Exam Cheat Sheet · Quick Reference
Oklahoma HVACR Contractor Unlimited
Oklahoma · PSI Services Contractor
Oklahoma State Portion 90 questions
Key Distinctions
A pictorial diagram shows all connections plus the approximate physical location of devices in the unit (and often wire colors), while a ladder diagram shows the order of component operation as horizontal rungs (individual circuits with loads and controls) between two vertical power lines, with devices near the top energized first.
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 heat passing through one square foot of material one inch thick in one hour per 1°F difference (Btu·in/hr·ft²·°F), while C-value is based on the material's actual measured thickness rather than a standardized 1-inch thickness.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 32 - Heating and Cooling Loads, Page 717 (20th ed.: Page 986)BRI is a specifically diagnosable illness from airborne agents whose symptoms do NOT disappear when the occupant leaves, whereas SBS has no specific cause or identifiable illness and its symptoms disappear when occupants leave the building.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 29 - Air Quality, Page 647 (20th ed.: Page 730)The subcooling method is used to check charge on systems with thermostatic expansion valves (low subcooling suggests undercharge, high suggests overcharge), while the superheat method is used on fixed-orifice systems such as capillary tubes.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 12 - Working with Refrigerants, Page 221 (20th ed.: Page 236)An in-line ammeter must be wired in series with the load (in dc, black lead to negative and red lead to positive before energizing), while a clamp-on ammeter clamps around only a single conductor — clamping both hot and neutral makes the opposing magnetic fields cancel and the meter reads zero.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 18 - Electrical Troubleshooting Fundamentals, Page 362 ("In-Line Ammeters") (20th ed.: Chapter 17, Section 17.1.3 — In-Line Ammeters)Turning the range screw clockwise raises both cut-in and cut-out an equal amount (differential unchanged), while the differential screw turned clockwise raises only the cut-out pressure and does not affect cut-in.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 17 - Electrical Control Systems, Page 333 (Section 17.4.4) (20th ed.: Chapter 16, Section 16.3.4 — Low-Pressure Motor Controls)In an HSI system the glow coil only ignites and cannot sense flame; a separate flame rod proves flame by rectification (a dirty/oxidized rod makes the module close the gas valve seconds after ignition), and thermocouples are used in standing-pilot systems, not HSI.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 33 - Gas-Fired Heating Systems, Page 765 (20th ed.: Pages 1126-1128 (Sections 41.5.4 and 41.6.1))Three-phase supply is already phase-split and its voltages peak at different times, so torque is produced without start windings or capacitors and power is essentially constant (never an instant when all voltages are zero), unlike single-phase, which needs phase-splitting components.
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)A current-limiting fuse uses temperature-sensitive resistors that open on overcurrent and close again when cooled (keeping current below rated locked rotor amperage), while a multipurpose fuse combines fast-acting and time-delay traits — riding through brief small overloads such as motor starting but blowing immediately above about 500% of rated current.
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 17 - Electrical Control Systems, Page 347 (Section 17.5.1) (20th ed.: Chapter 16, Section 16.4.1)The code covers natural gas piping systems operating at 125 psig or less, but LP-gas piping systems only at 20 psig (140 kPa gauge) or less, except as provided in Section 402.7.
IFGC 2024 Section 101.2.3 (2018 ed.: Section 101.2.2)Per the IECC, maximum leakage is 3 cfm per 100 ft² of conditioned space when tested before the air handler is installed, versus 4 cfm per 100 ft² when tested with the air handler installed (ducts pressurized to +25 Pa WRT the conditioned space).
Modern Refrigeration and Air Conditioning (22nd ed.), Chapter 31 - Ventilation System Service, Page 703 (Section 31.1.5) (20th ed.: Page 793)Per IMC 2024 §603.2, Manual D (or the manufacturer's instructions or other approved methods) is an accepted duct sizing method within a single dwelling unit, while Manual J is for load calculations and Manual S for equipment selection; other buildings use the ASHRAE Handbook of Fundamentals.
IMC 2024 Section 603.2Key Terms
Formulas to Know
R = E ÷ I (e.g., 240 V ÷ 10 A = 24 Ω); I = E ÷ R (e.g., 24 V ÷ 8 Ω = 3 A, the same current through every load in a series circuit)P = I × E (20 A × 120 V = 2400 W; 15 A × 240 V = 3,600 W; 8 A × 240 V = 1920 W)V = 4005 × √Pv, with Pv in inches of water (4005 × √0.25 = 2002.5 ≈ 2003 ft/min); for a swinging-vane anemometer (calibrated at 68°F) correct readings by velocity = instrument reading × √((460 + T) ÷ (460 + 68))ACH = (cfm × 60 min) ÷ room volume in ft³; ACH50 = (cfm50 × 60) ÷ conditioned volume (1,500 cfm × 60 ÷ 12,000 ft³ = 7.5)1 kW = 3,413 Btu/hr; kW required = Btu/hr ÷ 3,413 (50,000 ÷ 3,413 = 14.65 kW)COP = refrigeration effect ÷ heat of compression = (compressor-in h − evaporator-in h) ÷ (compressor-out h − compressor-in h) = (179 − 116) ÷ (198 − 179) = 63 ÷ 19 = 3.32:1Q = U × area × temperature difference; HTM = U × HTD (wall: 0.07 × 50 = 3.50 Btuh/ft², 136 ft² × 3.50 = 476 Btuh; glass: 0.30 × 50 = 15.00, 60 ft² × 15.00 = 900 Btuh; net wall = gross wall − window and door areas = 216 − 80 = 136 ft²; single glass example: 1.13 × 70°F TD × 12 ft² ≈ 949 Btu/hr)Slab-on-grade: perimeter length (ft) × 18 Btu/hr per linear foot (0°F outdoor design), added to above-ground losses; infiltration (heating, sensible only): Qs = 0.018 × CFH × TD (latent excluded for heating; both sensible and latent are used for cooling)TEL = greatest supply-run effective length + greatest return-run effective length (180 + 120 = 300 ft); Friction Rate = (ASP ÷ TEL) × 100 = (0.20 ÷ 250) × 100 = 0.08 in. WC per 100 ft; total equivalent length = straight duct + elbow allowances (28 + 22 + 15 = 65 ft straight; + 2 elbows × 25 ft = 115 ft)NS = (120 × frequency) ÷ number of poles; four-pole at 60 Hz = (120 × 60) ÷ 4 = 1800 RPM, and with 4%–5% slip actual full-load speed ≈ 1750 RPM