Free PE Civil — WRE practice problems

Ten real questions from our PE Civil: Water Resources and Environmental bank, drawn from 10 of its 12 study topics, with fully worked solutions and the same figures our members get. No signup, no email. Try each one first, then reveal the solution.

5 with figures10 topicsFull worked solutions

1
medium
A. Storms, IDF, time of concentration & runoff
A commercial parking catchment of $A=12\,\text{acres}$ has a runoff coefficient $C=0.65$ and a time of concentration $T_c=18\,\text{min}$ . The local 10-yr IDF curve is fitted as $i=a/(t_d+b)^c$ with $a=64.1$ , $b=11$ , $c=0.81$ ( $t_d$ in minutes, $i$ in in/hr). The 10-yr peak discharge from the catchment (cfs) is most nearly:
Figure — drag or pinch to move around
1. 14.8 cfs
2. 15.8 cfs
3. 23.4 cfs
4. 32.7 cfs
Show the answer & worked solution
Answer: D.32.7 cfs
Rational method (PE Civil Reference Handbook §6.5.2.1): the design intensity is read from the IDF curve at a duration equal to the time of concentration, $i=\dfrac{64.1}{(18+11)^{0.81}}=4.19$ in/hr at $t_d=T_c=18$ min. Then $Q=C\,i\,A=0.65(4.19)(12)=32.7$ cfs, taking 1 acre-in./hr as 1.0 cfs as the handbook does. Distractors: applied the exponent to $T_c$ alone, as the handbook's §6.5.3.1 equations do, $i=a/(T_c^{\,c}+b)=3.00$ in/hr and $Q=23.4$ cfs; divided first and raised the whole quotient to the power, $i=\left(a/(T_c+b)\right)^{c}=1.90$ in/hr and $Q=14.8$ cfs; and read the curve at the 1-hr storm ( $t_d=60$ min) instead of at $T_c$ , $i=2.03$ in/hr and $Q=15.8$ cfs.
2
medium
A. Soil classification, properties, concrete & piping
A nuclear-gauge field density test on a compacted lift of lean clay (CL) embankment fill gives a dry unit weight $\gamma_d=109.5$ pcf at a water content $w=16\%$ . Laboratory tests give a specific gravity of solids $G_s=2.70$ . The degree of saturation $S$ of the compacted soil is most nearly:
Figure — drag or pinch to move around
1. 28.1%
2. 43.2%
3. 55.0%
4. 80.2%
Show the answer & worked solution
Answer: D.80.2%
The dry unit weight gives the void ratio (PE Civil Reference Handbook §3.8.3, Dry Unit Weight: $\gamma_d=G_s\gamma_w/(1+e)$ , $\gamma_w=62.4$ pcf): $e=\dfrac{(2.70)(62.4)}{109.5}-1=0.539$ . Then $Se=wG_s$ : $S=\dfrac{(0.16)(2.70)}{0.539}=0.802=80.2\%$ . Distractors: LEFT OUT the $-1$ , taking $e=G_s\gamma_w/\gamma_d=1.54$ ( $S=28.1\%$ ); wrote $Se=wG_s$ and never divided by $e$ ( $wG_s=43.2\%$ ); and took the gauge's DRY unit weight for a moist one and divided it by $(1+w)$ again, $109.5/1.16=94.4$ pcf, so $e=0.785$ ( $S=55.0\%$ ).
3
medium
A. Uniform flow, specific energy & critical depth
An earthen trapezoidal irrigation canal has a bottom width $b=8\,\text{ft}$ , side slopes $z=2$ (H:V), and runs at a normal depth $y=3\,\text{ft}$ on a slope $S=0.0005$ with Manning $n=0.025$ . The uniform discharge is most nearly:
Figure — drag or pinch to move around
1. 34.4 cfs
2. 87.5 cfs
3. 91.5 cfs
4. 116 cfs
Show the answer & worked solution
Answer: B.87.5 cfs
Manning's equation (§6.4.5.1), with the trapezoid's area and wetted perimeter from the Geometric Elements of Channel Sections table (§6.4.5.4): $A=(b+zy)y=(8+2\cdot3)3=42$ ft², $P=b+2y\sqrt{1+z^2}=8+2(3)\sqrt{1+2^2}=21.416$ ft, $R=A/P=1.961$ ft. $Q=\frac{1.486}{n}AR^{2/3}S^{1/2}=\frac{1.486}{0.025}(42)(1.961)^{2/3}(0.0005)^{1/2}=87.5$ cfs. Distractors: used the RECTANGLE area $b y$ , dropping the side triangles ( $34.4$ ); used $P=b+2y$ , omitting $\sqrt{1+z^2}$ on the slanted sides ( $116$ ); and took the wetted perimeter as $b+2zy=20$ ft, the horizontal run of the side slopes (the top width), instead of their slant length ( $91.5$ ).
