Treat PE Mechanical review as a matching exercise between problem givens and named methods. For every topic, write a one-line trigger: what data must be given, what may remain unknown, and which equation form follows. Rehearse that trigger with worked scenarios and a unit-and-assumption log rather than by rereading formula sheets.
Closed System or Control Volume: Decide Before You Write the First Law
A closed system exchanges energy but no mass; a control volume exchanges both. Identify whether material crosses the boundary you drew before writing any energy equation, because the first law takes a different form for each case.
The distinction is about what crosses the boundary you chose. A piston-cylinder device with a fixed charge of gas is a closed system: energy enters as heat and work, but the mass inside never changes. A pump, turbine, nozzle, or duct section is a control volume: fluid enters and leaves continuously, so the energy balance must include the energy carried by that flow. An unsteady case, such as a tank being filled from a supply line, is a control volume with an accumulation term.
Applying a closed-system energy balance to a flow device silently drops the flow work, and the numbers come out wrong in a way that is hard to notice because the arithmetic still looks clean. Build the habit in this order: state the boundary, state what crosses it, then select the equation form. Worked Exercise — the assumption and unit log: for ten problems across thermodynamics and fluids, record five entries before computing anything. Rubric for each entry: (1) boundary classified as closed system or control volume; (2) governing equation named; (3) property basis justified (internal energy versus enthalpy); (4) units shown to cancel into the target quantity; (5) every simplifying assumption written down. Expected observation: classification is slow at first, then becomes nearly automatic; the log also exposes repeated unit slips, which are the errors most likely to survive unchecked.
Steady-Flow Devices Demand Enthalpy, Not Internal Energy
For pumps, compressors, turbines, and nozzles, energy accounting uses enthalpy, which bundles internal energy with flow work. Using internal energy alone omits the pv term that flowing fluid carries across the control surface.
Worked Scenario 1 — the mispriced compressor: air enters a compressor at 300 K and leaves at 500 K. Model air as an ideal gas with cv = 0.718 kJ/kg·K and cp = 1.005 kJ/kg·K. A tempting error is to compute the work as cv·ΔT = 0.718 × 200 = 143.6 kJ/kg, reusing a closed-system heating result. The correct path is the steady-flow energy equation: for an adiabatic compressor with negligible kinetic and potential energy changes, work per unit mass equals h2 − h1, and for an ideal gas that is cp·ΔT = 1.005 × 200 = 201 kJ/kg. The internal-energy shortcut is about 29% low — equivalently, the correct required work is roughly 40% higher than the shortcut value — an error large enough to translate directly into an undersized drive selection.
The trigger follows from Scenario 1: if mass crosses the boundary, the flow work term is real and must appear, and for ideal gases it converts the property basis from cv to cp. Contrast the closed-system case: heating the same air in a rigid tank from 300 K to 500 K gives ΔU = m·cv·ΔT, because no flow work exists there and cv is correct. Practicing both versions of the same temperatures back to back fixes the boundary between the two property bases far better than memorizing which constant belongs to which chapter heading.
LMTD or Effectiveness-NTU: Let the Givens Choose the Heat Exchanger Method
Use the log mean temperature difference method when both outlet temperatures are known; use effectiveness-NTU when capacity rates and inlet temperatures are known but outlets are not. The problem's unknowns select the method.
The LMTD method computes a driving temperature difference from the terminal differences, using a logarithmic mean of ΔT1 and ΔT2, with the pairing of terminals set by the flow arrangement: counterflow pairs hot-inlet with cold-outlet, parallel flow pairs inlet with inlet. Crossflow and shell-and-tube geometries then require a correction factor applied to the counterflow LMTD. The method is direct and accurate only when every terminal temperature is available.
