Free Certified Energy Manager Audit & Billing Questions and Answers — Questions and Answers
Question 1: Kilowatts are a unit of measurement for energy consumption, whereas kilowatt-hours are a unit of measurement for power consumption.
- TRUE
- FALSE (Correct answer)
Correct answer: FALSE
This statement is false. Kilowatts (kW) are a unit of power, which measures the rate at which energy is consumed or produced at any given moment. Kilowatt-hours (kWh) are a unit of energy, representing the total amount of energy consumed over a period of time (power multiplied by time). Therefore, the definitions in the statement are reversed.
Question 2: How many steam traps have failed in your steam system may be ascertained using a flue gas analyzer.
- True
- False (Correct answer)
Correct answer: False
This statement is false. A flue gas analyzer is used to measure the composition of combustion gases (e.g., oxygen, carbon monoxide, nitrogen oxides) exiting a boiler or furnace to assess combustion efficiency and emissions. It cannot detect failed steam traps, which are components in a steam distribution system. Failed steam traps are typically identified using methods like ultrasonic detection, infrared thermography, or visual inspection for temperature differences.
Question 3: Which device may be used to spot uninsulated steam lines:
- Psychrometer
- Tachymeter
- Bourdon Gauge
- Infrared Camera (Correct answer)
Correct answer: Infrared Camera
An infrared camera, also known as a thermal imager, detects and visualizes heat radiation. Uninsulated steam lines will emit significantly more heat than their surroundings, appearing as distinct hot spots in a thermal image. This allows for quick and non-invasive identification of areas with excessive heat loss, making it an ideal tool for energy audits and maintenance checks.
Question 4: How much more will you pay (for the following 11 months) if you had a 700kW additional spike (beyond usual demand) during the previous month if you pay $10 per kW per month and have an 80% demand ratchet?
- $0
- $6,160 per year
- $61,600 per year (Correct answer)
- $77,000 per year
Correct answer: $61,600 per year
A demand ratchet means that a percentage of the highest demand spike in a billing period will be charged for subsequent months. The additional spike was 700 kW. With an 80% demand ratchet, the additional demand charged is 0.80 * 700 kW = 560 kW. This charge applies for the following 11 months. So, the additional cost is 560 kW * $10/kW/month * 11 months = $61,600 per year.
Question 5: A system of electricity has 52.9 kVA and 50.5 kW. To get the power factor of the entire load to 95%, how many kVARs of capacitance are needed?
- 0 kVARs (Correct answer)
- 10.8 kVARs
- 35 kVARs
- 20 kVARs
- 75 kVARs
Correct answer: 0 kVARs
First, calculate the initial power factor (PF) using the given real power (kW) and apparent power (kVA): PF = kW / kVA = 50.5 kW / 52.9 kVA = 0.9546, or 95.46%. The target power factor is 95%. Since the existing power factor (95.46%) is already higher than the target power factor (95%), no additional kVARs of capacitance are needed. Adding capacitance would overcorrect the power factor.
Question 6: Take into account the inverted block rate structure shown below for a monthly bill: <br> Assume that a facility utilizes 200 kW and 86,400 kWh in total for all energy loads over the course of a month. If the lighting system is upgraded to use 25 kW less lighting power and is used for 300 hours a month during the busiest time of the day, how much money would they save each month?
- $675
- $600
- $925 (Correct answer)
- $850
- $1,150
Correct answer: $925
The total monthly savings are the sum of demand and energy savings. Energy savings are calculated as 25 kW * 300 hours/month = 7,500 kWh. This reduction falls within the 50,001-100,000 kWh block, priced at $0.07/kWh, resulting in $525 in energy savings. For demand, a 25 kW reduction from the 200 kW total, considering the block rate structure, leads to $400 in demand savings. Summing these, $525 (energy) + $400 (demand) equals $925 in total monthly savings.
Question 7: The power factor is high when an electric AC induction motor is 15% loaded.
- True
- False (Correct answer)
Correct answer: False
This statement is false. AC induction motors are designed to operate with a high power factor when they are loaded near their rated capacity. When an induction motor is lightly loaded, such as at 15% of its full load, the reactive power component (magnetizing current) remains relatively high while the real power component significantly decreases. This imbalance leads to a much lower power factor, indicating inefficient use of electrical power.
Free Certified Energy Manager Audit & Billing Questions and Answers
Kilowatts are a unit of measurement for energy consumption, whereas kilowatt-hours are a unit of measurement for power consumption.