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FE Chemical formula sheet

185 key equations from 17 study topics, each with what it is for and where to look it up: the FE Reference Handbook 10.6 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 FE Chemical study handbook; the full handbook covers every topic at this depth, with worked examples and the traps that cost points.

185

Equations

17

Study topics

17

Concepts

Mathematics
6% of the exam

Analytic geometry, logarithms, and trigonometry, single-variable and integral calculus, ordinary and partial differential equations and Laplace transforms, matrix algebra and systems of equations, and numerical methods including error propagation, Taylor series, curve fitting, and Newton-Raphson.

Numerical Methods, Error and Significant Figures

Newton-Raphson roots, trapezoidal and Simpson integration, Taylor truncation, relative-error propagation, and significant-figure discipline for FE-Chem.

Newton-Raphson iteration
xj+1=xj−f(xj)f′(xj)x_{j+1} = x_j - \dfrac{f(x_j)}{f'(x_j)}
Next root estimate; f′f' is the derivative at the current guess. Quadratic convergence near a simple root; needs f′≠0f'\neq 0 and a good seed.
Taylor series about a
f(x)=∑k=0∞f(k)(a)k!(x−a)kf(x) = \sum_{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k
Basis of truncation-error analysis. Truncating after the linear term gives the Newton tangent. a=0a=0 is the Maclaurin series.
Trapezoidal rule (n strips)
∫abf dx≈Δx2[f0+2f1+2f2+⋯+2fn−1+fn]\int_a^b f\,dx \approx \dfrac{\Delta x}{2}\Big[f_0 + 2f_1 + 2f_2 + \cdots + 2f_{n-1} + f_n\Big]
Straight-chord approximation; error ∝Δx2\propto \Delta x^2. Δx=(b−a)/n\Delta x=(b-a)/n.
Simpson's 1/3 rule
∫abf dx≈Δx3[f0+4f1+2f2+⋯+4fn−1+fn]\int_a^b f\,dx \approx \dfrac{\Delta x}{3}\Big[f_0 + 4f_1 + 2f_2 + \cdots + 4f_{n-1} + f_n\Big]
Parabolic fit; exact for cubics, error ∝Δx4\propto \Delta x^4. nn must be even (odd number of points).
Step width
Δx=b−an\Delta x = \dfrac{b-a}{n}
Uniform spacing for nn subintervals between limits aa and bb.
Relative error
εrel=∣xapprox−xtrue∣∣xtrue∣\varepsilon_{rel} = \dfrac{|x_{approx}-x_{true}|}{|x_{true}|}
Dimensionless; multiply by 100 for percent. Use the previous iterate as the reference when the true value is unknown.
Error propagation (products)
∣Δyy∣=∑i∣ai∣∣Δxixi∣\left|\dfrac{\Delta y}{y}\right| = \sum_i |a_i|\left|\dfrac{\Delta x_i}{x_i}\right|
Worst-case relative error for y=∏xiaiy=\prod x_i^{a_i}; fractional errors add, weighted by exponents.
Error propagation (RSS)
Δyy=∑iai2(Δxixi)2\dfrac{\Delta y}{y} = \sqrt{\sum_i a_i^2\left(\dfrac{\Delta x_i}{x_i}\right)^2}
Probable error when inputs are independent and random; always smaller than the linear sum.
Error propagation (sums)
y=∑icixi  ⇒  Δy=∑ici2 (Δxi)2y = \sum_i c_i x_i \;\Rightarrow\; \Delta y = \sqrt{\sum_i c_i^2\,(\Delta x_i)^2}
For additive combinations the absolute errors (not relative) combine.

Where it lives: NCEES FE Reference Handbook — Mathematics (Numerical Methods, Numerical Integration, Taylor's Series) · NCEES FE Reference Handbook — Units and Conversion Factors (Significant Figures, p.2) · NCEES FE Reference Handbook — Engineering Probability and Statistics (Measurement Uncertainty / Propagation of Error, RSS form) · Chapra & Canale, Numerical Methods for Engineers

Read the Mathematics chapter

Probability and Statistics
4% of the exam

Discrete, continuous, normal, and binomial distributions, expected value in decision making, hypothesis testing and design of experiments (t-test, ANOVA, outlier testing), measures of central tendency and dispersion with confidence intervals, regression and curve fitting, and statistical process-control limits.

Distributions, Confidence Intervals and Control Limits

Discrete and continuous distributions, the binomial and normal, mean/median/mode and spread, the confidence interval for a mean, and X-bar/R control limits.

Sample mean
Xˉ=1n∑i=1nXi\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i
Arithmetic mean (balance point) of nn observations; the statistic every interval and chart centers on.
Sample variance and standard deviation
s2=1n−1∑i=1n(Xi−Xˉ)2,s=s2s^2=\frac{1}{n-1}\sum_{i=1}^{n}\left(X_i-\bar{X}\right)^2,\qquad s=\sqrt{s^2}
Spread of a sample; divide by n−1n-1 (use NN for a full population). Units of ss match the data; s2s^2 is squared units.
Binomial mass function
P(x)=C(n,x) px(1−p) n−xP(x)=C(n,x)\,p^{x}(1-p)^{\,n-x}
Probability of exactly xx successes in nn independent trials, each with success probability pp. C(n,x)=n!/[x!(n−x)!]C(n,x)=n!/[x!(n-x)!].
Binomial mean and variance
μ=np,σ2=npq,q=1−p\mu=np,\qquad \sigma^{2}=npq,\quad q=1-p
Expected count and its variance (a count, not a fraction). Standard deviation σ=npq\sigma=\sqrt{npq}.
Normal probability density
f(x)=1σ2π e−12(x−μσ)2f(x)=\frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}
Gaussian curve, mean μ\mu, standard deviation σ\sigma. Never integrated by hand on the FE; standardize instead.
Standard-normal transform
z=x−μσz=\frac{x-\mu}{\sigma}
Converts any normal value to the unit normal so the handbook table gives F(z)F(z) (left area), R(z)R(z) (right area), W(z)W(z) (within ±z\pm z).
Within-band probability
P(−z≤Z≤z)=W(z)=2F(z)−1P(-z\le Z\le z)=W(z)=2F(z)-1
Fraction of a normal population inside μ±zσ\mu\pm z\sigma; e.g. W(1.96)≈0.95W(1.96)\approx0.95.
Standard error of the mean
σXˉ=σn    (or s/n)\sigma_{\bar X}=\frac{\sigma}{\sqrt{n}}\;\;(\text{or }s/\sqrt{n})
Spread of the sample mean; shrinks only as n\sqrt{n}. Drives both the confidence interval and the X-bar chart width.
Confidence interval, sigma known
Xˉ±zα/2 σn\bar{X}\pm z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}}
Interval for μ\mu when the population σ\sigma is known; z0.025=1.960z_{0.025}=1.960 for 95%, 1.6451.645 for 90%.
Confidence interval, sigma unknown
Xˉ±tα/2, n−1 sn\bar{X}\pm t_{\alpha/2,\,n-1}\,\frac{s}{\sqrt{n}}
Interval for μ\mu using sample ss; tt has n−1n-1 degrees of freedom and is wider than zz for small samples.
X-bar chart limits
UCL=X‾‾+A2Rˉ,LCL=X‾‾−A2Rˉ\text{UCL}=\overline{\overline{X}}+A_2\bar{R},\qquad \text{LCL}=\overline{\overline{X}}-A_2\bar{R}
Three-sigma limits on subgroup averages; X‾‾\overline{\overline{X}} is the grand mean, Rˉ\bar{R} the mean range, A2A_2 from the handbook table for subgroup size nn.
R chart limits and process sigma
UCLR=D4Rˉ,LCLR=D3Rˉ,σ^=Rˉd2\text{UCL}_R=D_4\bar{R},\quad \text{LCL}_R=D_3\bar{R},\qquad \hat{\sigma}=\frac{\bar{R}}{d_2}
Limits on subgroup range; D3,D4,d2D_3,D_4,d_2 from the table. σ^=Rˉ/d2\hat{\sigma}=\bar{R}/d_2 estimates the process standard deviation.

Where it lives: NCEES FE Reference Handbook — Engineering Probability and Statistics · NCEES FE Reference Handbook — Statistical Quality Control (Average and Range Charts) · Montgomery & Runger, Applied Statistics and Probability for Engineers

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Engineering Sciences
4% of the exam

Basic dynamics (friction, force, mass, acceleration, momentum), work, energy, and power for particles and rigid bodies, and electricity fundamentals (charge, current, voltage, power, Ohm's law, and Kirchhoff's laws).

Newtonian Dynamics, Work, Energy and Power

Newton's second law with friction, the work-energy theorem, and impulse-momentum for particles — the three bookkeeping methods that solve almost every FE dynamics question.

