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Good luck on the exam! -Justin Kauwale, PE, President of Engineering Pro Guides Link: https://drive.google.com/drive/folders/1qGrnnJh3ux9pDCXWJOtm9wpApbt1rVB3 Section 1.0 – Most Common Units This table shows a list of the popular unit conversions applicable for HVAC & Refrigeration 1 1 1 1 1 1 1 Btu/h Btu/h Btu/h EER SEER feet lb = = = = = = = 8.33 x10-5 2.93 x10-4 0.293 0.083 0.083 03048 0.4536 tons kW W kW/ton kW/ton meter kg 1 1 1 1 1 1 1 ton kW W kW/ton kW/ton meter kg = = = = = = = 12,000 3,412 3.412 12 12 3.2808 2.205 Btu/h Btu/h Btu/h EER SEER feet lb 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ft2 gal ft3 fpm gpm cfm in. wg ft. hd psi psi psi lb/ft3 Btu Btu/h HP = = = = = = = = = = = = = = = 0.093 3.79 0.1337 5.08x10-3 0.063 0.472 249.09 2.99 6.89 x10-3 27.68 2.31 16.018 0.293 3.93x10-4 0.7457 m2 liters gal m/s l/s l/s Pa Pa MPa in. wg ft. hd kg/m3 watthour HP kW 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 m2 liters ft3 m/s l/s l/s Pa Pa MPa in. wg ft. hd kg/m3 watthour HP kW = = = = = = = = = = = = = = = 10.7639 0.227 7.481 195.85 15.85 2.119 4.01 x10-3 3.34 x10-4 145.04 0.036 0.43 0.062 3.412 2,545 1.34 ft2 gal gal fpm gpm cfm in. wg ft. hd psi psi psi lb/ft3 Btu Btu/h HP πππ ππ = 16.02 π ππ‘ ππ πππ 1 = 0.0624 π ππ‘ πππ π 1 = 27.7 ππ ππ 1 ππ = 25.4 ππ 1 ππ = 2.54 ππ 1 ππ‘ = 0.3048 π 1ππ‘ = 0.093 π 1 ππ = 6.452 ππ 1ππ‘ = 0.0283 π 1 ππ = 16.39 ππ 1 π΅ππ = 1054 π½ 1 πππ − ππ‘ = 1.356 π½ 1 Density (π) Length (π) Area (π΄) Volume (π) Energy (π) 1 π΅ππ = 778 πππ − ππ‘ 1 http://www.engproguides.com ππ πππ = 0.0624 π ππ‘ π πππ 1 = 62.4 ππ ππ‘ π πππ 1 = 0.0361 ππ ππ 1 ππ = 0.0394 ππ 1 ππ = 0.394 ππ 1 π = 3.28 ππ‘ 1 π = 10.76 ππ‘ 1 ππ = 0.1550 ππ 1 π = 35.32ππ‘ 1 ππ = 0.0610 ππ 1 π½ = 9.48 π₯ 10 π΅ππ 1 π½ = 0.738 πππ − ππ‘ ππ − π 1π½ = 1 =1π−π π 1 Power (π, π) Temperature (π) Pressure (π) 1 π΅πππ» = 0.293 π π΅ππ 1 π΅πππ» = 1 π»π 12,000 π΅πππ» = 1 πππ 9 π(β) = (π(°πΎ) − 273) + 32 5 9 π(β) = π(°πΆ) + 32 5 1 ππ π€π = 0.0360912 ππ π 2 http://www.engproguides.com 1 π = 3.414 π΅πππ» π½ 1π =1 π 5 π(°πΆ) = (π(°πΉ) − 32) 9 π(°πΎ) = π(°πΆ) + 273 1 ππ π = 27.7076 ππ π€π Section 2.0 – Basic Engineering Practice Conversion Formula Present Value to Future Value πΉπ = ππ π₯ (1 + π) ππ = Future Value to Present Value Present