[2025] Practice with these CEM dumps Certification Sample Questions
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NEW QUESTION # 79
If the temperature difference across an external wall is 20°C and the heat flow across the wall is 2.5 Watts/m², calculate the insulation value of the wall (including inside and outside air films)?
- A. R = 8 m²-°C/W
- B. R = 12 m²-°C/W
- C. R = 4 m²-°C/W
- D. R = 6 m²-°C/W
- E. R = 10 m²-°C/W
Answer: A
NEW QUESTION # 80
The temperature in an unconditioned mechanical room is found to typically be around 23°C warmer than ambient air. If the air intake of an air compressor located in the mechanical room is routed through a nearby wall so that ambient air is fed directly to the air compressor's intake, what impact can be expected on the air compressor's energy requirement?
- A. No measureable change in energy use
- B. Decrease energy use by 3.8%
- C. Increase energy use by 3.8%
- D. Decrease energy use by 7.6%
- E. Increase energy use by 7.6%
Answer: D
NEW QUESTION # 81
A heat pump water heater has an annual average coefficient of performance (COP) of 2.1 and heats water from 15°C to 50°C. Hot water use is 100 liters per day, 365 days/yr. The electricity cost is $0.10/kWh. What is the annual water heating cost? (The specific heat of water (C) is 4.2 kJ/kg-°C.)
- A. $44.88/year
- B. $88.29/year
- C. $70.97/year
- D. $52.20/year
- E. $64.66/year
Answer: E
NEW QUESTION # 82
At the end of a performance contract, who owns the equipment installed by the project?
- A. The energy service company
- B. The bank
- C. The equipment manufacturer
- D. The company that paid the contract payments
Answer: D
NEW QUESTION # 83
Which of the following International Performance Measurement and Verification Protocol (IPMVP) options would you use to verify annual energy savings in a project that upgraded lighting in a large manufacturing plant?
- A. Option B - Retrofit isolation: all parameter measurement
- B. Option A - Retrofit isolation: key parameter measurement
- C. Option D - Calibrated simulation
- D. Option C - Whole facility measurement
- E. Option E - Stipulated savings
Answer: A
NEW QUESTION # 84
Which of the following best describes a thermally-light building?
- A. An underground storage facility
- B. A single-story office building where the load on the heating and cooling system is proportional to the outside air (ambient) temperature
- C. A computer data center housing multiple internet server racks
- D. A building filled with manufacturing and process equipment
Answer: B
NEW QUESTION # 85
Why do electric utility companies transmit power at high voltages?
SELECT THE CORRECT ANSWER
- A. Higher voltages are more stable
- B. Higher voltages are safer
- C. Transmission power losses (12R) are less
- D. Electrical frequency is more stable
Answer: C
Explanation:
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.
NEW QUESTION # 86
You can buy air compressor A for $15,000 or air compressor B for $6,000. Compressor A will cost $10,000 per year to operate. Compressor B will cost $11,500 per year to operate. If both air compressors have a 10- year life and your required return on investment is 12%, which air compressor has the lowest total life-cycle cost?
- A. Both air compressors have the same life-cycle cost
- B. Compressor A
- C. Compressor B
Answer: C
NEW QUESTION # 87
How much energy (kJ) is required to heat 2 kg of water from 25°C to 2 kg of water at 100°C? (The specific heat of water (C) is 4.2 kJ/kg-°C. Assume no system losses.)
- A. 510 kJ
- B. 420 kJ
- C. 840 kJ
- D. 630 kJ
Answer: C
NEW QUESTION # 88
Using the natural gas combustion efficiency tables in your workbook, find the combustion efficiency (based on higher-heating value) of a natural-gas-fired boiler if the excess air level is measured at 31.9% and the stack temperature rise is 187.8°C.
- A. 80%
- B. 76%
- C. 78%
- D. 82%
- E. 84%
Answer: D
NEW QUESTION # 89
A 50-Hz alternating current (AC) induction motor has 3 pole pairs (6 poles). What is the synchronous speed of the motor?