Newton's second law (particle)
∑F=ma=mdvdt\sum \mathbf{F} = m\mathbf{a} = m\dfrac{d\mathbf{v}}{dt}
Net external force equals mass times acceleration. SI: FF in N, mm in kg, aa in m/s2^2. Apply per axis from a free-body diagram.
Newton's law in USCS (gc form)
F=m agc,gc=32.174 lbm⋅ftlbf⋅s2F = \dfrac{m\,a}{g_c}, \quad g_c = 32.174\ \tfrac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}
Required whenever mass is in lbm and force in lbf; gcg_c reconciles the two. Weight W=mg/gcW = mg/g_c.
Coulomb friction
Ff≤μsN,Ff=μkN (sliding)F_f \le \mu_s N, \qquad F_f = \mu_k N \ (\text{sliding})
Friction opposes relative motion. NN = actual normal force (mgcos⁡θmg\cos\theta on an incline). μs,μk\mu_s,\mu_k dimensionless.
Constant-acceleration kinematics
v=v0+at,v2=v02+2a(x−x0)v = v_0 + at, \quad v^2 = v_0^2 + 2a(x-x_0)
Links velocity, acceleration, time, and displacement for constant aa. Use with ∑F=ma\sum F = ma to bridge force and motion.
Kinetic energy
T=12mv2T = \tfrac{1}{2}mv^{2}
Energy of motion of a particle. vv = speed (m/s), result in J. Note vv is squared — never linear.
Work of a force
U=∫F⋅dr=Fscos⁡θ (constant F)U = \int \mathbf{F}\cdot d\mathbf{r} = Fs\cos\theta\ (\text{constant } F)
θ\theta = angle between force and displacement; forces perpendicular to motion do zero work.
Work-energy theorem
U1→2=12mv22−12mv12U_{1\to 2} = \tfrac{1}{2}mv_2^{2} - \tfrac{1}{2}mv_1^{2}
Net work equals change in kinetic energy. Time-free — ideal for force-and-distance to speed problems.
Conservation of energy
T1+V1+U1→2nc=T2+V2T_1 + V_1 + U_{1\to 2}^{nc} = T_2 + V_2
V=mgh+12ks2V = mgh + \tfrac{1}{2}ks^2. U1→2ncU_{1\to 2}^{nc} = work of nonconservative forces only (negative for friction; zero for conservative systems). Distinct from the work-energy theorem U1→2U_{1\to 2}, which is the net work of all forces — do not equate the two.
Linear impulse-momentum
mv2=mv1+∫t1t2F dtm\mathbf{v}_2 = m\mathbf{v}_1 + \int_{t_1}^{t_2}\mathbf{F}\,dt
Impulse of net force equals change in momentum. For constant FF: F Δt=m ΔvF\,\Delta t = m\,\Delta v.
Conservation of momentum (impact)
m1v1+m2v2=m1v1′+m2v2′m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
Holds when external impulse is negligible during impact. KE conserved only if e=1e=1.
Coefficient of restitution
e=(v2′−v1′)n(v1−v2)n, 0≤e≤1e = \dfrac{(v_2' - v_1')_n}{(v_1 - v_2)_n}, \ 0\le e\le 1
e=1e=1 perfectly elastic, e=0e=0 perfectly plastic (bodies stick). Applied along the line of impact.
Power and efficiency
P=F⋅v,η=PoutPinP = \mathbf{F}\cdot\mathbf{v}, \qquad \eta = \dfrac{P_{out}}{P_{in}}
Power in W (or ft·lbf/s); 1 hp=746 W1\,\text{hp}=746\,\text{W}. Force and velocity must be collinear.

Where it lives: NCEES FE Reference Handbook — Dynamics · NCEES FE Reference Handbook — Dynamics (Laws of Friction) · Hibbeler, Engineering Mechanics: Dynamics

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Materials Science
4% of the exam

Chemical, electrical, mechanical, and physical properties and the effects of temperature, pressure, stress, and strain, material types and compatibilities for ferrous, nonferrous, and engineered materials, corrosion mechanisms and control, and polymers, ceramics, and composites.

Material Properties and Corrosion Control

Read the tensile curve, apply Hooke's law and the elastic constants, account for temperature and time, and set up the electrochemical cell that drives corrosion.

Engineering stress and strain
σ=FA0,ε=ΔLL0\sigma = \dfrac{F}{A_0}, \qquad \varepsilon = \dfrac{\Delta L}{L_0}
Based on ORIGINAL area A0A_0 and length L0L_0. σ\sigma in Pa (or MPa), ε\varepsilon dimensionless.
Hooke's law
σ=E ε\sigma = E\,\varepsilon
Linear elastic region only. EE = elastic (Young's) modulus, Pa; slope of the tensile curve's straight portion.
Axial deformation
δ=PLAE\delta = \dfrac{PL}{AE}
Elastic elongation of a bar of length LL, area AA, modulus EE under axial load PP. Units m.
Shear and bulk modulus
G=E2(1+ν),K=E3(1−2ν)G = \dfrac{E}{2(1+\nu)}, \qquad K = \dfrac{E}{3(1-2\nu)}
Isotropic elastic links. ν\nu = Poisson's ratio (≈0.3\approx 0.3 for steel). Only two of E,G,K,νE,G,K,\nu are independent.
Poisson's ratio
ν=−εlateralεlongitudinal\nu = -\dfrac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{longitudinal}}}
Negative ratio of transverse to axial strain. Dimensionless; 0≤ν≤0.50 \le \nu \le 0.5 for stable isotropic solids.
True stress and true strain
σT=FA,εT=ln⁡(1+ε)\sigma_T = \dfrac{F}{A}, \qquad \varepsilon_T = \ln(1+\varepsilon)
Use instantaneous area AA; needed in plastic/necking analysis. Differs from engineering values beyond yield.
Thermal deformation and stress
δt=αL (T−T0),σthermal=E α ΔT\delta_t = \alpha L\,(T - T_0), \qquad \sigma_{\text{thermal}} = E\,\alpha\,\Delta T
Free expansion (first) vs fully restrained stress (second). α\alpha = expansion coefficient, 1/K or 1/°C.
Thermal expansion coefficient
α=εΔT\alpha = \dfrac{\varepsilon}{\Delta T}
Engineering strain per degree of temperature change. Units 1/K (equivalently 1/°C for a difference).
Fracture toughness
KIC=Y σπaK_{IC} = Y\,\sigma\sqrt{\pi a}
Critical stress intensity; aa = crack length, YY = geometry factor (11 interior, 1.11.1 surface). Units MPa·m1/2^{1/2}.
Hardness–strength estimate
TS (MPa)≈3.5 BHN\text{TS (MPa)} \approx 3.5\,\text{BHN}
Plain-carbon steel approximation (or TS(psi) ≈ 500·BHN). BHN = Brinell hardness number.
Resistivity and capacitance
ρ=RAL,C=εAd\rho = \dfrac{RA}{L}, \qquad C = \dfrac{\varepsilon A}{d}
ρ\rho = resistivity (Ω·m); CC = parallel-plate capacitance, ε=κε0\varepsilon=\kappa\varepsilon_0, ε0=8.85×10−12 \varepsilon_0=8.85\times10^{-12}\,F/m.
Diffusion coefficient (Arrhenius)
D=D0 e−Q/(RT)D = D_0\,e^{-Q/(RT)}
Temperature dependence of diffusion/creep. QQ = activation energy, R=8.314 R=8.314\,J/(mol·K), TT in K.
Galvanic cell EMF
Ecell=Eox, anode∘+Ered, cathode∘E_{cell} = E^\circ_{ox,\,\text{anode}} + E^\circ_{red,\,\text{cathode}}
Positive EcellE_{cell} means the couple corrodes spontaneously. Use handbook oxidation potentials (reverse sign for the cathode).
Nernst equation (25 °C)
E=E∘−0.0592nlog⁡10QE = E^\circ - \dfrac{0.0592}{n}\log_{10} Q
Concentration correction to the standard potential; nn = electrons transferred, QQ = reaction quotient.

Where it lives: NCEES FE Reference Handbook — Materials Science/Structure of Matter · NCEES FE Reference Handbook — Mechanics of Materials · NCEES FE Reference Handbook — Chemistry and Biology · Callister & Rethwisch, Materials Science and Engineering: An Introduction

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Chemistry and Biology
7% of the exam

Inorganic chemistry (molarity, normality, acids and bases, redox, solubility product, pH and pK, electrochemistry), organic chemistry (nomenclature, structure, reactions, synthesis), analytical chemistry, biochemistry and microbiology (cell function, glycolysis and the Krebs cycle, enzymes, genetics), and bioprocessing (fermentation and aerobic/anaerobic treatment).

