FREE Certified Energy Manager Audit & Billing Questions and Answers
Take into account the inverted block rate structure shown below for a monthly bill:
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?
This problem provides a lot of extra information (much like the “real world”). The dollar savings has two components: Demand and kWh Savings:
Demand Savings:
= (25 kW)($10/kW per month)
= $250 per month
kWh Savings (Note the savings would be in the 2nd Tier, so the price would be $.09/kWh):
= (25 kW)(300 hours)($0.09/kWh)
= $675 per month
Total Savings:
$250 + $675
=$925 per month
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?
How many steam traps have failed in your steam system may be ascertained using a flue gas analyzer.
The effectiveness of combustion can be determined with a flue gas analyzer, but the amount of steam produced or really any other information about what's going farther down the steam system cannot.
Which device may be used to spot uninsulated steam lines:
You can locate uninsulated surfaces with the use of an infrared camera (if they are hot). RPM is measured with a tachymeter. A psychometer is used to assess humidity. To gauge static pressure, use a Bourdon gauge (usually in a tank).
Kilowatts are a unit of measurement for energy consumption, whereas kilowatt-hours are a unit of measurement for power consumption.
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?
Ratchet penalties are effective measures to encourage energy consumers to reduce kW spikes because if you have a ratchet penalty, you will pay the ratchet percent for the next 11 months (even if you don’t use any demand during that period). In this case, if you set a new peak 700kW above your normal operating load (old peak), you will pay:
= (700 kW)(0.8)(11 months)($10/month*kW)
= $61,600 in extra charges (over the next 11 months)
The power factor is high when an electric AC induction motor is 15% loaded.
If the load is less than 20% for this motor type, the power factor and resultant efficiency suffer significantly.