Reference Data

Conversion Factors: MULTIPLY BY TO OBTAIN MULTIPLY BY TO OBTAIN Atmospheres 14.70 Lbs./sq. inch (Absolute) Atmospheres 2116.8 Lbs./sq. inch (Absolute) B.T.U/min. 0.02356 Horsepower B.T.U/min. 0.01757 Kilowatts Feet 30.48 Centimeters Feet of Water 0.4335 Lbs./sq. inch (62F) Gallons 231 Cubic inches

Gallons of Water 8.34 Pounds Horsepower (Boiler) 33.479 B.T.U./hr. Horsepower (Boiler) 9.803 Kilowatt-hrs. Inches 2.54 Centimeters Inches of Water 0.03613 Lbs./sq. inch Liters 61.02 Cubic inches Liters 0.264 Gallons Tons of Refrigeration 12000 B.T.U./hr

_____________________________________________________________________ Heating and Cooling Calculations: CFM…………………………………=Cubic feet of air per minute passing through the coil Weight per cu ft……………………..=Weight of 1 pound (.075) Sp ht…………………………………=BTU required to raise the temperature of 1 lb of air 1 degree F (.24) To……………………………………=Temperature of air entering coil in degrees F. T……………………………………..=Temperature of air leaving coil in degrees F. T1……………………………………=Temperature of water leaving coil in degrees F. T2……………………………………=Temperature of water entering coil in degrees F. H1……………………………………=Enthalpy of entering air. H2……………………………………=Enthalpy of leaving air. Capacity of Hot Water Coil GPM = CFM X 1.08 X (T – To) (T2 – T1) X 500

Capacity of Chilled Water Coil GPM = (H1 – H2) X CFM X .075 X 60 (T2 – T1) X 500

Chiller Tonnage TONS = GPM X (CHWR – CHWS) 24

Chiller Coefficient of Performance COP = (CHWR – CHWS) X GPM X 0.0417 0.28433 X KW

DC Current Kilowatts KW = {Amps X Volts}/1000 AC Single Phase Kilowatts KW ={Amps X Volts X Power Factor}/1000 AC Three Phase Kilowatts KW = {Amps X Volts X 1.73 X Power Factor}/1000

VAV Box Air Flow Rate (CFM) CFM…………………………………=Cubic feet of air per minute passing through the duct A……………………………………..=Duct area in sq. ft.

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V…..…………………………………=Velocity of the air Pv…………………………………….= Pressure in inches of H2O from PV3 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))

Linear Reset Schedule Formula The linear reset schedule is used to reset a particular setpoint based on a particular parameter. Some examples are: • Boiler hot water supply temperature reset based on the outside air temperature. In this instance, the hot

water bypass valve is modulated to control the temperature of the water flow. • Supply air temperature reset based on the space zone calling for the most cooling or heating. • Chilled water supply temperature reset based on the chilled water return temperature.

Y……………………………………..=The variable that is reset m…………………………………..…=The slope of the reset line x.……………………………………..=The parameter that the reset is based upon b.…..…………………………………=System constant that defines the equation of the line Y1…………………………………….=Minimum value of the reset variable Y2…………………………………….=Maximum value of the reset variable X1…………………………………….=Minimum value of the dependent value X2…………………………………….=Maximum value of the dependent value STEP 1: Determine the minimum and maximum values of the system, i.e. Y1, Y2, X1, and X2. Example: Boiler hot water reset schedule. When the outside air temperature is 50 degrees F or below, the hot water supply temperature should be reset to 140 degrees F. When the outside air temperature is 90 degrees F or above, the hot water supply temperature should be reset to 110 degrees F. From the sequence of operation above, Y1 = 140, Y2 = 110, X1 = 50, and X2 = 90. STEP 2: Determine the slope of the line. m = Y2 – Y1 = 110 – 140 = -30 = -0.75 X2 – X1 90 – 50 40 STEP 3: Determine b. Plug in the known parameters into the equation of the line and solve for b. Y = mx + b ⇒ b = Y – mx = 140 – (-0.75)(50) = 72.5 b = 177.5

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STEP 4: Plug in the calculated values for m and b into the equation for a line.

Y = mx + b = (-0.75)x + 177.5 Here, Y is the HOT WATER SUPPLY TEMPERATURE and x is the OUTSIDE AIR TEMPERATURE. Although CBAS 2000 calculates linear reset equations automatically with the RESET SCHEDULE function, a sample calculation point in another program for this example would look like: HOT WATER SUPPLY TEMPERATURE = -0.75 X OUTSIDE AIR TEMPERATURE + 177.5

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