Solutions, pH, Buffers and Electrochemistry

Concentration units, weak-acid and buffer pH, the solubility product, and the Nernst equation — the aqueous-chemistry workhorses the FE rewards for speed.

Molarity and normality
M=nsoluteVsoln(L),N=M zM = \dfrac{n_{solute}}{V_{soln}(\text{L})}, \qquad N = M\,z
MM in mol/L; zz = equivalents per mole (protons for acids, electrons for redox). Normality is solution-context dependent.
Titration equivalence
N1V1=N2V2N_1 V_1 = N_2 V_2
Equal equivalents at the endpoint. Volumes in any consistent unit; valid only in normality, not molarity (unless 1:1). Not printed in the FE Reference Handbook 10.6.
pH and water ion product
pH=−log⁡10[H+],pH+pOH=14,Kw=[H+][OH−]=10−14\mathrm{pH} = -\log_{10}[\mathrm{H^+}], \quad \mathrm{pH}+\mathrm{pOH}=14, \quad K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 10^{-14}
At 25 ∘C25\,^\circ\mathrm{C}. [H+][\mathrm{H^+}] in mol/L. Acids pH < 7, bases pH > 7.
Weak-acid dissociation
Ka=[H+][A−][HA],pKa=−log⁡10KaK_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}, \qquad \mathrm{p}K_a = -\log_{10}K_a
Equilibrium constant for HA⇌H++A−\mathrm{HA}\rightleftharpoons \mathrm{H^+}+\mathrm{A^-}. Smaller KaK_a (larger pKa\mathrm{p}K_a) = weaker acid.
Weak-acid pH approximation
[H+]≈Ka C[\mathrm{H^+}] \approx \sqrt{K_a\,C}
Valid when dissociation <5%<5\% of formal concentration CC (mol/L). Verify the fraction [H+]/C[\mathrm{H^+}]/C afterward.
Henderson-Hasselbalch
pH=pKa+log⁡10[A−][HA]\mathrm{pH} = \mathrm{p}K_a + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}
Buffer pH from the conjugate-base to acid ratio. Independent of dilution; pH=pKa\mathrm{pH}=\mathrm{p}K_a when the ratio is 1.
Solubility product
AmBn⇌mAn++nBm−,Ksp=[An+]m[Bm−]n\mathrm{A}_m\mathrm{B}_n \rightleftharpoons m\mathrm{A}^{n+}+n\mathrm{B}^{m-}, \quad K_{sp} = [\mathrm{A}^{n+}]^m[\mathrm{B}^{m-}]^n
Solids and pure liquids omitted (activity = 1). Compare QQ to KspK_{sp} to test precipitation.
Equilibrium constant (general)
Keq=[C]c[D]d[A]a[B]bK_{eq} = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}
For aA+bB⇌cC+dDa\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}. Brackets are activities; = molar concentration in dilute solution, partial pressure for gases, 1 for pure solids/liquids.
Standard cell potential
Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}
Both values are reduction potentials (V). Ecell∘>0E^\circ_{cell}>0 means spontaneous as written.
Nernst equation (25 °C)
E=E∘−0.0592nlog⁡10QE = E^\circ - \dfrac{0.0592}{n}\log_{10}Q
nn = electrons transferred; QQ = reaction quotient (products over reactants). Full form: E=E∘−(RT/nF)ln⁡QE=E^\circ-(RT/nF)\ln Q, F=96,485 C/molF=96{,}485\,\mathrm{C/mol}.
Faraday's law of electrolysis
m=Q MzFm = \dfrac{Q\,M}{zF}
mm = mass deposited (g); QQ = charge (C = A·s); MM = molar mass (g/mol); zz = electrons per ion; F=96,485 C/molF=96{,}485\,\mathrm{C/mol}.

Where it lives: NCEES FE Reference Handbook — Chemistry and Biology · NCEES FE Reference Handbook — Environmental Engineering · Felder & Rousseau, Elementary Principles of Chemical Processes

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Fluid Mechanics and Dynamics
8% of the exam

Fluid properties and dimensionless numbers (Reynolds), the mechanical energy balance with pipe, valve, fitting, and packed-bed losses, the Bernoulli equation and hydrostatics, laminar and turbulent flow, flow measurement (orifices and Venturi meters), pumps, compressors, and vacuum systems, and compressible and non-Newtonian flow.

The Mechanical Energy Balance and Bernoulli

Write extended Bernoulli as heads, add pipe, valve and fitting friction losses, account for elevation, and size pump head or pressure drop with confidence.

Bernoulli equation (no friction, no machine)
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
Conservation of mechanical energy per unit weight along a streamline; each term in meters (or feet). γ=ρg\gamma=\rho g.
Extended (mechanical energy) balance
P1γ+v122g+z1+hp=P2γ+v222g+z2+hf\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_f
Adds pump head hph_p (energy in) and total head loss hfh_f (energy out, always positive). The form used for pump and pressure-drop sizing.
Continuity
m˙=ρQ=ρAv,Q=A1v1=A2v2\dot{m}=\rho Q=\rho A v,\qquad Q=A_1 v_1=A_2 v_2
Mass and (for constant density) volumetric flow conserved; A=π4D2A=\tfrac{\pi}{4}D^2 for round pipe. Units m3/s\mathrm{m^3/s}, m/s\mathrm{m/s}.
Darcy-Weisbach major loss
hf=f LD v22gh_f=f\,\frac{L}{D}\,\frac{v^{2}}{2g}
Straight-pipe friction head; ff from Moody chart vs Re\mathrm{Re} and ε/D\varepsilon/D. LL, DD in m, vv in m/s.
Fanning–Darcy relation
fDarcy=4 fFanningf_{\text{Darcy}}=4\,f_{\text{Fanning}}
Chemical-engineering charts often give Fanning fFf_F; the Darcy form needs the factor of four (or use hf=4fFLDv22gh_f=4f_F\tfrac{L}{D}\tfrac{v^2}{2g}).
Minor (fitting) losses
hf,minor=∑C v22gh_{f,\text{minor}}=\sum C\,\frac{v^{2}}{2g}
CC (or KK) is the dimensionless loss coefficient per fitting/valve/entrance/exit; sum over the run.
Hydrostatic pressure
P2−P1=−γ(z2−z1)=ρghP_2-P_1=-\gamma(z_2-z_1)=\rho g h
Static-fluid pressure variation with depth hh; basis of manometers. Use absolute or gauge consistently.
Pressure drop in a straight horizontal pipe
P1−P2=γ hf=f LD ρv22P_1-P_2=\gamma\,h_f=f\,\frac{L}{D}\,\frac{\rho v^{2}}{2}
Special case with z1=z2z_1=z_2, v1=v2v_1=v_2, no pump; converts head loss directly to a pressure drop in Pa.
Hydraulic (fluid) power
W˙fluid=ρgQhp=γQhp\dot{W}_{\text{fluid}}=\rho g Q h_p=\gamma Q h_p
Useful power delivered to fluid by a pump of head hph_p; in watts when QQ in m3/s\mathrm{m^3/s}, hph_p in m.
Brake and electrical power
W˙brake=W˙fluidηpump,W˙elec=W˙brakeηmotor\dot{W}_{\text{brake}}=\frac{\dot{W}_{\text{fluid}}}{\eta_{\text{pump}}},\qquad \dot{W}_{\text{elec}}=\frac{\dot{W}_{\text{brake}}}{\eta_{\text{motor}}}
Shaft power and electrical draw; efficiencies are fractions ≤1\le 1.
Torricelli (frictionless efflux)
v=2ghv=\sqrt{2gh}
Free jet velocity from a head hh when friction is negligible; the friction case puts 1+fLD+∑C1+f\tfrac{L}{D}+\sum C in the denominator inside the root (the leading 11 is the exit velocity head).

Where it lives: NCEES FE Reference Handbook — Fluid Mechanics · NCEES FE Reference Handbook — Fluid Mechanics (Energy Equation, Bernoulli, Darcy-Weisbach, Minor Losses) · Felder & Rousseau, Elementary Principles of Chemical Processes

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Thermodynamics
8% of the exam

Thermodynamic properties of pure components and mixtures, property data and phase diagrams (steam tables, P-h, T-s, x-y), the first and second laws, isothermal, adiabatic, and isentropic processes, power and refrigeration cycles, phase equilibrium (Raoult's law, fugacity, activity coefficients), chemical equilibrium, and heats of reaction and mixing.

First and Second Law: Processes, Cycles and Efficiency

Closed- and open-system energy balances, entropy and the second law, the isothermal/adiabatic/isentropic process family, and Carnot ceilings on efficiency and COP.

