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Formal CEM Test - Test CEM Simulator Online

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AEE Certified Energy Manager (CEM) Sample Questions (Q119-Q124):

NEW QUESTION # 119
An energy project implemented today saves $10,000 per year for the next 20 years. Using the 5% interest table, determine the total value of savings at the end of the 20th year?

A. $496,991
B. $124,620
C. $330,660
D. $146,907
E. $200,000

Answer: A

NEW QUESTION # 120
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. The project has a life of 10 years with no salvage value. The minimum attractive rate of return (MARR) is 10%. Using end-of-year cash flows, calculate the net- present value (NPV) of the project.

A. $311,556
B. $289,575
C. $456,254
D. $154,657

Answer: B

Explanation:
A screenshot of a math formula AI-generated content may be incorrect.

NEW QUESTION # 121
In a typical compressed air system, energy can be saved by which of the following actions?

A. Raising the air pressure
B. Reducing the air distribution pipe size
C. Move the inlet air filter to a warmer location
D. Reducing the capacity of the compressed air receiver tanks
E. Decreasing supply air pressure

Answer: E

NEW QUESTION # 122
How can you tell if steam traps in a steam distribution system may NOT be working correctly?

A. No sound from inverted bucket steam traps
B. Robust flash steam rising from condensate receiver tank vent pipes
C. An increase in make-up water requirements relative to steam demand
D. Rapid chattering of disk steam traps
E. All of the above

Answer: E

NEW QUESTION # 123
A 75-kW induction motor operates at 80% partial load and 88% efficiency. A new high-efficiency 75-kW motor can be installed that will operate at 95% efficiency (also at 80% partial load). What is the reduction in electric power demand?

A. 8 kW
B. 2 kW
C. 1 kW
D. 4 kW
E. 5 kW

Answer: D

NEW QUESTION # 124
......

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