Worked Scenario 2 — the looping oil cooler: an oil stream must be cooled from 90 °C to 60 °C by water entering at 20 °C, and the water outlet is not stated. A candidate starts the LMTD method by guessing the water outlet, computing the LMTD, checking the energy balance, guessing again, and iterating without a clear end. The better decision is effectiveness-NTU: compute the capacity rates, form the capacity ratio and the number of transfer units, read the effectiveness for the stated exchanger arrangement, and obtain the outlet temperatures directly. A second, quieter error appears when outlets are known: replacing the log mean with the arithmetic mean. With terminal differences of 70 °C and 40 °C, the arithmetic mean of 55 °C exceeds the log mean of about 53.6 °C, overstating the driving force and therefore the calculated duty. The log-mean correction is small at these values and grows sharply when the terminal differences are far apart, which is exactly when the error is most consequential.
The decision is systematic, not stylistic. The table below compresses the choice into the inputs each method needs.
| Aspect | LMTD method | Effectiveness-NTU method |
|---|---|---|
| Use when | All four terminal temperatures are known | Inlet temperatures and capacity rates known; outlets unknown |
| Required inputs | Terminal temperature differences, flow arrangement, correction factor | Capacity rates, overall U and area, arrangement effectiveness relation |
| Primary output | Heat duty or required area | Effectiveness, then duty and outlet temperatures |
| Typical pitfall | Arithmetic mean used instead of log mean; wrong terminal pairing for the arrangement | Guessing outlets and iterating instead of using capacity ratio and NTU |
| Fits problems worded as | Given both outlets, find duty or size | Given size and inlets, find performance |
Fatigue Criteria Under Mean Stress: Goodman, Gerber, and Soderberg Diverge
Mean stress separates the fatigue criteria: Soderberg guards against yielding, Goodman is linear in mean stress, and Gerber is parabolic and least conservative. Match the criterion to the design margin the problem states.
Fluctuating loading splits stress into an alternating component and a mean component, and only the alternating component is compared against the endurance limit directly. When a mean stress exists, a criterion is needed to combine the two. Soderberg compares mean stress against yield strength, making it the most conservative; Goodman uses ultimate strength in a straight-line relation; Gerber fits a parabola through the same data and allows the highest mean stress. Also keep the material properties distinct: an endurance limit applies to long life for steels that exhibit one, while finite-life fatigue strength at a stated cycle count is a different property used when infinite life is not the target.
Worked mini-case: a shaft sees alternating bending stress of 150 MPa and steady mean stress of 100 MPa, with an endurance limit of 200 MPa, ultimate strength of 600 MPa, and yield strength of 400 MPa. Goodman gives 150/200 + 100/600 = 0.75 + 0.167 = 0.917, an acceptable result. Soderberg gives 150/200 + 100/400 = 0.75 + 0.25 = 1.00, exactly at the limit, so the verdict flips to marginal. Gerber would clear it with room to spare. The same numbers produce three different engineering conclusions, which is the point: the criterion is a design decision the problem must specify, and switching criteria silently changes the answer. In practice problems, read whether the statement asks for a conservative check or a specific named criterion before combining terms.
The pairing habit generalizes: name the criterion, name the two strength properties it uses, and only then insert numbers. Mixing, for example, yield strength into a Goodman relation, produces a plausible-looking result with no meaning attached to it.
Psychrometric Mixing: Mass-Weight the Properties, Not the Volume Flows
Mixing two airstreams conserves mass and energy: humidity ratio and enthalpy are mass-flow-weighted, and the mixed state lies on the straight line between the two inlet states. Volume flows cannot be averaged directly.
The governing relations are simple: the mixture humidity ratio equals the mass-flow-weighted average of the inlet humidity ratios, and the mixture enthalpy is weighted the same way, so the mixed state falls on the segment joining the two inlet states on a psychrometric chart. Dry-bulb temperature of the mixture then follows from the mass and heat balances; it is not an independent input and it is not a conserved quantity in general.
