1.2 Engineering Units, Dimensional Analysis & Physical Property Lookups

Key Takeaways

  • HVAC engineering calculations rely heavily on Inch-Pound (IP) units, where 1 Ton of Refrigeration equals exactly 12,000 Btu/hr (200 Btu/min = 3.51685 kW), 1 Boiler Horsepower equals 33,475 Btu/hr (9.810 kW), and 1 Mechanical Horsepower equals 2,545 Btu/hr (745.7 W).
  • The standard air sensible heat constant (1.08) derives from standard air density (rho = 0.075 lbm/ft3), dry air specific heat (cp = 0.240 Btu/lbm-F), and the 60 min/hr time multiplier; non-standard conditions require calculating Cs = 14.4 * rho_actual.
  • Latent and total moist air heat equations rely on standard air derivations: Latent Heat q_l = 4840 * CFM * Delta_W (where W is in lbm_w/lbm_da) or 0.69 * CFM * Delta_w_grains (where 1 lbm = 7000 grains), and Total Enthalpy q_t = 4.5 * CFM * Delta_h.
  • Hydronic liquid heat transfer for standard water (density = 8.337 lbm/gal, cp = 1.000 Btu/lbm-F) uses q = 500 * GPM * Delta_T; for glycol solutions, derate using C_fluid = 500 * SG * cp_glycol.
  • Pressure conversions must distinguish between absolute and gauge references: 1 psi = 2.3087 ft w.g. = 27.70 in. w.g., and air duct velocity pressure relates to velocity by FPM = 4005 * sqrt(P_v in in. w.g.).
Last updated: August 2026

1.2 Engineering Units, Dimensional Analysis & Physical Property Lookups

Precise unit tracking and dimensional analysis are essential skills for passing the PE Mechanical: HVAC and Refrigeration exam. HVAC engineering calculations routinely combine thermodynamic, psychrometric, fluid dynamic, and electrical parameters across both Inch-Pound (IP) and International System (SI) units. Most foundational HVAC shortcut formulas (such as $1.08$, $4840$, $4.5$, and $500$) embed specific physical properties of standard air or standard liquid water. When systems operate at non-standard conditions—such as high-altitude installations in Denver or hydronic loops utilizing ethylene/propylene glycol—applying these shortcut constants without proper derating introduces substantial engineering errors.


1. Primary HVAC Unit Systems & Fundamental Equivalencies

+-----------------------------------------------------------------------------------------+
| FUNDAMENTAL HVAC UNIT CONVERSION BENCHMARKS                                             |
+-----------------------------------------------------------------------------------------+
| 1 Ton of Refrigeration (TR) = 12,000 Btu/hr = 200 Btu/min = 3.51685 kW = 4.715 BHP (mech)|
| 1 Boiler Horsepower (BHP)   = 33,475 Btu/hr = 9.8095 kW = 34.5 lbm steam/hr at 212°F    |
| 1 Mechanical Horsepower (hp)= 2,544.43 Btu/hr = 745.7 Watts = 550 ft·lbf/s             |
| 1 Kilowatt (kW)             = 3,412.14 Btu/hr = 1,000 Watts = 1.341 hp                  |
| 1 Gallon of Water (at 60°F) = 8.337 lbm = 0.13368 ft³ = 231.0 in³ = 3.7854 Liters       |
| 1 Cubic Foot of Water       = 62.37 lbm = 7.4805 Gallons                                |
| 1 Therm                     = 100,000 Btu = 105.5 MJ = 29.3 kWh                         |
| 1 Quad                      = 10¹⁵ Btu = 1.055 × 10¹⁸ J                                 |
+-----------------------------------------------------------------------------------------+

