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1.3 HVAC Fundamentals & Math Review

Key Takeaways

  • Temperature conversions between Fahrenheit and Celsius require standard linear formulas, while thermodynamic gas law calculations require absolute temperature scales: Rankine (°R = °F + 460) and Kelvin (K = °C + 273).

  • Pressure is expressed as gauge pressure (psig), absolute pressure (psia = psig + 14.7 at sea level), or inches of water column (in. w.c., where 1 psi = 27.7 in. w.c. and 1 in. Hg = 13.6 in. w.c.).

  • Ohm's law (E = I × R) and the electric power formula (P = E × I) govern all electrical troubleshooting and circuit analysis on the Core exam.

  • The sensible heat formula (Q = 1.08 × CFM × ΔT) and total heat formula (Q = 4.5 × CFM × Δh) represent the foundational equations for evaluating airflow and thermal capacity in air distribution systems.

  • Converting between units (such as Watts to Btu/hr via the multiplier 3.412) enables field verification of electric furnace heating output and system operating CFM.

Last updated: September 2026

1.3 HVAC Fundamentals & Math Review

Temperature Scales and Absolute Temperature Conversions

Temperature is the measurement of the average molecular kinetic energy of a substance. In HVAC/R work, technicians must routinely convert between temperature scales and understand when absolute temperature scales are mandatory.

The Four Standard Scales

  1. Fahrenheit (°F): The primary imperial scale used in North American HVAC/R. At standard sea-level atmospheric pressure (14.696 psia), pure water freezes at 32∘F32^\circ\text{F} and boils at 212∘F212^\circ\text{F} (a span of 180 degrees).
  2. Celsius (°C): The metric scale. Pure water freezes at 0∘C0^\circ\text{C} and boils at 100∘C100^\circ\text{C} (a span of 100 degrees). The ratio between Fahrenheit and Celsius degree increments is 180100=95=1.8\frac{180}{100} = \frac{9}{5} = 1.8.
  3. Rankine (°R): The imperial absolute temperature scale. Absolute zero—the theoretical point where all molecular motion ceases—is 0∘R0^\circ\text{R}, corresponding to −459.67∘F-459.67^\circ\text{F}. In field calculations, this is rounded to 460460.
  4. Kelvin (K): The metric absolute temperature scale. Absolute zero is 0 K0\text{ K}, corresponding to −273.15∘C-273.15^\circ\text{C} (rounded to 273273).

Conversion Formulas

∘F=(∘C×1.8)+32or∘F=(∘C×95)+32\mathbf{^\circ F = (^\circ C \times 1.8) + 32} \quad \text{or} \quad ^\circ\text{F} = \left(^\circ\text{C} \times \frac{9}{5}\right) + 32 ∘C=∘F−321.8or∘C=(∘F−32)×59\mathbf{^\circ C = \frac{^\circ F - 32}{1.8}} \quad \text{or} \quad ^\circ\text{C} = (^\circ\text{F} - 32) \times \frac{5}{9} ∘R=∘F+460(exact: 459.67)\mathbf{^\circ R = ^\circ F + 460} \quad (\text{exact: } 459.67) K=∘C+273(exact: 273.15)\mathbf{K = ^\circ C + 273} \quad (\text{exact: } 273.15)

CRITICAL EXAM RULE: Whenever you apply the ideal gas laws (Boyle's Law, Charles's Law, Gay-Lussac's Law, or P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}), temperature MUST ALWAYS be converted to an absolute scale (Rankine in imperial units, Kelvin in metric units). Inserting Fahrenheit or Celsius directly into gas law equations results in fatal mathematical errors.

Worked Example: Temperature Conversion to Absolute

A cylinder of nitrogen gas sits in an unconditioned service van overnight at 40∘F40^\circ\text{F}. During the afternoon, the van temperature rises to 95∘F95^\circ\text{F}. What are the initial and final temperatures in degrees Rankine?