- A. 750 rpm
- B. 1,000 rpm
- C. 600 rpm
- D. 1,500 rpm
Answer: D
NEW QUESTION # 90
A desk lamp is located inside a windowless office. The office is heated during the winter and cooled during the summer. When the lamp is switched on, how much of the energy consumed by the lamp becomes heat in the office space?
- A. 80%
- B. 25%
- C. 100%
- D. 120%
Answer: C
Explanation:
* All electrical energy consumed by the lamp is converted into heatwithin the space, either as direct heat or as waste heat from the light source.
* In a windowless office, no energy escapes as light.The light energy that is emitted eventually turns into heat through absorption by surfaces.
Thus, the correct answer isC. 100%.
NEW QUESTION # 91
A new lighting system costs $60,000. The new lighting system has a 15-year life and the minimum required rate of return (MARR) is 15%. How much must the new lighting system save to be life-cycle cost effective?
- A. $27,090/year
- B. $15,650/year
- C. $33,200/year
- D. $12,556/year
- E. $10,260/year
Answer: D
NEW QUESTION # 92
Using a degree-day base of 18°C, calculate the number of cooling degree days (CDD) if the outside temperature is a uniform 25°C throughout the year (365 days/year).
- A. 1,845 CDD/yr
- B. 1,955 CDD/yr
- C. 2,250 CDD/yr
- D. 2,555 CDD/yr
- E. 2,975 CDD/yr
Answer: E
NEW QUESTION # 93
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. Sensible heat
- B. No heat energy has been removed
- C. Latent heat
- D. Both sensible and latent heat
Answer: D
NEW QUESTION # 94
Which of the following steam turbines will likely not provide steam output at a useable temperature for industrial processes?
- A. Condensing turbine
- B. Extraction turbine
- C. Back-pressure turbine
- D. Answers A and B
Answer: A
NEW QUESTION # 95
A 50-kW induction motor (1,500 rpm synchronous speed) has a nameplate full-load rotational speed of 1,455 rpm. Field measurements show the actual rotational speed is 1,475 rpm. Using the slip method, calculate the partial load factor of the motor.
- A. 80%
- B. 98%
- C. 101%
- D. 46%
- E. 56%
Answer: A
NEW QUESTION # 96
Which problems can be caused by poor power quality?
- A. Answers B and C
- B. Equipment overheating
- C. All of the above
- D. Induction motors running backwards
- E. Circuit breakers tripping
Answer: A
NEW QUESTION # 97
A high-pressure steam system passes 180°C condensate (saturated liquid) through a steam trap to an atmospherically vented condensate receiver tank. If the steam trap is working properly, how much of the condensate mass can be lost out the vent of the receiver tank in the form of flash steam?
- A. 66%
- B. 6%
- C. 25%
- D. 15%
- E. 100%
Answer: C
NEW QUESTION # 98
An energy-saving project saves $30,000 per year. The project life is 10 years and the company minimum annual rate of return (MARR) is 15%. How much can the project cost and still be cost effective?
- A. $188,343
- B. $201,123
- C. $150,570
- D. $165,903
- E. $198,992
Answer: D
NEW QUESTION # 99
If installed without ancillary equipment, photovoltaic power cells generate what type of power?
- A. Direct current
- B. Sinusoidal alternating current
- C. Square-wave alternating current
Answer: A
NEW QUESTION # 100
A new energy-efficient boiler costs $60,890 to buy and install. The new energy-efficient boiler will save
$15,400 per year and last for 20 years. What is the internal rate of return (IRR) of the project?
- A. 20%
- B. 25%
- C. 10%
- D. 15%
- E. 30%
Answer: A
NEW QUESTION # 101
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. 4.0 years
- C. 5.0 years
- D. 3.0 years
Answer: B
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 # 102
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