4
medium
A. Lateral earth pressure & bearing capacity
A continuous (strip) footing of width $B=7\,\text{ft}$ is founded at depth $D_f=3\,\text{ft}$ in soil with cohesion $c=400\,\text{psf}$ , friction angle $\phi=32^\circ$ , and unit weight $\gamma=120\,\text{pcf}$ . Assume the same soil above and below the base, no surface surcharge, and no groundwater within the zone of influence. Using the bearing capacity factors in the PE Civil Reference Handbook, the ultimate (gross) bearing capacity $q_{ult}$ is most nearly:
Figure — drag or pinch to move around
1. 26,900 psf
2. 35,200 psf
3. 39,100 psf
4. 47,900 psf
Show the answer & worked solution
Answer: B.35,200 psf
Strip-footing bearing capacity equation (§3.4.2), with the Bearing Capacity Factors table row at $\phi=32^\circ$ : $N_c=35.5$ , $N_q=23.2$ , $N_\gamma=30.2$ ; $q=\gamma D_f=120(3)=360$ psf. $q_{ult}=cN_c+qN_q+\tfrac12\gamma B N_\gamma=400(35.5)+360(23.2)+\tfrac12(120)(7)(30.2)=14{,}200+8{,}352+12{,}684=35{,}236$ psf. Distractors: DROPPED the ½ on the $\gamma B N_\gamma$ term ( $47{,}900$ ); SWAPPED $B$ and $D_f$ between the $N_q$ and $N_\gamma$ terms ( $39{,}100$ ; reading a textbook's Terzaghi factors in place of the handbook's table lands here too); and omitted the $qN_q$ surcharge term, as if the footing sat at the ground surface ( $26{,}900$ ).
5
medium
A. Collection, primary & secondary treatment
Raw wastewater passes through primary settling tanks and then an activated-sludge process, and the discharge permit limits the final effluent to $S_e=30\,\text{mg/L}$ BOD $_5$ . The primary unit removes 35% of the BOD and the secondary unit removes 85% of the BOD reaching it. The highest raw-wastewater BOD $_5$ (mg/L) the plant can receive and still meet the limit is most nearly:
Figure — drag or pinch to move around
1. 75 mg/L
2. 200 mg/L
3. 270 mg/L
4. 308 mg/L
Show the answer & worked solution
Answer: D.308 mg/L
Each unit's removal is its process efficiency (§6.8.5.3) on the BOD that reaches that unit, not on the raw influent, so work back from the limit one unit at a time. The secondary unit passes $1-r_2=0.15$ of the BOD reaching it, so that BOD may be $S_e/(1-r_2)=30/0.15=200$ mg/L; the primary unit passes $1-r_1=0.65$ , so the raw wastewater may carry $200/0.65=307.7$ mg/L (overall removal $1-(1-r_1)(1-r_2)=90.3\%$ ). Distractors: applied the AVERAGE of the two removals once, $S_e/\left(1-\tfrac{r_1+r_2}{2}\right)=75$ mg/L; stopped at the BOD that may reach the secondary unit, $S_e/(1-r_2)=200$ mg/L, without working back through the primary unit; and worked back through the secondary unit, then scaled up by $1+r_1$ instead of dividing by $1-r_1$ , $200\times1.35=270$ mg/L.
6
medium
A. Mass balance, hydraulic & solids loading
A plant's raw influent flows at a metered average of 4,200 gpm and carries 220 mg/L of BOD. The BOD mass loading (lb/day) is most nearly:
1. 462 lb/day
2. 1,330 lb/day
3. 9,950 lb/day
4. 11,100 lb/day
Show the answer & worked solution
Answer: D.11,100 lb/day
A day has 1,440 minutes: $Q=4{,}200\times1440/10^6=6.048$ MGD. The handbook's concentration-to-mass conversion (PE Civil Reference Handbook §6.7.1) gives $M=C\,Q\,(8.34)=220\times6.048\times8.34=11{,}100$ lb/day. Distractors: DROPPED the 8.34 conversion ( $1{,}330$ lb/day); converted gpm with 60 min/hr but FORGOT the 24 hr/day, $Q=0.252$ MGD ( $462$ lb/day); and used 7.48, the gallons in a cubic foot, in place of 8.34, the pounds in a gallon of water ( $9{,}950$ lb/day).
7
medium
A. Energy/continuity & pipe losses
A 5-year-old cast-iron distribution main carries $Q=700\,\text{gpm}$ over $L=2{,}000\,\text{ft}$ of 8-in pipe. The friction head loss (ft) is most nearly:
1. 9.44 ft
2. 15.0 ft
3. 18.9 ft
4. 21.9 ft
Show the answer & worked solution
Answer: D.21.9 ft
For 5-year-old cast iron the handbook's table (§6.3.1.4) gives $C=120$ . $Q=700/448.9=1.56$ cfs, $D=8/12=0.667$ ft; Hazen-Williams head loss (PE Civil Reference Handbook §6.3.1.2): $h_f=\dfrac{4.73(2{,}000)(1.56)^{1.852}}{(120)^{1.852}(0.667)^{4.87}}=21.89$ ft. Distractors: used a LINEAR flow term $Q^{1}$ instead of $Q^{1.852}$ ( $15.0$ ft); took the pressure form (§6.3.1.3, gpm and inches), $P\cdot L=9.44$ psi, and reported it as feet without multiplying by $144/\gamma$ (2.31 ft/psi); and read the table row for NEW cast iron, $C=130$ , instead of the 5-year-old row ( $18.9$ ft).