Closed-system first law
Q−W=ΔU+ΔKE+ΔPEQ - W = \Delta U + \Delta KE + \Delta PE
Energy balance with no mass flow. QQ positive when added to the system, WW positive when done by the system; energies in kJ or Btu.
Steady-flow energy equation
Q˙in−W˙net=∑em˙e ⁣(he+Ve22+gZe)−∑im˙i ⁣(hi+Vi22+gZi)\dot{Q}_{in} - \dot{W}_{net} = \sum_e \dot{m}_e\!\left(h_e + \tfrac{V_e^2}{2} + gZ_e\right) - \sum_i \dot{m}_i\!\left(h_i + \tfrac{V_i^2}{2} + gZ_i\right)
Open-system rate balance at steady state. hh = specific enthalpy (kJ/kg), m˙\dot m = mass flow (kg/s); flow work is already inside h=u+Pvh = u + Pv.
Ideal-gas relations
Pv=RT,Δu=cv ΔT,Δh=cp ΔT,cp−cv=RPv = RT, \quad \Delta u = c_v\,\Delta T, \quad \Delta h = c_p\,\Delta T, \quad c_p - c_v = R
RR = specific gas constant (kJ/kg·K). Property changes depend only on ΔT\Delta T for an ideal gas, independent of path.
Isentropic (reversible adiabatic) relations
Pvk=const,T2T1=(P2P1)k−1k=(v1v2)k−1Pv^{k} = \text{const}, \qquad \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} = \left(\frac{v_1}{v_2}\right)^{k-1}
k=cp/cvk = c_p/c_v. Use absolute TT and PP. Valid for an ideal gas with constant specific heats.
Isentropic ideal-gas work
wclosed=R(T2−T1)1−k,wopen=kR(T2−T1)1−kw_{closed} = \frac{R(T_2 - T_1)}{1 - k}, \qquad w_{open} = \frac{kR(T_2 - T_1)}{1 - k}
Closed-system (boundary) work versus open-system (shaft) work for an isentropic process; the two differ by the factor kk.
Isothermal ideal-gas work
w=RTln⁡v2v1=RTln⁡P1P2w = RT\ln\frac{v_2}{v_1} = RT\ln\frac{P_1}{P_2}
Constant-temperature ideal-gas process; for a closed system q=wq = w since Δu=0\Delta u = 0.
Ideal-gas entropy change
Δs=cpln⁡T2T1−Rln⁡P2P1=cvln⁡T2T1+Rln⁡v2v1\Delta s = c_p\ln\frac{T_2}{T_1} - R\ln\frac{P_2}{P_1} = c_v\ln\frac{T_2}{T_1} + R\ln\frac{v_2}{v_1}
Entropy change between two states of an ideal gas (kJ/kg·K). Absolute TT throughout.
Increase-of-entropy principle
ΔStotal=ΔSsys+ΔSsurr≥0\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \ge 0
Total entropy never decreases; equality only for a reversible process. Sets the direction of spontaneous change.
Thermal efficiency
η=WQH=1−QLQH\eta = \frac{W}{Q_H} = 1 - \frac{Q_L}{Q_H}
Fraction of supplied heat converted to net work. Denominator is the heat supplied QHQ_H, never WW.
Carnot efficiency
ηCarnot=1−TLTH\eta_{Carnot} = 1 - \frac{T_L}{T_H}
Maximum efficiency between reservoirs at absolute temperatures THT_H, TLT_L (K or °R). Upper bound no real engine exceeds.
Coefficients of performance
COPref=QLW,COPHP=QHW=COPref+1\mathrm{COP}_{ref} = \frac{Q_L}{W}, \qquad \mathrm{COP}_{HP} = \frac{Q_H}{W} = \mathrm{COP}_{ref} + 1
Cooling values QLQ_L; heating values QHQ_H. The two differ by exactly one because QH=QL+WQ_H = Q_L + W.
Carnot COP ceilings
COPHP,max⁡=THTH−TL,COPref,max⁡=TLTH−TL\mathrm{COP}_{HP,\max} = \frac{T_H}{T_H - T_L}, \qquad \mathrm{COP}_{ref,\max} = \frac{T_L}{T_H - T_L}
Maximum COP of a reversed-Carnot cycle between the two reservoirs; absolute temperatures only.
Reservoir entropy change
ΔSres=QTres\Delta S_{res} = \frac{Q}{T_{res}}
Entropy change of a thermal reservoir exchanging heat QQ at constant absolute temperature TresT_{res} (kJ/K).

Where it lives: NCEES FE Reference Handbook — Thermodynamics · Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics · Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering Thermodynamics

Read the Thermodynamics chapter

Material and Energy Balances
10% of the exam

Steady-state and unsteady-state mass balances, steady-state and unsteady-state energy balances, recycle and bypass processes, and reactive systems including combustion, with conversion, yield, and extent of reaction.

Material Balances and Degrees of Freedom

The steady-state total and component mass balance, degrees-of-freedom analysis, tie components, and a disciplined recipe for solving multi-unit flowsheets.

General material balance
dmdt=m˙in−m˙out+m˙gen\dfrac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out} + \dot{m}_{gen}
Accumulation = in − out + generation. Per unit time for a defined control volume; m˙\dot m in kg/s (or mol/s for a species).
Steady-state total balance
∑m˙in=∑m˙out\sum \dot{m}_{in} = \sum \dot{m}_{out}
Total mass is conserved even with reaction; accumulation and net generation of total mass are zero.
Component balance (no reaction)
∑inFk xj,k=∑outPl xj,l\sum_{\text{in}} F_k\,x_{j,k} = \sum_{\text{out}} P_l\,x_{j,l}
For each species jj: mass (or mole) of jj in = out. xj,kx_{j,k} is the fraction of jj in stream kk.
Composition closure
∑jxj=1\sum_j x_j = 1
Mass or mole fractions in any single stream sum to unity — one usable equation per stream.
Tie-component relation
F xtie,F=P xtie,PF\,x_{\text{tie},F} = P\,x_{\text{tie},P}
An inert that enters and leaves in one stream each directly links the two stream flows FF and PP.
Degrees of freedom
DOF=Nunknowns−Nindep. equationsDOF = N_{\text{unknowns}} - N_{\text{indep. equations}}
DOF=0DOF=0 solvable; >0>0 underspecified (set a basis); <0<0 over-specified. Independent equations = balances + specs + relations.
Independent balances
Nindep. balances=SN_{\text{indep. balances}} = S
For SS species, only SS of the (total + SS component) balances are independent.
Mass–mole conversion
nj=mjMj,yj=nj∑knkn_j = \dfrac{m_j}{M_j}, \qquad y_j = \dfrac{n_j}{\sum_k n_k}
Convert a stream between mass basis (mjm_j, mass fraction) and mole basis (njn_j, mole fraction yjy_j) using molar masses MjM_j.

Where it lives: NCEES FE Reference Handbook — Chemical Engineering · NCEES FE Reference Handbook — Thermodynamics · Felder & Rousseau, Elementary Principles of Chemical Processes

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Heat Transfer
8% of the exam

Conductive, convective (natural and forced), and radiation heat transfer, overall, local, and fouling heat-transfer coefficients, and heat-transfer equipment design (double-pipe and shell-and-tube exchangers, log-mean temperature difference, the effectiveness-NTU method, and flow configuration).

Conduction, Thermal Resistance and the Overall U

Fourier conduction through walls and cylinders, the series-parallel resistance network, fouling, and the overall heat-transfer coefficient U.

