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PE Civil: Water Resources and Environmental formula sheet

115 key equations from 12 study topics, each with what it is for and where to look it up: the PE Civil Reference Handbook section, or the code or textbook when the handbook does not print it. No signup. These are the equations from the free chapters of our PE Civil: Water Resources and Environmental study handbook; the full handbook covers every topic at this depth, with worked examples and the traps that cost points.

115

Equations

12

Study topics

12

Concepts

Project Planning
6% of the exam

Quantity take-off, cost estimating, CPM schedules and activity sequencing, and economic/lifecycle comparison of alternatives (present worth).

Present Worth and Comparing Alternatives

Use time-value-of-money factors to compare alternatives by present worth, equivalent uniform annual cost, rate of return, benefit-cost ratio, payback, life-cycle cost, and the triple bottom line.

Single-payment compound amount / present worth
F=P(1+i)n,P=F(1+i)−nF = P(1+i)^n, \qquad P = F(1+i)^{-n}
(F/P,i,n)(F/P,i,n) and (P/F,i,n)(P/F,i,n). Move a lump sum forward or back nn periods at rate ii.
Uniform series present worth (P/A)
(P/A,i,n)=(1+i)n−1i (1+i)n(P/A,i,n) = \dfrac{(1+i)^n - 1}{i\,(1+i)^n}
Present worth of a level annual amount AA for nn periods. Inverse is the capital-recovery factor (A/P)(A/P).
Capital recovery factor (A/P)
(A/P,i,n)=i (1+i)n(1+i)n−1(A/P,i,n) = \dfrac{i\,(1+i)^n}{(1+i)^n - 1}
Converts a present cost PP into an equivalent uniform annual amount; the workhorse of EUAC.
Sinking-fund factor (A/F)
(A/F,i,n)=i(1+i)n−1(A/F,i,n) = \dfrac{i}{(1+i)^n - 1}
Converts a future amount (e.g., salvage) into an equivalent annual amount; credited in EUAC.
Equivalent uniform annual cost
EUAC=P (A/P,i,n)+AO&M−S (A/F,i,n)\text{EUAC} = P\,(A/P,i,n) + A_{\text{O\&M}} - S\,(A/F,i,n)
Annualized total cost; compare alternatives of different lives by lowest EUAC. SS = salvage.
Net present worth
PW=−P0+∑t=1nAt(1+i)t+S(1+i)n\text{PW} = -P_0 + \sum_{t=1}^{n}\dfrac{A_t}{(1+i)^t} + \dfrac{S}{(1+i)^n}
Discount all cash flows to time zero; choose least-cost (or greatest net-benefit) PW over equal lives.
Benefit-cost ratio
BC=PW or AW of benefitsPW or AW of costs≥1\dfrac{B}{C} = \dfrac{\text{PW or AW of benefits}}{\text{PW or AW of costs}} \ge 1
Project justified when ≥1\ge 1; for alternatives use incremental ΔB/ΔC\Delta B/\Delta C.
Rate of return condition
PW(i∗)=0,accept if i∗≥MARR\text{PW}(i^*) = 0,\quad \text{accept if } i^* \ge \text{MARR}
i∗i^* = interest rate that zeroes present worth; compare to the minimum attractive rate of return.
Capitalized cost (perpetual)
P=AiP = \dfrac{A}{i}
Present worth of a perpetual annual cost AA (infinite life), e.g., permanent maintenance of a structure.
Inflation-adjusted rate
d=i+f+i fd = i + f + i\,f
Combined rate when cash flows are in actual (then-current) dollars and inflation ff applies (handbook §1.7.4\S1.7.4).
Simple payback period
Payback=P0Asavings\text{Payback} = \dfrac{P_0}{A_{\text{savings}}}
Years to recover first cost from annual savings; ignores time value — a screen, not a decision rule.

Where it lives: NCEES PE Civil Reference Handbook — §1.7 Engineering Economics · NCEES PE Civil Reference Handbook — §1.7.9 Benefit-Cost Analysis · Newnan, Lavelle & Eschenbach, Engineering Economic Analysis · Envision / ISI Rating System (triple-bottom-line sustainability)

Read the Project Planning chapter

Soil Mechanics
4% of the exam

Lateral earth pressure, consolidation and compaction, bearing capacity, settlement, and slope stability for water-resources structures.

Lateral Earth Pressure: Active, At-Rest and Passive

Earth-pressure coefficients (active, at-rest, passive) by Rankine and Coulomb, the pressure triangle and resultant thrust, plus surcharge, water, and wall friction.