Value to Annual Value Annual Value to Present Value Multiply PV by (F/P, i, n) πΉπ (1 + π) Multiply FV by (P/F, i, n) π ∗ (1 + π) π΄ = ππ ∗ ( ) (1 + π) − 1 Multiply PV by (A/P, i, n) 1 − (1 + π) ππ = π΄ ∗ ( π Multiply AV by (P/A, i, n) Future Value to Annual Value Annual Value to Future Value Factor Value ) π π΄ = πΉπ( ) (1 + π) − 1 Multiply FV by (A/F, i, n) (1 + π) − 1 ) π Multiply A by (F/A, i, n) πΉπ = π΄ ∗ ( Ohm’s Law πΌ= πΌ = ππ’πππππ‘ [ππππ ] π = π£πππ‘πππ [π£πππ‘π ] π = πππ ππ π‘πππ‘ππ [ππππ ] π π =π +π +π +π Resistors in series π Resistors in parallel 1 1 1 1 1 = + + + π π π π π , π =πΌ∗π π π= π π =πΌ ∗π Power Equations π Pump Water Horsepower Equations π , , [ ,[ ∗ (ππΊ) β ∗π ; 3956 ∗ (ππΊ) π ∗π ; ] = 1,714 ] = ππΊ = 1.0 πππ π€ππ‘ππ ππ‘ 39 πΉ ππ πππ = 1,000 ; 62.4 π ππ‘ π 3 http://www.engproguides.com π = π£πππ’πππ‘πππ ππππ€ πππ‘π [πππ] β = ππππ π π’ππ [ππππ‘ ππ βπππ] π = πππ€ππ [βππππ ππππ€ππ] ππΊ = π πππππππ ππππ£ππ‘π¦ π = ππππ π π’ππ [ππ π] Fan Mechanical Horsepower Equation Pump or Fan Brake Horsepower Equation π = ∗ ππ 6356 π π Motor Horsepower Equation Electrical Power Supplied to Motor , / π π [ ; ] = [ = π = π£πππ’πππ‘πππ ππππ€ πππ‘π ππ πππ [ππ’πππ ππππ‘ πππ ππππ’π‘π] ππ = π‘ππ‘ππ ππππ π π’ππ [πππβππ π€ππ‘ππ πππ’ππ] π , [ ] = πππ πππβππππππ βπππ ππππ€ππ ] π [ π π / [ ]] π [ ] = ]] π ; ; [ ] ππΉ Real Power Single Phase → P = IV ∗ PF Three Phase → P = √3 ∗ IV ∗ PF PF = power factor; units → kW, W, MW Apparent Power Single Phase → S = IV Three Phase → S = √3 ∗ IV PF = power factor; units → KVA, VA, MVA Apparent Power P = IV PF = power factor; units → KVA, VA, MVA 4 http://www.engproguides.com Section 3.0 – Thermodynamics Term Equation [ππ π] = π [ππ π] + 14.7 ππ π π Pressure Absolute temperature Temperature °π = β + 460 Water π΅ππππ‘π ππ‘ 212 °πΉ πππ 1 ππ‘π π΅ππ 0.18505 πππ ) β = π’ + ππ£ ∗ ( 1 ππ π ∗ ππ‘ Enthalpy Description Water has a specific heat of 1.0 Heat Capacity π π =π ∗ Isentropic transition from Pressure State 1 to Pressure State 2 π , π΅π‘π’ πππ β π, π΅π‘π’ = 0.240 πππ °π π΅π‘π’ = 0.171 πππ °π β π π π π π π =π ∗ π K = 1.4 for air π = π ∗ (β − β ) π = π ∗ π ∗ (π − π ) π =π ∗ π π π π =π ∗ π π π£ =π£ ∗ π Transition π =π ∗ Isobaric transition with heat gain or heat rejection Other transitions Relation between the Enthalpy of Saturated Vapor and Liquid