  • Initial Temperature: T1=40∘F+460=500∘RT_1 = 40^\circ\text{F} + 460 = 500^\circ\text{R}
  • Final Temperature: T2=95∘F+460=555∘RT_2 = 95^\circ\text{F} + 460 = 555^\circ\text{R}

Pressure Fundamentals: Atmospheric, Gauge, Absolute, and Water Column

Pressure is defined as force per unit area:

P=ForceArea(e.g., pounds per square inch, or psi)P = \frac{\text{Force}}{\text{Area}} \quad \left(\text{e.g., pounds per square inch, or psi}\right)

1. Atmospheric Pressure

The earth's atmosphere exerts pressure due to the weight of air molecules. At sea level under standard conditions (59∘F/15∘C59^\circ\text{F} / 15^\circ\text{C}), standard atmospheric pressure equals:

  • 14.696 psia14.696\text{ psia} (commonly rounded to 14.7 psia14.7\text{ psia})
  • 29.92 in. Hg29.92\text{ in. Hg} (inches of mercury)
  • 760 mm Hg760\text{ mm Hg} (millimeters of mercury or Torr)
  • 101.325 kPa101.325\text{ kPa} (kilopascals)
  • 407 in. w.c.407\text{ in. w.c.} (inches of water column)

As altitude increases, atmospheric pressure decreases because there is less atmospheric air mass above the measurement point. For example, at Denver, Colorado (~5,280 feet), atmospheric pressure drops to approximately 12.1 psia12.1\text{ psia} (24.6 in. Hg24.6\text{ in. Hg}).

2. Gauge Pressure (psig) vs. Absolute Pressure (psia)

  • Gauge Pressure (psig\text{psig}): Calibrated to read 0 psig0\text{ psig} in open atmospheric air. A standard refrigeration manifold gauge measures pressure relative to the surrounding atmosphere.
  • Absolute Pressure (psia\text{psia}): Measures pressure relative to a perfect, complete vacuum (0 psia0\text{ psia}).
  • The Mathematical Bridge:
Ppsia=Ppsig+Patmospheric⟶Ppsia=Ppsig+14.7(at sea level)\mathbf{P_{\text{psia}} = P_{\text{psig}} + P_{\text{atmospheric}}} \quad \longrightarrow \quad \mathbf{P_{\text{psia}} = P_{\text{psig}} + 14.7} \quad (\text{at sea level}) Ppsig=Ppsia−14.7\mathbf{P_{\text{psig}} = P_{\text{psia}} - 14.7}

If your refrigeration gauge reads 118.3 psig118.3\text{ psig} on an R-410A system at sea level, the absolute pressure inside the circuit is:

Ppsia=118.3+14.7=133.0 psiaP_{\text{psia}} = 118.3 + 14.7 = 133.0\text{ psia}

3. Vacuum Measurements: Inches of Mercury and Microns

When measuring pressures below atmospheric pressure (deep evacuation), standard positive gauge scales lack adequate resolution. Technicians use two specialized scales:

  • Inches of Mercury Vacuum (in. Hg vac\text{in. Hg vac}): Ranges from 0 in. Hg0\text{ in. Hg} (atmospheric pressure) to 29.92 in. Hg29.92\text{ in. Hg} (perfect vacuum). Compound gauges display this scale on the retard side.
  • Microns: The definitive scientific unit for deep vacuum measurement. A micron is one-millionth of a meter of mercury column (0.001 mm Hg0.001\text{ mm Hg}).
    • Standard atmospheric pressure = 760,000 microns760,000\text{ microns}.
    • 1 in. Hg≈25,400 microns1\text{ in. Hg} \approx 25,400\text{ microns}.
    • Target system evacuation (EPA standard / NATE best practice) = below 500 microns500\text{ microns} (with an isolated vacuum rise test not exceeding 1,000 microns1,000\text{ microns}).

4. Inches of Water Column (in. w.c.)

In residential and light commercial HVAC, many critical pressures are far too subtle to be measured in pounds per square inch (psi). These include:

  • Gas manifold pressures on natural gas and propane furnaces.
  • Air handler external static pressure (ESP) across blower coils and ductwork.
  • Flue gas draft pressures in venting systems.

Technicians measure these using a digital manometer or inclined liquid manometer in inches of water column (in. w.c.\text{in. w.c.})—the pressure required to support a column of water one inch high.

