CEM Certified Energy Manager (CEM) Questions and Answers

Electric utility companies transmit power at high voltages primarily to reduce transmission power losses, which are proportional to the square of the current (I²) multiplied by the resistance (R) of the transmission lines. This relationship is expressed by the formula for power loss: P_loss = I²R.​

Key Points:

Power Loss and Current Relationship:Power losses in transmission lines are directly proportional to the square of the current flowing through them. Reducing the current decreases these losses significantly.

High Voltage Transmission:By increasing the transmission voltage, the current required to deliver the same amount of power decreases. This is because power (P) is the product of voltage (V) and current (I): P = VI. For a given power level, increasing voltage allows for a corresponding decrease in current.

Reduced I²R Losses:Lower current results in reduced I²R losses, enhancing the efficiency of power transmission over long distances.

Conclusion:

Transmitting power at high voltages minimizes transmission losses, making it the most efficient method for long-distance electrical power delivery. Therefore, the correct answer is C. Transmission power losses (I²R) are less.

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To determine which advantage makes an ice Thermal Energy Storage (TES) system more preferred over a water TES system for a load shifting strategy, we need to evaluate each option based on the principles of thermal energy storage as outlined in the Association of Energy Engineers (AEE) Certified EnergyManager (CEM) training materials. Load shifting involves storing energy (cooling capacity) during off-peak periods and releasing it during peak demand, making storage efficiency and capacity critical. Let’s analyze each option step-by-step.

Step 1: Understand Ice TES vs. Water TES in Load Shifting

Ice TES: Uses the latent heat of fusion of water (ice melting) to store cooling energy. Ice is formed during off-peak hours (e.g., overnight) and melted during peak hours to provide cooling.

Water TES: Uses the sensible heat capacity of water, storing chilled water (typically 4–6°C) to provide cooling.

Load Shifting Goal: Maximize cooling storage in minimal space and cost, shifting electrical demand from peak to off-peak periods.

CEM Reference: CEM materials in the "Thermal Energy Storage" section highlight ice TES for its high energy density and compact storage, contrasted with water TES for simpler operation but larger volume requirements.

Step 2: Evaluate Each Option

Option A: Ice-storage systems operate with a higher coefficient of performance (COP)

Analysis:

COP Definition: COP = (Cooling Output) / (Energy Input). For TES, this relates to the chiller’s efficiency.

Ice TES: Requires chillers to operate at lower temperatures (e.g., -5°C to 0°C) to freeze water, which typically reduces chiller COP (e.g., 3–4) compared to water TES chillers operating at 4–6°C (COP ~5–6).

Reality: Ice TES systems often have a lower COP due to the additional energy needed for phase change, though total system efficiency may improve with load shifting benefits.

CEM Reference: CEM notes that ice TES energy input is higher per unit of cooling due to lower evaporating temperatures, contradicting a "higher COP" claim.

Conclusion: This statement is incorrect and not an advantage for ice TES in load shifting.

Option B: Ice-storage systems require smaller storage tanks since ice has a higher energy storage density

Analysis:

Energy Storage Density:

Ice TES: Relies on latent heat of fusion = 334 kJ/kg (80 kcal/kg or ~144 Btu/lb). This is the energy absorbed/released when water freezes/melts, far exceeding sensible heat.

Water TES: Relies on sensible heat = cp⋅ΔT c_p \cdot \Delta T cp​⋅ΔT, where cp=4.18 kJ/kg\cdotp°C c_p = 4.18 \, \text{kJ/kg·°C} cp​=4.18kJ/kg\cdotp°C (1 Btu/lb·°F). For a typical ΔT=10°C\Delta T = 10°CΔT=10°C (e.g., 4°C to 14°C), energy stored = 4.18×10=41.8 kJ/kg 4.18 \times 10 = 41.8 \, \text{kJ/kg} 4.18×10=41.8kJ/kg (~20 Btu/lb).

Comparison: Ice stores ~8 times more energy per kg than water for a 10°C range (334 vs. 41.8 kJ/kg).

Volume Impact: Ice’s density (~917 kg/m³) is slightly less than water (~1000 kg/m³), but the latent heat advantage dominates, reducing required tank volume significantly.

Load Shifting: Smaller tanks mean less space and potentially lower capital costs, a key advantage for peak load management.

CEM Reference: CEM training emphasizes ice TES’s high energy density as a primary reason for its preference in space-constrained load shifting applications.

Conclusion: This statement is correct and a clear advantage for ice TES.

Option C: Water-storage systems require smaller storage tanks since water has a higher density than ice

Analysis:

Density: Water = 1000 kg/m³; Ice = 917 kg/m³. Water is denser, but density alone doesn’t determine storage size in TES.

