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
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.
Kilowatts are a unit of measurement for energy consumption, whereas kilowatt-hours are a unit of measurement for power consumption.
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).
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 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.