At-rest coefficient (Jaky)
K0=1−sin⁡ϕ′K_0 = 1-\sin\phi'
Normally consolidated soil, non-yielding wall. ϕ′\phi' = effective friction angle. For overconsolidated soil multiply by OCR Ω\text{OCR}^{\,\Omega}, Ω≈sin⁡ϕ′\Omega\approx\sin\phi'.
Rankine active coefficient
Ka=tan⁡2 ⁣(45∘−ϕ′2)=1−sin⁡ϕ′1+sin⁡ϕ′K_a = \tan^2\!\left(45^\circ-\tfrac{\phi'}{2}\right) = \frac{1-\sin\phi'}{1+\sin\phi'}
Smooth vertical wall, horizontal cohesionless backfill. Dimensionless; Ka<1K_a<1.
Rankine passive coefficient
Kp=tan⁡2 ⁣(45∘+ϕ′2)=1+sin⁡ϕ′1−sin⁡ϕ′K_p = \tan^2\!\left(45^\circ+\tfrac{\phi'}{2}\right) = \frac{1+\sin\phi'}{1-\sin\phi'}
Maximum resistance state; Kp=1/KaK_p = 1/K_a. Mobilizing full KpK_p requires large wall movement.
Active / passive pressure with cohesion
pa=Kaγz−2c′Ka,pp=Kpγz+2c′Kpp_a = K_a\gamma z - 2c'\sqrt{K_a}, \quad p_p = K_p\gamma z + 2c'\sqrt{K_p}
Lateral pressure at depth zz (psf). For cohesionless soil drop the 2c′K2c'\sqrt{K} term. Use γ′\gamma' below the water table.
Active / passive thrust
Pa=12KaγH2,Pp=12KpγH2P_a = \tfrac12 K_a\gamma H^2, \qquad P_p = \tfrac12 K_p\gamma H^2
Resultant per unit wall length (lb/ft) for a triangular distribution; acts at H/3H/3 above the base.
Tension crack depth
zc=2c′γKaz_c = \dfrac{2c'}{\gamma\sqrt{K_a}}
Depth over which active pressure is tensile (ignored in design) for a cohesive backfill.
Surcharge contribution
Δσh=K q\Delta\sigma_h = K\,q
Constant added horizontal pressure (psf) from a uniform surface surcharge qq; resultant Kq HKq\,H at mid-height.
Hydrostatic water pressure
u=γwzwu = \gamma_w z_w
Acts on the wall in addition to the soil pressure; KK does NOT apply to water. γw=62.4 pcf\gamma_w = 62.4\,\text{pcf}.
Coulomb active coefficient
Ka=cos⁡2(ϕ′−θ)cos⁡2θ cos⁡(δ+θ)[1+sin⁡(ϕ′+δ)sin⁡(ϕ′−β)cos⁡(δ+θ)cos⁡(θ−β)]2K_a = \dfrac{\cos^2(\phi'-\theta)}{\cos^2\theta\,\cos(\delta+\theta)\left[1+\sqrt{\frac{\sin(\phi'+\delta)\sin(\phi'-\beta)}{\cos(\delta+\theta)\cos(\theta-\beta)}}\right]^2}
δ\delta = wall friction, θ\theta = wall batter from vertical, β\beta = backfill slope. Reduces to Rankine when δ=θ=β=0\delta=\theta=\beta=0.

Where it lives: NCEES PE Civil Reference Handbook — §3.1 Lateral Earth Pressures · NCEES PE Civil Reference Handbook — §3.3 Effective and Total Stresses · FHWA-NHI-06-088/089 Soils and Foundations Reference Manual

Read the Soil Mechanics chapter

Materials
6% of the exam

Soil classification and boring-log interpretation, soil properties and phase relationships, concrete, piping materials, and test-method conformance.

Soil Classification and Boring Logs (USCS)

Run a soil through the USCS decision tree from sieve gradation and Atterberg limits, assign group symbol and name, and read SPT, water table, and strata off a boring log.

Uniformity coefficient
Cu=D60D10C_u = \dfrac{D_{60}}{D_{10}}
Spread of grain sizes. D60D_{60}, D10D_{10} = sieve openings (mm) passing 60% and 10%. Cu≥4C_u \ge 4 (gravel) or ≥6\ge 6 (sand) is needed for 'well graded.'
Coefficient of curvature
Cc=(D30)2D10 D60C_c = \dfrac{(D_{30})^2}{D_{10}\,D_{60}}
Shape of the gradation curve; well graded requires 1≤Cc≤31 \le C_c \le 3 together with the CuC_u test.
Plasticity index
PI=LL−PLPI = LL - PL
Range of water content over which soil is plastic (%). LLLL = liquid limit, PLPL = plastic limit.
A-line
PI=0.73 (LL−20)PI = 0.73\,(LL - 20)
Casagrande chart divider. On or above = clay (CL/CH); below = silt or organic (ML/MH/OL/OH). The A-line runs horizontal at PI=4PI=4 for low LLLL (below ≈25.5\approx 25.5, where 0.73(LL−20)=40.73(LL-20)=4), then rises along 0.73(LL−20)0.73(LL-20).
U-line (upper bound)
PI=0.9 (LL−8)PI = 0.9\,(LL - 8)
Empirical limit no natural soil exceeds; a plotted point above it signals a test or transcription error.
Liquidity index
LI=w−PLPILI = \dfrac{w - PL}{PI}
Where in-situ water content ww sits between limits. LI≤0LI \le 0 brittle/stiff, 0<LI<10<LI<1 plastic, LI≥1LI \ge 1 behaves as a viscous liquid (sensitive).
AASHTO group index
GI=(F−35)[0.2+0.005(LL−40)]+0.01(F−15)(PI−10)GI = (F-35)\big[0.2 + 0.005(LL-40)\big] + 0.01(F-15)(PI-10)
FF = % passing No. 200. The handbook prints the bare formula, but AASHTO M145 caps the partial products: (F−35)≤40(F-35)\le 40 (F≤75F\le75), (LL−40)≤20(LL-40)\le 20 in the first term, and (F−15)≤40(F-15)\le 40 (F≤55F\le55), (PI−10)≤20(PI-10)\le 20 in the second. Report as integer in parentheses; if negative, use 0. Higher = poorer subgrade.
Partial group index (A-2-6, A-2-7)
GI=0.01 (F−15)(PI−10)GI = 0.01\,(F-15)(PI-10)
Only the plasticity term is used for the A-2-6 and A-2-7 subgroups.
SPT energy correction (handbook form)
N60=Eeff60 N=Em0.60 NN_{60} = \dfrac{E_{eff}}{60}\,N = \dfrac{E_m}{0.60}\,N
Handbook §3.8.1.1: EeffE_{eff} = measured hammer efficiency (%) or EmE_m as a fraction (~0.45-0.55 donut, ~0.6 safety, ~0.8-0.9 automatic/trip). Correct before any correlation. The Skempton (1986) refinement N60=(EmCbCsCr/0.60)NN_{60} = (E_m C_b C_s C_r/0.60)N adds borehole CbC_b, sampler CsC_s, and rod CrC_r factors but is NOT in the NCEES handbook.
Relative density (granular state)
Dr=emax−eemax−emin×100%D_r = \dfrac{e_{max}-e}{e_{max}-e_{min}}\times 100\%
State of a cohesionless soil between its loosest (emaxe_{max}) and densest (emine_{min}) packing; the density class read from N60N_{60} on a boring log corresponds to a DrD_r range (e.g. medium dense ≈\approx 35-65%).