Primary Physical Quantities in IP and SI Units

Physical DimensionInch-Pound (IP) UnitSI Metric UnitExact Mathematical Conversion
LengthFoot ($\text{ft}$) / Inch ($\text{in.}$)Meter ($\text{m}$) / Millimeter ($\text{mm}$)$1\text{ ft} = 0.3048\text{ m}$; $1\text{ in.} = 25.4\text{ mm}$
MassPound-mass ($\text{lbm}$)Kilogram ($\text{kg}$)$1\text{ lbm} = 0.45359237\text{ kg}$
ForcePound-force ($\text{lbf}$)Newton ($\text{N}$)$1\text{ lbf} = 4.448222\text{ N}$
Pressure$\text{psi}$ / $\text{in. w.g.}$ / $\text{ft H}_2\text{O}$Pascal ($\text{Pa}$) / $\text{kPa}$ / $\text{bar}$$1\text{ psi} = 6,894.76\text{ Pa} = 6.895\text{ kPa} = 0.06895\text{ bar}$
Energy / HeatBritish Thermal Unit ($\text{Btu}$)Joule ($\text{J}$) / Kilojoule ($\text{kJ}$)$1\text{ Btu} = 1,055.056\text{ J} = 1.05506\text{ kJ}$
Power / Heat Rate$\text{Btu/hr}$ / $\text{Ton}$ / $\text{hp}$Watt ($\text{W}$) / Kilowatt ($\text{kW}$)$1\text{ Ton} = 12,000\text{ Btu/hr} = 3.51685\text{ kW}$
Volumetric Air FlowCubic feet per minute ($\text{CFM}$)$\text{m}^3/\text{s}$ / $\text{L/s}$$1\text{ CFM} = 0.471947\text{ L/s} = 0.00047195\text{ m}^3/\text{s}$
Volumetric Liquid FlowGallons per minute ($\text{GPM}$)$\text{m}^3/\text{hr}$ / $\text{L/s}$$1\text{ GPM} = 0.06309\text{ L/s} = 0.22712\text{ m}^3/\text{hr}$
Dynamic Viscosity$\text{lbm/(ft}\cdot\text{s)}$ / $\text{cP}$$\text{Pa}\cdot\text{s}$ / $\text{kg/(m}\cdot\text{s)}$$1\text{ cP} = 0.001\text{ Pa}\cdot\text{s} = 6.7197 \times 10^{-4}\text{ lbm/(ft}\cdot\text{s)}$
Kinematic Viscosity$\text{ft}^2/\text{s}$ / $\text{cSt}$$\text{m}^2/\text{s}$$1\text{ cSt} = 10^{-6}\text{ m}^2/\text{s} = 1.0764 \times 10^{-5}\text{ ft}^2/\text{s}$

2. Pressure Datums: Absolute, Gauge & Fluid Head Conversions

Thermodynamic equations (such as ideal gas laws and psychrometric saturation pressure evaluations) strictly require absolute pressure ($P_{\text{abs}}$), whereas mechanical gauges and pressure transducers measure gauge pressure ($P_{\text{gauge}}$):

Pabs=Pgauge+PatmP_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}

Standard Atmospheric Pressure ($P_{\text{atm}}$ at Sea Level, $59^\circ\text{F}$ / $15^\circ\text{C}$)

Patm=14.696 psia=29.921 in. Hg=760.0 mm Hg=101.325 kPa=1.01325 bar=407.2 in. w.g.=33.93 ft H2OP_{\text{atm}} = 14.696\text{ psia} = 29.921\text{ in. Hg} = 760.0\text{ mm Hg} = 101.325\text{ kPa} = 1.01325\text{ bar} = 407.2\text{ in. w.g.} = 33.93\text{ ft H}_2\text{O}

Liquid Head vs. Static Pressure Relations

Fluid static head ($h$) expressed in feet of fluid column relates to pressure ($P$) in pounds per square inch ($\text{psi}$) through fluid density ($\rho$, in $\text{lbm/ft}^3$):

h (ft)=P (psi)×144 in.2/ft2ρ (lbm/ft3)=144Pρh\text{ (ft)} = \frac{P\text{ (psi)} \times 144\text{ in.}^2/\text{ft}^2}{\rho\text{ (lbm/ft}^3\text{)}} = \frac{144 \cdot P}{\rho}

For liquid water at standard temperature ($60^\circ\text{F}$, $\rho = 62.37\text{ lbm/ft}^3$):

hwater (ft)=144P62.37=2.3087P    1 psi=2.3087 ft of water headh_{\text{water}}\text{ (ft)} = \frac{144 \cdot P}{62.37} = 2.3087 \cdot P \quad \implies \quad 1\text{ psi} = 2.3087\text{ ft of water head}

P (psi)=hwater (ft)2.3087=0.43315×hwater (ft)P\text{ (psi)} = \frac{h_{\text{water}}\text{ (ft)}}{2.3087} = 0.43315 \times h_{\text{water}}\text{ (ft)}

Air Duct Static Pressure: Inches of Water Gauge (in. w.g.)