Pressure Conversion ConstantsMathematical Value
1 psi1\text{ psi} in Inches of Water Column27.7 in. w.c.27.7\text{ in. w.c.} (exact: 27.71 in. w.c.27.71\text{ in. w.c.})
1 in. Hg1\text{ in. Hg} in Inches of Water Column13.6 in. w.c.13.6\text{ in. w.c.} (mercury is 13.6 times denser than water)
1 in. w.c.1\text{ in. w.c.} in psi0.0361 psi0.0361\text{ psi}

Standard HVAC Field Pressure Applications:

  • Natural Gas Furnace Manifold Pressure: Typically 3.5 in. w.c.3.5\text{ in. w.c.} (equivalent to 3.527.7≈0.126 psi\frac{3.5}{27.7} \approx 0.126\text{ psi}).
  • Liquefied Petroleum (LP/Propane) Manifold Pressure: Typically 10.5 to 11.0 in. w.c.10.5\text{ to } 11.0\text{ in. w.c.} (approx. 0.38 to 0.40 psi0.38\text{ to } 0.40\text{ psi}).
  • External Static Pressure (ESP): Total external static pressure across an air handler typically targets 0.50 in. w.c.0.50\text{ in. w.c.} for standard residential blowers.
  • Natural Draft Chimney Draft: Typically −0.02 to −0.04 in. w.c.-0.02\text{ to } -0.04\text{ in. w.c.} (negative draft pulling combustion gases upward).

Fundamental Electrical Formulas: Ohm's Law and Electric Power

Electrical diagnostics represents 26% of the Core exam. Technicians must be thoroughly fluent in the mathematical relationships governing voltage, current, resistance, and power.

Ohm's Law

Ohm's law defines the relationship between electromotive force (Voltage, EE), intensity of current (Amperage, II), and opposition to current flow (Resistance, RR):

E=I×RI=ERR=EI\mathbf{E = I \times R} \qquad \mathbf{I = \frac{E}{R}} \qquad \mathbf{R = \frac{E}{I}}
  • EE (Voltage): Measured in Volts (V\text{V}). The electrical pressure driving charge through a conductor.
  • II (Current): Measured in Amperes or Amps (A\text{A}). The rate of electron flow (1 Ampere=1 Coulomb/second1\text{ Ampere} = 1\text{ Coulomb/second}).
  • RR (Resistance): Measured in Ohms (Ω\Omega). The opposition to current flow.

Electric Power (Watt's Law)

Power is the rate at which electrical energy is converted into work or heat. In direct current (DC) circuits and single-phase alternating current (AC) pure resistive circuits (such as electric heating elements), power is given by:

P=E×IP=I2×RP=E2R\mathbf{P = E \times I} \qquad \mathbf{P = I^2 \times R} \qquad \mathbf{P = \frac{E^2}{R}}
  • PP (Power): Measured in Watts (W\text{W}) or Kilowatts (1 kW=1,000 W1\text{ kW} = 1,000\text{ W}).

Power Conversion to Thermal Energy

In HVAC calculations, technicians must frequently bridge the gap between electrical input and thermal output:

1 Watt=3.412 Btu/hr1 kW=3,412 Btu/hr\mathbf{1\text{ Watt} = 3.412\text{ Btu/hr}} \qquad \mathbf{1\text{ kW} = 3,412\text{ Btu/hr}}
  • A 5 kW5\text{ kW} electric heating element produces:
5 kW×3,412 Btu/kW=17,060 Btu/hr5\text{ kW} \times 3,412\text{ Btu/kW} = 17,060\text{ Btu/hr}
  • A 10 kW10\text{ kW} heating package produces:
10 kW×3,412 Btu/kW=34,120 Btu/hr10\text{ kW} \times 3,412\text{ Btu/kW} = 34,120\text{ Btu/hr}

The Fundamental Airflow Equations: Sensible and Total Heat

Air distribution is the vehicle through which conditioning is delivered to the space. The two most critical equations tested on the NATE Core exam are the sensible heat equation and the total heat equation.

1. The Sensible Heat Equation

Sensible heat is heat that causes a measurable change in dry-bulb temperature without changing the moisture content (humidity ratio) of the air.

