4EX Water
4EX Water
4EX Water
_________________________________________________________________
A) B) C)
The Answers is D
According to the fluid mechanic principles, the pressure is not related to the shape, so at
the depth of “h” the amount of pressure is the same for all shapes.
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Practice Problems PE Style Exam (AM)
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34) A 3h storm over a 150 km2 area produces a total runoff volume of 7*106 m3
With a peak discharge of 360 m3/Sec. What is the total excess precipitation?
A) 1.4 cm
B) 2.6 cm
C) 3.6 cm
D) 4.6 cm
The Answers is D
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Practice Problems PE Style Exam (AM)
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35) A 3h storm over a 150 km2 area produces a total runoff volume of 7*106 m3 with
a peak discharge of 360 m3/Sec. find the unit hydrograph discharge?
A) 78 m3/s.cm
B) 120 m3/s.cm
C) 210 m3/s.cm.
D) 260 m3/s.cm
The Answers is A
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Practice Problems PE Style Exam (AM)
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36) What is the flow rate for a street V channel finished (clean) concrete channel with
a width of 1’, channel slope of 0.5%, with a “normal” water depth of 0.5’?
A) 0.55 cfs
B) 1.20 cfs
C) 0.25 cfs
D) 2.41 cfs
The Answers is A
n is 0.015,
hydraulic radius= (d Cos α)/2
1’
α = Arch tan (0.5/0.5) =45o
So, HR = 0.5* (cos 45) / 2 = 0.176
d=0.5’
S is 0.005 ft/ft, so
V = 2.2 ft/sec
Q = V*A= 2.2 ft/sec*(0.5*0.5)/2*2 sq.ft. = 0.55 cfs
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
The Answers is C
58
Practice Problems PE Style Exam (AM)
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38) What is the definition According to the US environment protection (EPA) which
area needs permit for the land disturbing and it will called as “disturb”?
The Answers is B
US environment protection (EPA) effective March 10, 2003 any activity in the area of 1
or more acres needs NPDES (National Pollutant Discharge Elimination System) permit.
59
Practice Problems PE Style Exam (AM)
_________________________________________________________________
39) 8 MGD (million gallon per day) of water flows into the new schedule-40 steel
pipe network as shows below. Find the rate of flow in the upper branch.
A) 1.1 MGD
B) 6.2 MGD
C) 5.0 MGD
D) 1.95 MGD
The Answers is B
Parallel pipes have three principles that govern the distribution of flow between the two
branches. 1- head loss are the same for each branch 2- head loss at each junction is the
same as each branches, 3- the total flow rate is the sum of the flow rates in the two
branches. According to the 3rd principle, Vt = Va + Vb
So:
Diameter = 8” flow area = 50.24 in2
Diameter = 4” flow area = 12.56 in2
Using the Hazen-Williams expression for the velocity of flow in the pipe:
V= (0.55CD0.63hf0.54)/L0.54
V1= 0.55 * 80 * (8/12)0.63 * hf0.54 / 2000.54 = 1.95 hf0.54
The same for lower branch:
V2=2.28 hf0.54
. hf1=hf2
V1*A1/V2*A2 = (1.95 * 50.24) / (2.28*12.56) = 3.42, V2=0.29 V1
Vt =V1+V2 = V1.A1+0.29 V1.A2 = 1.29 V1.A1=1.29 Q1, then Q1= 5MGD/1.29 =
6.2MGD
60
Practice Problems PE Style Exam (AM)
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40) In a drainage project for an underground subway station if the required amount of
well drawdown is 3 ft. in the 50 ft. depth of the aquifer. The hydraulic conductivity is
given equal to 120 gal/(day-ft2), well radius is equal to 0.3 ft. and the water table recover
at radius of 1000 ft. find the required pumping flow rate ?
A) 13524 gal/day
B) 71020 gal/day
C) 18520 gal/day
D) 5624 gal/day
The Answers is A
( − )
=
1
( )
2
Where, K= hydraulic conductivity = 120 gal/ (day-ft2)
For the maximum drawdown of well we need to check it at the well center, so:
y1= 5 ft. and, y2=50-3 = 47 ft.