Where it lives: NCEES PE Civil Reference Handbook — §3.7 Soil Classification and Boring Log Interpretation · NCEES PE Civil Reference Handbook — §3.8 Material Test Methods (Atterberg Limits, Index Testing) · ASTM D2487 — Unified Soil Classification System (USCS) · AASHTO M 145 — Classification of Soils and Soil-Aggregate Mixtures

Read the Materials chapter

Analysis and Design
9% of the exam

Mass balance, hydraulic and solids (sediment/sludge) loading, and hydraulic flow measurement with weirs, flumes, and meters.

Steady-State Mass Balance

Control-volume bookkeeping — accumulation = in − out ± reaction — that, killed at steady state, solves blending, splitting, and CSTR/PFR reactor problems across the WRE exam.

General mass balance
dMdt=m˙in−m˙out±m˙rxn\frac{dM}{dt} = \dot m_{in} - \dot m_{out} \pm \dot m_{rxn}
Accumulation = in − out ± reaction. M=CVM=CV (mass in the control volume); each transport rate m˙=QC\dot m = QC. Foundation of every WRE balance (§6.7\S6.7).
Steady-state, conservative
∑QiCi∣in=∑QjCj∣out\sum Q_i C_i\big|_{in} = \sum Q_j C_j\big|_{out}
When dM/dt=0dM/dt=0 and no reaction: mass in equals mass out. Use for blending, splitting, and tracer balances.
Continuity (volumetric balance)
Q=vA,∑Qin=∑QoutQ = vA, \qquad \sum Q_{in} = \sum Q_{out}
Flow balance, independent of the mass balance. QQ = volumetric flow, vv = velocity, AA = cross-sectional area.
Flow-weighted blend concentration
Cmix=∑iQiCi∑iQiC_{mix} = \dfrac{\sum_i Q_i C_i}{\sum_i Q_i}
Concentration after streams merge (conservative, steady state). Result must fall between the lowest and highest inlet concentration.
Reaction rate
r=kCnr = kC^{n}
nn = reaction order (0 or 1 on the exam); kk = rate constant (mg/L⋅d−1\text{mg/L}\cdot\text{d}^{-1} for n=0n{=}0, d−1\text{d}^{-1} for n=1n{=}1); CC = concentration.
Hydraulic detention time
τ=θ=VQ\tau = \theta = \dfrac{V}{Q}
Mean residence time in a reactor or tank; VV = volume, QQ = flow. Keep units consistent (e.g. days).
CSTR, first-order
CC0=11+kτ\dfrac{C}{C_0} = \dfrac{1}{1+k\tau}
Steady-state effluent of a complete-mix reactor with first-order decay. Reaction evaluated at exit (= bulk) concentration.
CSTR, zero-order
C=C0−kτC = C_0 - k\tau
Complete-mix reactor with constant-rate (zero-order) removal. Valid only while C≥0C \ge 0.
PFR, first-order
CC0=e−kτ\dfrac{C}{C_0} = e^{-k\tau}
Steady-state effluent of an ideal plug-flow reactor with first-order decay. Removes more than a CSTR at equal τ\tau.
N CSTRs in series, first-order
CNC0=1(1+kτi)N\dfrac{C_N}{C_0} = \dfrac{1}{(1+k\tau_i)^{N}}
NN equal-volume complete-mix tanks, each with detention τi\tau_i; approaches PFR as N→∞N\to\infty (§6.1.5.10\S6.1.5.10).
Mass-to-load conversion (USCS)
M (lb/day)=C (mg/L) Q (MGD) (8.34)M\,(\text{lb/day}) = C\,(\text{mg/L})\, Q\,(\text{MGD})\,(8.34)
8.348.34 = lb per Mgal per mg/L. The most-used conversion in WRE; QQ MUST be in MGD.
Mass-to-load conversion (SI)
M (kg/d)=C (g/m3) Q (m3/d)1000M\,(\text{kg/d}) = \dfrac{C\,(\text{g/m}^3)\, Q\,(\text{m}^3/\text{d})}{1000}
Uses mg/L=g/m3\text{mg/L}=\text{g/m}^3. No 8.34 needed in SI — divide by 1000 to reach kg.

Where it lives: NCEES PE Civil Reference Handbook — §6.7 Water Quality (Mass Conservation and Continuity) · NCEES PE Civil Reference Handbook — §6.1.5.10 Flow Reactors, Steady State (CSTR/PFR) · Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery

Read the Analysis and Design chapter

Hydraulics — Closed Conduit
10% of the exam

Energy/continuity (Bernoulli, EGL/HGL), Hazen-Williams and Darcy-Weisbach losses, pump application, NPSH and cavitation, and pipe-network analysis.

