PE Chemical common mistakes
PE Chemical · Understand the traps and how to avoid them.
Plugging a gauge pressure (e.g., 100 psig) directly into the ideal gas law, a compressibility chart, Raoult's law, or a relief-valve sizing equation instead of converting to absolute.
Plant experience and most equipment datasheets quote gauge pressure, and the word 'psig' disappears once the number is copied from the problem statement into a formula; absolute-temperature errors get caught more often than absolute-pressure errors because the units look different.
What to do instead
Before substituting any pressure, write it as P = X + 14.7 psia (or + 101.3 kPa) on the same line as the equation, and keep a running 'absolute only' column for every pressure and temperature in the stream table.
Treating the fresh feed as the reactor feed in a recycle process, so the single-pass conversion is used where the overall conversion (or vice versa) is being asked for, and the inert purge stream is left out of the balance entirely.
The flowsheet is drawn with a recycle loop but the solver mentally collapses it into a straight line to save time; the inert component also feels irrelevant because it does not react.
What to do instead
Draw separate envelopes around the reactor and around the whole process, do a degrees-of-freedom count for each, and write the inert balance around the overall envelope first - the purge flow usually falls out immediately. Track conversion with an explicit extent of reaction variable per envelope.
Using a standard-state Gibbs free energy of reaction (ΔG° at 298 K) to compute the equilibrium constant at a reactor temperature of 400–600 K, without a van't Hoff or Kirchhoff correction.
ΔG°298 and ΔH°298 are the values that appear in the appendix tables, and the temptation is to insert them straight into K = exp(−ΔG°/RT) using the reactor T.
What to do instead
Always write the temperature path explicitly: ΔH°(T) = ΔH°298 + ∫ΔCp dT, then ΔG°(T) = ΔH°(T) − TΔS°(T), or use ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1) when ΔH° is roughly constant. If the problem gives only one temperature, check that the tabulated data are at that same temperature before using them.
Mixing Darcy and Fanning friction factors - using an f read off a Moody chart (Darcy) inside h_f = 4fLv²/(2gD) (Fanning), which produces a head loss that is off by a factor of four.
Both symbols are 'f', both charts look alike, and different references (Crane vs. textbook vs. Perry) define them differently; under time pressure the formula is recalled without its definition.
What to do instead
Write the defining equation next to the symbol every time: Darcy f gives h_f = f(L/D)(v²/2g); Fanning f gives h_f = 2f(L/D)(v²/2g). Sanity-check the answer against f ≈ 0.02–0.03 for turbulent flow - if the computed head loss implies f outside that band, the definition was swapped.
Using the arithmetic mean temperature difference instead of the log mean, and omitting the F correction factor for multipass or crossflow exchangers, when the terminal temperature approaches differ by more than about 20%.
Arithmetic mean is easier to compute mentally and is nearly correct for close approaches, so it survives early practice problems and is never revisited; F is easy to forget because it lives in a separate chart.
What to do instead
Compute ΔT1 and ΔT2 first and compare them; if their ratio exceeds about 2, the arithmetic mean is not acceptable. Then always check: for a multipass or crossflow unit, look up or compute F and confirm F > 0.8, otherwise the configuration is impractical and the problem is signalling an error.
Using mole fractions with an overall mass-transfer coefficient correlated in mole ratios, or applying the Kremser/absorption-factor equation without first confirming that the operating and equilibrium lines are straight over the range being integrated.
The shortcut equations are memorized as 'plug in y_in, y_out, L/V' and the assumptions that justify them (dilute solution, linear equilibrium, constant L and V) are never checked against the problem data.
What to do instead
State the assumptions in one line before using any shortcut: dilute (<10% solute), isothermal, constant L and V. If any fails, set up the differential balance and integrate numerically or graphically between the operating and equilibrium curves rather than forcing the closed-form equation.
Substituting the feed concentration into the CSTR design equation V = F_A0·X/(−r_A) while evaluating the rate at the outlet concentration - or, equivalently, using C_A0 where C_A at the exit belongs.
The CSTR is perfectly mixed, so the outlet composition IS the reaction composition, but the feed flow F_A0 appears in the same equation, and the two 'A' subscripts get conflated during a fast substitution.
What to do instead
Write the design equation in the form V = (F_A0 − F_A)/(−r_A at exit conditions) and evaluate −r_A explicitly with the exit concentration computed from the conversion. For a CSTR, if the rate used and the concentration used are not the same point, the answer is wrong.
Applying a purchased-equipment cost to a whole plant, or applying the six-tenths rule across capacity ratios well above about 10, or omitting working capital and salvage when computing NPV or payback.
Lang factors, the six-tenths rule, and NPV all appear as short formulas, and under time pressure only the visible term (equipment cost, capacity ratio, annual cash flow) is carried through while the modifiers are dropped.
What to do instead
Build a one-line cost ladder and tick each rung: purchased cost → installed cost (× Lang factor) → total capital including working capital → annual cash flow after tax and depreciation → NPV including salvage and working-capital recovery. Restrict the six-tenths rule to capacity ratios under roughly 10 and note the assumption.
Specifying controller action (direct/reverse) without first determining the valve's fail-safe position, or assuming an air-to-open valve when the loop requires fail-closed.
Controller action and valve action are taught as separate topics, and the exam question usually supplies only one of the two pieces of information, so the candidate fills in the other from habit rather than from a safety analysis.
What to do instead
Work the loop in order every time: what happens to the process variable if the signal to the valve is lost → which valve position is safe → is that air-to-open or air-to-close → does the process gain invert across the valve → what controller action makes the loop negative feedback. Write each arrow down; do not skip to the last step.
Sizing a relief device for a single scenario without checking whether the flow is choked at the relief pressure, and without comparing the blocked-outlet, fire, and thermal-expansion cases to find the controlling one.
The sizing equation is applied as soon as a relieving flow rate is given, and the critical-pressure ratio check is a separate step that is easy to omit; the fire case is often the governing one but is rarely the first case considered.
What to do instead
Before using any relief equation, compute the critical pressure ratio and confirm whether flow is sonic or subsonic, then select the corresponding equation. Enumerate every credible relief scenario in a short table, size each one, and take the maximum orifice - never size against the first case that comes to mind.