Fourier's law (1-D)
Q˙=−kA dTdx\dot{Q} = -kA\,\dfrac{dT}{dx}
Conduction heat rate (W). kk = thermal conductivity [W/(m·K)], AA = area normal to flow (m²); minus sign = flow down the gradient.
Plane-wall conduction
Q˙=kAL (T1−T2)\dot{Q} = \dfrac{kA}{L}\,(T_1 - T_2)
Steady 1-D slab of thickness LL (m); T1>T2T_1>T_2 are the two face temperatures (K or °C — a difference is the same in both).
Thermal-resistance (Ohm) form
Q˙=ΔTtotal∑R\dot{Q} = \dfrac{\Delta T_{\text{total}}}{\sum R}
Heat rate equals total temperature drop over the sum of series resistances (K/W). The same Q˙\dot{Q} flows through each.
Plane-wall resistance
R=LkAR = \dfrac{L}{kA}
Conduction resistance of a slab (K/W). LL = thickness, AA = face area.
Convection resistance
R=1hAR = \dfrac{1}{hA}
Film resistance (K/W). hh = convection coefficient [W/(m²·K)], AA = wetted area on that side.
Cylindrical-wall resistance
R=ln⁡(r2/r1)2πkLR = \dfrac{\ln(r_2/r_1)}{2\pi k L}
Radial conduction through a tube of length LL between radii r1r_1 (inner) and r2r_2 (outer).
Critical insulation radius
rcr=kinshor_{cr} = \dfrac{k_{\text{ins}}}{h_o}
Outer radius that maximizes heat loss from an insulated cylinder; below it, adding insulation increases loss.
Fouling resistance
Rfoul=RfAR_{\text{foul}} = \dfrac{R_f}{A}
RfR_f is the tabulated per-area fouling factor [(m²·K)/W]; divide by the fouled-side area to get K/W.
Overall coefficient (UA)
1UA=1hiAi+Rf,iAi+ln⁡(Do/Di)2πkL+Rf,oAo+1hoAo\dfrac{1}{UA} = \dfrac{1}{h_i A_i} + \dfrac{R_{f,i}}{A_i} + \dfrac{\ln(D_o/D_i)}{2\pi k L} + \dfrac{R_{f,o}}{A_o} + \dfrac{1}{h_o A_o}
Series sum of all tube-side resistances. UAUA (W/K) carries the area, so Q˙=UA ΔT\dot{Q}=UA\,\Delta T needs no reference-area choice.
Overall U vs reference area
UiAi=UoAo=1∑RU_i A_i = U_o A_o = \dfrac{1}{\sum R}
UU must be quoted against a stated area (inside or outside); only the product UAUA is unambiguous.
Interface temperature
Tn+1=Tn−Q˙ RnT_{n+1} = T_n - \dot{Q}\,R_n
March the same Q˙\dot{Q} through each resistance to recover any intermediate temperature.
Biot number (lumped check)
Bi=hVkAs=hLckBi = \dfrac{h V}{k A_s} = \dfrac{h L_c}{k}
If Bi<0.1Bi < 0.1 a body is nearly isothermal and the lumped-capacitance transient model applies; Lc=V/AsL_c=V/A_s.

Where it lives: NCEES FE Reference Handbook — Heat Transfer · Incropera & DeWitt, Fundamentals of Heat and Mass Transfer · Perry's Chemical Engineers' Handbook — Heat Transfer

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Mass Transfer and Separation
8% of the exam

Molecular and convective mass transfer with diffusion and mass-transfer coefficients, separation systems (distillation, absorption, extraction, membranes, adsorption), equilibrium-stage methods (McCabe-Thiele, stage efficiency), continuous-contact methods (NTU, HTU, HETP), and humidification, drying, and evaporation.

Molecular Diffusion and Convective Mass Transfer

Fick's law, equimolar counter-diffusion versus diffusion through stagnant B, the two-film model and its coefficients, and the momentum-heat-mass analogy.

Fick's first law
JA=−D dCAdxJ_A = -D\,\dfrac{dC_A}{dx}
Steady 1-D molar flux (mol⋅m−2s−1\text{mol·m}^{-2}\text{s}^{-1}) of A; DD = diffusivity (m2/s\text{m}^2/\text{s}), CAC_A = concentration (mol/m3\text{mol/m}^3). Minus sign: flux goes down the gradient.
Fick's second law
∂C∂t=D ∂2C∂x2\dfrac{\partial C}{\partial t} = D\,\dfrac{\partial^2 C}{\partial x^2}
Unsteady 1-D diffusion (constant DD). The mass-transfer twin of the transient heat-conduction equation.
Semi-infinite solution (erf)
C−CsC0−Cs=erf⁡ ⁣(x2Dt)\dfrac{C - C_s}{C_0 - C_s} = \operatorname{erf}\!\left(\dfrac{x}{2\sqrt{Dt}}\right)
Surface held at CsC_s, bulk initially C0C_0. Form z=x/(2Dt)z = x/(2\sqrt{Dt}), then read erf⁡(z)\operatorname{erf}(z) from the handbook table.
Equimolar counter-diffusion (gases)
NA=DmRT Δz (pA1−pA2)N_A = \dfrac{D_m}{RT\,\Delta z}\,(p_{A1}-p_{A2})
Use when NB=−NAN_B=-N_A (no net molar flow), e.g., binary distillation. Δz\Delta z = path length.
Diffusion through stagnant B
NA=Dm PRT Δz (pB)lm (pA1−pA2)N_A = \dfrac{D_m\,P}{RT\,\Delta z\,(p_B)_{lm}}\,(p_{A1}-p_{A2})
Use when B is inert and motionless (gas absorption). The (pB)lm(p_B)_{lm} term adds the Stefan drift; always gives a larger flux than equimolar.
Log-mean partial pressure of B
(pB)lm=pB2−pB1ln⁡(pB2/pB1),pB=P−pA(p_B)_{lm} = \dfrac{p_{B2}-p_{B1}}{\ln(p_{B2}/p_{B1})}, \quad p_B = P - p_A
Log mean of the inert's partial pressures at the two film faces.
Two-film flux
NA=kG′(pAG−pAi)=kL′(CAi−CAL)N_A = k_G'(p_{AG}-p_{Ai}) = k_L'(C_{Ai}-C_{AL})
Gas-film and liquid-film rate expressions; equal at steady state. kG′k_G', kL′k_L' are individual coefficients, interface in equilibrium.
Overall coefficients
1KG′=1kG′+HkL′,1KL′=1HkG′+1kL′\dfrac{1}{K_G'} = \dfrac{1}{k_G'} + \dfrac{H}{k_L'}, \quad \dfrac{1}{K_L'} = \dfrac{1}{H k_G'} + \dfrac{1}{k_L'}
Series resistances. Large HH (insoluble) -> liquid film controls; small HH (soluble) -> gas film controls.
Henry's law (interface)
pA∗=H CAL,CA∗=pAG/Hp_A^{*} = H\,C_{AL}, \qquad C_A^{*} = p_{AG}/H
Links a liquid concentration to the equilibrium gas partial pressure; HH has units of pressure per concentration.
Sherwood correlation (turbulent tube)
Sh=kmDDm=0.023 Re0.8 Sc1/3\mathrm{Sh} = \dfrac{k_m D}{D_m} = 0.023\,\mathrm{Re}^{0.8}\,\mathrm{Sc}^{1/3}
Here DD = tube diameter (NOT the diffusivity DmD_m); Sc=μ/(ρDm)\mathrm{Sc}=\mu/(\rho D_m), Re=ρVD/μ\mathrm{Re}=\rho V D/\mu. Mass-transfer twin of Dittus-Boelter.
Chilton-Colburn analogy
jH=NuRe Pr1/3=jM=ShRe Sc1/3=f8j_H = \dfrac{\mathrm{Nu}}{\mathrm{Re}\,\mathrm{Pr}^{1/3}} = j_M = \dfrac{\mathrm{Sh}}{\mathrm{Re}\,\mathrm{Sc}^{1/3}} = \dfrac{f}{8}
Estimates kmk_m from a friction factor or heat coefficient when a direct correlation is missing. The f/8f/8 form uses the DARCY friction factor (with the Fanning factor it is f/2f/2). FE Reference Handbook 10.6 prints it without the name, p. 255 (Use of Friction Factor to Predict Heat-Transfer and Mass-Transfer Coefficients): jH=(Nu/Re Pr) Pr2/3=f/8j_H = (\mathrm{Nu}/\mathrm{Re\,Pr})\,\mathrm{Pr}^{2/3} = f/8 and jM=(Sh/Re Sc) Sc2/3=f/8j_M = (\mathrm{Sh}/\mathrm{Re\,Sc})\,\mathrm{Sc}^{2/3} = f/8.

Where it lives: NCEES FE Reference Handbook — Chemical Engineering (Mass Transfer) · Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat, and Mass Transfer · Geankoplis, Transport Processes and Separation Process Principles

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Solids Handling
3% of the exam

Particle properties and size distributions, surface and bulk forces, solids processing (crushing, grinding, and crystallization), and transportation and storage (belt and pneumatic conveying, slurries, tanks, and hoppers).

Particle Size, Distributions and Comminution

How to describe a particle-size distribution, compute the right mean diameter, relate surface area to size, and apply the Rittinger/Kick/Bond grinding-energy laws and crystallization basics.

