FREE Certified Energy Auditor Certification Trivia Questions and Answers
There are 50 employees working in an office building, which is occupied 250 days a year. Each employee is thought to use an average of 1 gallon of hot water per day for hand washing. City water is 60°F while the required hot water temperature is 120°F. Determine the annual energy consumption for heating this water:
Explanation:
To calculate the annual energy required to heat the water, we can use the following formula:
Energy (MMBtu/year) = (Volume x Density x Specific Heat x Temperature Rise) / (Efficiency x 1,000,000)
where:
Volume: the volume of water used per year, in gallons
Density: the density of water, in pounds per gallon (8.33 lb/gal)
Specific Heat: the specific heat of water, in Btu/lb°F (1 Btu/lb°F)
Temperature Rise: the temperature rise required to heat the water, in °F (120°F - 60°F = 60°F)
Efficiency: the efficiency of the water heater (we will assume 90% for this calculation)
Plugging in the values, we get:
Volume = 50 employees x 1 gallon/employee/day x 250 days/year = 12,500 gallons/year
Energy = (12,500 gallons/year x 8.33 lb/gal x 1 Btu/lb°F x 60°F) / (0.9 x 1,000,000) = 6.25 MMBtu/year
Therefore, the amount of annual energy required to heat this water is 6.25 MMBtu.
13 amps at 240 volts are required to run a three-phase induction motor. There is a 0.9 power factor. Identify the kW.
Explanation:
To calculate the kW (kilowatt) for a three-phase induction motor, we can use the following formula:
kW = sqrt(3) x V x I x PF / 1000
where:
sqrt(3) is the square root of 3 (which is approximately 1.732)
V is the voltage, in volts
I is currently, in amps
PF is the power factor (which is given as 0.9 in this case)
1000 is a conversion factor to convert watts to kilowatts
Plugging in the values, we get:
kW = 1.732 x 240 volts x 13 amps x 0.9 / 1000 = 4.86 kW
Therefore, the kW for this three-phase induction motor is 4.86.
Ground source heat pumps may not achieve predicted levels of efficiency because of
Explanation:
Ground source heat pumps (GSHPs) can be highly efficient heating and cooling systems, but their performance can be affected by many factors, including the design and installation of the ground loop. The ground loop is the underground piping system that circulates the heat exchange fluid between the heat pump and the ground, and it is a critical component of the GSHP system.
Every year, a hospital uses 400,000 gallons of water for showers alone. The showers have 4.5 gpm, older-style showerheads. What is the annual water savings if they switch to new showerheads that adhere to the maximum flow rates permitted by the Energy Policy Act of 1992?
Explanation:
The Energy Policy Act of 1992 mandates maximum flow rates for showerheads at 2.5 gallons per minute (gpm). If the hospital replaces the old showerheads that use 4.5 gpm with new ones that comply with the maximum flow rates, the annual amount of water saved can be calculated as follows:
Annual water usage with old showerheads = 400,000 gallons/year
Annual water usage with new showerheads = (400,000 gallons/year) x (2.5 gpm / 4.5 gpm) = 222,222 gallons/year
Water saved per year = Annual water usage with old showerheads - Annual water usage with new showerheads
= 400,000 gallons/year - 222,222 gallons/year
= 177,778 gallons/year
Therefore, replacing the old showerheads with new ones that comply with the Energy Policy Act of 1992 would save 177,778 gallons of water per year.
The phrase "lumens per watt" refers to lighting efficiency.
Explanation:
True. The lighting efficiency term is lumens per watt (lm/W), which measures the amount of light (in lumens) produced by a light source per unit of power (in watts) consumed. It is a measure of how effectively a light source converts electrical energy into visible light. The higher the lumens per watt, the more efficient the light source.
The minimal requirements are specified in ASHRAE standard 62-2010 for:
Explanation:
ASHRAE Standard 62-2010, "Ventilation for Acceptable Indoor Air Quality," establishes the minimum requirements for ventilation and indoor air quality in commercial and institutional buildings.
The standard defines the acceptable levels of indoor air quality for various pollutants, including carbon dioxide, carbon monoxide, volatile organic compounds (VOCs), and particulate matter. It also establishes the minimum ventilation rates required to maintain these acceptable levels of indoor air quality.
The full load rating of a chiller is 0.7 kW/ton. Given that this unit has a 200 ton rating, what is the full load kW?
Explanation:
The full load rating of the chiller is given as 0.7 kW/ton, which means that for every ton of cooling capacity, the chiller consumes 0.7 kW of power at full load.
To calculate the full load kW for a 200-ton chiller, we can use the following formula:
Full load kW = Tons x kW/ton
Substituting the values given in the problem, we get:
Full load kW = 200 tons x 0.7 kW/ton = 140 kW
Therefore, the full load kW for a 200-ton chiller is 140 kW.