the allowable soil pressure is 3500 psf. Other than
the soil above the footing, there is no surcharge. The soil unit weight is 100 lbf/ft3. The footing
is located at the corner of a building and is loaded through a concentric 14 in square column.
The column transmits a 125,000 lbf service dead load and a 175,000 lbf service live load to the
footing. The critical (plan) area contributing to two-way punching shear is assumed to be
92.44ft2.Th
ede
a
dl
oa
di
nc
l
ude
st
hec
ol
umn’
swe
i
g
htbutdoe
snoti
nc
l
udet
hef
oot
i
ng’
swe
i
ght.
The compressive strength for all of the concrete used is 3000 psi.
The ultimate two-way punching shear, in lbf, is most nearly?
(A)
(B)
(C)
(D)
350,000
450,000
550,000
650,000
Solution
The footing area is
A B 2 (10 ft )(10 ft ) 100 ft 2
The ultimate load carried by the footing is
P 1.4 Pd 1.7 Pl
(1.4)(125, 000 lbf)+(1.7)(175,000 lbf)
472,500 lbf
The factored soil pressure acting on the footing is
P 472, 500 lbf
pu  u 
A
100 ft 2
4725 lbf/ft 2
The ultimate two-way punching shear is
lbf 

Vu pu Acritical 4725 2 
92.44ft 2 

ft 

436, 779 lbf
THE CORRECT ANSWER IS (B)
Practice Exam 1
AM (A)
Solutions
Q. 5
A 6.0 MGD water treatment plant is being planned that will use a local river as its water source.
The raw water characteristics of the river are shown in Table 1.
Tracer studies done on the disinfection tank of a drinking water treatment plant after construction
produced the plot shown in the figure below. To satisfy the surface water treatment rule for
disinfection of Giardia (Table 2) on a day when the peak hourly flow rate is 5 MGD, the water
temperature is 10
C and the pH is 7.0, the residual chlorine concentration (mg/L) needed is most
nearly:
NOTE: The chlorine dosage must not exceed 1/8 mg/L to minimize THM formation.
(A) 0.60
(B) 0.52
(C) 0.46
(D) 0.42
200
150
CONTACT
TIME, t10
100
(min)
50
0
1
2
3
4
5
6
PEAK HOURLY FLOW RATE
TABLE 1
Parameter
Turbidity
pH
Alkalinity
Calcium
Temperature
Value
17 NTU
6.8-7.1 S.U.
150 mg/L as CaCO3
51 mg/L as CaCO3
10
C to 26
C
7
Practice Exam 1
AM (A)
Solutions
TABLE 2
CT values (mg/L
min) for inactivation of Giardia cysts by free chlorine at 10
C
Chlorine
Concentration
(mg/L)
0.5
1.0
pH=7.0
Log Inactivations
1.5
2.0
2.5
3.0
18
33
33
70
82
105
0.4
19
38
56
75
94
113
0.6
20
40
59
79
99
119
0.8
21
41
62
82
103
123
1.0
21
42
64
85
106
127
1.2
22
44
65
87
109
131
1.4
22
45
67
89
112
134
1.6
23
46
68
91
114
137
1.8
23
46
70
93
116
139
2.0
24
47
71
95
118
142
2.2
24
48
72
96
120
144
2.4
24
49
73
97
122
146
2.6
25
49
74
99
123
148
2.8
25
50
75
100
125
150
3.0
Source: U.S. Environmental Protection Agency, Guidance Manual for Compliance with
Filtration and Disinfection Requirements for Public Water Systems Using Surface Water
Sources, Criteria and Standards Division, Office of Drinking Water (U.S.E.P.A. NTIS
Publication NO. PB 90-148016), Washington, D.C: U.S. Government Printing Office, October,
1979.
Solution
Surface Water Treatment Rule for disinfection of Giardia; 3 log inactivation required. 2.5 log
inactivation allowed for treatment prior to disinfection. Therefore, 3-2.5 = 0.5 log inactivation
required by disinfection.
From the figure, at a peak hourly flow rate of 5 MGD, t10 = 50 min contact time.
CT 23 mg / L min; CT concentration x time
Ref: Davis 7 Cornwell, Environmental Engineering, McGraw Hill, 1998, p. 244, Table 3-20.
C = CT/T (23 mg / L min) / 50 min 0.46 mg / L residual chlorine concentration needed.
THE CORRECT ANSWER IS: (C)
Practice Exam 1
AM (A)
Solutions
Q. 6
An eight phase intersection looses 2 seconds per phase and has an all-red duration of 3 seconds
on phases 1 and 4. If the sum of the ratios of approach flow to saturation flows for all phases is
0.
7(
i
.
e
.t
hef
l
owt
os
a
t
ur
a
t
i
onf
l
owsf
ora
l
lpha
s
e
si
s0.
7(
i
.
e
.t
he"
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ue
)
,t
he
nt
heopt
i
mum
cycle time, in seconds, is most nearly:
(A)
(B)
(C)
(D)
80
95
105
125
Solution
1 5 L 5
Co  o
1 y
i 1
where
Co = optimum cycle time (sec)
L = total lost time per cycle (sec)
yi = maximum value of approach flows to saturation flows for all lane groups using
phase i, qij/s;