Number-length (mean-length) diameter
XML=∫01x dFN=∑i=1mxi ΔFNiX_{ML}=\int_0^1 x\,dF_N=\sum_{i=1}^{m} x_i\,\Delta F_{Ni}
Plain number-average particle size. xix_i = mean size of sub-range ii (μ\mum or m); ΔFNi\Delta F_{Ni} = number fraction in that sub-range (dimensionless, summing to 1).
Sauter (surface-mean) diameter
XSM=∑ixi3 ΔFNi∑ixi2 ΔFNi=1∑i1xi ΔFViX_{SM}=\dfrac{\sum_i x_i^{3}\,\Delta F_{Ni}}{\sum_i x_i^{2}\,\Delta F_{Ni}}=\dfrac{1}{\sum_i \frac{1}{x_i}\,\Delta F_{Vi}}
Diameter of a sphere with the same surface-to-volume ratio as the sample; use for surface-rate processes (dissolution, drying, catalysis). ΔFVi\Delta F_{Vi} = volume/mass fraction in sub-range ii.
Mean-volume diameter
XMV=(∑ixi3 ΔFNi)1/3X_{MV}=\left(\sum_i x_i^{3}\,\Delta F_{Ni}\right)^{1/3}
Volume-weighted mean; governs mass distribution. Always XML≤XMV≤XSMX_{ML}\le X_{MV}\le X_{SM} for the same powder.
Specific surface area of a sphere
av=6d,am=6ρp da_v=\dfrac{6}{d}, \qquad a_m=\dfrac{6}{\rho_p\,d}
Surface area per unit volume (ava_v, m2/m3\text{m}^2/\text{m}^3) or per unit mass (ama_m, m2/kg\text{m}^2/\text{kg}). Scales as 1/d1/d: halving size doubles area. ρp\rho_p = particle density.
Specific surface area with sphericity
av=6Φs da_v=\dfrac{6}{\Phi_s\,d}
Non-spherical correction. Φs\Phi_s = sphericity (0–1): area of equal-volume sphere divided by actual surface area; the same Φs\Phi_s used in the Ergun equation.
General comminution law
dEdx=−C x−n\dfrac{dE}{dx}=-C\,x^{-n}
Energy per unit mass to reduce size; n=2n=2 Rittinger, n=1n=1 Kick, n=1.5n=1.5 Bond. xx = particle size; CC = material constant. Not printed in the FE Reference Handbook 10.6.
Rittinger's law
E=KR(1x2−1x1)E=K_R\left(\dfrac{1}{x_2}-\dfrac{1}{x_1}\right)
Energy ∝\propto new surface created; best for fine grinding. x1x_1 feed, x2x_2 product size; KRK_R Rittinger constant. Not printed in the FE Reference Handbook 10.6.
Kick's law
E=KK ln⁡x1x2E=K_K\,\ln\dfrac{x_1}{x_2}
Energy depends only on size reduction ratio; best for coarse crushing. KKK_K = Kick constant. Not printed in the FE Reference Handbook 10.6.
Bond's law (size form)
E=KB(1x2−1x1)E=K_B\left(\dfrac{1}{\sqrt{x_2}}-\dfrac{1}{\sqrt{x_1}}\right)
Intermediate law for rod/ball mills. KBK_B = Bond constant tied to the work index. Not printed in the FE Reference Handbook 10.6.
Bond work-index equation
W=10 Wi(1P80−1F80)W = 10\,W_i\left(\dfrac{1}{\sqrt{P_{80}}}-\dfrac{1}{\sqrt{F_{80}}}\right)
Specific energy in kWh/short ton; WiW_i = Bond work index (kWh/short ton), F80,P80F_{80},P_{80} = 80%-passing feed and product sizes in μ\mum. Mill power =W×=W\times mass throughput. Not printed in the FE Reference Handbook 10.6.
Supersaturation
S=CC∗,ΔC=C−C∗S=\dfrac{C}{C^{*}}, \qquad \Delta C=C-C^{*}
Crystallization driving force. CC = actual concentration, C∗C^{*} = equilibrium solubility at the operating temperature. S>1S>1 required for nucleation and growth.
Stokes terminal settling velocity
vt=g (ρp−ρf) d218 μv_t=\dfrac{g\,(\rho_p-\rho_f)\,d^{2}}{18\,\mu}
Laminar (Re≤1Re\le1) settling/classification of a particle of diameter dd; ρp,ρf\rho_p,\rho_f = particle and fluid density, μ\mu = fluid viscosity. Ties particle size to separation.

Where it lives: NCEES FE Reference Handbook — Chemical Engineering (Solids Processing: particle-size distributions and mean diameters; it prints no comminution laws) · NCEES FE Reference Handbook — Fluid Mechanics (sphericity, packed beds, drag/terminal velocity) · Perry's Chemical Engineers' Handbook · McCabe, Smith & Harriott, Unit Operations of Chemical Engineering

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Chemical Reaction Engineering
7% of the exam

Reaction rates and order, the Arrhenius rate constant, conversion, yield, and selectivity, series, parallel, homogeneous, heterogeneous, and biological reactions, reactor types (batch, semibatch, CSTR, plug flow, gas and liquid phase), and catalysis.

Rate Laws, Integrated Forms and Arrhenius

Reaction order and rate laws, the zero/first/second-order integrated forms and half-life, and the Arrhenius temperature dependence — the kinetic backbone of every reactor problem.

Rate of reaction (constant volume)
−rA=−1VdNAdt=−dCAdt-r_A = -\dfrac{1}{V}\dfrac{dN_A}{dt} = -\dfrac{dC_A}{dt}
Moles of AA consumed per volume per time; positive for a disappearing reactant. CAC_A in mol/L\text{mol/L}, tt in s\text{s} or min\text{min}.
Power-law rate and order
−rA=kCAxCBy,n=x+y-r_A = kC_A^{x}C_B^{y}, \quad n = x+y
Order x,yx,y are empirical; overall order nn sets the units of kk.
Fractional conversion
CA=CA0(1−XA)C_A = C_{A0}(1-X_A)
Links concentration and conversion at constant VV; XAX_A = mol AA reacted per mol AA fed.
Zero-order integrated form
CA=CA0−kt,t1/2=CA02kC_A = C_{A0}-kt, \quad t_{1/2}=\dfrac{C_{A0}}{2k}
kk in mol⋅L−1time−1\text{mol·L}^{-1}\text{time}^{-1}. CAC_A linear in tt; half-life grows with CA0C_{A0}.
First-order integrated form
ln⁡CA0CA=kt,CA=CA0e−kt\ln\dfrac{C_{A0}}{C_A}=kt,\quad C_A=C_{A0}e^{-kt}
kk in time−1\text{time}^{-1}. Equivalent: −ln⁡(1−XA)=kt-\ln(1-X_A)=kt. ln⁡CA\ln C_A linear in tt.
First-order half-life
t1/2=ln⁡2kt_{1/2}=\dfrac{\ln 2}{k}
Independent of CA0C_{A0} — the diagnostic signature of first-order kinetics.
Second-order integrated form
1CA−1CA0=kt,t1/2=1kCA0\dfrac{1}{C_A}-\dfrac{1}{C_{A0}}=kt,\quad t_{1/2}=\dfrac{1}{kC_{A0}}
kk in L⋅mol−1time−1\text{L·mol}^{-1}\text{time}^{-1}. 1/CA1/C_A linear in tt; half-life shrinks with CA0C_{A0}.
Arrhenius equation
k=A e−Ea/RTk = A\,e^{-E_a/RT}
AA = pre-exponential factor (same units as kk), EaE_a = activation energy (J/mol\text{J/mol}), TT in K, R=8.314 J/(mol⋅K)R=8.314\ \text{J/(mol·K)}.
Two-temperature Arrhenius
ln⁡k2k1=EaRT2−T1T1T2\ln\dfrac{k_2}{k_1}=\dfrac{E_a}{R}\dfrac{T_2-T_1}{T_1 T_2}
Solve for EaE_a from two (k,T)(k,T) pairs; AA cancels. TT in K.
Arrhenius linear form
ln⁡k=ln⁡A−EaR1T\ln k = \ln A - \dfrac{E_a}{R}\dfrac{1}{T}
Slope −Ea/R-E_a/R, intercept ln⁡A\ln A on a ln⁡k\ln k vs 1/T1/T plot.
Differential method for order
ln⁡(−rA)=ln⁡k+nln⁡CA\ln(-r_A)=\ln k + n\ln C_A
Slope of ln⁡(−rA)\ln(-r_A) vs ln⁡CA\ln C_A gives the order nn.

Where it lives: NCEES FE Reference Handbook — Chemical Engineering · NCEES FE Reference Handbook — Chemistry and Biology · Fogler, Elements of Chemical Reaction Engineering

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Engineering Economics
4% of the exam

Time value of money (present, annual, and future worth, rate of return), economic analyses (break-even, benefit-cost, optimal economic life), uncertainty with expected value and risk, and project selection with unequal lives, depreciation, and discounted cash flow.

Time Value of Money and Interest Factors

The six interest factors, gradients, nominal-vs-effective rate, and rate of return — the toolkit that converts cash flows across time on the FE.