The Energy Equation, EGL and HGL

Continuity and the steady-flow energy equation between two points, the velocity/pressure/elevation heads, pump and turbine head, and how to sketch the energy and hydraulic grade lines.

Continuity (incompressible)
Q=A1v1=A2v2Q = A_1 v_1 = A_2 v_2
Discharge QQ (cfs or m³/s) is constant along a single pipe; AA = cross-sectional area, vv = mean velocity. Velocity rises where area falls.
Steady-flow energy equation
p1γ+v122g+z1+hp=p2γ+v222g+z2+ht+hL\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 + h_p = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2 + h_t + h_L
Head balance between sections 1 and 2. hph_p = pump head added, hth_t = turbine head removed, hLh_L = total head loss, γ\gamma = specific weight.
Bernoulli equation (ideal)
p1γ+v122g+z1=p2γ+v222g+z2\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2
Energy equation with no machine and no loss (steady, inviscid, incompressible). Total head is constant.
Total head at a section
H=pγ+v22g+zH = \frac{p}{\gamma} + \frac{v^2}{2g} + z
Sum of pressure, velocity, and elevation heads (ft or m). This is the height of the EGL above the datum.
Energy and hydraulic grade lines
EGL−HGL=v22g\text{EGL} - \text{HGL} = \frac{v^2}{2g}
EGL plots total head; HGL plots piezometric head p/γ+zp/\gamma + z. They are separated by the velocity head everywhere.
Pressure head conversion
h=pγ,1 psi=2.31 ft H2Oh = \frac{p}{\gamma}, \qquad 1\ \text{psi} = 2.31\ \text{ft H}_2\text{O}
Converts gauge pressure to feet of water with γ=62.4 lbf/ft3\gamma = 62.4\,\text{lbf/ft}^3. In SI, p/γp/\gamma with γ=9.81 kN/m3\gamma = 9.81\,\text{kN/m}^3.
Fluid power of a machine
W˙fluid=γ Q h\dot{W}_{\text{fluid}} = \gamma\,Q\,h
Power added by a pump or extracted by a turbine; hh = head change across the machine. Divide by η\eta for pump shaft power.
Velocity head
hv=v22gh_v = \frac{v^2}{2g}
Kinetic energy per unit weight (ft or m). Grows with the square of velocity; the gap between EGL and HGL.

Where it lives: NCEES PE Civil Reference Handbook — §6.2.1 Principles of One-Dimensional Fluid Flow (Bernoulli §6.2.1.3, Energy Line §6.2.1.5) · NCEES PE Civil Reference Handbook — §6.3 Closed Conduit Flow and Pumps · Munson, Young & Okiishi, Fundamentals of Fluid Mechanics

Read the Hydraulics — Closed Conduit chapter

Hydraulics — Open Channel
10% of the exam

Manning uniform flow, specific energy and critical depth, sub-/supercritical flow and hydraulic jumps, stormwater collection, gutter/inlet flow, and culverts.

Manning Uniform Flow and Normal Depth

Manning's equation in both SI and USCS forms, hydraulic radius and conveyance, finding normal depth by iteration, best hydraulic sections, and partially full circular pipes.

Manning's equation (discharge)
Q=kn A RH2/3 S1/2Q = \dfrac{k}{n}\,A\,R_H^{2/3}\,S^{1/2}
Steady uniform flow. k=1.486k = 1.486 USCS (ft, cfs), 1.01.0 SI (m, m³/s); nn = roughness; AA = flow area; RHR_H = hydraulic radius; S=S0S = S_0 = bed slope (ft/ft or m/m).
Manning's equation (velocity)
v=kn RH2/3 S1/2v = \dfrac{k}{n}\,R_H^{2/3}\,S^{1/2}
Mean velocity (ft/s or m/s). Equal to Q/AQ/A; same kk convention.
Hydraulic radius
RH=APR_H = \dfrac{A}{P}
Flow area AA over wetted perimeter PP (the boundary in contact with water, excluding the free surface). For a full circular pipe, RH=D/4R_H = D/4.
Trapezoidal geometry
A=(b+zy) y,P=b+2y1+z2,T=b+2zyA = (b + zy)\,y, \quad P = b + 2y\sqrt{1 + z^2}, \quad T = b + 2zy
Bottom width bb, depth yy, side slope zz (H:V). TT = top width. Set z=0z = 0 for a rectangle.
Conveyance
K=kn A RH2/3,Q=K S1/2K = \dfrac{k}{n}\,A\,R_H^{2/3}, \qquad Q = K\,S^{1/2}
Geometry-and-roughness factor (cfs) independent of slope; useful for compound channels (sum KiK_i) and backwater work.
Normal-depth iteration target
Q nk=A RH2/3 S1/2\dfrac{Q\,n}{k} = A\,R_H^{2/3}\,S^{1/2}
Left side is a known constant; iterate yy until the right side matches. The solution is the normal depth yny_n.
Best hydraulic sections
Half-hexagon: RH=y2, A=3 y2;Best rectangle: b=2y, RH=y2\text{Half-hexagon: } R_H = \tfrac{y}{2},\ A=\sqrt3\,y^2; \quad \text{Best rectangle: } b=2y,\ R_H=\tfrac{y}{2}
Sections that minimize wetted perimeter for a given area (maximum efficiency). Semicircle is the overall optimum.
Minimum full-flow diameter
Dmin=(2.16 Q nS1/2)3/8D_{min} = \left(\dfrac{2.16\,Q\,n}{S^{1/2}}\right)^{3/8}
USCS; smallest circular pipe (ft) that conveys QQ (cfs) just full at slope SS. Round up to the next commercial size.

