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[General] CEM Reliable Exam Camp & CEM Training Solutions

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【General】 CEM Reliable Exam Camp & CEM Training Solutions

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AEE Certified Energy Manager (CEM) Sample Questions (Q125-Q130):NEW QUESTION # 125
A compressed air system has a total volume of 247,500 liters. During a 30-minute break in production (air compressor system turned off and no production demand for compressed air), the system pressure drops from
7 bar, gauge (700 kPa, gauge) to 5 bar, gauge (500 kPa, gauge). Calculate the leakage rate in the compressed air system.
  • A. 8,250 liters/minute
  • B. 22,750 liters/minute
  • C. 16,500 liters/minute
  • D. 32,000 liters/minute
  • E. 48,250 liters/minute
Answer: C

NEW QUESTION # 126
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. 2.0 years
  • B. 5.0 years
  • C. 4.0 years
  • D. 3.0 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 # 127
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?
  • A. 20 Watt/(m²-°C)
  • B. 2.20 Watt/m²
  • C. 20 (m²-°C)/Watt
  • D. 1.25 (m²-°C)/Watt
  • E. 0.8 (m²-°C)/Watt
Answer: D

NEW QUESTION # 128
What will commonly be the most appropriate expected lamp life for a fluorescent lamp fitted with an occupancy sensor, that is operational in a school?
SELECT THE CORRECT ANSWER
  • A. 50,000 operational hours
  • B. 1,000 operational hours
  • C. 20,000 calendar hours
  • D. 18,000 operational hours
  • E. 50,000 calendar hours
Answer: D

NEW QUESTION # 129
In a PID control system, what is the purpose of the Proportional Gain?
  • A. Reacts to change
  • B. Senses high load rate
  • C. Control point
  • D. Reduces/eliminates offset
Answer: A

NEW QUESTION # 130
......
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