Worked example with the classic mistake: stream A supplies 2 kg/s of dry air at 10 °C with humidity ratio 0.006; stream B supplies 3 kg/s at 30 °C with humidity ratio 0.015. The mixture humidity ratio is (2 × 0.006 + 3 × 0.015) / 5 = 0.0114, and the mixed dry-bulb works out to about 22 °C on a mass basis. A plausible error is averaging the two temperatures to 20 °C while ignoring the flow imbalance, or averaging the two volume flow rates when the streams are quoted in volumetric units, which weights the mixture by the wrong quantity because the streams differ in density and humidity. The better decision is to convert every stream to mass flow of dry air first, then apply the weighted balances. The same weighting logic extends to coil processes: the sensible heat ratio fixes the slope of the process line on the chart, so a stated SHR tells you how the coil's total capacity splits between sensible and latent portions before any numbers are placed.
Pipe Losses: Each Term Must Use the Velocity at Its Own Component
Major losses come from wall friction through Darcy-Weisbach; minor losses come from fittings expressed as a loss coefficient times the velocity head. Every loss term must reference the velocity in its own pipe or fitting.
Darcy-Weisbach writes head loss as f times (L/D) times the velocity head V²/2g, with the friction factor determined by Reynolds number and relative roughness; in laminar flow the factor reduces to 64/Re. Fittings, valves, entrances, and exits contribute minor losses of the form K times V²/2g. Total loss in a series line is the sum of the individual terms, and the summation is only valid when each term uses the velocity that its coefficient was defined against.
Worked mini-case: a line contracts from a 100 mm pipe to a 50 mm pipe. Since velocity scales inversely with diameter squared, the velocity head in the small pipe is four times that in the large pipe. A candidate who computes a fitting loss using the large-pipe velocity head because the calculation started there understates that loss by a factor of four. The better decision is to annotate each diameter on the sketch, mark which velocity belongs to each loss term, and only then sum. The same discipline separates the two loss families: friction term uses length, diameter, and the pipe's friction factor; fitting terms use their K values and their local velocity. Conflating them — for instance, applying a pipe-length multiplier to a fitting — is easy to do and difficult to catch, because the result still carries plausible units of head.
Economics Factors: Match the Cash-Flow Pattern, Then Compute
Equal end-of-period series use the (A/P) and (P/A) factors; arithmetic gradients require the (A/G) or (P/G) factor added to a base amount; single future sums use (F/P). Pattern recognition precedes computation.
Each compound-interest factor encodes one cash-flow shape: a uniform series, a single future amount, or a gradient that starts at zero in the first period and grows by a fixed increment G each period thereafter. A recurring cost that starts at A1 in year 1 and grows by G per year is not a uniform series plus a shifted series; it equals A1 plus G applied through the gradient factor: A = A1 + G·(A/G, i, n). Applying the gradient factor to the first payment, or treating the increment as though it were already in the base, misprices every year of the analysis.
Worked mini-case: maintenance costs 10,000 in year 1 and rises by 500 each year for 10 years at i = 8 percent. The uniform-series equivalent is A = 10,000 + 500·(A/G, 8%, 10). With (A/G, 8%, 10) ≈ 3.87, the result is about 11,936 per year. A tempting error is to average 10,000 and the final payment of 14,500 into a uniform series, which ignores the timing of every intermediate year and shifts weight toward the later, more heavily discounted payments. Present worth and annual worth give the same ranking when both are computed correctly, so a mismatch between them signals a pattern error, not a rounding issue.
Adaptable preparation sequence: (1) map the concept pairs in each syllabus area — energy models, exchanger methods, fatigue criteria, mixing rules, loss terms, cash-flow patterns — into one trigger line each; (2) run the assumption-and-unit log from the first section across all six areas; (3) work topic-mixed scenario sets where classifying the problem is part of the task; (4) finish with timed mixed sets and recheck the log entries. Readiness checks: you can classify a system boundary in under a minute; you can state which exchanger method fits a problem from its unknowns alone; you can name the criterion and strength properties before inserting fatigue numbers; and your log shows consistent unit cancellation. Administrative matters such as registration and availability belong with NCEES at ncees.org; this guide covers subject preparation only.
References and further reading
Use these references to explore the concepts and check the latest information from the relevant organizations.