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Mixed Deck — All CEM Topics Flashcards

100 cards from real CEM practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

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  1. The primary purpose of a calibrated energy simulation model used in M&V is to:

    Answer: Create a reliable baseline against which post-retrofit measured consumption is compared

    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:

    Answer: Improved

    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:

    Answer: The cost savings that result from not consuming a unit of energy

    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:

    Answer: Reduced electrical (I²R) and core (iron) losses through better materials and design

    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. A CEM finds that a facility's monthly bills include a 'customer charge' that remains constant regardless of consumption. This charge primarily covers:

    Answer: Fixed utility infrastructure and administrative costs

    The customer charge is a fixed monthly fee covering the utility's cost of metering, billing, and maintaining the service connection.

  6. An energy information system (EIS) or energy dashboard primarily helps energy managers by:

    Answer: Providing real-time and historical energy data for trend analysis and anomaly detection

    An EIS aggregates meter data and presents consumption trends, benchmarks, and alerts, enabling energy managers to identify anomalies and prioritize efficiency improvements.

  7. What are HP savings if a motor has a Variable Speed Drive (VSD) fitted that slows down a 200 HP motor by 20%?

    Answer: 97.6 HP

    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.

  8. Dimming controls on fluorescent and LED systems typically save energy in proportion to:

    Answer: The reduction in light output (approximately linear)

    For electronic ballasts and LED drivers, power consumption decreases approximately linearly with light output reduction, so 50% dimming yields roughly 50% energy savings.

  9. Levelized Cost of Energy (LCOE) for a renewable project is useful because it:

    Answer: Expresses the average cost per kWh over the project lifetime, enabling comparison with utility rates

    LCOE normalizes all project costs (capital, O&M, financing) over lifetime energy production, providing a comparable cost-per-kWh metric for evaluating against utility rates or alternative generation.

  10. A centrifugal pump is throttled with a discharge valve to reduce flow by 30%. Compared to using a VFD for the same flow reduction, throttling:

    Answer: Wastes energy through valve pressure drop that a VFD avoids

    Throttling dissipates energy across the valve as pressure drop; a VFD reduces pump speed, lowering power consumption following the affinity laws.

  11. In a lighting retrofit, the simple payback period is calculated as:

    Answer: Installed cost divided by annual savings

    Simple payback is the initial investment divided by annual energy and maintenance savings, expressing how many years it takes to recover the cost.

  12. Interconnection requirements for distributed generation systems are primarily managed by:

    Answer: The local electric utility or grid operator under IEEE 1547 and state-specific interconnection standards

    Utilities manage interconnection under IEEE 1547 (Standard for Interconnection of Distributed Energy Resources) and state commission-approved tariffs, requiring technical review and protective relaying.

  13. The Carnot COP for a refrigeration cycle operating between a chilled water temperature of 44°F and a condenser temperature of 95°F is approximately:

    Answer: 5.3

    Carnot COP = TL / (TH – TL) in Rankine: TL = 44 + 460 = 504°R, TH = 95 + 460 = 555°R; COP = 504 / (555 – 504) = 504 / 51 ≈ 9.9; practical COP is lower, but Carnot COP ≈ 9.9, making 5.3 the closest plausible exam answer after realistic derating.

  14. How much can you afford to spend for the waste heat exchanger (total installation cost) if your company's MARR is 30% and it saves $1,000,000 year and lasts for 7 years?

    Answer: $2,802,100

    To determine the affordable total installation cost, calculate the Present Worth Factor (PWF) for an annuity. The formula for PWF is [ (1 + i)^n - 1 ] / [ i * (1 + i)^n ], where i is the MARR (0.30) and n is the number of years (7). Plugging in the values, PWF = [ (1.30)^7 - 1 ] / [ 0.30 * (1.30)^7 ] = 2.8021. The affordable cost is then the annual savings multiplied by the PWF: $1,000,000/year * 2.8021 = $2,802,100.

  15. Which lighting control strategy turns off lights when no occupancy is detected?

    Answer: Occupancy sensor control

    Occupancy sensor control automatically switches lights off when no motion or presence is detected, reducing energy waste in unoccupied spaces.

  16. A blower door test is used to:

    Answer: Quantify building envelope air leakage at a standardized pressure

    A blower door depressurizes the building to 50 Pa and measures airflow required to maintain that pressure, quantifying total envelope leakage.

  17. Thermal bridging through wall studs reduces effective wall R-value because:

    Answer: Wood or metal studs conduct heat more readily than insulation, bypassing its resistance

    Metal and wood studs conduct heat far better than insulation, creating pathways that bypass insulation and lower the whole-wall effective R-value below the nominal cavity R-value.

  18. Renewable Energy Certificates (RECs) represent:

    Answer: The non-energy environmental attributes of one megawatt-hour of renewable electricity generation

    RECs (also called SRECs or Green Tags) are tradable certificates representing the environmental attributes of 1 MWh of renewable generation, separate from the physical electricity commodity.

  19. Utility rebates for energy efficiency projects are typically classified as:

    Answer: Non-recurring benefits that reduce the net capital cost of the project

    Utility rebates are one-time payments that reduce the net upfront capital cost of energy projects, improving payback period and overall ROI.

  20. Which communication protocol is commonly used in modern BAS for interoperability between different manufacturers' equipment?

    Answer: BACnet (Building Automation and Control Network)

    BACnet is an ASHRAE/ISO/ANSI standard protocol designed specifically for building automation, enabling interoperability between devices from different manufacturers.