Qsensible=1.08×CFM×ΔT\mathbf{Q_{\text{sensible}} = 1.08 \times \text{CFM} \times \Delta T}

Where:

  • QsensibleQ_{\text{sensible}}: Sensible heat transfer rate in Btu/hr\text{Btu/hr}.
  • CFM\text{CFM}: Airflow volume in Cubic Feet per Minute.
  • ΔT\Delta T: Temperature difference between entering and leaving air in ∘F^\circ\text{F} (ΔT=Tsupply−Treturn\Delta T = T_{\text{supply}} - T_{\text{return}} or Treturn−TsupplyT_{\text{return}} - T_{\text{supply}}).
  • 1.081.08: The sensible heat constant for standard air at sea level (70∘F70^\circ\text{F}, 29.92 in. Hg29.92\text{ in. Hg}, density ρ=0.075 lb/ft3\rho = 0.075\text{ lb/ft}^3, specific heat cp=0.24 Btu/lb⋅∘Fc_p = 0.24\text{ Btu/lb}\cdot^\circ\text{F}):
Constant=60minhr×0.075lbft3×0.24Btulb⋅∘F=1.08Btu⋅minft3⋅hr⋅∘F\text{Constant} = 60\frac{\text{min}}{\text{hr}} \times 0.075\frac{\text{lb}}{\text{ft}^3} \times 0.24\frac{\text{Btu}}{\text{lb}\cdot^\circ\text{F}} = 1.08\frac{\text{Btu}\cdot\text{min}}{\text{ft}^3\cdot\text{hr}\cdot^\circ\text{F}}

Algebraic Variations of the Sensible Heat Formula:

  • Solving for Airflow (CFM\text{CFM}):
CFM=Qsensible1.08×ΔT\mathbf{\text{CFM} = \frac{Q_{\text{sensible}}}{1.08 \times \Delta T}}
  • Solving for Temperature Difference (ΔT\Delta T):
ΔT=Qsensible1.08×CFM\mathbf{\Delta T = \frac{Q_{\text{sensible}}}{1.08 \times \text{CFM}}}

2. The Total Heat Equation (Enthalpy Equation)

In cooling and dehumidification, air undergoes both a dry-bulb temperature drop (sensible cooling) and moisture condensation (latent cooling). Total cooling capacity must therefore account for both sensible and latent heat removal using specific enthalpy (hh), measured in Btu per pound of dry air\text{Btu per pound of dry air}:

Qtotal=4.5×CFM×Δh\mathbf{Q_{\text{total}} = 4.5 \times \text{CFM} \times \Delta h}

Where:

  • QtotalQ_{\text{total}}: Total heat transfer rate in Btu/hr\text{Btu/hr}.
  • Δh\Delta h: Change in specific enthalpy between entering and leaving air (hentering−hleavingh_{\text{entering}} - h_{\text{leaving}}) in Btu/lb\text{Btu/lb}.
  • 4.54.5: The total heat air constant for standard air:
Constant=60minhr×0.075lbft3=4.5lb⋅minft3⋅hr\text{Constant} = 60\frac{\text{min}}{\text{hr}} \times 0.075\frac{\text{lb}}{\text{ft}^3} = 4.5\frac{\text{lb}\cdot\text{min}}{\text{ft}^3\cdot\text{hr}}

3. Cooling Tonnage Equivalents

Technicians must remember standard cooling capacity ratings:

1 Ton of Refrigeration=12,000 Btu/hr=200 Btu/min\mathbf{1\text{ Ton of Refrigeration} = 12,000\text{ Btu/hr} = 200\text{ Btu/min}}
  • A nominal residential system operates at approximately 400 CFM per ton400\text{ CFM per ton} of cooling under standard humid climate design conditions (varying between 350 CFM/ton350\text{ CFM/ton} for high latent removal and 450 CFM/ton450\text{ CFM/ton} for dry, arid climates).
  • A 3-ton system typically requires: 3 tons×400 CFM/ton=1,200 CFM3\text{ tons} \times 400\text{ CFM/ton} = 1,200\text{ CFM}.

Worked Engineering & Diagnostic Calculations

Calculation 1: Determining Airflow via Electric Heat Strip Testing

Scenario: A technician is commissioning a residential air handler with an electric heat strip. The technician measures:

  • Line Voltage: E=240 VE = 240\text{ V}
  • Total Current Draw of Heaters: I=40 AI = 40\text{ A} (blower motor amperage is isolated and excluded)
  • Return Air Dry-Bulb Temperature: Treturn=68∘FT_{\text{return}} = 68^\circ\text{F}
  • Supply Air Dry-Bulb Temperature: Tsupply=104∘FT_{\text{supply}} = 104^\circ\text{F}

