CEM Study Guide 2026
Everything you need to pass the CEM exam in one place: the exam format, every topic to study, real practice questions with explanations, flashcards, and full-length practice tests. Free, no sign-up needed.
📋 CEM Exam Format at a Glance
📚 CEM Topics to Study (21)
✍️ Sample CEM Questions & Answers
1. The primary purpose of a calibrated energy simulation model used in M&V is to:
A calibrated simulation is adjusted until its outputs closely match actual measured pre-retrofit data, creating a credible baseline model for calculating savings under Option D of IPMVP.
2. In general, if cooler air enters a compressor, the compressor's efficiency is:
Cooler air is denser, meaning a greater mass of air can be compressed per unit of volume. When a compressor takes in denser air, it can deliver more compressed air for the same amount of volumetric displacement and energy input. This increases the compressor's mass flow rate and overall efficiency, as less energy is expended on compressing a smaller mass of less dense air.
3. The 'avoided cost' of energy is most accurately described as:
Avoided cost represents the economic value of energy that does not need to be purchased due to conservation or efficiency measures.
4. NEMA Premium efficiency motors save energy compared to standard efficiency motors primarily because they have:
Premium efficiency motors use higher grades of silicon steel, more copper in windings, closer tolerances, and improved designs to reduce core and copper losses, achieving 2-8% better efficiency than standard motors.
5. An energy information system (EIS) or energy dashboard primarily helps energy managers by:
An EIS aggregates meter data and presents consumption trends, benchmarks, and alerts, enabling energy managers to identify anomalies and prioritize efficiency improvements.
6. What are HP savings if a motor has a Variable Speed Drive (VSD) fitted that slows down a 200 HP motor by 20%?
For centrifugal loads like pumps and fans, power consumption is proportional to the cube of the speed (affinity laws). If the motor speed is reduced by 20%, the new speed is 80% (0.8) of the original. The new power consumed will be (0.8)^3 * 200 HP = 0.512 * 200 HP = 102.4 HP. Therefore, the HP savings are 200 HP (original) - 102.4 HP (new) = 97.6 HP.