L i R
i 1
2 8 2 3
= 22 sec
(1.5 22) 5
Co 
1 0.7
= 126.7 sec
THE CORRECT ANSWER IS (D)
Practice Exam 1
AM (A)
Solutions
Q. 7
I
ft
hede
pt
hoff
l
owi
ngwa
t
e
ri
nadi
t
c
hwi
t
hMa
nni
ng
’
sr
oug
hne
s
sof0.
02a
nda
na
ve
r
a
g
ebe
d
slope of 0.5% is 2 feet, the velocity (fps) of the water would be most nearly:
(A) 3.5
(B) 3.8
(C) 5.6
(D) 6.8
1
1
3
3
2' - 0"
DITCH SECTION
NOT TO SCALE
Solution
Determine the water velocity for the trapezoidal ditch. For a depth = 2.0 ft.
A 
2.0 3x2.0 
2.0 16.0 ft 2
b zy y 2.0 3x2.0 2.0 1.092 ft
R
b 2 y 1 z 2 2.0 2x2.0 1 32
2
1
1.486 2 3 1 2 1.486
V
R s 
x
1.092 3 
0.005 2 5.6 ft / s
n
0.02
THE CORRECT ANSWER IS: (C)
Practice Exam 1
AM (A)
Solutions
Q. 8
Analyze the truss shown.
h
2
h
2
The force in member AH is most nearly
(A)
(B)
(C)
(D)
300 lbf (compression)
600 lbf (compression)
670 lbf (compression)
670 lbf (tension)
Hint: Solve by using the method of joints.
Solution
From the sum of the moments about E, the vertical reaction at support A is
(300 lbf)(2h)+(200 lbf)(3h)
RA,v 
4h
=300 lbf
Draw the free-body diagram of joint A.
By summing the forces in the horizontal direction, determine that the horizontal reaction at A is
zero.
The sum of the forces in the vertical direction is
AH v RA,v 0 lbf
Practice Exam 1
AM (A)
Solutions
AHv = 300 lbf (compression)
From the geometry of the truss, the horizontal component of AH must be twice the vertical
component.
AHh = (2)(300 lbf)
= 600 lbf (compression)
The resultant force in member AH is
AH  AH v2 AH h2
 (300 lbf) 2 
600 lbf 
2
= 670.8 lbf (670 lbf (compression))
THE CORRECT ANSWER IS (C)
Why Other Options Are Wrong
(A) This incorrect solution finds only the vertical component of the force in AH.
(B) This incorrect solution finds the horizontal component of the force in AH.
(D) This incorrect solution identifies the resultant force in AH as a tensile force.
Practice Exam 1
AM (A)
Solutions
Q. 9
A city wants to design a sludge dewatering system for their 20-MGD secondary, activated sludge
treatment plant shown below.
The sludge volume reduction (%) achieved by the thickener is most nearly:
(A) 50
(B) 58
(C) 65
(D) 70
BOD f =5mg/L
TSS1 =200mg/L
TSSPE =80mg/L
TSSf =10mg/L
10, 000 lb/d@7500
Solution
Assuming specific gravity of sludge is 1.00
Sludge quantity and volume to be dewatered.
Primary Sludge
lb / d (200 mg / L 80 mg / L)(8.34)(20 MGD ) 20, 016 lb / d
20, 016 lb / d
gpd
68.571 gpd
(8.34 lb / gal )(0.035)
Secondary Sludge (was)
10, 000 lb / d
gpd
0.1599 MGD 0.159,872 gpd
(8.34 lb / gal )(7,500 mg / L)
Before Thickening
Total quantity = 20, 016 10, 000 30, 016 lb / d
Total volume = 68,571+159,872 = 228,443 gpd
Practice Exam 1
AM (A)
Solutions
After Thickening
Total quantity = 30,016 lb/d
Total volume
30, 016

