HVAC FORMULAS TON OF REFRIGERATION - The amount of heat required to melt a ton (2000 lbs.) of ice at 32°F 288,000 BTU/24 hr. 12,000 BTU/hr. APPROXIMATELY 2 inches in Hg. (mercury) = 1 psi WORK = Force (energy exerted) X Distance Example: A 150 lb. man climbs a flight of stairs 100 ft. high Work = 150 lb. X 100 ft. Work = 15,000 ft.-lb. ONE HORSEPOWER = 33,000 ft.-lb. of work in 1 minute ONE HORSEPOWER = 746 Watts CONVERTING KW to BTU: 1 KW = 3413 BTU’s Example: A 20 KW heater (20 KW X 3413 BTU/KW = 68,260 BTU’s CONVERTING BTU to KW: 3413 BTU’s = 1 KW Example: A 100,000 BTU/hr. oil or gas furnace (100,000 ¸ 3413 = 29.3 KW) COULOMB = 6.24 X 1018 (1 Coulomb = 1 Amp) E = voltage (emf) I = Amperage (current) R = Resistance (load) WATTS (POWER) = volts x amps or P = E x I P(in KW) = E x I 1,000 U FACTOR = reciprocal of R factor Example: 1 R = .05U 19 = BTU’s transferred / 1 Sq.Ft. / 1ºF / 1 HourVA (how the secondary of a transformer is rated) = volts X amps Example: 24V x .41A = 10 VA ONE FARAD CAPACITY = 1 amp. stored under 1 volt of pressure MFD (microfarad) = 1 Farad 1,000,000 LRA (Locked rotor amps) = FLA (Full Load Amps) 5 LRA = FLA x 5 TXV (shown in equilibrium) 46.7 Bulb Pressure _______________ Spring Pressure 9.7 37 Evaporator Pressure Bulb Pressure = opening force Spring and Evaporator Pressures = closing forces RPM of motor = 60Hz x 120_ No. of Poles 1800 RPM Motor – slippage makes it about 1750 3600 RPM Motor – slippage makes it about 3450 DRY AIR = 78.0% Nitrogen 21.0% Oxygen 1.0% Other Gases WET AIR = Same as dry air plus water vapor SPECIFIC DENSITY = 1_______ Specific Volume SPECIFIC DENSITY OF AIR = __1__ = .075 lbs./cu.ft. 13.33 STANDARD AIR = .24 Specific Heat (BTU’s needed to raise 1 lb. 1 degree)SENSIBLE HEAT FORMULA (Furnaces): BTU/hr. – Specific Heat X Specific Density X 60 min./hr. = X CFM X DT .24 X .075 X 60 X CFM X DT = 1.08 X CFM X DT ENTHALPHY = Sensible heat and Latent heat TOTAL HEAT FORMULA (for cooling, humidifying or dehumidifying) BTU/hr. = Specific Density X 60 min./hr. X CFM X DH = 0.75 x 60 x CFM x DH = 4.5 x CFM x DH RELATIVE HUMIDITY = __Moisture present___ Moisture air can hold SPECIFIC HUMIDITY = grains of moisture per dry air 7000 GRAINS in 1 lb. of water DEW POINT = when wet bulb equals dry bulb TOTAL PRESSURE (Ductwork) = Static Pressure plus Velocity Pressure CFM = Area (sq. ft.) X Velocity (ft. min.) HOW TO CALCULATE AREA Rectangular Duct Round Duct A = L x W A = pD 2__ OR pr 2 4 RETURN AIR GRILLES – Net free area = about 75% 3 PHASE VOLTAGE UNBALANCE = 100 x maximum deg. from average volts Average Volts NET OIL PRESSURE = Gross Oil Pressure – Suction PressureCOMPRESSION RATIO = Discharge Pressure Absolute Suction Pressure Absolute HEAT PUMP AUXILIARY HEAT – sized at 100% of load ARI HEAT PUMP RATING