Single-payment compound amount (F/P)
F=P (1+i)nF = P\,(1+i)^n
Future worth of a present lump sum. ii = rate per period, nn = number of periods (same period as ii).
Single-payment present worth (P/F)
P=F (1+i)−nP = F\,(1+i)^{-n}
Discounts a single future amount FF to the present. The base building block of all discounting.
Uniform-series present worth (P/A)
P=A (1+i)n−1i(1+i)nP = A\,\dfrac{(1+i)^n-1}{i(1+i)^n}
Present worth of nn equal end-of-period payments AA. Reciprocal is capital recovery (A/P)(A/P).
Capital recovery (A/P)
A=P i(1+i)n(1+i)n−1A = P\,\dfrac{i(1+i)^n}{(1+i)^n-1}
Equal payment that repays a present amount PP (loan payment); also used to annualize a present cost.
Uniform-series compound amount (F/A)
F=A (1+i)n−1iF = A\,\dfrac{(1+i)^n-1}{i}
Future worth of nn equal deposits AA. Reciprocal is the sinking-fund factor (A/F)(A/F).
Sinking fund (A/F)
A=F i(1+i)n−1A = F\,\dfrac{i}{(1+i)^n-1}
Equal deposit needed to accumulate a target future fund FF in nn periods.
Gradient present worth (P/G)
P=G[(1+i)n−1i2(1+i)n−ni(1+i)n]P = G\left[\dfrac{(1+i)^n-1}{i^2(1+i)^n}-\dfrac{n}{i(1+i)^n}\right]
Present worth of an arithmetic gradient GG (first increment at end of period 2). Add to base-annuity PP.
Gradient to uniform series (A/G)
(A/G,i,n)=1i−n(1+i)n−1(A/G,i,n)=\dfrac{1}{i}-\dfrac{n}{(1+i)^n-1}
Converts a gradient GG to an equivalent level annuity: Aeq=Abase+G (A/G,i,n)A_\text{eq}=A_\text{base}+G\,(A/G,i,n).
Effective interest rate
ie=(1+rm)m−1i_e=\left(1+\dfrac{r}{m}\right)^m-1
Annual effective rate from nominal annual rr with mm compoundings/yr. Continuous limit (not printed in the handbook): ie=er−1i_e=e^r-1.
Equivalence among PW, AW, FW
AW=PW (A/P,i,n),FW=PW (F/P,i,n)AW = PW\,(A/P,i,n),\quad FW = PW\,(F/P,i,n)
The three worth measures are interconvertible; sign and accept/reject conclusion are identical.
Capitalized cost (perpetuity)
P∞=AiP_\infty=\dfrac{A}{i}
Present worth of an infinite uniform series — capitalized cost of perpetual service or maintenance.
Rate of return condition
PW(i∗)=∑t=0nCFt(1+i∗)t=0PW(i^*)=\sum_{t=0}^{n}\dfrac{CF_t}{(1+i^*)^t}=0
i∗i^* = internal rate of return. Accept if i∗≥MARRi^*\ge MARR. Beware multiple roots with sign changes.

Where it lives: NCEES FE Reference Handbook — Engineering Economics · Newnan, Eschenbach & Lavelle, Engineering Economic Analysis · Blank & Tarquin, Engineering Economy

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Process Design
7% of the exam

Process flow diagrams and piping and instrumentation diagrams, equipment selection, sizing, and scale-up, equipment and facilities cost estimation with cost indices, process design and optimization (sustainability, efficiency, green engineering, inherently safer design), and design standards (regulatory, ASTM, ISO, OSHA).

PFDs, P&IDs and Cost Estimation

Read a PFD versus a P&ID, then turn an equipment list into a capital-cost number with the six-tenths rule, a cost index, and Lang factors.

Cost index update
C2=C1(I2I1)C_{2} = C_{1}\left(\dfrac{I_{2}}{I_{1}}\right)
Escalate a past cost C1C_1 (year-1 index I1I_1) to a present cost C2C_2 (current index I2I_2). CEPCI base 1957-59=1001957\text{-}59=100. Time only — never size.
Capacity scaling (power law)
CACB=(SASB)n\dfrac{C_{A}}{C_{B}} = \left(\dfrac{S_{A}}{S_{B}}\right)^{n}
SS is a capacity measure (area, flow, volume, power); nn is the cost-capacity exponent. Interpolate within a size class only.
Six-tenths rule
CA=CB(SASB)0.6C_{A} = C_{B}\left(\dfrac{S_{A}}{S_{B}}\right)^{0.6}
Default when the specific exponent is unknown: cost rises as capacity to the 0.60.6 power.
Back-out the exponent
n=ln⁡ ⁣(C2/C1)ln⁡ ⁣(S2/S1)n = \dfrac{\ln\!\left(C_{2}/C_{1}\right)}{\ln\!\left(S_{2}/S_{1}\right)}
Fit nn from two cost-capacity data points; slope of the cost line on log-log axes.
Combined escalate-and-scale
C=CB(SASB)n(I2I1)C = C_{B}\left(\dfrac{S_{A}}{S_{B}}\right)^{n}\left(\dfrac{I_{2}}{I_{1}}\right)
Move a base cost across both size and time in one expression.
Lang factor (installed/fixed-capital cost)
CFCI=fL Cp,deliveredC_{\text{FCI}} = f_{L}\,C_{p,\text{delivered}}
fLf_L: fluid plant 5.05.0 (fixed) / 6.06.0 (total), solid-fluid 4.3/5.04.3/5.0, solid 4.0/4.74.0/4.7. CpC_p = total delivered purchased-equipment cost.
Module (factored) estimate
CTM=∑iCp,i FBM,iC_{\text{TM}} = \sum_{i} C_{p,i}\,F_{BM,i}
Each item's purchased cost times its bare-module factor FBMF_{BM} (raises with alloy, pressure); summed for the total module cost.
Total capital investment
CTCI=CFCI+CWCC_{\text{TCI}} = C_{\text{FCI}} + C_{\text{WC}}
Fixed capital (built, depreciable) plus working capital (inventory, receivables; typically ∼15%\sim 15\% of TCI), recovered at end of life.

Where it lives: NCEES FE Reference Handbook — Chemical Engineering (Cost Estimation: cost indexes, scaling of equipment costs, capital cost / Lang factors) · NCEES FE Reference Handbook — Engineering Economics (cost indices, capital recovery, working capital) · Peters, Timmerhaus & West, Plant Design and Economics for Chemical Engineers · ISA-5.1 Instrumentation Symbols and Identification

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Process Control
4% of the exam

Process dynamics (first- and second-order processes, gains and time constants, stability, damping, transfer functions), control strategies (feedback, feedforward, cascade, ratio, PID tuning), and control-loop design and hardware (sensors, control valves, interlocks, and conceptual DCS and PLC programming).

Process Dynamics: First- and Second-Order Response

Process gain and time constant, the 63.2% first-order step, second-order damping and overshoot, transfer functions, and what makes a loop stable.

First-order transfer function
Y(s)R(s)=Kτs+1\dfrac{Y(s)}{R(s)} = \dfrac{K}{\tau s + 1}
KK = steady-state gain (output units / input units); τ\tau = time constant (s). Single pole at s=−1/τs=-1/\tau.
First-order step response
y(t)=y0e−t/τ+KM(1−e−t/τ)y(t) = y_0 e^{-t/\tau} + KM\left(1 - e^{-t/\tau}\right)
Response to a step of magnitude MM. In deviation variables y0=0y_0=0. Reaches 63.2%63.2\% of KMKM at t=τt=\tau.
Percent-complete landmarks
63.2, 86.5, 95.0, 98.2, 99.3 % at t=1,2,3,4,5 τ63.2,\ 86.5,\ 95.0,\ 98.2,\ 99.3\,\% \text{ at } t = 1,2,3,4,5\,\tau
Fraction of the final change reached after each time constant; practically settled by 5τ5\tau.
First-order plus dead time (FOPDT)
Y(s)R(s)=Ke−θsτs+1\dfrac{Y(s)}{R(s)} = \dfrac{K e^{-\theta s}}{\tau s + 1}
θ\theta = dead time (transport lag, s). The step response is delayed by θ\theta before the exponential begins.
Second-order standard form
Y(s)R(s)=K ωn2s2+2ζωns+ωn2\dfrac{Y(s)}{R(s)} = \dfrac{K\,\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}
ωn\omega_n = undamped natural frequency (rad/s); ζ\zeta = damping ratio (dimensionless); KK = steady-state gain.
Second-order (process form)
Y(s)R(s)=Kτ2s2+2ζτs+1\dfrac{Y(s)}{R(s)} = \dfrac{K}{\tau^2 s^2 + 2\zeta\tau s + 1}
Equivalent form with τ=1/ωn\tau = 1/\omega_n. Common in the chemical-process literature.
Damping classification
ζ>1 (over),ζ=1 (critical),0<ζ<1 (under)\zeta > 1\ \text{(over)}, \quad \zeta = 1\ \text{(critical)}, \quad 0<\zeta<1\ \text{(under)}
Only the underdamped band overshoots and oscillates; critically damped is fastest with no overshoot.
Damped natural frequency
ωd=ωn1−ζ2\omega_d = \omega_n\sqrt{1-\zeta^2}
Actual ringing frequency of an underdamped response (rad/s); always less than ωn\omega_n.
Percent overshoot
%OS=100 exp⁡ ⁣(−ζπ1−ζ2)\%\text{OS} = 100\,\exp\!\left(\dfrac{-\zeta\pi}{\sqrt{1-\zeta^2}}\right)
Overshoot above the final value for a unit step; depends on ζ\zeta only, not ωn\omega_n.
Peak time
tp=πωd=πωn1−ζ2t_p = \dfrac{\pi}{\omega_d} = \dfrac{\pi}{\omega_n\sqrt{1-\zeta^2}}
Time to the first (largest) peak of an underdamped step response (s).
Two-percent settling time
Ts=4ζωnT_s = \dfrac{4}{\zeta\omega_n}
Time for the envelope e−ζωnte^{-\zeta\omega_n t} to decay within 2%2\% of final value (s).
DC gain (Final Value Theorem)
dc gain=lim⁡s→0G(s)\text{dc gain} = \lim_{s\to 0} G(s)
Steady-state gain, valid only when all poles have negative real parts. Equals KK for the first-order form.

