Exception: Reduced I²R losses in cables can slightly lower kW — usually negligible. Define: Degree-Day method for estimating heating energy. Card 9 – Back A: Heating Energy (MMBtu) = (HDD × UA × 24 hr/day) / (System Efficiency × 10^6)
Subtitle: CEM Test Practice Questions & Review for the Certified Energy Manager Exam
Also relevant for CEM: ASHRAE/IESNA 100 (Existing Buildings), ASHRAE 62.1 (Ventilation). Practice Question: A plant has annual energy cost $2,000,000. A project saves 8% with a $250,000 cost. What is the ROI first year? Card 13 – Back A: ROI = (Annual Savings / Project Cost) × 100
The effective annual return from the project’s cash flows. Compare to company’s hurdle rate (e.g., 12%). If IRR > hurdle rate → accept.
If NPV > 0 → accept project (earns more than hurdle rate). If NPV < 0 → reject.
Isolated retrofit, stipulated assumption for unmeasured factors. Option B: All parameters measured. Card 8 – Front True or False: Power factor correction capacitors reduce kW demand charges. Card 8 – Back A: False. Capacitors reduce kvar , thus lowering kVA (apparent power), which may reduce kVA demand charges , but not kW demand. kW is real power, unchanged by capacitors.
Note: Power factor (0.85) not needed for kW calculation — only for kVA. Term: Chilled Water Reset Card 4 – Back A: A control strategy that increases chilled water supply temperature as cooling load decreases (e.g., from 44°F to 50°F at part load).
Each card features a question, key term, formula, or scenario on the front, and a detailed, exam-relevant answer on the back. Use these cards in 15-minute bursts to reinforce your weak areas and test your recall under exam conditions. Card 1 – Front Q: What is the formula for Simple Payback Period (SPP), and what is its primary limitation in energy project evaluation? Card 1 – Back A: SPP = Initial Investment / Annual Savings