Since we want to find the Q at the center of well we do not need the information about
the radius of well and radial distance and r1=r2
120(50 − 47 )
= = 13524.17 /
1000
( )
0.3
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
The Answers is B
A siphon is a bent or curved tube that carries fluid from a container at a high elevation to
another container at lower elevation however, the fluid seems to flow “uphill” in a
siphon. See the following picture.
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
34) A 2h storm over a 111 km2 area produces a total runoff volume of 4*106 m3
With a peak discharge of 260 m3/Sec. What is the total excess precipitation?
A) 1.4 cm
B) 2.6 cm
C) 3.6 cm
D) 4.0 cm
The Answers is C
119
Practice Problems PE Style Exam (AM)
_________________________________________________________________
35) A 2h storm over a 111 km2 area produces a total runoff volume of 4*106 m3 with
a peak discharge of 260 m3/Sec., find the unit hydrograph discharge?
A) 72 m3/s.cm
B) 120 m3/s.cm
C) 210 m3/s.cm.
D) 260 m3/s.cm
The Answers is A
120
Practice Problems PE Style Exam (AM)
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36) What is the flow rate for a rectangular finished (clean) concrete channel with a
base width of 8’, channel slope of 0.5%, with a “normal” water depth of 2’?
A) 140 cfs
B) 8.5 cfs
C) 100 cfs
D) 200 cfs
The Answers is A
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Practice Problems PE Style Exam (AM)
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37) The ratio of the depth of flow to the hydraulic radius for the most economical
trapezoidal section, in open channel flow is
A) 0.5
B) 1
C) 2
D) 1.2
The Answers is C
The most efficient open channel cross section will maximize the flow for the given
Manning coefficient, slope, and flow area. The most efficient trapezoid channel is always
one which the flow depth is twice the hydraulic radius. If the side slope is adjustable, the
sides of the most efficient trapezoid should be inclined at 60 degree.
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Practice Problems PE Style Exam (AM)
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38) When does the silt fence barrier (below picture) be used for the sediment control?
The Answers is B
Choice “A” is the definition for bale slope barrier, “C” is the definition of a spillway, “D”
is the ditch with silt fencing. The barrier should be used at the toe of a slope when the
ditch does not exist.
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
39) 5 MGD (million gallon per day) of water flows into the new schedule-40 steel
pipe network as shows below. Find the rate of flow in the upper branch.
a. MGD
B) 3.9 MGD
C) 5.0 MGD
D) 1.95 MGD
The Answers is B
.
Parallel pipes have three principles that govern the distribution of flow between the two
branches. 1- head loss are the same for each branch 2- head loss at each junction is the
same as each branches, 3- the total flow rate is the sum of the flow rates in the two
branches. According to the 3rd principle, Vt = Va + Vb
So:
Diameter = 8” flow area = 50.24 in2
Diameter = 4” flow area = 12.56 in2
Using the Hazen-Williams expression for the velocity of flow in the pipe:
V= (0.55CD0.63hf0.54)/L0.54
V1= 0.55 * 80 * (8/12)0.63 * hf0.54 / 2000.54 = 1.95 hf0.54
The same for lower branch:
V2=2.28 hf0.54
. hf1=hf2
V1*A1/V2*A2 = (1.95 * 50.24) / (2.28*12.56) = 3.42, V2=0.29 V1
Vt =V1+V2 = V1.A1+0.29 V1.A2 = 1.29 V1.A1=1.29 Q1, then Q1= 5MGD/1.29 =
3.87MGD
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Practice Problems PE Style Exam (AM)
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20) A gutter at the side of a street has a curb height of 8 inches, slope along the length of the
street of 0.7%, and Manning n=0.032. The pavement slopes from the center of the street to
the gutter at 5%. What is the flow capacity of this gutter?
A) 4.9 cfs
B) 8.0 cfs
C) 11.1 cfs
D) 14.7 cfs
The Answers is B
Step 1: When the gutter is full, the water extends a distance =8/0.05=160 inches=13.3 feet from
the curb. The cross sectional area A=0.5(13.3)(8/12)=4.44 ft2, and wetted perimeter P=
8/12+13.3 = 14 ft, hydraulic radius Rh=(4.44)/(14)=0.317 ft.