Where it lives: NCEES FE Reference Handbook — Instrumentation, Measurement, and Control (Control Systems; First- and Second-Order Control System Models) · Seborg, Edgar, Mellichamp & Doyle — Process Dynamics and Control · Coughanowr & LeBlanc — Process Systems Analysis and Control

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Safety, Health, and Environment
5% of the exam

Hazardous material properties and safety data sheets, industrial hygiene (toxicity, noise, PPE, ergonomics), process safety and hazard analysis (LOPA, HAZOP, fault and event trees, dispersion modeling), overpressure and underpressure protection (relief and inherently safer design), waste minimization, treatment, and regulation (RCRA, CWA, EPA, OSHA), and reactivity hazards (inerting, runaway reactions, compatibility).

Process Safety: HAZOP, LOPA and Risk Assessment

Generate deviations with HAZOP guide words, multiply independent-protection-layer PFDs in LOPA, and combine fault- and event-tree gates with the AND-multiply, OR-add rules.

Risk definition
Risk=Hazard×Probability\text{Risk} = \text{Hazard}\times\text{Probability}
Risk combines the inherent capacity to harm with the likelihood (and severity) of the harmful event.
HAZOP deviation
Deviation=Guide word×Parameter\text{Deviation} = \text{Guide word}\times\text{Parameter}
Guide words (NO, MORE, LESS, REVERSE, etc.) crossed with flow, pressure, temperature, level, composition generate credible deviations.
LOPA mitigated frequency
fmit=finit∏jPFDjf_{\text{mit}} = f_{\text{init}}\prod_{j}\mathrm{PFD}_j
finitf_{\text{init}} in events/yr; each independent IPL contributes a dimensionless PFD. Result in events/yr is compared to a tolerable target.
Risk reduction factor
RRF=1PFD\mathrm{RRF} = \dfrac{1}{\mathrm{PFD}}
Demand-failure reciprocal; an RRF=100\mathrm{RRF}=100 layer has PFD=0.01\mathrm{PFD}=0.01, the bottom of the SIL 1 band — to guarantee a factor of 100 you specify SIL 2 (PFD<10−2\mathrm{PFD}<10^{-2}).
Required risk reduction
RRFreq=funmitigatedftolerable\mathrm{RRF}_{\text{req}} = \dfrac{f_{\text{unmitigated}}}{f_{\text{tolerable}}}
Total reduction the protection layers must collectively supply; product of all IPL RRFs must meet or exceed it.
Fault-tree AND gate
PAND=∏ipiP_{\text{AND}} = \prod_i p_i
Top event needs all inputs to fail (redundant safeguards). Independent probabilities multiply.
Fault-tree OR gate
POR=1−∏i(1−pi)≈∑ipiP_{\text{OR}} = 1-\prod_i(1-p_i)\approx \sum_i p_i
Top event needs any one input to fail (series). Approximate sum valid for small, independent pip_i.
Event-tree outcome frequency
foutcome=finit∏kpkf_{\text{outcome}} = f_{\text{init}}\prod_k p_k
Multiply the initiating frequency by the branch probability (success or failure) along each path.
Gaussian plume centerline
C=m˙π u σyσzC = \dfrac{\dot{m}}{\pi\,u\,\sigma_y \sigma_z}
Ground-level downwind concentration for a continuous neutral release; inverse in wind speed uu — calm air is worst case.

Where it lives: NCEES FE Reference Handbook — Safety · CCPS (AIChE), Guidelines for Hazard Evaluation Procedures and Layer of Protection Analysis · IEC 61511 Functional Safety — Safety Instrumented Systems for the Process Industry

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Ethics and Professional Practice
3% of the exam

Codes of ethics of professional and technical societies, agreements, contracts, and contract law (noncompete, nondisclosure, memoranda of understanding), public health, safety, and welfare with licensing and professional liability, and intellectual property (copyrights, trade secrets, patents, and trademarks).

Codes of Ethics and Public Welfare

The NCEES Model Rules, the paramount duty to public health and safety, competence and conflict-of-interest obligations, licensure, and a method for resolving ethical dilemmas.

Priority of obligations
Public  ≫  Employer/Client  ≳  Other licensees\text{Public} \;\gg\; \text{Employer/Client} \;\gtrsim\; \text{Other licensees}
Only the public-paramount rule is codified: the duty to public health, safety, and welfare outranks every other duty (Model Rules, Obligations to the Public). The employer/client-over-other-licensee ordering is a practical exam heuristic, not an explicit Model Rules ranking.
Paramount duty
First responsibility=safeguard public health, safety, and welfare\text{First responsibility} = \text{safeguard public health, safety, and welfare}
Model Rules §240.15(A)(1): the licensee's first and foremost responsibility when serving clients and employers.
Sealing rule
Seal  ⟺  (prepared by you or under your responsible charge)  ∧  (within your competence)\text{Seal} \iff (\text{prepared by you or under your responsible charge}) \;\wedge\; (\text{within your competence})
A licensee may sign and seal a document only if both conditions hold; sealing work outside competence or not in responsible charge is a violation.
Conflict-of-interest test
Disclose if it could influence or appear to influence professional judgment\text{Disclose if it could influence } \textbf{or appear to influence} \text{ professional judgment}
Both actual and apparent conflicts must be disclosed to the employer or client; appearance alone triggers the duty.
Dual compensation rule
Accept pay from ≥2 parties on one project  ⟺  full disclosure∧written agreement of all\text{Accept pay from } \ge 2 \text{ parties on one project} \iff \text{full disclosure} \wedge \text{written agreement of all}
Compensation from more than one party for the same project requires disclosure to and written consent of every interested party.
Escalation duty
Judgment overruled  ∧  public endangered⇒notify employer/client AND appropriate authority\text{Judgment overruled} \;\wedge\; \text{public endangered} \Rightarrow \text{notify employer/client AND appropriate authority}
When a safety-critical judgment is overruled, the obligation extends beyond the employer to an outside authority.
Licensure path
ABET degree→FE exam (EI)→∼4 yr experience→PE exam→PE license\text{ABET degree} \to \text{FE exam (EI)} \to \sim 4\text{ yr experience} \to \text{PE exam} \to \text{PE license}
Typical Model Law route to a Professional Engineer license; experience is four years after a qualifying bachelor's degree.
Competence boundary
Undertake assignment  ⟺  qualified by education or experience in that specific field\text{Undertake assignment} \iff \text{qualified by education or experience in that specific field}
Model Rules: accept only work for which you are technically qualified.

Where it lives: NCEES FE Reference Handbook — Ethics and Professional Practice · NCEES Model Rules, Section 240.15 — Rules of Professional Conduct · NCEES Model Law, Sections 110.20, 130.10, 150.10 — Definitions, Licensure, Disciplinary Action · NSPE Code of Ethics for Engineers

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Ten free FE Chemical questions with full worked solutions and figures where needed, no account required. Then the full bank, timed mock exams and the complete study handbook when you are ready.

Many equations follow the FE Reference Handbook 10.6 as NCEES prints it; the rest come from the code, design standard or textbook named in their “Where it lives” line or note. The on-screen reference NCEES supplies on exam day is the copy that counts, so check the edition for your sitting. Independent study resource; not affiliated with, endorsed by, or sponsored by NCEES.