Where it lives: NCEES PE Civil Reference Handbook — §6.4 Open-Channel Flow · NCEES PE Civil Reference Handbook — §6.4.5 Steady Uniform Flow (Manning's Equation) · Chow, Open-Channel Hydraulics

Read the Hydraulics — Open Channel chapter

Hydrology
11% of the exam

Storm frequency and IDF, time of concentration, Rational and SCS/NRCS runoff, unit and synthetic hydrographs, routing, depletions, and stormwater BMPs.

The Rational Method, Time of Concentration and IDF

Q = CiA for peak discharge on small catchments: composite runoff coefficient, time of concentration by flow segment, and selecting intensity from an IDF curve.

Rational formula
Q=C i AQ = C\,i\,A
QQ = peak discharge (cfs); CC = runoff coefficient; ii = intensity at duration tct_c (in/hr); AA = area (acres). 1 ac-in/hr≈1 cfs1\,\text{ac-in/hr}\approx1\,\text{cfs}.
Composite runoff coefficient
Cw=∑CiAi∑AiC_w = \dfrac{\sum C_i A_i}{\sum A_i}
Area-weighted CC for a mixed-cover catchment.
Time of concentration
tc=Tti+tsc+tcht_c = T_{ti} + t_{sc} + t_{ch}
Sum of sheet, shallow-concentrated, and channel travel times along the longest hydraulic path.
Sheet-flow travel time
Tti=Kui0.4(nLS)0.6T_{ti} = \dfrac{K_u}{i^{0.4}}\left(\dfrac{nL}{\sqrt{S}}\right)^{0.6}
Kinematic-wave overland time (min), handbook §6.5.4.2 form — exponent 0.60.6 on nL/SnL/\sqrt{S}, 0.40.4 on ii; Ku=0.933K_u=0.933 USCS, 6.926.92 SI; nn = overland roughness, L≤300 ftL\le300\,\text{ft}, SS = slope, ii = intensity.
Shallow concentrated velocity
V=kSV = k\sqrt{S}
k≈20.3k\approx20.3 (paved), 16.116.1 (unpaved), VV in ft/s, SS in ft/ft; then tsc=L/(60V)t_{sc}=L/(60V) in minutes.
Channel travel time
tch=L60 V,V=1.486nR2/3S1/2t_{ch} = \dfrac{L}{60\,V}, \quad V = \dfrac{1.486}{n}R^{2/3}S^{1/2}
Manning velocity for the channel/pipe segment (USCS); tcht_{ch} in minutes with LL in ft, VV in ft/s.
IDF intensity
i=c Tm(tc+f)ei = \dfrac{c\,T^{m}}{(t_c + f)^{e}}
Design intensity (in/hr); enter at duration =tc=t_c (minutes) and the design return period TT (yr).

Where it lives: NCEES PE Civil Reference Handbook — §6.5.2.1 Rational Formula Method · NCEES PE Civil Reference Handbook — §6.5.4 Time of Concentration · FHWA HEC-22, Urban Drainage Design Manual

Read the Hydrology chapter

Groundwater and Wells
5% of the exam

Aquifer properties, Darcy groundwater flow, and steady/transient well drawdown analysis (Thiem, Theis, Dupuit) for confined and unconfined aquifers.

Darcy's Law and Aquifer Properties

Confined vs unconfined aquifers, hydraulic conductivity and transmissivity, storativity and specific yield, and the difference between Darcy and seepage velocity.

Darcy's law (discharge)
Q=K i A=K ΔhΔL AQ = K\,i\,A = K\,\dfrac{\Delta h}{\Delta L}\,A
Volumetric flow QQ (ft3/s\text{ft}^3/\text{s}, m3/s\text{m}^3/\text{s}) from conductivity KK, gradient i=Δh/ΔLi=\Delta h/\Delta L (taken positive), and gross area AA. Flow is directed toward lower head.
Specific (Darcy) velocity
q=QA=K iq = \dfrac{Q}{A} = K\,i
Bulk/superficial flux across the full cross-section (length/time). Not the velocity of a water particle.
Seepage (pore) velocity
vs=qne=K inev_s = \dfrac{q}{n_e} = \dfrac{K\,i}{n_e}
Average linear velocity through the pores; nen_e = effective (drainable) porosity. Use for travel time and contaminant transport.
Transmissivity
T=K bT = K\,b
Conductivity times saturated thickness bb (ft2/s\text{ft}^2/\text{s}, m2/s\text{m}^2/\text{s}, or gpd/ft\text{gpd/ft}). Used directly in well equations.
Conductivity vs intrinsic permeability
K=k γμ=k ρ gμK = \dfrac{k\,\gamma}{\mu} = \dfrac{k\,\rho\,g}{\mu}
kk = intrinsic permeability (area, darcys); γ,μ\gamma,\mu = fluid specific weight and dynamic viscosity. Separates soil and fluid effects.
Storativity (confined)
S=Ss bS = S_s\,b
Volume released per unit area per unit head drop; SsS_s = specific storage (ft−1\text{ft}^{-1}). Dimensionless, 5×10−55\times10^{-5} to 5×10−35\times10^{-3}.
Porosity split (unconfined)
n=Sy+Sr,S≈Syn = S_y + S_r, \qquad S \approx S_y
Total porosity = specific yield (drainable) + specific retention (held). Unconfined storativity ≈Sy\approx S_y (0.10.1–0.30.3).
Layered equivalent conductivity
Kx=∑Kibi∑bi,Kz=∑bi∑(bi/Ki)K_x=\dfrac{\sum K_i b_i}{\sum b_i}, \quad K_z=\dfrac{\sum b_i}{\sum (b_i/K_i)}
Arithmetic (thickness-weighted) mean parallel to layers; harmonic mean perpendicular. Kx≥KzK_x \ge K_z always.