79,979 gpd
(8.34 lb / gal )(0.045 percentage)
% R eduction in sludge volume=
CORRECT ANSWER IS: (C)
228.443-79,979
65%
228.443
Practice Exam 1
AM (A)
Solutions
Q. 10
Groundwater monitoring wells have been installed at the site of a proposed sanitary landfill such
that Well B is located 1,500 feet north and 300 feet west of Well A.
A proposed containment cell, having bottom dimensions of 500 feet (north to south) and 100 feet
(east to west), is to be located such that the southeast corner is 100 feet west and 500 feet north
of Well A.
The bottom of the landfill cell must be a minimum of 5 feet above the groundwater elevation.
The elevations that were determined at each well location are shown in the table below.
The minimum bottom elevation for the proposed landfill cell if the bottom is to be level is most
nearly:
(A) 227
(B) 232
(C) 237
(D) 242
Well
A
B
Groundwater
Elevation(ft)
229.75
222.25
Ground Surface
Elevation (ft)
248.75
243.75
Solution
Determine minimum bottom elevation of the containment cell.
The highest groundwater elevation within cell area = groundwater elevation at point A less the
groundwater drop to closest point of the cell, Point C highest groundwater elevation within cell
area = groundwater elevation at Point A less the (distance A to C)*I
Distance Well A to cell [(500) *(500) (100)*(100)0.5 ] 510 ft
Highest groundwater elevation within cell area 229.75 510* 0.0049 227.25
Minimum bottom elevation of containment cell = highest groundwater elevation within cell area
plus 5 feet = 227.25 + 5 feet
Minimum bottom elevation of containment cell = 232 feet
CORRECT ANSWER IS: (B)
Practice Exam 1
AM (A)
Solutions
Q 11
A soil sample, taken from a borrow pit has a specific gravity of soil solids of 2.66. The sample
was taken to a materials laboratory and tested. The results of a standard Proctor test are
tabulated below.
Weight of soil (lb) Moisture content (%)
3.20
12.8%
3.78
13.9%
4.40
15.0%
4.10
15.7%
3.70
16.6%
3.30
18.1%
The maximum dry density, in lb/ft3, is most nearly:
(A) 85
(B) 90
(C) 100
(D) 115
Solution
Step 1: Derive unit weight (of wet soil) from:


= weight of soil/volume of proctor mold,
where volume of proctor mold = 1/30 ft3.
Weight Moist unit
of soil weight
lb
lb/ft3
W


