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New Jersey Master HVACR Contractor Exam
New Jersey · PSI Services Contractor
New Jersey State Portion 50 questions
Key Distinctions
A pictorial diagram shows all device connections plus the approximate physical location of devices in the unit (and may show wire colors), while a ladder diagram shows the order of component operation using horizontal rungs between two vertical power lines.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 18 - Electrical Troubleshooting Fundamentals, Page 369 (Section 18.2.2) (20th ed.: Chapter 16, Section 16.1.1)K-value is based on a standardized one-inch thickness of material, while C-value is based on the material's actual measured thickness.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 32 - Heating and Cooling Loads, Page 717 (20th ed.: Page 986)ACR tubing size is designated by its actual outside diameter, while copper water tubing nominal size equals the outside diameter minus 1/8″.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 5 - Working with Tubing and Piping, Page 71 (Section 5.1.1) (20th ed.: Chapter 8, Section 8.1.1)A hard lockout shuts the system down for an unspecified time after a failed trial for ignition and requires a service call or power interruption to reset, while a soft lockout automatically retries after a waiting period.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 33 - Gas-Fired Heating Systems, Page 770 (20th ed.: Page 1129)Category III is positive-pressure, noncondensing and typically vented with stainless steel pipe, while Category IV is positive-pressure, condensing (flue gas below 140°F/60°C, below its dew point) with a secondary heat exchanger and PVC vent piping.
Modern Refrigeration and Air Conditioning (22nd ed.), Page 776 (Section 33.8)The coiled water tubing of a shell-and-coil condenser cannot be accessed for mechanical cleaning and must be cleaned with chemicals, while a shell-and-tube condenser has removable end caps that allow mechanical cleaning of its straight water tubes.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 44 - Commercial Evaporators, Condensers, and Other Heat Exchangers, Page 1069 (Section 44.2.2) (20th ed.: Page 537)The subcooling method is generally used to check the charge on systems with thermostatic expansion valves, while the superheat method is used on systems with fixed orifice metering devices such as capillary tubes.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 12 - Working with Refrigerants, Page 221 (20th ed.: Page 236)HSI systems require a separate flame rod to prove flame by flame rectification because the glow coil only ignites and cannot sense, while DSI systems can use the high-voltage electrode for both ignition and sensing.
Modern Refrigeration and Air Conditioning (22nd ed.), Pages 765-766 (Section 33.5.4)Per the IECC, maximum leakage is 3 cfm per 100 ft² of conditioned space when tested before the air handler is installed and 4 cfm per 100 ft² when tested with the air handler installed.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 31 - Ventilation System Service, Page 703 (Section 31.1.5) (20th ed.: Page 793)Per IFGC 2021 Section 305.5, appliances in private garages require a minimum clearance of 6 feet above the floor while public garages require 8 feet; an exception applies where the appliance is protected from motor vehicle impact and installed per Section 305.3.
IFGC 2021 Section 305.5A three-phase supply is already phase-split (each phase peaks at a different time), so the motor develops starting torque without start windings or capacitors, which single-phase motors need for phase splitting.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 16 - Electric Motors, Page 308 (Section 16.2.2) (20th ed.: Chapter 15, Section 15.2.2)Heating load calculations consider only the sensible heat loss from infiltration, while cooling load calculations must include both sensible and latent heat gains from infiltration.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 32 - Heating and Cooling Loads, Page 733 (20th ed.: Page 1004)Low-temperature evaporators have fins spaced further apart than medium- and high-temperature evaporators so air can still flow around the fins when some frost has built up.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 44 - Commercial Evaporators, Condensers, and Other Heat Exchangers, Page 1050 (20th ed.: Page 521)Apparent power is calculated as I × E, while true power is the wattmeter reading; their ratio ×100 is the power factor, which falls below 100% when inductive or capacitive components shift voltage and current out of phase.
Modern Refrigeration and Air Conditioning (22nd ed.), Page 275 (Section 14.1.3)Key Terms
Formulas to Know
I = E ÷ R (also Amperes = Watts ÷ Volts); R = E ÷ I; E = I × R. Examples: 1500 W ÷ 120 V = 12.5 A; 240 V ÷ 8 A = 30 Ω; 240 V ÷ 10 A = 24 Ω; 10 A × 12 Ω = 120 V.P = E × I (volts × amperes); when voltage and resistance are known, P = E² ÷ R. Examples: 20 A × 120 V = 2400 W; 15 A × 240 V = 3600 W; (240 V)² ÷ 20 Ω = 2880 W. Rearranged for current: I = P ÷ E (3600 W ÷ 240 V = 15 A; 3600 W ÷ 120 V = 30 A).Apparent power = I × E (VA); true power = wattmeter reading; Power factor = (true power ÷ apparent power) × 100. Example: (2880 W ÷ 3600 VA) × 100 = 80%.Total current equals the current in any other part of the circuit (IT = I1 = I2 = I3); total resistance is the sum of the resistances; I = E ÷ RT; voltage drop across a component = I × R and the individual drops sum to the applied voltage. Examples: 24 V ÷ 8 Ω = 3 A; (0.1+0.1+0.3+23.5) = 24 Ω, 120 V ÷ 24 Ω = 5 A, 5 A × 23.5 Ω = 117.5 V.kW = Btu/hr ÷ 3,413. Example: 50,000 ÷ 3,413 = 14.65 kW minimum capacity.COP = refrigeration effect ÷ heat of compression. Example: (179 − 116) ÷ (198 − 179) = 63 ÷ 19 = 3.32:1.ACH = (airflow in cfm × 60 min) ÷ volume of room in cubic feet.Latent ventilation load = 0.68 × ACF × CFMvent × grains difference (example: 0.68 × 1.0 × 50 × 28 = 952 Btuh); sensible ventilation load is a separate line item = 1.1 × ACF × CFMvent × CTD (example: 1.1 × 1.0 × 50 × 18 = 990 Btuh).Envelope infiltration Cfm = (ACH × above-grade volume) ÷ 60; add the fireplace adjustment Cfm; load = 1.1 × ACF × total Cfm × HTD. Example: (0.40 × 16,000)/60 = 106.7 + 12 = 118.7 Cfm; 1.1 × 1.0 × 118.7 × 60 ≈ 7,832 Btuh.Wall heating HTM = U × HTD and load = HTM × NET area (gross panel area − window and door area). Example: 0.090 × 55 = 4.95 Btuh/SqFt × (400 − 60 − 20) = 1,584 Btuh. Shade line Z = overhang length × shade-line multiplier; shaded glass height S = Z − Y (example: 2.0 × 1.6 = 3.2; 3.2 − 1.0 = 2.2 ft).