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 |
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 | |
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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) |