PE Mechanical HVAC common mistakes
PE Mechanical HVAC · Understand the traps and how to avoid them.
Mixing two airstreams by averaging dry-bulb temperature and humidity ratio with a 50/50 or CFM-weighted split, then reading the mixed enthalpy straight off the chart at that averaged temperature.
Psychrometric mixing is a mass balance on dry air and water vapor, but both streams are almost always given to you in CFM, so volume weighting feels natural and the chart makes the averaged temperature look plausible. The error is small when the two specific volumes are similar and large when one stream is hot or at altitude.
What to do instead
Convert each stream to dry-air mass flow (CFM x 60 / specific volume at that stream's own conditions), mass-weight W and h, then back out the mixed dry-bulb from the mixed h and W. Reserve volume weighting for reheat/coil work where the problem explicitly fixes the airflow mass.
Solving an entire psychrometric problem on the sea-level chart and using 1.08 and 4.5 for sensible and total heat when the problem statement gives a site elevation of 3,000-6,000 ft.
The standard chart and the memorized constants are what everyone practices with, and the elevation is buried in the first sentence of the problem rather than in the data table. At 5,000 ft the density is roughly 17% low, so sensible loads and fan power are off by the same fraction.
What to do instead
Scan for elevation before opening the chart, and write the density ratio next to your constants: sensible factor = 1.08 x (rho/0.075), total factor = 4.5 x (rho/0.075). Rebuild W from the local barometric pressure if no altitude chart is available.
Equating the coil load to the space load, skipping fan heat, duct gain, and the outdoor-air load, and ignoring a stated bypass factor when finding the leaving-air condition.
A simple room sketch shows a supply airflow removing a room load, so the two numbers seem interchangeable. In real systems the coil sees more heat than the space does, and the bypass factor is the only thing that relates the apparatus dew point to the leaving air.
What to do instead
Draw the load path explicitly: space sensible and latent, then add duct gain, fan heat, and the outdoor-air load to get coil load, and state which one the question asks for. When a bypass factor is given, leaving condition = ADP + BF x (entering - ADP) for both dry-bulb and humidity ratio, and check that the ADP is below the leaving dew point.
Computing compressor power as refrigerant mass flow times the enthalpy difference between the evaporator exit and the condensing-pressure enthalpy, without applying the isentropic efficiency, and deriving mass flow from compressor displacement rather than from the evaporator duty.
The vapor-compression diagram is drawn with the discharge state already on the condensing line, which is the ideal case, and the efficiency is one line of text that is easy to skip. Displacement also looks like the obvious place to get flow when it appears in the problem.
What to do instead
Compute h2s at s2s = s1, then h2 = h1 + (h2s - h1)/eta_isen, and power = mass flow x (h2 - h1). Get refrigerant mass flow from evaporator capacity divided by the refrigeration effect (h1 - h4), and only use displacement x volumetric efficiency when the problem asks for swept volume or states the compressor is the given constraint.
Using the entering water temperature as the condensing or evaporating temperature, or swapping approach and range when sizing a condenser or cooling tower.
All four temperatures are single numbers in the problem statement and none of them is labeled in a way that forces you to think about which side of the heat exchanger it belongs to. Approach and range both look like 'temperature difference across the equipment.'
What to do instead
Write the temperature ladder for each exchanger before computing: entering water, leaving water, refrigerant saturation temperature, ambient wet bulb. Approach = saturation temperature minus leaving water (condenser) or leaving chilled water minus evaporating temperature (evaporator); range = entering minus leaving water. Cooling tower: range = hot water minus cold water, approach = cold water minus wet bulb.
Applying the fan or pump laws directly to predict flow at a new speed on a system that has substantial fixed pressure drop from filters, coils, or static head, and reporting the result as the new operating point.
Fan laws are drilled as three simple ratios, and the system curve is usually not given, so the laws get applied to any speed-change question. They only describe the fan, not the intersection with a system that has a constant-loss component.
What to do instead
Sketch the system curve as fixed loss plus flow-squared loss and find the intersection with the new fan curve. Only scale flow linearly with speed when the problem states the system is entirely dynamic (all losses vary with the square of flow) or the equipment is a centrifugal pump with no static lift; otherwise solve the intersection.
Reading a duct friction chart or computing velocity pressure with V = 4005 x sqrt(VP) for air that is not at 0.075 lb/ft3, or sizing at standard density and then specifying a fan static pressure without a density correction.
The 4005 constant and the standard duct charts are the default tools, and the fact that the air is at 120 F or at altitude is treated as a note rather than a correction. Duct velocity is also often given as if it were independent of density.
What to do instead
For non-standard air, use VP = rho x V^2 / (2 x gc) and correct the friction chart by the density ratio; scale fan static pressure by rho/0.075 when the fan selection is at a different density than the catalog. Keep ACFM for velocity and SCFM (or mass flow) for the fan curve.
Substituting an arithmetic mean temperature difference for LMTD in a coil or heat exchanger, or using LMTD for a cross-flow arrangement without the correction factor F, and mixing U-values computed on inside versus outside area.
Arithmetic mean is much faster and gives nearly the right answer when the two terminal differences are close, so it survives practice problems with balanced flow. It fails badly on condensers and on chilled water coils where the terminal differences differ by a factor of two or more.
What to do instead
Compute both terminal differences first; if they differ by more than about 20%, use the logarithmic mean without exception. For cross-flow or multipass arrangements apply the correction factor F and confirm it is above roughly 0.8. Keep the area basis consistent: Q = U_o A_o LMTD requires U referenced to the outside area, with fouling and both film resistances included.
Summing each zone's required outdoor airflow to get system outdoor intake, or averaging the zone outdoor-air fractions, on a multiple-zone recirculating system.
Zone-by-zone calculation is how the code table is applied, so summing the zone results is the natural next step. But recirculation means the critical zone sets the fraction for the whole system, and the uncorrected sum over- or under-ventilates.
What to do instead
Compute V_ou as the sum of zone requirements, find the critical zone fraction Z_d as the largest V_oz/V_pz, and use V_ot = V_ou / (1 + X_s - Z_d) with X_s = V_ou/V_ps. Identify the critical zone explicitly by name before finishing the calculation.
Dropping unit conversions mid-calculation, such as dividing a Btu/h load by a gpm flow directly, mixing lb/h with cfm, or using 500, 1.08, and 4.5 without knowing where each comes from and when it must be scaled.
The exam is full of mixed customary units and the memorized constants hide their own dimensions, so a missing 60 or a missing 8.33 never shows up as a suspicious answer choice. The result is an answer off by exactly the factor in one of the distractors.
What to do instead
Carry units on every line and cancel them before substituting numbers. Write the derivations once on your scratch sheet: 500 = 60 x 8.33 (gpm to lb/h for water), 1.08 = 60 x 0.075 x 0.24, 4.5 = 60 x 0.075, and note that 1.08 and 4.5 both scale with density. After solving, check the magnitude against a rough rule of thumb (about 400 cfm/ton, 2.4 gpm/ton) before selecting an answer.