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CERTIFIED ENERGY MANAGER (CEM)Verified Exam
Actual Test Questions and Correct Answers With
Rationales LATEST THIS YEAR||NEWEST EXAM!!
An energy saving device will save $25,000 per year for 8
years. How much can a company pay for this device if the
interest rate (discount rate) is 15%?
A. $10,000
B. $77,000
C. $112,000
D. $173,000 - Answer-C. $112,000
P = present value
A = single payment or interest received in a series of n
equal annual payments
i = Annual interest rate = MARR
n = the number of annual interest periods
P = A * [(1+i)^n - 1 / i(1+i)^n ]
P = 25,000 * [(1+0.15)^8 - .15(1+0.15)^8]
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What would be used to find hot spots or phase imbalances
in an AC circuit?
A. Ohmmeter
B. Infrared Camera
C. Wattmeter
D. All of the above - Answer-Answer:
B. Infrared Camera
Descriptions:
Ohmmeter: an instrument for measuring electrical
resistance.
Wattmeter: a meter for measuring electric power in watts.
An audit for one firm showed that the power factor is
almost always 70% and that the demand is 1000kw. What
capacitor size is needed to correct power factor to 90%?
A. 266 kVAR
B. 536 kVAR
C. 618 kVAR
D. 1000 kVAR - Answer-B. 536 kVAR
Look up tables
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Existing power factor = 0.7
Target power factor = 0.9
table value=0.536
kVAR=kW*table value=1000 kW * 0.536 = 536 kVAR
The amount of reactive power that must be supplied by
capacitors to correct a power factor of 84% to 95% in a
400 HP motor at 75% load and 98% efficiency is
A 72.4 kVAR
B. 82.5 kVAR
C. 90.04 kVAR
D. 93.4 kVAR
E. 123.5 kVAR - Answer-A 72.4 kVAR
Power Factor Chart: 0.84, 0.95 = 0.317
(400HP * 0.75)/0.98 = 306.12 HP
306.12 HP * 0.7457 kW/HP = 228.27 kW
kVAR= 228.27 kW * 0.317 = 72.4 kVAR
Power factor correcting capacitors may be located
A. At the inductive load
CERTIFIED ENERGY MANAGER (CEM)Verified Exam
Actual Test Questions and Correct Answers With
Rationales LATEST THIS YEAR||NEWEST EXAM!!
An energy saving device will save $25,000 per year for 8
years. How much can a company pay for this device if the
interest rate (discount rate) is 15%?
A. $10,000
B. $77,000
C. $112,000
D. $173,000 - Answer-C. $112,000
P = present value
A = single payment or interest received in a series of n
equal annual payments
i = Annual interest rate = MARR
n = the number of annual interest periods
P = A * [(1+i)^n - 1 / i(1+i)^n ]
P = 25,000 * [(1+0.15)^8 - .15(1+0.15)^8]
,2|Page
What would be used to find hot spots or phase imbalances
in an AC circuit?
A. Ohmmeter
B. Infrared Camera
C. Wattmeter
D. All of the above - Answer-Answer:
B. Infrared Camera
Descriptions:
Ohmmeter: an instrument for measuring electrical
resistance.
Wattmeter: a meter for measuring electric power in watts.
An audit for one firm showed that the power factor is
almost always 70% and that the demand is 1000kw. What
capacitor size is needed to correct power factor to 90%?
A. 266 kVAR
B. 536 kVAR
C. 618 kVAR
D. 1000 kVAR - Answer-B. 536 kVAR
Look up tables
, 3|Page
Existing power factor = 0.7
Target power factor = 0.9
table value=0.536
kVAR=kW*table value=1000 kW * 0.536 = 536 kVAR
The amount of reactive power that must be supplied by
capacitors to correct a power factor of 84% to 95% in a
400 HP motor at 75% load and 98% efficiency is
A 72.4 kVAR
B. 82.5 kVAR
C. 90.04 kVAR
D. 93.4 kVAR
E. 123.5 kVAR - Answer-A 72.4 kVAR
Power Factor Chart: 0.84, 0.95 = 0.317
(400HP * 0.75)/0.98 = 306.12 HP
306.12 HP * 0.7457 kW/HP = 228.27 kW
kVAR= 228.27 kW * 0.317 = 72.4 kVAR
Power factor correcting capacitors may be located
A. At the inductive load
