Free Certified Energy Manager Motors and HVAC Questions and Answers — Questions and Answers
Question 1: The US industries and structures with the highest installation rates of electric motors are:
- Direct Generation
- Cogeneration
- AC Induction (Correct answer)
- DC
- AC Synchronous
Correct answer: AC Induction
AC induction motors are the most widely installed type of electric motor in industrial and commercial applications globally. Their popularity stems from their robust construction, reliability, relatively low cost, and ease of maintenance. They are extensively used in pumps, fans, compressors, conveyors, and various other machinery across almost all sectors, making them ubiquitous in modern industries and structures.
Question 2: Pick the most accurate estimate of electric motor efficiency (in percent) when motors are appropriately loaded in the absence of additional data:
- 100
- 90 (Correct answer)
- 33
- 50
- 72
Correct answer: 90
When electric motors are appropriately sized and loaded, especially modern high-efficiency or premium-efficiency AC induction motors, their efficiency typically falls within the range of 85% to 95%. In the absence of specific data, 90% serves as a reasonable and commonly accepted estimate for a well-operating motor under optimal load conditions. This value reflects the significant improvements in motor technology aimed at reducing energy losses.
Question 3: What are HP savings if a motor has a Variable Speed Drive (VSD) fitted that slows down a 200 HP motor by 20%?
- 160 HP
- 102 HP
- 40 HP
- 97.6 HP (Correct answer)
- None of the above
Correct answer: 97.6 HP
For centrifugal loads like pumps and fans, power consumption is proportional to the cube of the speed (affinity laws). If the motor speed is reduced by 20%, the new speed is 80% (0.8) of the original. The new power consumed will be (0.8)^3 * 200 HP = 0.512 * 200 HP = 102.4 HP. Therefore, the HP savings are 200 HP (original) - 102.4 HP (new) = 97.6 HP.
Question 4: There are roughly how many cooling degree days in a calendar year if the outside temperature stays constant at 75°F:
- 1,825 CDD
- 8,760 CDD
- 0 CDD
- 3,650 CDD (Correct answer)
Correct answer: 3,650 CDD
Cooling Degree Days (CDD) are calculated as the difference between the average daily temperature and a base temperature, typically 65°F. If the outside temperature remains constant at 75°F, each day contributes (75°F - 65°F) = 10 CDD. Over a calendar year of 365 days, the total CDD would be 10 CDD/day * 365 days = 3,650 CDD.
Question 5: The COP of a 250-ton chiller is 4.2. What is the chiller's load (in kW) while it is operating at full capacity?
- 841 kW
- Cannot be determined
- 1050 kW
- 209 kW (Correct answer)
Correct answer: 209 kW
First, convert the chiller's cooling capacity from tons to kilowatts: 250 tons * 3.517 kW/ton = 879.25 kW. The Coefficient of Performance (COP) is defined as the cooling output divided by the electrical input (COP = Output/Input). Rearranging for input, Electrical Input (kW) = Cooling Capacity (kW) / COP. Therefore, the chiller's electrical load is 879.25 kW / 4.2 = 209.345 kW, which rounds to 209 kW.
Question 6: At a flow rate of 6750 cubic feet per minute, air with a relative humidity of 50% and a dry bulb temperature of 69°F is heated to 90°F. How much BTU per hour is needed for this process? Don't consider duct losses.
- 153 kBtu/year (Correct answer)
- 73,857 Btu/year
- 637,875 Btu/year
- Cannot be determined
Correct answer: 153 kBtu/year
To calculate the sensible heat needed, use the formula Q = Flow Rate (CFM) * Specific Heat of Air * Delta T * 60 min/hr. The specific heat of air is approximately 0.018 Btu/(ft³·°F). So, Q = 6750 CFM * 0.018 Btu/(ft³·°F) * (90°F - 69°F) * 60 min/hr. This calculates to 6750 * 0.018 * 21 * 60 = 153,090 Btu/hour. Converting to kBtu/hour (assuming the 'year' in the answer is a typo and should be 'hour'), this is 153.09 kBtu/hour, which rounds to 153 kBtu/hour.
Question 7: How much can you afford to spend for the waste heat exchanger (total installation cost) if your company's MARR is 30% and it saves $1,000,000 year and lasts for 7 years?
- $2,802,100 (Correct answer)
- $3,300,000
- $8,172,200
- $356,900
- None of the above
Correct answer: $2,802,100
To determine the affordable total installation cost, calculate the Present Worth Factor (PWF) for an annuity. The formula for PWF is [ (1 + i)^n - 1 ] / [ i * (1 + i)^n ], where i is the MARR (0.30) and n is the number of years (7). Plugging in the values, PWF = [ (1.30)^7 - 1 ] / [ 0.30 * (1.30)^7 ] = 2.8021. The affordable cost is then the annual savings multiplied by the PWF: $1,000,000/year * 2.8021 = $2,802,100.
Free Certified Energy Manager Motors and HVAC Questions and Answers
The US industries and structures with the highest installation rates of electric motors are: