HVAC Formulas
Dewpoint and Wetbulb Temperature
The following equations are used to calculate the wetbulb temperature of air given the drybulb temperature and relative humidity %. The equation assumes that the ambient barometric pressure is constant at a value of 29.15 “Hg since the change in wetbulb temperature is very insignificant with changes in the ambient barometric pressure.
| Input Variables | System Variables | Output Variables | |||
|---|---|---|---|---|---|
| RH | Relative Humidity % | e | Ambient vapor pressure in kPa | Td | Dewpoint temperature in degrees C |
| T | Drybulb temperature in degrees C | GAMMA | Constant based upon ambient barometric pressure | Tw | Wetbulb temperature |
| DELTA | Constant | ||||
| Equations | |||||
| e | (RH / 100) * 0.611*EXP(17.27*T/(T+237.3)) | ||||
| Td | [116.9 + 237.3 ln(e)] / [16.78 – ln(e)] | ||||
| GAMMA | 0.00066*P (Use P = 98.642 kPa. This is equal to 29.15 “Hg… about the pressure we normally experience.) | ||||
| DELTA | 4098*(e / Td + 237.3)^2 | ||||
| Wetbulb Temperature in Degrees F Equals: | |||||
| Tw | 1.8 * [[(GAMMA*T) + (DELTA*Td)] / (GAMMA + DELTA)] + 32 | ||||
| Dewpoint Temperature in Degrees F Equals: | |||||
| Td | 1.8 * [[116.9 + 237.3 ln(e)] / [16.78 – ln(e)]] + 32 | ||||
Air Handling Unit Tonnage Output
The following equation calculates the refrigeration output in Tonns of a coil.
| Input Variables | Output Variables | ||
|---|---|---|---|
| T1 | Entering air temperature of the coil in degrees F | TONNS | Dewpoint temperature in degrees F |
| T2 | Leaving air temperature of the coil in degrees F | ||
| CFM | Volume of air passing through the coil | ||
| Equation | |||
| TONNS | 1.08*(T1 – T2)*CFM | ||
Chiller Tonnage Output
The following equation calculates the refrigeration output in Tonns of a chiller.
| Input Variables | Output Variables | ||
|---|---|---|---|
| T1 | Chilled water return temperature in degrees F | TONNS | Energy output of the chiller |
| T2 | Chilled water supply temperature in degrees F | ||
| GPM | Volume of water passing through the chiller | ||
| Equation | |||
| TONNS | GPM*(T1 – T2) / 24 | ||
Chiller Coefficient of Performance
The following equation calculates the ratio of energy used to the energy output of a chiller.
| Input Variables | |
|---|---|
| T1 | Chilled water return temperature in degrees F |
| T2 | Chilled water supply temperature in degrees F |
| GPM | Volume of water passing through the chiller |
| KW | Kilowatts |
| Output Variables | |
|---|---|
| COP | Energy output of the chiller |
| Equation | |
|---|---|
| COP | (T1 – T2) * GPM * 0.0417 / (0.28433 * KW) |
VAV Box Air Flow Rate (CFM)
| Input Variables | |
|---|---|
| A | Duct area in sq. ft |
| Pv | Pressure in inches of H2O from PV3 |
| Output Variables | |
|---|---|
| V | Velocity of the air |
| CFM | Cubic feet of air per minute |
| Equation | |
|---|---|
| Q | AV |
| 0.0763 is the density of dry air at 60o F The duct diameter units are in ft. | |
| CFM | 1096Π(Duct Diameter/2)2(√(Pv/.0763)) |
Heat Index Calculation
The following equation calculates the heat index of the outside air.
| Input Variables | |
|---|---|
| Tf | Outside air temperature in degrees F |
| RH | Outside air relative humidity % (enter 50 for 50%, etc.) |
| Output Variables | |
|---|---|
| HI | Heat index |
| Equation | |
|---|---|
| HI |  |
Wind Chill Temperature Calculation
The following equation calculates the wind chill temperature of the outside air.
| Input Variables | |
|---|---|
| V | Outside air velocity in Miles per Hour |
| T | Outside air temperature in degrees F |
| Output Variables | |
|---|---|
| WC | Wind chill temperature |
| Equation | |
|---|---|
| WC | 0.0817(3.71(V)^0.5 + 5.81 - 0.25V)(T - 91.4) + 91.4 |
Pressure Measurement
| Velocity Pressure | |
|---|---|
 Where V = Air Velocity (FPM) Pv = Velocity Pressure (in. w.g.) | |
| Equivalent Measures of Pressure | |
|---|---|
| 1lb. per square inch | = 144lbs. per sq. ft. = 2.036in. Mercury at 32°F = 2.311ft. Water at 70°F = 27.74in. Water at 70°F |
| 1 inch Water at 70°F | = .03609lb. per sq. in. = .5774oz. per sq. in. = 5774oz. per sq. in. = 5.196lbs. per sq. ft. |
| 1 ounce per sq. in. | = 1272in. Mercury at 32°F = 1.733in. Water at 70°F |
| 1ft. Water at 70°F | = .433lbs. per sq. in. = 62.31lbs. sq. ft. |
| 1 Atmosphere | = 14.696lbs. per sq. in. = 2116.3lbs. per sq. ft. = 33.96ft. Water at 70°F = 29.92in. Mercury at 32°F |
| 1in. Mercury at 32°F | = .491lbs. per sq. in. = 7.86oz. per sq. in. = 1.136ft. Water at 70°F = 13.63in. Water at 70°F |
| Compression Ratio | |
|---|---|
| Compression Ratio | = Absolute Discharge Pressure / Absolute Suction Pressure |
| Absolute Discharge Pressure | = gauge reading + 15psi |
| Absolute Suction Pressure | = gauge reading + 15psi |
| Refrigerant Mass Flow Rate | |
|---|---|
| Mass Flow Rate (Pounds/Minute) | = Piston Displacement X Refrigerant Density = (Cubic Feet/Minute) X (Pounds/Cubic Feet) |