Energy Storage: As calculated, water’s sensible heat capacity (e.g., 41.8 kJ/kg for 10°C) is much lower than ice’s latent heat (334 kJ/kg). To store the same cooling capacity, water TES requires ~8 times more mass and thus larger tanks (even accounting for density differences).

Implication: Water TES tanks are larger, not smaller, contradicting the statement.

CEM Reference: CEM materials note water TES’s larger volume requirements as a disadvantage compared to ice TES.

Conclusion: This statement is incorrect and not an advantage for ice TES (it favors water TES incorrectly).

Option D: Ice-storage systems require lower maintenance due to lower pumping volume

Analysis:

Pumping Volume: Ice TES often uses glycol or brine solutions to transfer heat at lower temperatures, requiring pumps sized for smaller volumes due to concentrated coolingcapacity. Water TES circulates larger volumes of chilled water. However, "lower pumping volume" doesn’t directly translate to "lower maintenance."

Maintenance: Ice TES systems are more complex (ice-making equipment, heat exchangers), potentially increasing maintenance (e.g., defrost cycles, corrosion from brine). Water TES is simpler, often with lower maintenance needs.

CEM Reference: CEM discusses ice TES complexity as a trade-off for its density advantage, not a maintenance benefit.

Conclusion: This statement is questionable and not a primary advantage for load shifting.

Step 3: Identify the Key Advantage for Load Shifting

Load Shifting Context: The goal is to store maximum cooling capacity efficiently during off-peak hours. Option B (smaller tanks due to higher energy storage density) directly supports this by reducing space and installation costs, a critical factor in TES design per CEM guidelines.

Elimination:

A: Incorrect (lower COP, not higher).

C: Incorrect (water TES tanks are larger).

D: Weak (maintenance isn’t clearly lower; not the primary driver).

B: Correct and relevant.

Renewable energy resources are naturally replenished on a human timescale. Let's evaluate each option:​

A.Saw grass:A biomass resource, renewable through regrowth.

B.Shale gas:A fossil fuel extracted from shale formations, non-renewable.

C.Geothermal heat:Energy from Earth's internal heat, renewable.

D.Ocean waves:Mechanical energy from ocean surface waves, renewable.

E.Crop residue:Organic materials from agriculture, renewable.

Conclusion:

Shale gas is not a renewable energy resource. Therefore, the correct answer is B.

When conducting internal audits of an Energy Management System (EnMS) in accordance with ISO 50001, auditors are expected to perform several key activities to ensure the system's effectiveness and continual improvement. These activities include:​

Focus on Areas of Significant Energy Use:Auditors should prioritize auditing areas that have substantial energy consumption, as improvements in these areas can lead to significant energy performance enhancements.

Review Written Procedures Related to the Area Under Audit:A thorough examination of documented procedures ensures that they align with ISO 50001 requirements and organizational practices. This review helps in identifying any discrepancies between documented procedures and actual practices.

Assess Actual Activities Against Written Procedures:Auditors compare on-ground activities with documented procedures to verify compliance and identify deviations. This assessment ensures that operations are performed as intended and in line with established protocols.

Look for Opportunities for Energy Improvement and Minimize Bureaucracy:Identifying opportunities for enhancing energy performance is a core objective of the audit.Auditors should also aim to streamline the EnMS by reducing unnecessary complexities, thereby making it more efficient and user-friendly.

By encompassing all these activities, internal auditors ensure a comprehensive evaluation of the EnMS, leading to effective energy management and continual improvement. Therefore, the correct answer is E. All of the above.

Instrumentation in energy audits serves multiple functions, but theprimary reasonis to determine energy flows.

Measuring Energy Flows:

Instrumentation helpsquantify energy usein different systems, such as electricity, HVAC, and industrial processes.

It helps identifywhere energy is used, lost, or wasted.

Supporting Energy Conservation Efforts:

By measuring energy flows, auditors canidentify inefficienciesand recommend solutions.

Other Options Analysis:

Saving energy (A):A result of energy audits, not the main purpose.

Validating utility bills (C):Important but secondary to understanding energy flows.

Supporting the local economy (D):Not relevant.

Thus, the correct answer isB. Determine energy flows.

Step 1: Purpose of Each Instrument

Hot-wire anemometer→ Measures air velocity, NOT temperature.

Bimetallic thermometer→ Measures temperature using metal expansion.

Thermistor→ Measures temperature using resistance change.

Thermometer→ Directly measures temperature.

Optical pyrometer→ Measures high temperatures using infrared radiation.

Since ahot-wire anemometer measures airspeed, not temperature, the correct answer isA. Hot-wire anemometer.

The Energy Utilization Index (EUI) is measured astotal energy consumption per unit area.

Step 1: Compare Energy Inputs for Both Systems

Existing Natural Gas System:

Heating output:10,000 kJ/h

Efficiency:80%

Required input energy

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