Where it lives: NCEES PE Civil Reference Handbook — §6.6 Groundwater and Wells · NCEES PE Civil Reference Handbook — §3.16 Groundwater and Seepage · Fetter, Applied Hydrogeology · Freeze and Cherry, Groundwater

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Surface Water & Groundwater Quality
7% of the exam

Stream oxygen dynamics and the Streeter-Phelps DO sag, TMDL and load allocation, and biological and chemical contaminant fate.

The Streeter-Phelps Dissolved-Oxygen Sag

Track dissolved oxygen downstream of a discharge: BOD deoxygenation versus reaeration, the deficit equation, and the critical point that decides whether a stream stays alive.

Dissolved-oxygen deficit
D=DOsat−DOD = DO_{sat} - DO
DD (mg/L) is the state variable of the model: how far below saturation the water sits. DOsatDO_{sat} is read from the saturation table at the stream temperature.
Streeter-Phelps deficit equation
D=kdLakr−kd(e−kdt−e−krt)+Da e−krtD = \dfrac{k_d L_a}{k_r - k_d}\left(e^{-k_d t} - e^{-k_r t}\right) + D_a\,e^{-k_r t}
Deficit at travel time tt (days). kdk_d, krk_r base-ee (day−1^{-1}); LaL_a = mixed ultimate BOD (mg/L); DaD_a = initial deficit after mixing (mg/L).
Mixed ultimate BOD and DO at outfall
La=QwLw+QrLrQw+Qr,DOmix=QwDOw+QrDOrQw+QrL_a = \dfrac{Q_w L_w + Q_r L_r}{Q_w + Q_r}, \quad DO_{mix} = \dfrac{Q_w DO_w + Q_r DO_r}{Q_w + Q_r}
Flow-weighted mixing of waste (ww) and river (rr). Initial deficit Da=DOsat−DOmixD_a = DO_{sat} - DO_{mix}.
Critical time (worst sag location)
tc=1kr−kdln⁡ ⁣[krkd(1−Da(kr−kd)kdLa)]t_c = \dfrac{1}{k_r - k_d}\ln\!\left[\dfrac{k_r}{k_d}\left(1 - \dfrac{D_a(k_r - k_d)}{k_d L_a}\right)\right]
Travel time (days) to the minimum DO. Valid for kr≠kdk_r \neq k_d; the bracket must be positive.
Critical deficit
Dc=kdkr La e−kdtcD_c = \dfrac{k_d}{k_r}\,L_a\,e^{-k_d t_c}
Maximum deficit (mg/L). Minimum dissolved oxygen is DOmin=DOsat−DcDO_{min} = DO_{sat} - D_c.
BOD remaining and exerted
Lt=L0 e−kdt,BODt=L0(1−e−kdt)L_t = L_0\,e^{-k_d t}, \qquad BOD_t = L_0\left(1 - e^{-k_d t}\right)
First-order BOD decay. L0L_0 = ultimate BOD (mg/L); exerted BOD is the oxygen consumed up to time tt.
Base-e and base-10 rate constants
kd=2.303 Kdk_d = 2.303\,K_d
Lab BOD data give base-10 KK (day−1^{-1}); the e−kte^{-kt} form of Streeter-Phelps needs base-ee kk, while the handbook's 10−Kt10^{-Kt} form takes KK directly. Convert once, for the form you use.
Arrhenius temperature correction
kT=k20 θ T−20k_T = k_{20}\,\theta^{\,T - 20}
Correct rate constants to stream temperature TT (°C). Handbook §6.8.3.3 banded BOD values: θ=1.135\theta = 1.135 (T=4T = 4–20∘20^\circC) and θ=1.056\theta = 1.056 (T=21T = 21–30∘30^\circC) for kdk_d; θ=1.024\theta = 1.024 for reaeration krk_r.
Oxygen saturation (Henry's law)
DOsat=H PO2DO_{sat} = H\,P_{O_2}
Saturation DO is proportional to the partial pressure of oxygen; in practice read it from the temperature/salinity table. ≈9.17 mg/L\approx 9.17\,\text{mg/L} at 20∘20^\circC, freshwater, 1 atm1\,\text{atm}.
Travel time from distance
t=xvt = \dfrac{x}{v}
Convert downstream distance xx to travel time tt (days) using stream velocity vv; mind unit conversion to days.

Where it lives: NCEES PE Civil Reference Handbook — §6.7.4 Oxygen Dynamics (Streeter-Phelps) · NCEES PE Civil Reference Handbook — §6.7.3 Biochemical Oxygen Demand · NCEES PE Civil Reference Handbook — §6.8.3.3 Kinetic Temperature Corrections · Davis & Cornwell, Introduction to Environmental Engineering

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Drinking Water Distribution & Treatment
9% of the exam

Demand and storage, distribution systems, sedimentation, coagulation/flocculation, filtration and membranes, disinfection (CT), softening, and adsorption.

Sedimentation and Overflow Rate

The ideal-basin theory of discrete (Type I) settling: why overflow rate alone decides removal, how settling velocity compares to it, and how weir loading and detention time bound the design.