3.2
96
3.8
113
4.4
132
4.1
123
3.7
111
3.3
99
Practice Exam 1
AM (A)
Solutions
Step 2: Derive dry density 
d
d / (1+w)
Moist unitMoisture Dry
weight content density
lb/ft3
%
lb/ft3



w
d
96
11%
86
113
13% 100
132
15% 115
123
17% 105
111
18%
94
99
19%
83
Step 3: Plot Moisture content (w) vs dry density (
d)
115
110
Dry unit weight,
d
(lb/ft3)
120
105
100
95
90
85
80
10%
12%
14%
16%
18%
20%
Moisture Content, w (% )
Step 4: Identify maximum dry density (
d) and optimum moisture content (w) from the plot:
Optimum moisture content = w = 15% and 
d(max) = 115 pcf
THE CORRECT ANSWER IS (D)
Practice Exam 1
AM (A)
Solutions
Q. 12
A horizontal curve for a section of a highway has a design speed of 60 mph. The terrain restricts
the radius of the curve to 1200 ft, which therefore requires the super elevation at the curve, in
percent, to be most nearly:
(A) 4
(B) 6
(C) 8
(D) 10
Solution
u2
Radius ( ft ) 
15(e f s )
Given the radius of 1200 feet and the design speed of 60 mph, the side friction factor (fs) comes
from Exhibit 3-14onpa
ge145oft
he2001AASHTO“
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ome
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e
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pe
e
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ti
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a
t
e
st
ha
tf
ora60mphde
s
i
g
n
speed the side friction factor is 0.12
fs = 0.12
Substituting the known values into the radius equation, the only unknown is the super elevation,
e. By solving for e, the super elevation rate is found.
(60mph) 2
1200 ft 
15(e 0.12)
602
3600
e 0.12 
e
0.12 0.08 8%
1200(15)
1200(15)
THE CORRECT ANSWER IS ( C )
Practice Exam 1
AM (A)
Solutions
Q. 13
A horizontal curve is defined by its radius of 851 feet, as shown below. The length of the chord
AB, in feet, is most nearly:
A
B
12%
36o
(A)
(B)
(C)
(D)
505
515
525
535
Solution
The chord is the straight-line distance from A to B.
AD
sin 36 
2
AO
AD C / 2
 
Practice Exam 1
AM (A)
Solutions
AO R 851
C 2(851)sin(18) 525.95
Arc ACB=
2851
534.70
360
THE CORRECT ANSWER IS (C)
Q. 14
A concrete column and footing caries the loads shown.
The footing is 6 ft wide.
column dead load
column live load
moment due to wind
compressive strength of concrete
maximum allowable soil pressure
80 kips
100 kips
50 ft-kips
3000 lbf/in2
3000 lbf/in2
What is the minimum footing length, L, required for the entire footing to be considered effective
in carrying these loads?
(A)
(B)
(C)
(D)
1.7 ft
1.8 ft
7.7 ft
10 ft
Hint: For the entire footing to be effective, avoid tensile stress in the soil.
Practice Exam 1
AM (A)
Solutions
Solution
Limit eccentricity, e, to one-sixth of the footing length to keep the soil resultant within the kern
limit and avoid tension in the soil.
M
M
e   wind
P PD PL
50 ft kips

80 kips 100 kips
= 0.28 ft
L
e
6
L
0.28 ft 
6
Solving for L,
L 
0.28 ft 
6 1.68 ft

