Certified Energy Auditor Certification Certified Energy Auditor HVAC Systems and Performance 1 — Questions and Answers
Question 1: A heat pump in heating mode is measured to have a coefficient of performance (COP) of 3.2. What does this value mean in practical terms?
- The system produces 3.2 BTUs of heat for every BTU of electrical energy consumed (Correct answer)
- The system converts 3.2 kW of heat into 1 kW of electrical power
- The system loses 3.2 units of energy for every unit of useful heat delivered
- The system's seasonal efficiency rating is 3.2 times the SEER baseline
Correct answer: The system produces 3.2 BTUs of heat for every BTU of electrical energy consumed
COP is the ratio of useful heat output to electrical energy input. A COP of 3.2 means the heat pump delivers 3.2 units of heat energy for every 1 unit of electrical energy consumed, making it far more efficient than resistance heating (COP = 1.0).
Question 2: An energy auditor inspects a commercial air handling unit with an enthalpy-based economizer. Under which condition should the economizer damper be commanded to 100% open?
- Outdoor air dry-bulb temperature falls below 55°F regardless of humidity
- Outdoor air enthalpy is lower than return air enthalpy and mechanical cooling is active (Correct answer)
- CO2 concentration in the return air stream exceeds 1,000 ppm
- Supply air temperature rises above the cooling setpoint by more than 2°F
Correct answer: Outdoor air enthalpy is lower than return air enthalpy and mechanical cooling is active
An enthalpy-based economizer compares the total heat content (sensible + latent) of outdoor air versus return air. When outdoor enthalpy is lower, bringing in 100% outdoor air provides free cooling and reduces compressor runtime, saving energy.
Question 3: Applying the fan affinity laws, if a variable frequency drive (VFD) reduces a centrifugal fan's motor speed from 1,200 RPM to 900 RPM, the resulting power consumption will be approximately what fraction of the original?
- 75% of original power (linear with speed)
- 56% of original power (varies as the square of speed ratio)
- 42% of original power (varies as the cube of speed ratio) (Correct answer)
- 90% of original power (accounts for motor slip losses only)
Correct answer: 42% of original power (varies as the cube of speed ratio)
The fan affinity laws state that power varies as the cube of the speed ratio. (900/1200)³ = (0.75)³ = 0.422, so power drops to approximately 42% of original — a dramatic saving that makes VFDs one of the highest-ROI energy efficiency measures in HVAC.
Question 4: During an energy audit of a large commercial building, an auditor wants to quantify duct leakage to the outside. Which test method directly measures this?
- Pitot tube traverse across the main supply trunk to calculate total airflow
- Duct pressurization test using a calibrated fan to pressurize the duct system while supply and return registers are sealed (Correct answer)
- Blower door depressurization of the entire building envelope at 50 Pascals
- Infrared thermographic scan of supply duct surfaces during peak cooling operation
Correct answer: Duct pressurization test using a calibrated fan to pressurize the duct system while supply and return registers are sealed
A duct pressurization test (using a 'duct blaster') seals all registers and pressurizes the duct to a reference pressure (typically 25 Pa). The airflow required to maintain that pressure equals the duct leakage rate, expressed in CFM25. This directly quantifies distribution losses.
Question 5: A central chiller plant consumes 480 kW of electrical power while producing 1,200 tons of cooling. What is the plant's efficiency expressed in the metric most commonly used by energy auditors for chiller evaluation?
- 0.40 kW/ton, indicating a highly efficient plant
- 2.50 tons/kW, which is equivalent to a COP of 8.8
- 0.40 kW/ton and 2.50 tons/kW are both correct representations of the same efficiency (Correct answer)
- 1,200 EER, derived from multiplying tons by the input wattage
Correct answer: 0.40 kW/ton and 2.50 tons/kW are both correct representations of the same efficiency
Chiller efficiency is expressed as kW/ton (lower is better) or its reciprocal tons/kW (higher is better). Here: 480 kW ÷ 1,200 tons = 0.40 kW/ton, and 1,200 tons ÷ 480 kW = 2.50 tons/kW. Both values describe the same operating point and auditors may use either.
Question 6: When verifying refrigerant charge on a split-system air conditioner equipped with a fixed-orifice metering device, which combination of measurements does an auditor use as the primary diagnostic?
- Suction line pressure alone compared to manufacturer's pressure-temperature chart
- Superheat at the evaporator outlet and subcooling at the condenser outlet (Correct answer)
- Compressor amperage draw compared to nameplate RLA under current outdoor conditions
- Supply-to-return temperature split measured at the air handler grilles
Correct answer: Superheat at the evaporator outlet and subcooling at the condenser outlet
For fixed-orifice systems, the correct refrigerant charge produces specific superheat at the evaporator exit (ensuring no liquid flood-back) and specific subcooling at the condenser exit (ensuring full liquid to the metering device). Measuring both confirms charge accuracy under actual operating conditions.
Certified Energy Auditor Certification Certified Energy Auditor HVAC Systems and Performance 1
A heat pump in heating mode is measured to have a coefficient of performance (COP) of 3.2.
What does this value mean in practical terms?