Overflow rate (critical settling velocity)
v0=QAsurfacev_0 = \dfrac{Q}{A_{surface}}
Surface loading rate. The critical velocity: particles with vt≥v0v_t \ge v_0 are fully removed. Depends only on plan area, not depth. Units ft/s, m/d, or gpd/ft2^2.
Stokes' law settling velocity
vt=g (ρp−ρw) d218 μ=g (SG−1) d218 νv_t = \dfrac{g\,(\rho_p - \rho_w)\,d^{2}}{18\,\mu} = \dfrac{g\,(SG-1)\,d^{2}}{18\,\nu}
Terminal velocity of a discrete particle for NRe<1N_{Re}<1. dd = diameter, ν\nu = kinematic viscosity, SGSG = particle specific gravity.
Reynolds number check
NRe=vt dν<1N_{Re} = \dfrac{v_t\,d}{\nu} < 1
Validity bound for Stokes' law. Above ∼1\sim 1, use the general drag-coefficient form vt=4g(ρp−ρw)d/(3CDρw)v_t=\sqrt{4g(\rho_p-\rho_w)d/(3C_D\rho_w)}.
Removal ratio (partial)
r=vtv0(vt<v0)r = \dfrac{v_t}{v_0}\quad (v_t < v_0)
Fraction of a slower particle removed in an ideal basin; equals h/Hh/H in the geometric model (hh = limiting entry/fall height, HH = basin depth, so r≤1r \le 1). Equals 11 when vt≥v0v_t \ge v_0.
Detention time
t=VQ=Hv0t = \dfrac{V}{Q} = \dfrac{H}{v_0}
Hydraulic residence time. Bounds short-circuiting and sludge storage but does NOT set Type I removal.
Horizontal (approach) velocity
vH=QAxv_H = \dfrac{Q}{A_{x}}
Flow over basin cross-sectional (flow-normal) area. Kept <0.5 ft/min< 0.5\,\text{ft/min} in water treatment to avoid scouring settled sludge.
Weir overflow rate
WOR=QLweirWOR = \dfrac{Q}{L_{weir}}
Flow per unit weir length (gpd/ft). Water-treatment limit ≤20,000 gpd/ft\le 20{,}000\,\text{gpd/ft} to prevent floc liftover at the effluent.
Ideal-basin geometry relation
v0vH=HL\dfrac{v_0}{v_H} = \dfrac{H}{L}
Critical particle traverses depth HH while moving length LL; ties overflow rate, horizontal velocity, and basin proportions.
Flow-rate conversion
1 gpd/ft2=1.547×10−6 ft/s=4.72×10−7 m/s=0.0407 m/d1\,\text{gpd/ft}^2 = 1.547\times10^{-6}\,\text{ft/s} = 4.72\times10^{-7}\,\text{m/s} = 0.0407\,\text{m/d}
Convert a reported overflow rate to a velocity before comparing to a Stokes settling velocity.

Where it lives: NCEES PE Civil Reference Handbook — §6.9.6 Settling and Sedimentation (Type I Discrete Settling, Overflow Rate) · Recommended Standards for Water Works (Ten States Standards), 2018 · Davis & Cornwell, Introduction to Environmental Engineering

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Wastewater Collection & Treatment
10% of the exam

Collection systems and lift stations, preliminary/primary/secondary treatment, activated-sludge F/M and SRT, nutrient removal, solids handling, and disinfection.

Activated Sludge: F/M, SRT and Secondary Clarifiers

The F/M ratio, mean cell residence time (SRT), MLSS mass balance, hydraulic retention time, RAS/WAS flows, and the coupled secondary clarifier that together control a biological treatment plant.

Food-to-microorganism ratio
FM=Q0S0VX\frac{F}{M}=\frac{Q_0 S_0}{V X}
Q0Q_0 influent flow, S0S_0 influent BOD (mg/L), VV aeration volume, XX MLSS (mg/L). Units day−1^{-1}. Conventional target 0.2–0.40.2\text{–}0.4.
Solids retention time (SRT, θc\theta_c)
θc=VXQwXw+QeXe\theta_c=\frac{V X}{Q_w X_w + Q_e X_e}
Sludge age in days = solids inventory ÷ solids wasted per day. Master control variable; nitrification needs θc≳8–10 d\theta_c\gtrsim8\text{–}10\ \text{d}.
MLSS from kinetics
X=θc Y(S0−Se)θ(1+kdθc)X=\frac{\theta_c\,Y(S_0-S_e)}{\theta(1+k_d\theta_c)}
YY = yield (mg VSS/mg BOD, typical range 0.4–1.2), kdk_d = endogenous decay (day−1^{-1}, ~0.06), θ=V/Q0\theta=V/Q_0 = HRT. Predicts the MLSS the basin carries.
Hydraulic retention time
θ=VQ0\theta=\frac{V}{Q_0}
Liquid residence time (hours). Distinct from SRT: HRT is on influent flow; SRT is on the solids. Conventional aeration θ≈4–8 hr\theta\approx4\text{–}8\ \text{hr}.
Process efficiency
E=S0−SeS0×100%E=\frac{S_0-S_e}{S_0}\times100\%
BOD removal across secondary; conventional activated sludge 85–95%85\text{–}95\%.
Recycle (RAS) ratio
R=QrQ0=XXr−XR=\frac{Q_r}{Q_0}=\frac{X}{X_r-X}
From a solids balance at the reactor inlet. XX = MLSS, XrX_r = RAS concentration. Typical R≈0.25–1.0R\approx0.25\text{–}1.0.
SVI and limiting RAS concentration
SVI=SV×1000MLSS,Xr,max⁡=106SVI\text{SVI}=\frac{SV\times1000}{\text{MLSS}},\qquad X_{r,\max}=\frac{10^6}{\text{SVI}}
SVI (mL/g) from the 30-min settled volume SVSV (mL/L). Good sludge: 80–12080\text{–}120. Sets the maximum thickening of the underflow.
Secondary clarifier overflow rate
v0=Q0Av_0=\frac{Q_0}{A}
Surface overflow rate uses INFLUENT flow only (Q0Q_0), NOT Q0+QrQ_0+Q_r. RAS does not leave over the weir. Limit ~400–800 gpd/ft2^2 avg.
Secondary clarifier solids loading rate
SLR=8.34 (Q0+Qr) XA\text{SLR}=\frac{8.34\,(Q_0+Q_r)\,X}{A}
Thickening check uses Q0+QrQ_0+Q_r at MLSS XX. Units lb/day-ft2^2; QQ in MGD, XX in mg/L. Typical avg ~19–29.
Clarifier solids mass balance
(Q0+Qr)X=QeXe+QrXr+QwXw(Q_0+Q_r)X = Q_e X_e + Q_r X_r + Q_w X_w
Steady-state solids in = solids out. Closes the system; solve for an unknown flow or concentration.
Solids (system) loading mass
SL=8.34 Q XSL=8.34\,Q\,X
Mass of suspended solids per day (lb/day); QQ in MGD, XX in mg/L. General form of the 8.348.34 conversion for solids.