Step 1: Calculate electrical heat output in Watts and Btu/hr

P=E×I=240 V×40 A=9,600 W=9.6 kWP = E \times I = 240\text{ V} \times 40\text{ A} = 9,600\text{ W} = 9.6\text{ kW} Qsensible=9,600 W×3.412Btu/hrW=32,755.2 Btu/hrQ_{\text{sensible}} = 9,600\text{ W} \times 3.412\frac{\text{Btu/hr}}{\text{W}} = 32,755.2\text{ Btu/hr}

Step 2: Calculate temperature rise (ΔT\Delta T)

ΔT=104∘F−68∘F=36∘F\Delta T = 104^\circ\text{F} - 68^\circ\text{F} = 36^\circ\text{F}

Step 3: Solve for CFM using the sensible heat formula

CFM=Qsensible1.08×ΔT=32,755.21.08×36=32,755.238.88≈842.5 CFM\text{CFM} = \frac{Q_{\text{sensible}}}{1.08 \times \Delta T} = \frac{32,755.2}{1.08 \times 36} = \frac{32,755.2}{38.88} \approx 842.5\text{ CFM}

The blower is delivering approximately 843 CFM843\text{ CFM}.

Calculation 2: Determining Gas Furnace Temperature Rise

Scenario: A technician services an 80%80\% AFUE gas furnace with an input rating of 80,000 Btu/hr80,000\text{ Btu/hr}. The system is delivering a measured airflow of 1,200 CFM1,200\text{ CFM}. What is the expected temperature rise (ΔT\Delta T) across the heat exchanger?

Step 1: Calculate furnace heating output capacity (QQ)

Q=Input Rating×Efficiency=80,000 Btu/hr×0.80=64,000 Btu/hrQ = \text{Input Rating} \times \text{Efficiency} = 80,000\text{ Btu/hr} \times 0.80 = 64,000\text{ Btu/hr}

Step 2: Solve for ΔT\Delta T

ΔT=Q1.08×CFM=64,0001.08×1,200=64,0001,296≈49.38∘F\Delta T = \frac{Q}{1.08 \times \text{CFM}} = \frac{64,000}{1.08 \times 1,200} = \frac{64,000}{1,296} \approx 49.38^\circ\text{F}

The expected temperature rise across the furnace heat exchanger is approximately 49.4∘F49.4^\circ\text{F}. If the furnace nameplate lists an allowable rise of 40∘F to 70∘F40^\circ\text{F} \text{ to } 70^\circ\text{F}, this operating condition falls perfectly within manufacturer specifications.

Calculation 3: Manifold Gas Pressure Conversion

Scenario: A field technician measures a natural gas manifold pressure of 3.5 in. w.c.3.5\text{ in. w.c.} using a digital manometer. The customer's industrial engineering manager asks for this pressure in pounds per square inch (psi).

Calculation: Since 1 psi=27.7 in. w.c.1\text{ psi} = 27.7\text{ in. w.c.}:

Ppsi=3.5 in. w.c.27.7 in. w.c./psi≈0.126 psiP_{\text{psi}} = \frac{3.5\text{ in. w.c.}}{27.7\text{ in. w.c./psi}} \approx 0.126\text{ psi}

This calculation highlights why manometers are indispensable in HVAC work: a pressure difference of 0.126 psi0.126\text{ psi} would barely register off the pin of a standard 0–100 psig0\text{–}100\text{ psig} gauge.


