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2025 Reliable CEM Study Notes | Pass-Sure AEE CEM: Certified Energy Manager (CEM) 100% Pass

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AEE Certified Energy Manager (CEM) Sample Questions (Q87-Q92):

NEW QUESTION # 87
Thermal scans of electrical equipment are useful for:

A. Locating areas of high power factor
B. Locating a loose electrical connection
C. Reading current carrying capacity of installed conductors
D. Measuring the load on a refrigerant system

Answer: B

Explanation:
Step 1: Purpose of Thermal Scans
Thermal scans (infrared thermography) detecttemperature variationsin electrical components.
* Loose connectionsincrease resistance # causeoverheating, which isvisible in thermal scans.
Step 2: Analysis of Each Option
* A. Locating areas of high power factor#Incorrect#Power factor is measured with power meters.
* B. Reading current-carrying capacity of conductors#Incorrect#Measured using ammeters and engineering data.
* C. Locating a loose electrical connection#Correct#Loose connections causeheat buildup, easily detected with thermal scans.
* D. Measuring the load on a refrigerant system#Incorrect#Refrigerant loads are measured using pressure sensors and flow meters.
Thus, the correct answer isC. Locating a loose electrical connection.

NEW QUESTION # 88
Which of the advantages listed below, makes an ice TES system more preferred over a water TES system, when a load shifting strategy is considered?

A. Ice-storage systems require lower maintenance due to lower pumping volume
B. Ice-storage systems require smaller storage tanks since ice has a higher energy storage density
C. Water-storage systems require smaller storage tanks since water has a higher density than ice
D. Ice-storage systems operate with a higher coefficient of performance (COP)

Answer: B

Explanation:
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##T c_p cdot Delta T cp##T, where cp=4.18 kJ
/kgcdotp°C c_p = 4.18 , ext{kJ/kg °C} cp=4.18kJ/kgcdotp°C (1 Btu/lb °F). For a typical #T=10°CDelta T = 10°C#T=10°C (e.g., 4°C to 14°C), energy stored = 4.18×10=41.
8 kJ/kg 4.18 imes 10 = 41.8 , ext{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.

NEW QUESTION # 89
You are using a performance contract to finance the added cost of using more-efficient equipment in the construction of a new building. Which of the following International Performance Measurement and Verification Protocol (IPMVP) options would you use to verify annual energy savings?

A. Option D - Calibrated simulation
B. Option C - Whole facility measurement
C. Option A - Retrofit isolation: key parameter measurement
D. Option E - Stipulated savings
E. Option B - Retrofit isolation: all parameter measurement

Answer: B

NEW QUESTION # 90
An energy-saving project costs $540,000. The project will have maintenance costs of $25,000 per year. The energy savings from the project are $160,000 per year. What is the simple payback of the project?

A. 5.0 years
B. 3.0 years
C. 4.0 years
D. 2.0 years

Answer: C

Explanation:
To determine the simple payback period for the energy-saving project, we need to apply the standard formula used in energy management as per the Association of Energy Engineers (AEE) Certified Energy Manager (CEM) guidelines. The simple payback period is a widely used metric in energy efficiency projects to evaluate how long it takes for the initial investment to be recovered through net savings. Let's break this down step-by-step using the provided data and CEM-aligned methodology.
Step 1: Understand the Simple Payback Formula
* Formula: Simple Payback Period (years)=Initial Investment CostNet Annual Savings ext{Simple Payback Period (years)} = rac{ ext{Initial Investment Cost}}{ ext{Net Annual Savings}} Simple Payback Period (years)=Net Annual SavingsInitial Investment Cost
* Definition: The simple payback period represents the time (in years) required for the cumulative savings to equal the initial investment, without considering the time value of money (e.g., discount rates or inflation).
* CEM Reference: AEE CEM training materials emphasize this formula in the "Energy Economics" section, where simple payback is a fundamental tool for assessing project feasibility.
Step 2: Identify Given Data
* Initial Investment Cost: $540,000 (one-time cost of the project).
* Annual Energy Savings: $160,000 per year (benefit from the project).
* Annual Maintenance Costs: $25,000 per year (additional cost incurred due to the project).
* Net Annual Savings: This must account for both the savings and the costs incurred annually.
Step 3: Calculate Net Annual Savings
* Definition: Net annual savings is the difference between the annual energy savings and any additional annual costs (e.g., maintenance).
* Verification: The problem specifies maintenance costs as an ongoing expense tied to the project, which reduces the effective savings. CEM guidelines require including such costs in payback calculations unless explicitly stated otherwise.
Step 4: Compute the Simple Payback Period
* Apply the Formula: Simple Payback Period=Initial Investment CostNet Annual Savings ext{Simple Payback Period} = rac{ ext{Initial Investment Cost}}{ ext{Net Annual Savings}} Simple Payback Period=Net Annual SavingsInitial Investment Cost Simple Payback Period=540,
000135,000=4.0 years ext{Simple Payback Period} = rac{540,000}{135,000} = 4.0 , ext{years} Simple Payback Period=135,000540,000=4.0years
* Result: The payback period is exactly 4.0 years, meaning it takes 4 years for the net savings to recover the initial investment.
Step 5: Validate Against Options
* Options:A. 2.0 yearsB. 3.0 yearsC. 4.0 yearsD. 5.0 years
* Check:
* If we ignored maintenance costs (incorrectly), payback would be 540,000160,000=3.375 rac
{540,000}{160,000} = 3.375 160,000540,000=3.375 years, which rounds to 3.4-not an exact match for any option.
* With maintenance costs included, 540,000135,000=4.0 rac{540,000}{135,000} = 4.0
135,000540,000=4.0, which matches option C precisely.
* Conclusion: Option C (4.0 years) is correct based on the net savings approach.

NEW QUESTION # 91
An air conditioning unit cools make-up air from 25°C (dry bulb temperature) and 50% relative humidity down to 13°C. What type of heat has been removed from the air?

A. Latent heat
B. Both sensible and latent heat
C. Sensible heat
D. No heat energy has been removed

Answer: B

NEW QUESTION # 92
......

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