8
medium
A. Demand, storage & distribution
A trunk main will supply a new pressure zone of 12,000 people with an average-day demand of 150 gpcd. Referenced to the average day, the maximum-day factor is 1.8 and the peak-hour factor is 2.7. The needed fire flow is 3,500 gpm. The governing design flow (gpm) for sizing the main is most nearly:
1. 3,380 gpm
2. 4,750 gpm
3. 5,750 gpm
4. 6,080 gpm
Show the answer & worked solution
Answer: C.5,750 gpm
Average day: $Q_{avg}=12{,}000\times150=1{,}800{,}000$ gpd, and over the 1,440 min of a day $1{,}800{,}000/1440=1{,}250$ gpm. Both factors multiply the average day. Peak hour: $2.7\times1{,}250=3{,}375$ gpm. Maximum day with the fire flow: $1.8\times1{,}250+3{,}500=2{,}250+3{,}500=5{,}750$ gpm. The main is sized for the larger case, here the maximum day plus fire flow: $5{,}750$ gpm. Distractors: sized the main for the peak hour alone and never checked the fire case, $2.7\times1{,}250=3{,}375$ gpm; added the fire flow to the AVERAGE day instead of the maximum day, $1{,}250+3{,}500=4{,}750$ gpm; and compounded the factors, taking the peak hour as $1.8\times2.7\times1{,}250=6{,}075$ gpm (both factors are referenced to the average day).
9
medium
A. Aquifers, Darcy flow & well drawdown
Leachate from an unlined landfill cell has reached the water table in a fine sand aquifer. Slug tests and core samples give the aquifer a hydraulic conductivity of $900\,\text{gpd/ft}^2$ , a total porosity of 0.32 and an effective porosity of 0.26. The water-table elevations in two monitoring wells $1{,}250\,\text{ft}$ apart along the direction of flow are $732.6\,\text{ft}$ and $729.4\,\text{ft}$ . The seepage velocity of the groundwater carrying the leachate is most nearly:
1. 0.308 ft/day
2. 0.963 ft/day
3. 1.18 ft/day
4. 8.86 ft/day
Show the answer & worked solution
Answer: C.1.18 ft/day
Gradient $i=\Delta h/L=(732.6-729.4)/1{,}250=0.00256$ . $K=900\,\text{gpd/ft}^2\div7.48\,\text{gal/ft}^3=120.3\,\text{ft/day}$ , so the Darcy flux is $q=Ki=120.3\times0.00256=0.308\,\text{ft/day}$ . Water moves only through the connected pores, so the seepage velocity is $v_s=q/n_e=0.308/0.26=1.18\,\text{ft/day}$ (PE Civil Reference Handbook §6.6.2.1 Darcy's Law: the average seepage velocity is $q$ over the effective porosity, which it writes $S_y$ ; §6.6.2.4 Seepage Velocity prints only a bare porosity $\eta$ ). Distractors: left the porosity out, taking the Darcy flux $q=Ki$ as the velocity of the water ( $0.308$ ft/day); divided by the total porosity, 0.32, where only the effective porosity, 0.26, carries the flow ( $0.963$ ft/day); and used $K=900$ as if it were already in ft/day, skipping the 7.48 gal/ft³ that turns gpd/ft² into ft/day ( $8.86$ ft/day).
10
medium
A. Scheduling, quantities, cost & economics
The contract for 4,800 LF of 24-in. sanitary sewer replacement has a budget at completion of $1,920,000, spread uniformly over its length. At the month-7 progress review, 2,880 LF has been installed, the baseline schedule called for 2,400 LF by this date, and the actual cost of the work performed to date is $1,250,000. The cost performance index (CPI) is most nearly:
1. 0.77
2. 0.92
3. 1.20
4. 1.30
Show the answer & worked solution
Answer: B.0.92
By §2.4.1.3 Earned-Value Analysis, project completion is $2{,}880/4{,}800=60\%$ , so the budgeted cost of the work performed is $BCWP=0.6\times1{,}920{,}000=1{,}152{,}000$ dollars (the schedule called for $BCWS=0.5\times1{,}920{,}000=960{,}000$ dollars). The actual cost of that work is $ACWP=1{,}250{,}000$ dollars, so $CPI=BCWP/ACWP=1{,}152{,}000/1{,}250{,}000=0.92$ : under 1, so the work is costing more than its budget. Distractors: computed the SCHEDULE performance index, $SPI=BCWP/BCWS=1{,}152{,}000/960{,}000=1.20$ , instead of the cost index; put the PLANNED value where the earned value belongs, $BCWS/ACWP=960{,}000/1{,}250{,}000=0.77$ ; and compared the actual cost with the PLANNED value, $ACWP/BCWS=1{,}250{,}000/960{,}000=1.30$ , a spend-to-plan ratio that ignores the work actually done.

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