Air distribution ductwork pressures are small and expressed in inches of water gauge ($\text{in. w.g.}$ or $\text{in. w.c.}$):

1 in. w.g.=112 ft H2O=0.4331512 psi=0.03610 psi=5.202 lbf/ft2=249.08 Pa1\text{ in. w.g.} = \frac{1}{12}\text{ ft H}_2\text{O} = \frac{0.43315}{12}\text{ psi} = 0.03610\text{ psi} = 5.202\text{ lbf/ft}^2 = 249.08\text{ Pa}

1 psi=27.70 in. w.g.1\text{ psi} = 27.70\text{ in. w.g.}


3. Derivation of Fundamental HVAC Air Constants

In standard HVAC sensible, latent, and total heat calculations, engineers use the shortcut multipliers 1.08, 4840, and 4.5. These values are derived from standard dry air physical properties at sea level.

Standard Air Definition (ASHRAE / NCEES Standards)

  • Standard Temperature ($T_0$): $70^\circ\text{F}$ ($21.1^\circ\text{C}$)
  • Standard Barometric Pressure ($P_0$): $14.696\text{ psia}$ ($29.921\text{ in. Hg}$)
  • Standard Air Density ($\rho_{\text{std}}$): $0.075\text{ lbm/ft}^3$ ($1.204\text{ kg/m}^3$)
  • Standard Specific Volume ($v_{\text{std}}$): $1 / 0.075 = 13.333\text{ ft}^3/\text{lbm}$
  • Specific Heat of Dry Air ($c_p$): $0.240\text{ Btu/(lbm}\cdot{}^\circ\text{F)}$ ($1.005\text{ kJ/(kg}\cdot{}^\circ\text{C)}$)
  • Latent Heat of Vaporization at $70^\circ\text{F}$ ($h_{fg}$): $1054\text{ to }1076\text{ Btu/lbm}$

1. Sensible Heat Multiplier Derivation ($q_s$)

Sensible heat transfer changes air dry-bulb temperature without phase change:

q˙s=m˙aircpΔT\dot{q}_s = \dot{m}_{\text{air}} \cdot c_p \cdot \Delta T

Converting volumetric flow rate $\text{CFM}$ (cubic feet per minute) to mass flow rate $\dot{m}_{\text{air}}$ in $\text{lbm/hr}$:

m˙air=CFM(ft3min)×60(minhr)×ρair(lbmft3)=60×0.075×CFM=4.50(lbmhr)×CFM\dot{m}_{\text{air}} = \text{CFM}\left(\frac{\text{ft}^3}{\text{min}}\right) \times 60\left(\frac{\text{min}}{\text{hr}}\right) \times \rho_{\text{air}}\left(\frac{\text{lbm}}{\text{ft}^3}\right) = 60 \times 0.075 \times \text{CFM} = 4.50\left(\frac{\text{lbm}}{\text{hr}}\right) \times \text{CFM}

Substituting $\dot{m}_{\text{air}}$ and $c_p = 0.240\text{ Btu/(lbm}\cdot{}^\circ\text{F)}$ into the sensible equation:

q˙s=(4.50×CFM)×0.240×ΔT=(60×0.075×0.240)×CFM×ΔT\dot{q}_s = (4.50 \times \text{CFM}) \times 0.240 \times \Delta T = (60 \times 0.075 \times 0.240) \times \text{CFM} \times \Delta T

60×0.075×0.240=1.0860 \times 0.075 \times 0.240 = 1.08

q˙s (Btu/hr)=1.08×CFM×ΔT (F)\dot{q}_s\text{ (Btu/hr)} = 1.08 \times \text{CFM} \times \Delta T\text{ (}{}^\circ\text{F)}

Non-Standard Altitude & Temperature Derivation: At non-standard density $\rho_{\text{actual}}$ (e.g., in Denver at 5,000 ft elevation, where $P_{\text{atm}} \approx 12.23\text{ psia}$ and $\rho \approx 0.0624\text{ lbm/ft}^3$): Csensible, actual=60×ρactual×0.240=14.4×ρactualC_{\text{sensible, actual}} = 60 \times \rho_{\text{actual}} \times 0.240 = 14.4 \times \rho_{\text{actual}} For Denver: $C_{\text{sensible}} = 14.4 \times 0.0624 = 0.899 \approx 0.90$. Using 1.08 at 5,000 ft overpredicts coil capacity by 20%!