K
Step 2: Manning’s equation for the discharge Q ARh2 / 3 S 1/ 2 , where K=1 for metric units,
n
K=1.49 for USCS units, n = Manning roughness coefficient
Rh=A/P = hydraulic radius, A = cross sectional area, P = wetted perimeter, S = channel slope, so
Q=(1.49/0.032)(4.44)(0.317)2/3 (.7/100)1/2 = 8.0 cfs.
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Practice Problems PE Style Exam (AM)
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21) Water from a 175-ac light industrial watershed is collected and drained by a trapezoidal
open channel. The channel (Manning' roughness coefficient, = 0.02) has a 4.5-ft-wide
bottom and 1:1 sides. The channel direction is perpendicular to a road where twin, side-by-
side 54-in-diameter corrugated metal pipe (CMP) culverts take the water under the roadway.
The average slope of the channel and culverts is 0.75% (i.e., 0.0075 ft/ft). The time for
runoff from the farthest part of the watershed to begin contributing to the flow is 35 min. -
Using the rational method and assuming the intensity after 35 min is 2 in/hr, what is the
runoff?
A) 308 B) 180
C) 228 D) 340
The Answers is C
= =( . ) ( )=
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Practice Problems PE Style Exam (AM)
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22) A parking lot adjoining a shopping center has a surface area of 3.5 acres. A rainstorm that
delivers rainfall at a rate of 2.5 inches/hour occurs. The parking area has a runoff
coefficient C=0.7. What is the peak runoff from the parking lot during this rainstorm?
The Answers is C
Q=CiA
where Q is the discharge, C is the runoff coefficient, i is the rainfall intensity in inches/hour, and
A is the drainage area in acres.
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Practice Problems PE Style Exam (AM)
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23) A 10 acre basin stores approximately 7.0 inches of water. What is the runoff for the given
basin for a 2-hr storm with an average of 0.5 in/hr of rainfall?
The Answers is B
= .
= ∗ . =
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Practice Problems PE Style Exam (AM)
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24) A horizontal pipeline carries water at a discharge of 13.5 cfs. Upstream of a contraction the
pipe diameter is 24 inches and pressure is 14 psi, while downstream of the contraction the
diameter is 18 inches. Neglecting head loss, what is the pressure downstream?
A) 11.2 psi
B) 13.7 psi
C) 14.2 psi
D) 15.9 psi
The Answers is B
p1 V12 p V2
z1 z2 2 2
Step 1: The Bernoulli equation states that 2g 2 g , where z =
elevation, p = pressure, V = average velocity, g = acceleration of gravity , and = specific
weight of water. Let point 1 be upstream of the contraction, and point 2 downstream.
Step 2: Since pipe is horizontal, z1=z2. Area A1= D2/4 = 3.14 ft2, A2= 3.14(18/12)2/4=
1.77ft2. V1=Q1/A1= (13.5)/(3.14)= 4.3 ft/sec, V2=Q2/A2= (13.5)/(1.77)=7.63 ft/sec. Also
p1=14 psi = 14(144)=2020 lbs/ft2, p1/ =4900/62.4 = 32.3 ft.
p2 p1 V12 V22
Step 3: Rearranging the Bernoulli Eq gives 2 g 2 g =32.3 + (4.3)2/2/32.2 -
(7.63)2/2/32.2 = 31.7 ft; p2=31.7(62.4 lbs/ft3)/144=13.7 psi.
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
25) Which one has more pressure at the depth of h = 10 ft? (the inclined surfaces has the
measure of angle of 60 degree.)
The Answers is A
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
20) Three rain gages are located within or nearby a watershed, whose total area is 1100 acres.
Using the Thiessen polygon method, the portions of the watershed associated with gages A, B,
and C are 520, 310, and 270 acres, respectively. For a particular storm, it is determined that the
excess rainfall for these 3 portions of the watershed is 0.95, 1.23, and 1.44 inches, respectively.
What is the volume of runoff from this watershed for this storm?
A) 19.2 MG
B) 28.3 MG
C) 34.3 MG
D) 44.7 MG
The Answers is C
Step 1: The volume of runoff is equal to the product of the watershed area and the excess
rainfall depth.