Comprehensive HVAC Formula & Constant Reference Sheet

Formula NameEquationUnits & Key Constants
Fahrenheit to Celsius∘C=(∘F−32)÷1.8^\circ\text{C} = (^\circ\text{F} - 32) \div 1.8∘F^\circ\text{F}, ∘C^\circ\text{C}
Celsius to Fahrenheit∘F=(∘C×1.8)+32^\circ\text{F} = (^\circ\text{C} \times 1.8) + 32∘F^\circ\text{F}, ∘C^\circ\text{C}
Absolute Temperature (Rankine)∘R=∘F+460^\circ\text{R} = ^\circ\text{F} + 460Mandatory for imperial gas laws
Absolute Temperature (Kelvin)K=∘C+273\text{K} = ^\circ\text{C} + 273Mandatory for metric gas laws
Absolute Pressure (psia)Ppsia=Ppsig+14.7P_{\text{psia}} = P_{\text{psig}} + 14.7At standard sea level
Water Column to psipsi=in. w.c.÷27.7\text{psi} = \text{in. w.c.} \div 27.71 psi=27.7 in. w.c.1\text{ psi} = 27.7\text{ in. w.c.}
Mercury to Water Columnin. w.c.=in. Hg×13.6\text{in. w.c.} = \text{in. Hg} \times 13.6Mercury density ratio = 13.6
Ohm's Law (Voltage)E=I×RE = I \times RVolts, Amperes, Ohms
Electric Power (Watts)P=E×IP = E \times IPure resistive / DC loads
Electrical Thermal EquivalentQ=Watts×3.412Q = \text{Watts} \times 3.4121 kW=3,412 Btu/hr1\text{ kW} = 3,412\text{ Btu/hr}
Sensible Heat EquationQsensible=1.08×CFM×ΔTQ_{\text{sensible}} = 1.08 \times \text{CFM} \times \Delta TStandard air constant = 1.08
Total Heat EquationQtotal=4.5×CFM×ΔhQ_{\text{total}} = 4.5 \times \text{CFM} \times \Delta hStandard air constant = 4.5
Cooling Tonnage Equivalent1 Ton=12,000 Btu/hr1\text{ Ton} = 12,000\text{ Btu/hr}Nominal airflow ≈400 CFM/ton\approx 400\text{ CFM/ton}

Common Math Traps on the NATE Core Exam

  1. Forgetting to convert Watts to Btu/hr: When evaluating electric heat strip temperature rise, remember that multiplying Amps by Volts yields Watts, not Btu/hr. You must multiply Watts by 3.4123.412 before inserting into Q=1.08×CFM×ΔTQ = 1.08 \times \text{CFM} \times \Delta T.
  2. Using Input Rating instead of Output Capacity: On gas or oil heating calculations, furnaces are rated by input Btu/hr. The heat transferred to the air stream is the output (Qoutput=Input×AFUE EfficiencyQ_{\text{output}} = \text{Input} \times \text{AFUE Efficiency}). Using the gross input rating will overstate calculated airflow or temperature rise by 10% to 20%.
  3. Plugging Gauge Pressure into Gas Law Formulas: In any formula evaluating pressure changes due to temperature (e.g., P1/T1=P2/T2P_1/T_1 = P_2/T_2), both pressure and temperature must be absolute (PP in psia, TT in °R).
  4. Confusing Water Column (in. w.c.) with Inches of Mercury (in. Hg): Remember that water is much lighter than mercury. 1 in. Hg1\text{ in. Hg} equals 13.6 in. w.c.13.6\text{ in. w.c.}, and 1 psi1\text{ psi} equals 27.7 in. w.c.27.7\text{ in. w.c.}, whereas 1 psi1\text{ psi} equals only 2.036 in. Hg2.036\text{ in. Hg}.
Test Your Knowledge

An electric furnace operates at 240 V and draws 45 A across its resistive heating elements (excluding the blower motor). A technician measures a return air temperature of 68°F and a supply air temperature of 102°F. Using the sensible heat formula Q = 1.08 × CFM × ΔT, what is the approximate operating airflow of the blower system?

A

650 CFM

B

1,004 CFM

C

1,250 CFM

D

1,520 CFM

Test Your Knowledge

A recovery cylinder filled with nitrogen sits at an ambient temperature of 75°F, and its pressure gauge indicates 150 psig at sea level. If a technician needs to apply Charles's Law or the Ideal Gas Law to calculate pressure changes at higher temperatures, what are the initial absolute pressure (psia) and absolute temperature (°R)?

A

135.3 psia and 75°R

B

150.0 psia and 348°R

C

164.7 psia and 75°R

D

164.7 psia and 535°R

Test Your Knowledge

A technician is checking the manifold gas pressure on a natural gas furnace using a digital manometer and measures 3.5 in. w.c. How many pounds per square inch (psi) does this pressure represent, and why are inches of water column used instead of standard psig gauges?

A

Approximately 0.126 psi; inches of water column provide the high measurement resolution needed for low-pressure systems where 1 psi equals 27.7 in. w.c.

B

Approximately 1.26 psi; water column scales are designed specifically for liquid fuel lines such as No. 2 fuel oil.

C

Approximately 0.476 psi; 1 psi equals 7.35 in. w.c., which matches standard atmospheric pressure.

D

Approximately 3.50 psi; inches of water column and pounds per square inch are identical direct 1:1 units.

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