Where it lives: NCEES PE Civil Reference Handbook — §6.8.5.3 Activated Sludge Treatment (F/M, SRT, MLSS, RAS, SVI, solids loading) · Recommended Standards for Wastewater Facilities (Ten States Standards), 2014 · Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery

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Project Sitework
13% of the exam

Excavation and embankment (cut/fill), horizontal and vertical curve geometry, retaining walls, erosion and sediment control, construction safety, and methods.

Earthwork: Cut/Fill, Shrinkage, Swell and the Mass Diagram

Average-end-area volumes, the bank/loose/compacted state changes that govern borrow and haul, and how the mass-haul diagram sets balance points and overhaul.

Average end-area volume
V=A1+A22 LV = \dfrac{A_1+A_2}{2}\,L
Volume between two cross sections; A1,A2A_1,A_2 in ft2\text{ft}^2, LL in ft\text{ft}, VV in ft3\text{ft}^3 (÷27 for cy). Slightly overestimates for nonlinear sections.
Prismoidal volume
V=L6(A1+4Am+A2)V = \dfrac{L}{6}\left(A_1 + 4A_m + A_2\right)
More accurate volume using midsection area AmA_m; use when the problem supplies AmA_m or asks for prismoidal.
Weight conservation across states
γBVB=γLVL=γCVC\gamma_B V_B = \gamma_L V_L = \gamma_C V_C
Master identity. BB = bank (in-situ), LL = loose (hauled), CC = compacted (fill). Dry weight is conserved, volume is not.
Swell (percent)
swell (%)=(γBγL−1)×100\text{swell}\,(\%) = \left(\dfrac{\gamma_B}{\gamma_L}-1\right)\times 100
Bulking from bank to loose. γL<γB\gamma_L<\gamma_B so swell is positive; loose volume VL=VB(γB/γL)V_L = V_B(\gamma_B/\gamma_L).
Shrinkage (percent)
shrinkage (%)=(1−γBγC)×100\text{shrinkage}\,(\%) = \left(1-\dfrac{\gamma_B}{\gamma_C}\right)\times 100
Densification from bank to compacted. γC>γB\gamma_C>\gamma_B; compacted volume VC=VB(γB/γC)V_C = V_B(\gamma_B/\gamma_C).
Load factor
LF=γLγB,VL=VBLFLF = \dfrac{\gamma_L}{\gamma_B}, \qquad V_L = \dfrac{V_B}{LF}
Converts bank to loose for trucking; LF<1LF<1. Some texts define load factor as the reciprocal — confirm against weight conservation.
Relative compaction
RC=γd,fieldγd,max⁡×100RC = \dfrac{\gamma_{d,\text{field}}}{\gamma_{d,\max}}\times 100
Field acceptance ratio of field dry density to laboratory maximum (Proctor); specs commonly require ≥95%\ge 95\%.
Water to add for moisture
G=γd V (wdes−wborrow)8.33G = \dfrac{\gamma_{d}\,V\,(w_{des}-w_{borrow})}{8.33}
Gallons of water; γd\gamma_d in lb/ft3\text{lb/ft}^3, VV in ft3\text{ft}^3, ww as decimal moisture content, 8.33 lb/gal8.33\,\text{lb/gal}.
Borrow-pit grid square volume
V=Agrid4 (a+b+c+d)V = \dfrac{A_{grid}}{4}\,(a+b+c+d)
Cut volume of one full grid square from corner cut depths a,b,c,da,b,c,d; AgridA_{grid} is the plan area of the square.
Triangular spoil bank volume
V=H2Ltan⁡RV = \dfrac{H^2 L}{\tan R}
Loose spoil pile of length LL, height HH, angle of repose RR; base width B=2H/tan⁡RB = 2H/\tan R.
Overhaul quantity
Overhaul=Voh (dCG-CG−dfh)\text{Overhaul} = V_{oh}\,(d_{\text{CG-CG}}-d_{fh})
Pay quantity in station-yards: overhaul volume times the centroid-to-centroid haul beyond the freehaul distance dfhd_{fh}.
Average haul distance
davg=area under mass curveordinate heightd_{avg} = \dfrac{\text{area under mass curve}}{\text{ordinate height}}
Graphical haul distance for a balanced segment of the mass-haul diagram.

Where it lives: NCEES PE Civil Reference Handbook — §2.1 Earthwork Construction and Layout · NCEES PE Civil Reference Handbook — §2.1.2 Earthwork Volumes · NCEES PE Civil Reference Handbook — §2.1.4 Earthwork Balancing and Haul Distances · Caterpillar Performance Handbook

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