Free Certified Energy Manager Motors and HVAC Questions and Answers

Question 1

The US industries and structures with the highest installation rates of electric motors are:

Accepted Answer

AC Induction

Answer

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:

Accepted Answer

90

Answer

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%?

Accepted Answer

97.6 HP

Answer

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:

Accepted Answer

3,650 CDD

Answer

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?

Accepted Answer

209 kW

Answer

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.

CEM - Certified Energy Manager Practice Test

CEM - Certified Energy Manager Practice Test

Free Certified Energy Manager Motors and HVAC Questions and Answers