Enthalpy of Mixed Vapor-Liquid Relationship of Entropy of Vaporization, Entropy of Saturated Vapor and Liquid Water Entropy of Wet Steam K = 1.4 for air π = π ∗ (β − β ) π = π ∗ π ∗ (π − π ) Description Adiabatic Isothermal Isochoric β =β +β No change in energy No change in temperature No change in volume β = πππ‘βππππ¦ ππ π ππ‘π’πππ‘ππ π£ππππ; β = πππ‘βππππ¦ ππ π ππ‘π’πππ‘ππ ππππ’ππ β = πππ‘βππππ¦ ππ π£ππππππ§ππ‘πππ =β +π₯∗β β π =π +π π = πππ‘ππππ¦ ππ π ππ‘π’πππ‘ππ π£ππππ π = πππ‘ππππ¦ ππ π ππ‘π’πππ‘ππ ππππ’ππ π = πππ‘ππππ¦ ππ π£ππππππ§ππ‘πππ π =π +π₯∗π 5 http://www.engproguides.com π₯ = π£ππππ ππ’ππππ‘π¦ (Mixed Region) as a Function of Steam Quality π = πππ‘ππππ¦ ππ π€ππ‘ π π‘πππ (πππ₯ ππ ππππ’ππ & π£ππππ) π₯ = π π‘πππ ππ’ππππ‘π¦, ππππ¦ππ π πππππ‘πππ, % π£ππππ π =π∗β πππ π = πππ π ππππ€ πππ‘π [ ] βπ π΅π‘π’ π = ππππππ¦ [ ] βπ Heat Available from Condensing Steam Throttling: Irreversible Adiabatic [Constant Enthalpy] or Isenthalpic Tank Heating/Cooling: Isometric [Constant Volume] β =β π£ =π£ ππ‘ π£ = π πππππππ π£πππ’ππ [ ] ππ Turbine Expansion: Isentropic [Constant Entropy] or Reversible Adiabatic π =π Compressor: Isentropic [Constant Entropy] or Reversible Adiabatic π =π =π π π = ππππ π π’ππ [ππ ππ] Boiler Heating: Isobaric [Constant Pressure] Simplified Steam Heating Coil: Steam to Water Heat Transfer π ∗β = 500 ∗ πΊππ ∗ βπ Simplified Steam Heating Coil: Steam to Air Heat Transfer π ∗β = 1.08 ∗ πΆπΉπ ∗ βπ 6 http://www.engproguides.com ππππ Turbines, Pumps and Compressors Energy Balances π π Boilers, Condensers and Evaporators Energy Balances =π , ∗ β Nozzles, Diffusers Energy Balances Heat Exchangers Energy Balances Mixing Energy Balances ∗π , π π ∗ π ∗π , −β ∗β = ππππ =π ∗β +π ∗β =π =β + = π ∗π , π£ ∗β π£ β + , −π ∗β + π ∗π , ∗π =π ∗π , + π ∗β =π ∗β ∗π π −β +π / π ∗ β 2 , 2 ∗ π , −π ∗π π΄ ∗ πΆ π» + π΅ ∗ (π + 3.76π ) → π»πππ‘ + πΆ ∗ πΆπ + π· ∗ π» π π€βπππ π΄, π΅, πΆ & π· πππ ππ’πππππππ π£πππ’ππ π‘βππ‘ πππ πππππ π‘βπ πππ‘ππ ππ πππβ ππππππ’ππ ππ π‘βπ πππ’ππ‘πππ Combustion π΄ππ = (π + 3.76π ) % πΈπ₯πππ π π΄ππ = π΄ππ‘π’ππ π΄ππ − 100% πβπππππ‘ππππ π΄ππ πΆππ = πΆππ = Coefficient of Performance (COP) π π πΈπΈπ 3.412 (πππ‘β π£πππ’ππ ππ’π π‘ ππ π πππ π’πππ‘π ) πΆππ = π΅π‘π’ ( ) βπ π π 7 http://www.engproguides.com π΅π‘π’ ( ) βπ , Evaporator Net Refrigeration Effect [π΅π‘π’β] π π΅π‘π’ ππ ∗ (π πππππ πΉπππ€ π ππ‘π) ππ πππ = (π» − π» ) ∗ (60) πππ βπ π΅π‘π’ ;π» ππ π» = ππππ£πππ ππ£ππππππ‘ππ πππ‘βππππ¦ π΅π‘π’ = πππ‘πππππ ππ£ππππππ‘ππ πππ‘βππππ¦ [ ] ππ Compressor Work [π΅π‘π’β] π π΅π‘π’ ππ ∗ (π πππππ πΉπππ€ π ππ‘π) ππ πππ = (π» − π» ) ∗ (60) πππ βπ π» = ππππ£πππ πππππππ π ππ πππ‘βππππ¦ Refrigeration Cycle π΅π‘π’ ;π» ππ π΅π‘π’ = πππ‘πππππ πππππππ ππ πππ‘βππππ¦ [ ] ππ Net Condenser Effect [π΅π‘π’β] π π΅π‘π’ ππ ∗ (π πππππ πΉπππ€ π ππ‘π) ππ πππ = (π» − π» ) ∗ (60) πππ βπ π» = πππ‘πππππ πππππππ ππ πππ‘βππππ¦ π΅π‘π’ ;π» ππ π΅π‘π’ = ππππ£πππ πππππππ ππ πππ‘βππππ¦ [ ] ππ Net Condenser Effect [π΅π‘π’β] π = π [π΅π‘π’β] + π [π΅π‘π’β] = π ∗ (β − β ) π −π =π∗ ππ’ππ → ππππ π πππ πππ π΅π‘π’ ; π = [ ]; ππππ = π€βπππ π πππ π = [ππ π], π = ππ‘ βπ βπ π΅πππππ → π = π ∗ (β − β ) ππ’ππ → ππππ Rankine Cycle πΆππππππ ππ → π = π ∗ (β − β ) ππ’πππππ → π = π ∗ (β − β ) πΈπππππππππ¦ = ππππ 8 http://www.engproguides.com − ππππ π Section 4.0 – Psychrometrics Dry Bulb Temperature Wet Bulb Temperature Dew Point Humidity Ratio Relative Humidity Sensible Heat Latent Heat Enthalpy Dry bulb temperature indicates the amount of energy independent of the amount of water in the air. Measured with a thermometer. Units = [β] Wet bulb temperature indicates the amount of water in the air. Measured with a sling psychrometer or hygrometer. Units = [β] The temperature at which moist air must be cooled to, in order for water to condense out of the air. Units = [β] Humidity ratio or specific humidity is the measure of the amount of water in air. ππππ‘π = ] [ Relative Humidity indicates the amount of water in the air relative to the total amount of water the air can hold. Units = [%] Sensible heat indicates the amount of dry heat. It indicates the amount of energy either absorbed or released to change the dry bulb temperature of the air. Btu ] Units = [ lb of air Latent heat indicates the amount of energy in the air due to moisture. It is the amount of heat released when water in the air condenses out or the amount of heat absorbed by water in order to vaporize the water. Btu ] Units = [ lb of air Enthalpy is an indication of the total amount of energy in the air, both sensible and latent. Btu Units = [ ] lb of air T , =T , ∗% +T , ∗% T , ∗ CFM + T , ∗ CFM T , = CFM + CFM W Mixing of Air Streams W , h h =W , ∗% +W , ∗% W , ∗ CFM + W , ∗ CFM = CFM + CFM , , , π =h , ∗% +h , ∗% h , ∗ CFM + h , ∗ CFM = CFM + CFM = 0.68 ∗ βπ ∗ πΆπΉπ = π ∗ π»π π π»π = βπππ‘ ππ π£ππππππ§ππ‘πππ π΅π‘π’ ππ‘ πππππ ππππππ‘ππππ π»π = 1,060 βπ Latent Heat (Air at Sea Level and Ideal Conditions) βπ πππππ ππ π»20 = πβππππ ππ π πππππππ βπ’πππππ‘π¦ ππ ππ πππ¦ πππ π = 4,734 ∗ βπ 9 http://www.engproguides.com ∗ πΆπΉπ ππ ππ π»20 = πβππππ ππ π πππππππ βπ’πππππ‘π¦ ππ ππ πππ¦ πππ πππ π πππ π»20 πππ , ∗ 0.075 ∗ ∗ π = πΆπΉπ ∗ 60 βπ ππ‘ πππ πππ¦ πππ 7000 ππππππ π»20 or πππ π , πππ ∗ 0.075 ∗ π = πΆπΉπ ∗ 60 ππ‘ πππ πππ¦ πππ βπ βπ Latent Heat π΅π‘π’ [ ] = ππ βπ β π€βπππ π = 0.240 ; π Sensible Heat (Air at Sea Level and Ideal Conditions) . π= ( π Sensible Heat Adjusted for Different Elevations π β )*( )*( ) . π΅π‘π’ [ ] = 1.08 ∗ βπ ∗ πΆπΉπ β π΅π‘π’ = 1.08 ∗ βπ ∗ πΆπΉπ ∗ [1 − ππππ£ ∗ 6.8754π₯10 ] . β ; ππππ£ ππ ππππ‘ = 4.5 ∗ (βh) ∗ CFM π΅π‘π’ π = π‘ππ‘ππ βπππ‘ [ ] βπ βh = πβππππ ππ πππ‘βππππ¦ πππ‘π€πππ πππ‘πππππ πππ ππππ£πππ πΆπΉπ = π£πππ’πππ‘πππ ππππ€ πππ‘π, ππ’πππ ππππ‘ πππ ππππ’π‘π Q Total Heat Equation Total Heat Equation Relative Humidity as a Function of Humidity Ratio and Partial Pressures SHR π΅π‘π’ = 4.5 ∗ πΆπΉπ ∗ (βh) ∗ [1 − ππππ£ ∗ 6.8754π₯10 ] . ; ππππ£ ππ ππππ‘ β W p x 100% ≈ x 100% RH = p W RH = πππππ‘ππ£π βπ’πππππ‘π¦ p = ππππ‘πππ ππππ π π’ππ ππ π€ππ‘ππ π£ππππ ππ π‘βπ πππ π π‘ππππ p = π ππ‘π’πππ‘ππ π£ππππ ππππ π π’ππ ππ π€ππ‘ππ ππ‘ π‘βπ π‘πππππππ‘π’ππ ππ ππ’ππ π‘πππ W = βπ’πππππ‘π¦ πππ‘ππ ππ π‘βπ πππ π π‘ππππ W = βπ’πππππ‘π¦ πππ‘ππ ππ π‘βπ πππ π π‘ππππ ππ‘ π ππ‘π’πππ‘πππ ππ‘ π‘βπ π‘πππππππ‘π’ππ ππ ππ’ππ π‘πππ π Btu Btu sensible heat ( ) sensible heat ( ) h h SHR = = Btu Btu total heat ( sensible heat + latent heat ( ) ) h h 10 http://www.engproguides.com Section 5.0 – Heat Transfer Term Equation Convert U-Factor to R-Value 1 π= π Addition of RValues Addition of UFactors Thermal Conductivity Units 1 = πΏππ¦πππ ππ π πππππ 1 1 1 + + β―+ π π π π= πΏππ¦πππ ππ π πππππ π΅π‘π’ βπ ∗ ππ‘ ∗ β Convert Thermal Conductivity to RValue and U-Factor π‘ π = π Heat Transfer Equation π = π ∗ π΄ ∗ βπ Log Mean Temperature Difference (LMTD) π΅π‘π’ ] βπ ∗ ππ‘ ∗ β βπ ∗ ππ‘ ∗ β π = π‘βπππππ πππ ππ π‘ππππ [ ] π΅π‘π’ π = βπππ‘ π‘ππππ πππ πππππππππππ‘ [ = π + π +β―+π π π Description πΏπππ· = βπ − βπ βπ ln( ) βπ π‘ = π‘βππππππ π ππ πππ‘πππππ [ππ‘] π΅π‘π’ π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ βπ ∗ ππ‘ ∗ β π π= π‘ π΅π‘π’ ] π = ππ£πππππ βπππ‘ π‘ππππ πππ πππππππππππ‘[ βπ ∗ ππ‘ ∗ β π΄ = ππππ ππ βπππ‘ π‘ππππ πππ [ππ‘ ] βπ = π‘πππ. ππππππππππ πππ‘π€πππ βππ‘ & ππππ πππππ [β] βπ = π‘πππππππ‘π’ππ ππππππππππ ππ‘ πππ‘πππππ βπ = π‘πππππππ‘π’ππ ππππππππππ ππ‘ ππ₯ππ‘ 11 http://www.engproguides.com Counter-flow Heat Exchanger Parallel-flow Heat Exchanger π΅π‘π’ βπ π΅π‘π’ π ∗ π΄ ∗ (π − π ) π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ ππ πππ‘πππππ π= βπ ∗ ππ‘ ∗ β π‘ π −π = π‘πππ. ππππ. πππ‘π€πππ πππππππ /ππ’π‘πππππ [β] π‘ = π‘βππππππ π ππ πππ‘πππππ [ππ‘] π΄ = ππππ ππ βπππ‘ π‘ππππ πππ [ππ‘ ] π΅π‘π’ ] β = ππππ£πππ‘ππ£π βπππ‘ π‘ππππ πππ πππππππππππ‘[ βπ ∗ ππ‘ ∗ β π΄ = ππππ ππ βπππ‘ π‘ππππ πππ [ππ‘ ] π = β ∗ π΄ ∗ βπ βπ = π‘πππ. ππππ. πππ‘π€πππ βππ‘ & ππππ πππππ ππ βπππ‘ π‘ππππ πππ [β] π = ππ’πππ‘ππ‘π¦ ππ βπππ‘ π‘ππππ ππππππ Conduction Heat Transfer Equation, Through Wall Convection Heat Transfer Equation 12 http://www.engproguides.com Radiative Heat Transfer Equation Radiative Heat Transfer Between Two Objects π΅π‘π’ = ππππππ‘πππ βπππ‘ π‘ππππ πππ πππππππππππ‘[ ] βπ ∗ ππ‘ ∗ β π΄ = ππππ ππ βπππ‘ π‘ππππ πππ [ππ‘ ] βπ = π‘πππ. ππππ. πππ‘π€πππ βππ‘ & ππππ πππππ ππ βπππ‘ π‘ππππ πππ [β] β ∗ π΄ ∗ βπ π=β π = π ∗ π ∗ π΄ ∗ (π = π ln −π π π π ) π = ππππππ‘πππ βπππ‘ (π΅π‘π’β); π΄ = ππππ (ππ‘ ); π΅π‘π’ π = 0.1713 π₯ 10 ; β ∗ ππ‘ ∗ π π = π π’πππππ π‘πππππππ‘π’ππ (°π ); π = ππππ π ππ£ππ‘π¦ 2ππΏ ∗ (π −π π ln π 1 ++ + β π π ) + π 1 β π΅π‘π’ βπ π΅π‘π’ π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ ππ πππ‘πππππ βπ ∗ ππ‘ ∗ β π = ππππππππ‘π’ππ ππ‘ π‘βπ πππππ ππππ π€πππ [β] = ππππππππ‘π’ππ ππ‘ π‘βπ ππππ ππ₯π‘πππππ [β] π π = πππππ πππππ’π ππ ππππ [ππ‘] = ππ’π‘π‘ππ πππππ’π ππ ππππ [ππ‘] π πΏ = ππππ πππππ‘β [ππ‘] π΅π‘π’ β = ππππ£πππ‘ππ£π βπππ‘ π‘ππππ πππ πππππππππππ‘[ βπ ∗ ππ‘ ∗ β π = ππ’πππ‘ππ‘π¦ ππ βπππ‘ π‘ππππ ππππππ Heat Transfer Equation, Through Pipe π = π πππππππ βπππ‘ Prandtl Number Nusselt Number ππ = ππ’ = π∗π π π·∗β π π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ π΅π‘π’ ; πππ ∗ β π΅π‘π’ β ∗ ππ‘ ∗ πππ ) π = ππ¦πππππ π£ππ πππ ππ‘π¦ ( βπ − ππ‘ β ππ‘ π· = ππππππ‘ππ ππ ππππ (ππ‘), π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ ππ π‘βπ πππ’ππ = πΆ Turbulent flow inside circular pipe (heating) Nusselt Number ππ’ = π·∗β