Commissioning and Measurement & Verification (M&V)
What the exam expects
Commissioning verifies that systems meet the owner's project requirements and basis of design. Know design review, installation checks, functional performance testing, seasonal testing, training, O&M documentation, and ongoing commissioning. Retro-commissioning applies the process to existing buildings. M&V follows IPMVP options: A, retrofit isolation with stipulated operational parameters; B, retrofit isolation with measured parameters; C, whole-building metering; and D, calibrated simulation. Savings = adjusted baseline energy minus reporting-period energy, with adjustments for weather, occupancy, and scope.
Traps
Option A is not the same as Option B: one stipulates key parameters, the other measures them. Another trap is calling avoided cost a savings claim without adjusting for baseline changes. Watch measurement boundary and non-routine adjustments.
Worked example
A lighting retrofit baseline is 200 kW for 3,000 operating hours, or 600,000 kWh. After retrofit, demand is 120 kW for the same hours, or 360,000 kWh. Savings are 240,000 kWh. If operating hours rise to 3,300, the baseline must be adjusted to 200 x 3,300 = 660,000 kWh before claiming savings of 300,000 kWh.
Building Systems
What the exam expects
Building-systems questions cover HVAC, lighting, envelope, service water heating, controls, and plug loads. Know chiller efficiency in kW/ton and COP, boiler AFUE and combustion efficiency, packaged-unit EER and SEER, fan and pump laws, economizers, demand-controlled ventilation, heat recovery, and variable-frequency drives. Understand part-load performance and why full-load ratings may not predict annual energy. Controls sequences, scheduling, setpoints, and simultaneous heating and cooling are frequent audit targets. Envelope and internal gains affect load shape as well as total load.
Traps
Confusing SEER with EER, or AFUE with combustion efficiency, is common. Another trap is assuming a high-efficiency chiller saves energy if it runs at poor part-load conditions or fights an uncontrolled air handler. Watch units: kW/ton and COP are inversely related with a conversion factor.
Worked example
A 1,000-ton chiller runs 2,000 equivalent full-load hours. At 0.75 kW/ton it uses 1,500,000 kWh; at 0.60 kW/ton it uses 1,200,000 kWh. Savings are 300,000 kWh. At $0.08/kWh, that is $24,000 per year before demand savings. The estimate assumes equivalent full-load hours are unchanged.
Industrial Process and Utility Equipment
What the exam expects
Industrial questions emphasize compressed air, steam, pumps, fans, motors, process heating, refrigeration, and heat recovery. Compressed air is one of the most expensive utilities per unit of useful work; leaks, artificial demand, and inappropriate uses are prime targets. Steam systems lose energy through traps, insulation, condensate, and blowdown. Pumps and fans follow affinity laws. Motors should be sized to actual load; oversizing raises no-load losses. Understand specific energy, load profiles, and waste-heat recovery opportunities.
Traps
Assuming a leak is free because it is small, or measuring compressor kW without accounting for storage and controls, is common. Another trap is applying a pump curve without checking system curves and throttling losses. Watch steam-trap testing and insulation economics.
Worked example
A compressed-air leak flows 100 cfm continuously. At roughly 25 kW per 100 cfm, the leak draws about 25 kW. Over 8,000 hours, energy is 200,000 kWh. At $0.08/kWh, the annual cost is $16,000, before demand charges. Repairing the leak is often a low-cost measure.
Combined Heat and Power and On-site Generation
What the exam expects
CHP questions test prime movers, heat recovery, efficiency, sizing, and economics. Know reciprocating engines, combustion turbines, microturbines, and fuel cells, plus heat-to-power ratios and useful thermal output. CHP efficiency = (electricity generated x 3,412 + useful heat recovered) / fuel input, in consistent Btu. Economic feasibility depends on spark spread, avoided utility charges, thermal load duration, interconnection, and emissions. On-site renewable generation adds intermittency, net metering, and storage considerations. Standby and supplemental charges can erode savings.
Traps
Using nameplate capacity rather than net output, or counting recovered heat that has no use, overstates efficiency. Another trap is ignoring parasitic loads, maintenance, and standby charges. The spark spread is not the same as total project economics.
Worked example
A CHP unit uses 10 MMBtu/h of fuel, produces 1 MW and 4 MMBtu/h of useful heat. Electric output = 1,000 kW x 3,412 Btu/kWh = 3.412 MMBtu/h. Useful output = 3.412 + 4 = 7.412 MMBtu/h. Efficiency = 7.412 / 10 = 74.1%. If only 2 MMBtu/h of heat is used, efficiency falls to 54.1%.