CEM Certified Energy Manager (CEM) Questions and Answers

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.

We need to calculate the monthly electric bill based on the given pricing structure.

Step 1: Identify Charges from the Table

Customer charge:$50/month

Peak demand charge:$10/kW-month

Electric consumption charges:

First25,000 kWh@$0.06/kWh

Excess(75,000 - 25,000) = 50,000 kWh@$0.05/kWh

Tax:20% of total

A screenshot of a calculator AI-generated content may be incorrect.

Amanometeris used to measurepressure, particularly in gases and liquids.

Step 1: Analysis of Each Option

Pressure (A):✅Correct. Manometers are commonly used in HVAC, industrial, and lab applications.

Volume (B):❌Incorrect. Volume is measured using flow meters or tanks.

Distance (C):❌Incorrect. Distance is measured using rulers or lasers.

Voltage (D):❌Incorrect. Voltage is measured using voltmeters.

Thus, the correct answer isA. Pressure.

To determine the annual energy cost savings from switching to LED luminaires:

A table with numbers and symbols AI-generated content may be incorrect.

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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A screenshot of a math formula AI-generated content may be incorrect.

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\text{Simple Payback Period (years)} = \frac{\text{Initial Investment Cost}}{\text{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\text{Simple Payback Period} = \frac{\text{Initial Investment Cost}}{\text{Net Annual Savings}}Simple Payback Period=Net Annual SavingsInitial Investment Cost​ Simple Payback Period=540,000135,000=4.0 years\text{Simple Payback Period} = \frac{540,000}{135,000} = 4.0 \, \text{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 \frac{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 \frac{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.

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.