2. Latent Heat Multiplier Derivation ($q_l$)

Latent heat transfer changes moisture content (humidity ratio $W$, in $\text{lbm moisture / lbm dry air}$) at constant temperature:

q˙l=m˙airhfgΔW=(60×0.075×CFM)×1076×ΔW\dot{q}_l = \dot{m}_{\text{air}} \cdot h_{fg} \cdot \Delta W = (60 \times 0.075 \times \text{CFM}) \times 1076 \times \Delta W

60×0.075×1076=4842484060 \times 0.075 \times 1076 = 4842 \approx 4840

q˙l (Btu/hr)=4840×CFM×ΔW (lbm H2Olbm dry air)\dot{q}_l\text{ (Btu/hr)} = 4840 \times \text{CFM} \times \Delta W\text{ }\left(\frac{\text{lbm } \text{H}_2\text{O}}{\text{lbm dry air}}\right)

When humidity ratio is expressed in grains of moisture ($1\text{ lbm} = 7000\text{ grains}$):

Clatent, grains=48407000=0.69140.69C_{\text{latent, grains}} = \frac{4840}{7000} = 0.6914 \approx 0.69

q˙l (Btu/hr)=0.69×CFM×Δw (grains/lbm dry air)\dot{q}_l\text{ (Btu/hr)} = 0.69 \times \text{CFM} \times \Delta w\text{ (grains/lbm dry air)}

3. Total Heat (Enthalpy) Multiplier Derivation ($q_t$)

Total heat encompasses both sensible temperature and latent phase changes via moist air specific enthalpy ($h$, in $\text{Btu/lbm dry air}$):

q˙t=m˙airΔh=(60×ρair×CFM)×Δh=(60×0.075)×CFM×Δh\dot{q}_t = \dot{m}_{\text{air}} \cdot \Delta h = (60 \times \rho_{\text{air}} \times \text{CFM}) \times \Delta h = (60 \times 0.075) \times \text{CFM} \times \Delta h

60×0.075=4.560 \times 0.075 = 4.5

q˙t (Btu/hr)=4.5×CFM×Δh (Btu/lbm dry air)\dot{q}_t\text{ (Btu/hr)} = 4.5 \times \text{CFM} \times \Delta h\text{ (Btu/lbm dry air)}


4. Hydronic Heat Transfer & Antifreeze Glycol Solutions

For closed hydronic piping loops (chilled water, condenser water, heating hot water):

q˙hydronic=m˙watercp,waterΔT\dot{q}_{\text{hydronic}} = \dot{m}_{\text{water}} \cdot c_{p,\text{water}} \cdot \Delta T

Expressing water flow in Gallons per Minute ($\text{GPM}$):

m˙water=GPM(galmin)×60(minhr)×ρwater(lbmgal)\dot{m}_{\text{water}} = \text{GPM}\left(\frac{\text{gal}}{\text{min}}\right) \times 60\left(\frac{\text{min}}{\text{hr}}\right) \times \rho_{\text{water}}\left(\frac{\text{lbm}}{\text{gal}}\right)

At standard water conditions ($60^\circ\text{F}$, $\rho_{\text{water}} = 8.337\text{ lbm/gal}$, $c_p = 1.000\text{ Btu/(lbm}\cdot{}^\circ\text{F)}$):

Cwater=60×8.337×1.000=500.22500C_{\text{water}} = 60 \times 8.337 \times 1.000 = 500.22 \approx 500

q˙ (Btu/hr)=500×GPM×ΔT (F)\dot{q}\text{ (Btu/hr)} = 500 \times \text{GPM} \times \Delta T\text{ (}{}^\circ\text{F)}

Glycol Mixture Corrections

When ethylene or propylene glycol is added for freeze protection, the fluid density increases while the specific heat capacity decreases:

Cglycol=500×Specific Gravity (SG)×cp,glycolC_{\text{glycol}} = 500 \times \text{Specific Gravity (SG)} \times c_{p,\text{glycol}}

For example, a 30% by weight aqueous Propylene Glycol solution at $40^\circ\text{F}$ has $\text{SG} = 1.028$ and $c_p = 0.925\text{ Btu/(lbm}\cdot{}^\circ\text{F)}$:

Cmixture=500×1.028×0.925=475.45C_{\text{mixture}} = 500 \times 1.028 \times 0.925 = 475.45

q˙glycol (Btu/hr)=475.5×GPM×ΔT\dot{q}_{\text{glycol}}\text{ (Btu/hr)} = 475.5 \times \text{GPM} \times \Delta T


5. Dimensional Homogeneity & Gravitational Constant ($g_c$)

In the English Engineering unit system, Newton's second law of motion requires the gravitational conversion constant $g_c$ to equate force ($\text{lbf}$) and mass ($\text{lbm}$):