POINTS (SEER Ratings) 47° 17° NON-BLEND REFRIGERANTS: Constant Pressure = Constant Temperature during Saturated Condition BLENDS – Rising Temperature during Saturated Condition 28 INCHES OF WC = 1 psi NATURAL GAS COMBUSTION: Excess Air = 50% 15 ft.3 of air to burn 1 ft.3 of methane produces: 16 ft.3 of flue gases: 1 ft.3 of oxygen 12 ft.3 of nitrogen 1 ft.3 of carbon dioxide 2 ft.3 of water vapor Another 15 ft.3 of air is added at the draft hood GAS PIPING (Sizing – CF/hr.) = Input BTU’s Heating Value Example: ___ 80,000 Input BTU’s____________ 1000 (Heating Value per CF of Natural Gas) = 80 CF/hr. Example: _________ 80,000 Input BTU’s_________ 2550 (Heating Value per CF of Propane) = 31 CF/hr. FLAMMABILITY LIMITS Propane Butane_ Natural Gas 2.4-9.5 1.9-8.5 4-14COMBUSTION AIR NEEDED Propane Natural Gas (PC=Perfect Combustion) 23.5 ft.3 (PC) 10 ft.3 (PC) (RC=Real Combustion) 36 ft.3 (RC) 15 ft.3 (RC) ULTIMATE CO2 13.7% 11.8% CALCULATING OIL NOZZLE SIZE (GPH): _BTU Input___ = Nozzle Size (GPH) 140,000 BTU’s OR _______ BTU Output___________ 140,000 X Efficiency of Furnace FURNACE EFFICIENCY: % Efficiency = energy output energy input OIL BURNER STACK TEMPERATURE (Net) = Highest Stack Temperature minus Room Temperature Example: 520° Stack Temp. – 70° Room Temp. = Net Stack Temperature of 450° KELVIN TO CELSIUS: C = K – 273 CELSIUS TO KELVIN: K = C + 273 ABSOLUTE TEMPERATURE MEASURED IN KELVINS SINE = side opposite COSINE = side adjacent sin hypotenuse cos hypotenuse TANGENT = side opposite tan side adjacent PERIMETER OF SQUARE: P = 4s P = Perimeter s = side PERIMETER OF RECTANGLE: P = 2l + 2w P – Perimeter l = length w = width PERIMETER OF TRIANGLE: P = a + b + c P = Perimeter a = 1st side b = 2nd side c = 3rd side PERIMETER OF CIRCLE: C = pD C = Circumference C = 2pr p = 3.1416 D = Diameter r = radius AREA OF SQUARE: a = s2 A = Area s = side AREA OF RECTANGLE: A = l w A = Area l = length w = width AREA OF TRIANGLE: A = 1/2bh A = Area b = base h = height AREA OF CIRCLE: A = pr 2 A = Area p = 3.1416 A = p D2 r = radius 4 D = Diameter VOLUME OF RECTANGULAR SOLID:V = l wh V = Volume l = length w = width h = height VOLUME OF CYLINDRICAL SOLID: V = pr 2h V = Volume p = 3.1416 V = p D2h r = radius 4 D = Diameter h = height CAPACITANCE IN SERIES: C = ______1________________ 1 + 1 + . . . . . C1 C2 CAPACITANCE IN PARALLEL: C = C1 + C2 + . . . . . GAS LAWS: Boyle’s Law: P1 V1 = P2 V2 P = Pressure (absolute) V = Volume Charles’ Law: P1 = P2 P = Pressure (absolute) T1 T2 T = Temperature (absolute) General Gas Law: P1 V1 P2 V2 P = Pressure (absolute) _____ = _____ V = Volume T1 T2 T = Temperature (absolute) PYTHAGOREAN THEOREM: C 2 = a2 + b2 c = hypotenuse a & b = sides
Posted on: Fri, 28 Mar 2014 03:53:05 +0000
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