Step 2: Here the volume = (0.95 in)(1/12 ft/in)(520 acres)(43560 ft2/acre)+ (1.23 in)(1/12
ft/in)(310 acres)(43560 ft2/acre) + (1.44 in)(1/12 ft/in)(270 acres)(43560 ft2/acre)=4.59x106 ft3 =
34.3 MG (million gallons)
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Practice Problems PE Style Exam (AM)
_________________________________________________________________
21) At the design discharge, an existing single circular culvert has the following
characteristics: headwater elevation = 5.5 ft above invert, culvert diameter= 2.5 ft, critical depth
= 1.9 ft, uniform depth = 1.7 ft, and the outlet is free. Which of the following describes the
culvert flow under these conditions?
The Answers is B
Step 1: Since the critical depth is greater than the uniform depth, the culvert slope is steep.
Therefore the culvert is under inlet control.
Step 2: Since the headwater elevation above the invert is more than 1.2 times the pipe diameter,
the inlet will be submerged, and an orifice flow condition exists at the inlet.
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Practice Problems PE Style Exam (AM)
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22) Runoff from a 3-acre site is to be drained by a channel. The time of concentration for
this site is 40 minutes. The site has a runoff coefficient C=0.2. Rainfall quantities to be used for
design are 0.5 inches for a storm of duration 20 minutes, 0.7 inches for duration 40 minutes, and
0.9 inches for duration 60 minutes. For what discharge should this channel be designed?
The Answers is B
Q=CiA
where Q is the discharge in cfs, C is the runoff coefficient for the watershed, i is the
Step 2: In applying the rational method, rain falling over a time period equal to the time
of concentration of the watershed should be used. In this case, the time of concentration
is given as 40 minutes, or 0.67 hours, and the corresponding rainfall amount is 0.7 inches. So
the rainfall intensity is
Step 3: Solver for the discharge Q = C i A= (0.2) (1.04) (3) = 0.63 cfs
248
Practice Problems PE Style Exam (AM)
_________________________________________________________________
The Answers is B
The total precipitation is the sum of all the rainfall over a given period of time for a watershed.
249
Practice Problems PE Style Exam (AM)
_________________________________________________________________
24) All three pipes in the figure below have the same length and are connected in parallel.
The diameters are given. Determine the ratio of velocities in the branches. Assume the friction
factor for all the lines to be the same.
The Answers is C
fA( )( ) = f ( )( ) = f ( )( )
( ) = ( ) = ( )
( ) = ( ) = ( )
. .
2 2 2
1 VA = 1.2 VB = 1.5VC
Let Vc = 1
Then
1.5 (1)2 = 1.2 VB2
. ( )
VB = [ ]
.
VB =1.118
Also
1.5 (1)2 = VA2
. ( )
VA = [ ]
.
VA = 1.224
V to V toV = 1 to 1.118 to 1.224
250
Practice Problems PE Style Exam (AM)
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25) What is the approximate runoff for a watershed with the given hydrograph?
The Answers is C
Step 1: Divide the area under the hydrograph into easily calculable areas
. ∗( . . )
Triangle 1 = ∗ 3600 = 8190
. ∗( . . )
Triangle 2 = ∗ 3600 = 6300
. ∗( . . )
Triangle 3 = ∗ 3600 = 2700
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Practice Problems PE Style Exam (AM)
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26) Two pipelines carry water from a common starting point to a common end point. The
two pipes have the same friction factor f and diameter, but pipe 1 is twice as long as pipe 2.
What fraction of the discharge between the start and end points flows through pipe 1?
The Answers is B
Step 1: For 2 pipes in parallel, the head loss in each pipe is equal.
Step 2: Using the Darcy Weisbach equation, equating the head losses in pipes 1 and 2 gives f1
(L1/D1) V12/2g = f2 (L2/D2) V22/2g
Step 3: Given that f1 = f2, D1 = D2, we get that V12/V22 = (L2/L1)=0.5, or V1/V2 =(0.5)1/2
= 0.71. Since the diameters and pipe areas are the same Q1/Q2 = 0.71. Then Q1/(Q1+Q2)=
1/(1+1/0.71)=0.41 = 41%
252