F=magcwheregc=32.174 lbmftlbfs2F = \frac{m \cdot a}{g_c} \quad \text{where} \quad g_c = 32.174\text{ }\frac{\text{lbm}\cdot\text{ft}}{\text{lbf}\cdot\text{s}^2}

Derivation of Air Duct Velocity Pressure ($P_v$)

Fluid kinetic energy per unit volume defines velocity pressure:

Pv (lbf/ft2)=ρ (lbm/ft3)V2 (ft/s)22gc (lbmft/(lbfs2))P_v\text{ (lbf/ft}^2\text{)} = \frac{\rho\text{ (lbm/ft}^3\text{)} \cdot V^2\text{ (ft/s)}^2}{2 \cdot g_c\text{ (lbm}\cdot\text{ft/(lbf}\cdot\text{s}^2))}

For standard air ($\rho = 0.075\text{ lbm/ft}^3$), expressing velocity $V$ in feet per minute ($V_{\text{fpm}}$, where $V_{\text{fps}} = V_{\text{fpm}} / 60$) and converting pressure from $\text{lbf/ft}^2$ to $\text{in. w.g.}$ ($1\text{ in. w.g.} = 5.202\text{ lbf/ft}^2$):

Pv (in. w.g.)=0.075×(Vfpm/60)22×32.174×5.202=0.075×Vfpm23600×334.73=Vfpm216,043,000=(Vfpm4005.4)2P_v\text{ (in. w.g.)} = \frac{0.075 \times (V_{\text{fpm}} / 60)^2}{2 \times 32.174 \times 5.202} = \frac{0.075 \times V_{\text{fpm}}^2}{3600 \times 334.73} = \frac{V_{\text{fpm}}^2}{16,043,000} = \left(\frac{V_{\text{fpm}}}{4005.4}\right)^2

V (FPM)=4005Pv (in. w.g.)V\text{ (FPM)} = 4005 \sqrt{P_v\text{ (in. w.g.)}}


6. Unit Conversion Pitfall Matrix

Conversion PathCorrect Formula / MultiplierCommon Exam Pitfall
Tons of Refrig. $\rightarrow$ Btu/hr$\text{Btu/hr} = \text{Tons} \times 12,000$Confusing thermal cooling tons with mass short tons ($2000\text{ lbm}$)
GPM $\rightarrow$ lbm/hr (Water)$\text{lbm/hr} = \text{GPM} \times 500$Forgetting $60\text{ min/hr}$ multiplier ($8.33 \times 60 = 500$)
psi $\rightarrow$ ft of Water Head$\text{Head (ft)} = \text{psi} \times 2.3087$Inverting ratio (dividing by $2.3087$ instead of multiplying)
in. w.g. $\rightarrow$ psi$\text{psi} = \text{in. w.g.} \times 0.03610$Treating $\text{in. w.g.}$ as direct $\text{psi}$ (erroneous by factor of $27.7$)
Humidity Ratio $\rightarrow$ Grains$\text{grains/lbm} = W\text{ (lbm/lbm)} \times 7000$Mixing grains with $\text{lbm/lbm}$ in latent heat equations
Test Your Knowledge

An air handling unit operates at an elevation of 6,000 ft where the ambient barometric pressure is 11.78 psia and air density is 0.060 lbm/ft3. If the supply fan delivers 8,000 CFM across a heating coil with a 35°F temperature rise, what is the actual sensible heat output of the coil?

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Test Your Knowledge

A closed hydronic cooling loop circulates 120 GPM of a 30% by weight aqueous propylene glycol solution (Specific Gravity = 1.028, Specific Heat c_p = 0.925 Btu/lbm-°F) through a process heat exchanger with an entering temperature of 54°F and leaving temperature of 44°F (Delta_T = 10°F). What is the total cooling rate in tons of refrigeration?

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Test Your Knowledge

A differential pressure transducer across a high-efficiency particulate air (HEPA) filter bank reads 1.50 in. w.g. What is the equivalent pressure drop across the filter in pounds per square foot (lbf/ft2) and pounds per square inch (psi)?

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Test Your Knowledge

A dedicated outdoor air system (DOAS) unit dehumidifies 3,000 CFM of outside air from an entering humidity ratio of 105 grains/lbm dry air to a supply condition of 45 grains/lbm dry air. What is the latent cooling load removed by the cooling coil?

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