1.7 ft 

THE CORRECT ANSWER IS (A)
Why Other Options Are Wrong
(B) This incorrect solution calculates eccentricity using factored loads. Factored loads are used
in concrete design, but not for determining footing size. Footing size should be based on
unfactored soil pressure.
(C) This incorrect solution calculates the required size of a square footing (ignoring the footing
width given) based on the dead and live loads and allowable soil pressure. The problem
statement is asking for the minimum length required for the entire footing to be effective, not the
minimum size based on the soil pressure.
(D) This incorrect solution calculates the required length based on the dead and live loads and
allowable soil pressure. The problem statement is asking for the minimum length required for
the entire footing to be effective, not the minimum size based on the soil pressure.
Practice Exam 1
AM (A)
Solutions
Q. 15
The mechanical and plasticity tests of a soil under consideration as a fill material are shown
below:
A soil is under investigation. Mechanical and plasticity test results are shown below:
Mechanical Analysis
Sieve % passing by weight
10
19
40
25
200
66
Plasticity
Liquid Limit Plastic Limit
40
20
The soil may be classified, according to the Unified Soil Classification (USCS) system as:
(A)
(B)
(C)
(D)
GP
SW
CL
CH
Solution
Refer to the Unified Soil Classification System, Technical Memorandum No. 3-357, US Army
Engineers Waterways Experiment Station, Vicksburg, Mississippi, 1960. Note, this
classification table is reproduced in most geotechnical textbooks. You will find it typically titled
a
s“
TheUni
f
i
e
dSoi
lCl
a
s
s
i
f
i
c
a
t
i
onSy
s
t
e
m”
.
The Unified Soil Classification system has two analytical components:
1. Particle size
2. Plasticity
Particle size
Less than half of the material is larger than No. 200 sieve: the material is a Fine-grained soil.
Plasticity
TheAt
t
e
r
be
r
g
’
spl
oti
sa
bovet
he“
A”l
i
ne
,a
ndt
hePIi
sg
r
e
a
t
e
rt
ha
n7,t
heg
r
oups
y
mboli
sCL.
The soil can be classified as CL –Silty Clay.
THE CORRECT ANSWER IS (C)
Practice Exam 1
AM (A)
Solutions
Q. 16
A transition curve is to be used to implement a change in cross-section from a normally
crowned section to a fully superelevated section. The outer lane is to be gradually aligned
from the normally crowned section to a straight level section at the Tangent-to-Spiral (T.S.)
point. The full superelevation is rotated about the centerline.
Degree of Curve (D) = 2.5°
Two 11-foot lanes
Design superelevation = 0.08 ft/ft
Grade = +1.50%
Crown = 0.015 ft/ft
T.S. Station = 100 + 00.00
T.S. CL Elevation = 2,500.00 ft.
Length of spiral = 230 ft
The station (feet) where full superelevation is reached is most nearly:
(A) 100 + 00
(B) 100 + 25
(C) 101 + 00
(D) 102 + 50
Solution
S.C. Sta. = T.S. Sta. + LS
= (100+00) + (2 + 30) = 102 + 30
THE CORRECT ANSWER IS: (D)
Practice Exam 1
AM (A)
Solutions
Q. 17
Using the NDS, which of the following statements must be true?
I.
The temperature factor, Ct, applies to member subjected to extremely cold
temperatures.
II.
The volume factor, CV, applies only to glued laminated timber bending members.
III.
The bending design value, Fb, for a floor framed with 1 x 6 sawn lumber joists must
be multiplied by the repetitive member factor.
IV.
The load duration factor, CD, does not apply to the modulus of elasticity values.
(A)
(B)
(C)
(D)
I and II
II and III
II and IV
III and IV
Hint: Refer to NDS Sec 2.3 for adjustment of design values.
Solution
Section 2.3 and Table 2.3.1 of the NDS contain the adjustment factors and their applicability for
design values.
The volume factor, CV, is discussed in the footnotes to NDS Table 2.3.1. Footnote 3 states, "The
volume factor, CV, shall apply only to glued laminated timber bending members (see Sec.
5.3.2)." Statement II is true.
The load duration factor, CD, is discussed in NDS sec. 2.3.2 and NDS table 2.3.2. Both NDS
Sec. 2.3.2.1 and Footnote 1 to NDS Table 2.3.2 state that "Load duration factors shall not apply
t
omodul
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l
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s
t
i
c
i
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a
t
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me
ntI
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r
ue
.
THE CORRECT ANSWER IS (C)
Why Other Options Are Wrong
(A) Although statement II is true, statement I is false. The temperature factor, Ct, is discussed in
NDS Sec. 2.3.4. The temperature factor applies to members subjected to sustained exposure to
elevated (over 100 F) temperatures. Because the statement applies to members subjected to
extreme cold, not heat, statement I is false.
(B) Although statement II is true, statement III is false. The repetitive member factor, Cr, is
discussed in Footnote 5 of NDS Table 2.3.1 Footnote 5 states, "The repetitive member factor,
Cr, shall apply only to dimension lumber bending members 2 in to 4 in thick (see Sec. 4.3.4)."
Because the joists in this case are 1 in thick, statement III is false.
(D) Although statement IV is true, statement III is false because the repetitive member factor,
Cr, applies only to dimension lumber bending members 2 in to 4 in thick. The joists in this case
are 1 in thick.
Practice Exam 1
AM (A)
Solutions
Q.-18
A soil sample weighing 66.8 lb with a moisture content of 19 percent, volume of 0.55 cubic feet,
has a specific gravity of 2.72. The degree of saturation is most nearly:
(A)
(B)
(C)
(D)
0.5
0.6
0.7
0.8
Solution
W / V
= 66.83 / 0.55
= 121.5 pcf
d / (1 + w)
d = 121.5 / (1 + 0.19)
d = 102.1 pcf
d = (Gs * w ) / (1 + e)
e = ((Gs * w ) / d ) –1
e = ((2.72 * 62.4) / 102.1) –1
e = 0.66
S = (w * Gs) / e
S = ((0.19)*(2.72)) / 0.66
S = 0.78
THE CORRECT ANSWER IS: (D)
Practice Exam 1
AM (A)
Solutions
Q. 19
What is the average hydraulic detention time for a rectangular tank with dimensions of 2.5 m by
15 m by 3.0 m deep receiving a flow of 900 m3/d? The hydraulic efficiency of the tank is 83%.
(A)
(B)
(C)
(D)
2.3 h
2.5 h
3.0 h
3.6 h
Hint: The average detention time will be less than the theoretical detention time.
Solution
E
Q
t
ta
V
fractional efficiency
flow rate
theoretical hydraulic detention time
actual hydraulic detention time
volume
m3/d
d
d
m3
 h
(2.5m)(15m)(3.0m) 24 
V
 d
t 
3
m
Q
900
d
3.0 h
ta tE
83%

