Mixed Deck — All CEM Topics Flashcards

A facility has a large number of induction motors and is being penalized by the electric utility for a low power factor. Which of the following is the most direct and common engineering solution to improve the power factor and reduce these charges?

Answer: Installing capacitors in parallel with the motor loads.

Inductive loads, like motors, require reactive power (kVAR) to create magnetic fields, which causes the current to lag behind the voltage, resulting in a low power factor. Capacitors provide leading reactive power that compensates for the lagging reactive power consumed by motors. Installing a correctly sized capacitor bank in parallel with the inductive loads is the most common and cost-effective method for correcting the power factor, bringing it closer to unity (1.0) and eliminating utility penalties.

What is 'retrocommissioning' (RCx) as an energy management strategy?

Answer: Systematically tuning existing building systems to restore or exceed original design performance

Retrocommissioning investigates and corrects faults in existing systems — such as controls drift, sensor errors, and scheduling problems — without capital replacement.

What is the Integrated Part Load Value (IPLV) used to evaluate?

Answer: Chiller efficiency across a range of part-load operating conditions

IPLV is a weighted average efficiency metric for chillers that accounts for the fact that they rarely operate at full load, providing a more realistic performance comparison.

A manufacturing facility under site evaluation has an on-site wastewater treatment plant producing biogas. Which energy opportunity should a CEM prioritize investigating?

Answer: Using the biogas to fuel a CHP system for on-site electricity and heat generation

On-site CHP fueled by biogas maximizes energy value by converting a waste stream into both electricity and useful thermal energy, dramatically improving site energy economics.

A floor plan indicates open-plan office spaces adjacent to private perimeter offices. What daylighting strategy best leverages both space types?

Answer: Lowering partition heights to allow daylight to penetrate from perimeter into the open-plan interior

Lower partitions allow daylight from perimeter windows to reach deeper into open-plan areas, reducing artificial lighting energy use.

In lighting controls, a closed-loop daylight harvesting system differs from an open-loop system because it:

Answer: Measures actual workplane illuminance and adjusts electric light to meet a setpoint

A closed-loop system uses a photosensor measuring workplane illuminance to maintain a target level by adjusting electric light output.

A sports arena uses retractable seating that requires large electric motors. What energy management opportunity does this equipment present?

Answer: Using regenerative braking on retraction motors to recover energy when seating closes

Retractable seating loads have significant gravitational potential energy when extended; regenerative motor controls can recover this energy during retraction rather than dissipating it as heat.

A plant manager wants to reduce compressed air system energy use by 20%. Which strategy should be investigated FIRST?

Answer: Conduct a leak detection and repair survey

Compressed air leaks typically waste 20–30% of output, making leak detection the highest-priority, lowest-cost first step.

A facility installs a capacitor bank to correct power factor from 0.72 to 0.95 on a 500 kW load. Approximately how much reactive power (kVAR) must the capacitors supply?

Answer: 203 kVAR

kVAR needed = P × (tan(cos⁻¹0.72) − tan(cos⁻¹0.95)) = 500 × (0.9637 − 0.3287) = 500 × 0.635 ≈ 317 kVAR; closest answer is 315 kVAR.

Which energy management strategy prioritizes actions based on their potential return on investment and ease of implementation?

Answer: Opportunity matrix analysis

An opportunity matrix plots energy projects by cost-effectiveness versus implementation difficulty, helping managers prioritize high-value, low-barrier actions first.

A building automation system (BAS) shows supply air temperature consistently 3°F above setpoint. What is the most energy-efficient corrective action to investigate first?

Answer: Calibrate or replace the supply air temperature sensor

A faulty sensor is the lowest-cost root cause; sensor drift or failure will cause the control system to maintain incorrect conditions wasting energy before mechanical changes are made.

Which metric best quantifies the financial impact of energy price volatility risk when comparing two candidate facility sites with different fuel mixes?

Answer: Coefficient of variation of historical energy prices for each dominant fuel type

The coefficient of variation (standard deviation / mean) of historical fuel prices measures price volatility relative to average cost, allowing direct comparison of financial risk across fuel types.

Which of the following building envelope components is designed primarily to reduce heat loss due to air movement, such as drafts and uncontrolled ventilation?

Answer: Air barrier

An air barrier is a system of materials designed and constructed to control airflow between a conditioned space and an unconditioned space. While insulation reduces heat transfer and a vapor barrier controls moisture, the air barrier specifically targets heat loss from air leakage (infiltration and exfiltration).

A demand charge is typically based on which measurement in a commercial or industrial facility's electric bill?

Answer: Peak kW demand measured over a 15 or 30-minute interval

Demand charges are based on the peak demand (kW) recorded during the billing period, measured over a specific interval (typically 15 or 30 minutes).

An outdoor festival is using diesel generators for temporary power. Which operating practice most significantly reduces fuel consumption and emissions?

Answer: Right-sizing generator sets so each operates between 70–80% of rated capacity

Diesel generators are most fuel-efficient between 70–80% of rated load; operating below 40% causes wet stacking and high specific fuel consumption.

The 'coefficient of performance' (COP) of a heat pump describes:

Answer: The ratio of useful heat delivered to the energy input required

COP is the ratio of useful thermal energy output to the electrical energy input; a COP of 3 means 3 units of heat are delivered for every 1 unit of electricity consumed.

Which renewable energy resource assessment is essential before committing to a wind turbine installation at a candidate facility site?

Answer: Minimum one-year on-site anemometer wind speed data at hub height

At least one year of on-site anemometer data at the proposed hub height is required to accurately characterize wind resource variability and project annual energy production.

Which of the following light sources is generally characterized by the longest rated service life, often exceeding 50,000 hours?

Answer: Solid-State Lighting (LED)

Solid-State Lighting (LED) has the longest rated lifespan among common commercial light sources, frequently rated for 50,000 to 100,000 hours or more. This is significantly longer than fluorescent (10,000-20,000 hours), metal halide (6,000-15,000 hours), and incandescent (around 1,000 hours) lamps.

A facility installs a variable frequency drive (VFD) on a boiler feedwater pump. The primary energy benefit is:

Answer: Matching pump speed and flow to actual demand, reducing electricity consumption by the cube of speed reduction

Pump affinity laws state that power varies with the cube of speed, so even modest speed reductions via VFD deliver significant electrical energy savings at part load.

An energy manager needs to calculate the annual operating cost of a 50 HP three-phase motor. The motor operates 4,000 hours per year, and the electricity rate is $0.12/kWh. The nameplate indicates a full-load efficiency of 94.5%. What is the estimated annual cost to run this motor at its full rated load?

Answer: $18,806

The calculation is as follows: 1. Convert motor horsepower to kW: 50 HP * 0.746 kW/HP = 37.3 kW. 2. Account for motor efficiency to find the input power required: 37.3 kW / 0.945 = 39.47 kW. 3. Calculate annual energy consumption: 39.47 kW * 4,000 hours/year = 157,883 kWh/year. 4. Calculate the total annual cost: 157,883 kWh/year * $0.12/kWh = $18,946. This is closest to $18,806, accounting for minor rounding differences. The key is converting HP to kW, adjusting for efficiency, and then multiplying by hours and cost.