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| Section | Weight | Objectives |
|---|---|---|
| Topic 1: Heating, Ventilating, and Air Conditioning (HVAC) | 12-18% | - Psychrometrics - Equipment efficiency ratings - Indoor air quality - Controls and optimization - Heat transfer principles - HVAC system types and components - Load calculations |
| Topic 2: Electric Power Systems | 10-15% | - Power distribution systems - Harmonics and power quality - Motors and drives - Standby power systems - Lighting systems and controls - Power factor correction |
| Topic 3: Facilities Management | 8-12% | - Monitoring and targeting (M&T) - Operations and maintenance best practices - Commissioning and re-commissioning - Preventive maintenance programs - Building automation systems (BAS) |
| Topic 4: Building Envelope | 8-12% | - Windows and glazing systems - Air leakage and infiltration - Insulation types and R-values - Daylighting and shading - Thermal envelope analysis |
| Topic 5: Environmental and Regulatory Compliance | 5-10% | - Sustainability reporting - Green building standards (LEED, ENERGY STAR) - EPA regulations (Clean Air Act, Clean Water Act) - Carbon footprint calculation - Energy reporting requirements |
| Topic 6: Energy Accounting and Economics | 10-15% | - Energy price forecasting - Utility rate structures and tariffs - Financial analysis techniques (ROI, NPV, IRR) - Energy cost accounting and allocation - Life cycle cost analysis (LCCA) |
| Topic 7: Energy Auditing | 15-20% | - Audit procedures and methodologies - Audit report writing - Types of energy audits (preliminary, detailed, investment-grade) - Baseline establishment - Data collection and analysis - Metering and measurement equipment |
| Topic 8: Renewable Energy Technologies | 8-12% | - Solar photovoltaic systems - Biomass and bioenergy - Geothermal systems - Energy storage technologies - Solar thermal systems - Wind energy systems |
| Topic 9: Energy Project Implementation | 8-12% | - Energy service companies (ESCOs) - Implementation strategies - Performance contracting (ESPC) - Project risk management - Measurement and verification (M&V) - Project financing models |
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NEW QUESTION # 21
In a warehouse, lights are mounted 4 meters above the floor. This provides an average light level of 150 lux.
To increase the light level without installing more lights or higher Wattage lights, you lower the light fixtures with pendants so that the lights are located 3.25 meters above the floor. Calculate the new average light level on the floor.
Answer: E
NEW QUESTION # 22
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?
Answer: D
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\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.
NEW QUESTION # 23
Insulation has a thermal conductivity of 0.04 Watt/(m-°C). What is the R-value for this insulation when it is 5- cm thick?
Answer: A
NEW QUESTION # 24
A 100-kW electrical load has a power factor of 75%. What capacitance, in KVAR, will improve the power factor to 95%?
Answer: D
NEW QUESTION # 25
A concentrating solar thermal system collects 5 GJ/h of thermal heat, which is delivered to a 94% efficient hot water generator. The hot water generator is used to raise the water temperature from 15°C to 65°C. The specific heat of water (Cp) is 4.2 kJ/kg-°C. What flow rate (liter/second) of hot water can be delivered from the hot water generator?
Answer: B
Explanation:
Step 1: Compute Useful Thermal Energy Output
Given:
* Solar system delivers 5 GJ/h
* Hot water generator efficiency = 94%
NEW QUESTION # 26
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