3.0 h 
100%
=2.5 h
THE CORRECT ANSWER IS (B)
Why Other Options Are Wrong
(A)
This incorrect solution divides the flow rate by the volume to get the theoretical detention
time. The theoretical detention time is then divided by the percent efficiency instead of being
multiplied by the fractional efficiency. Other definitions and equations are unchanged from the
correct solution.
 m3 
 h
900


24 
d 
Q
 d

t 
V 
2.5m)(15m)(3.0m 
192 h/d 2
h
192 2
ta  d
83%
2.3 h
Units do not make sense.
Practice Exam 1
AM (A)
Solutions
(C) This incorrect solution calculates the theoretical detention time. The hydraulic efficiency
is ignored. Other definitions and equations are unchanged from the correct solution.
h
2.5m)(15m)(3.0m 
24 
 d
t
3
m
900
d
3.0 h
(D)
In this incorrect solution, the theoretical detention time is divided by the hydraulic
efficiency instead of being multiplied by it. Other definitions and equations are unchanged from
the correct solution.
h
2.5m)(15m)(3.0m 
24 
 d
t
3
m
900
d
3.0 h
100% 

ta 
3.0 h 


83% 
=3.6 h
Q. 20
Assume all soils in a drainage basin are in the Soil Conservation Service (SCS) hydrologic soil
Group B. Also assume that the vegetative covers are in good condition. The land use is parks
and open space. The SCS Runoff Curve Number (CN) for the entire area is most closely
approximated by:
(A) 43
(B) 54
(C) 61
(D) 81
Solution
Determine the SCS Runnoff Curve Number for the entire area. For soil Group B, with good
vegetative cover in urban, fully developed open space (parks, lawns), the appropriate SCS curve
number (Gupta, Hydrology and Hydraulic Systems, Prentice Hall, 1989, p. 101)
CN = 61
THE CORRECT ANSWER IS (C)