π = .023 ∗ π π . ∗ ππ . Turbulent flow inside circular pipe (cooling) Nusselt Number ππ’ = π·∗β π = .023 ∗ π π . ∗ ππ . Laminar flow inside circular pipe (heating) Nusselt Number ππ’ = π·∗β π = 4.36 Laminar flow inside circular pipe (cooling) Nusselt Number ππ’ = π·∗β π = 3.66 13 http://www.engproguides.com ; Section 6.0 – Fluid Mechanics Term Kinematic and Dynamic Viscosity Specific Gravity Equation ππ π[ ] ππ‘ ππ‘ ∗ π =[ ] π£= ππ π π[ ] ππ‘ π ππΊ = π π= Mach Number Speed of Sound Description π= π€βπππ π = ππππ ππ‘π¦, π = ππ¦πππππ π£ππ πππ ππ‘π¦; π£ = πππππππ‘ππ π£ππ πππ ππ‘π¦ π’ π π∗π∗π ∗π = 62.4 π ππ @ 60β ππ‘ π = πππβ ππ’ππππ [π’πππ‘πππ π ]; π’ = π£ππππππ‘π¦ = π ππππ ππ π ππ’ππ π = πππβ ππ’ππππ; π = ππππππππ πππ ππππ π‘πππ‘; π = π‘πππππππ‘π’ππ ππ π ππππππ π π = πππ‘ππ ππ π πππππππ βπππ‘π = , π π = π πππππππ βπππ‘ ππ πππ ππ‘ ππππ π‘πππ‘ ππππ π π’ππ πππ π = π πππππππ βπππ‘ ππ πππ ππ‘ ππππ π‘πππ‘ π£πππ’ππ ππ‘ π ππ Air Properties Nozzles & Diffusers π / β / π£ / π΄ Conservation of mass βπ = 0 βπ = 0 π ∗π΄ ∗π£ =π ∗π΄ ∗π£ Units ππ‘ − ππ π ππ’π − °π π 1.4 N/A = πππ¦ βπππ‘ πππ‘πππππ ππ ππ₯ππ‘πππ π‘βπ π π¦π π‘ππ = πππ‘βππππ¦ ππ‘ π‘βπ πππ‘πππππ/ππ₯ππ‘ ππ π‘βπ πππ§π§ππ = π£ππ ππ‘ π‘βπ πππ‘πππππ πππ ππ₯ππ‘ ππ π‘βπ πππ§π§ππ π 1 π +β + π£ 2 1 =π +β + π£ 2 ππ‘ ;π π ππ 1,716 = ππππ ππ‘ π‘βπ πππ‘πππππ /ππ₯ππ‘ ππ π‘βπ πππ§π§ππ = π£ππππππ‘π¦ ππ‘ π‘βπ πππ‘πππππ /ππ₯ππ‘ ππ π‘βπ πππ§π§ππ / π / πππ ; ππ‘ π = πΌππππ£πππ’ππ πΊππ πΆπππ π‘πππ‘ ππ‘ − πππ πππ πππ; = 53.35 πππ − π π = π‘πππππππ‘π’ππ ππ π ππππππ π = ππππ π π’ππ Ideal gas law Darcy Weisbach Equation π =π∗π ∗π β= ππΏπ£ 2π·π π€βπππ β = ππ‘ ππ βπππ; π = π·ππππ¦ πππππ‘πππ ππππ‘ππ; π£ ππ‘ , = π£ππππππ‘π¦ π ππ ππ‘ ] π· = πππππ ππππππ‘ππ [ππ‘], π = ππππ£ππ‘π¦ [32.2 π ππ 14 http://www.engproguides.com Convert GPM to Cubic Feet Multiply GPM by ft 1 to get . 448.83 sec ππ‘ ; sec ππ‘ π = π£ππππππ‘π¦ ππ πππ’ππ ; π ππ π· = ππππππ‘ππ ππ ππππππ βπ¦ππππ’πππ ππππππ‘ππ [ππ‘] π = πππππππ‘ππ π£ππ πππ ππ‘π¦ Reynolds Number π∗π· π ππ¦πππππ ππ’ππππ = π Lift πΉ =πΆ ∗π∗π΄∗ π£ 2 π€βπππ πΆ = πππππππππππ‘ ππ ππππ‘; π = πππ’ππ ππππ ππ‘π¦; π΄ = ππππ ππ ππππππ‘; π£ = π£ππππππ‘π¦ ππ πππ’ππ Drag πΉ =πΆ ∗π∗π΄∗ π£ 2 π€βπππ πΆ = πππππππππππ‘ ππ ππππ; π = πππ’ππ ππππ ππ‘π¦; π΄ = ππππ ππ ππππππ‘; π£ = π£ππππππ‘π¦ ππ πππ’ππ ∗ [π] π΅ = ππ’ππ¦ππππ¦ πππππ; (πππ ππ π) ππ πππ = πππ’ππ ππππ ππ‘π¦; ( ππ ) π π ππ‘ = π£πππ’ππ ππ πππ ππππππ πππ’ππ π πππ ππΌ π = 9.81 π Buoyancy π΅= π π = Fluid Power Bulk Modulus Viscosity ∗π π½= π πΉ π΄ πππ ; πΉ = πππππ (πππ ); ππ π΄ = ππ = ππππ (ππ ) π = ππππ π π’ππ βπ ; ππ π βπ/π πΆπππ‘ππ π‘ππππ = 1.0764 π₯ 10 ππ‘ π ππ ππ‘ − π ππ£ πππ§ ππ£ πππ§ + +β =π + + π + 2π π 2π π π = 32.2 ; π = 32.2 ππ‘ π£ = ; π§ = ππ‘; π = ππ/ππ‘ ; β = πππ/ππ‘ π ππππ π = 0.067197 Bernouli’s 15 http://www.engproguides.com Section 7.0 – Energy/Mass Balance Term Equation π Dehumidification/ Humidification π =π Mixing General Equations π , ∗ (π − π ) π Mixing Air based on Temperature Mixing Air based on Enthalpy Mixing Air based on Humidity Ratio Latent Heat of Vaporization Description +π , +π π , , +π T T , , , = π , ππ ππ π»20 = π πππππππ βπ’πππππ‘π¦ ππ ππ πππ¦ πππ π = πππ π ππππ€ πππ‘π ππ πππ; = πππ π ππππ€ πππ‘π ππ π€ππ‘ππ π = π , +π , , = π , =T , ∗% +T , ∗% T , ∗ CFM + T , ∗ CFM = CFM + CFM =h , ∗% +h , ∗% h , ∗ CFM + h , ∗ CFM h , = CFM + CFM W , =W , ∗% +W , ∗% W , ∗ CFM + W , ∗ CFM W , = CFM + CFM h , 970 Btu/lbm 16 http://www.engproguides.com Section 8.0 – Heating/Cooling Loads Term Equation Convert U-Factor to R-Value π= Addition of R-Values Addition of U-Factors Thermal Conductivity Units Convert Thermal Conductivity to RValue and U-Factor Heat Transfer Equation 1 π 1 = π΅π‘π’ ] βπ ∗ ππ‘ ∗ β βπ ∗ ππ‘ ∗ β π = π‘βπππππ πππ ππ π‘ππππ [ ] π΅π‘π’ π = βπππ‘ π‘ππππ πππ πππππππππππ‘ [ = π + π + π …+ π π π Description 1 1 1 1 + + …+ π π π π π= π΅π‘π’ βπ ∗ ππ‘ ∗ β π‘ π π π= π‘ π = π = π ∗ π΄ ∗ βπ π‘ = π‘βππππππ π ππ πππ‘πππππ [ππ‘] π΅π‘π’ π = π‘βπππππ πππππ’ππ‘ππ£ππ‘π¦ [ βπ ∗ ππ‘ ∗ β π΅π‘π’ ] π = ππ£πππππ βπππ‘ π‘ππππ πππ πππππππππππ‘[ βπ ∗ ππ‘ ∗ β π΄ = ππππ ππ βπππ‘ π‘ππππ πππ [ππ‘ ] βπ = π‘πππππππ‘π’ππ ππππππππππ πππ‘π€πππ βππ‘ πππ ππππ πππππ ππ βπππ‘ π‘ππππ πππ [β] Roof/Wall πΆππππ’ππ‘ππ£π πΏππππ → π = π ∗ π΄ ∗ πΆπΏππ· Skylight/Window πΆππππ’ππ‘ππ£π πΏππππ → π = π ∗ π΄ ∗ πΆπΏππ· Or πΆππππ’ππ‘ππ£π πΏππππ → π = π ∗ π΄ ∗ βπ π πππππ‘ππ£π πΏππππ → π = π΄ ∗ ππΆ ∗ ππΆπΏ ππΆ = π βπππππ πππππππππππ‘ ππΆπΏ = π ππππ πππππππ ππππ ππππ‘ππ People Sensible loads π = π ∗ ππ»πΊ ∗ πΆπΏπΉ π = ππ’ππππ ππ ππππππ ππ»πΊ = π πππ ππππ βπππ‘ ππππ, πππ‘ππ£ππ‘π¦ πππππππππ‘ πΆπΏπΉ = πππππππ ππππ ππππ‘ππ π = π ∗ πΏπ»πΊ ∗ πΆπΏπΉ π = ππ’ππππ ππ ππππππ People Latent loads πΏππ‘πππ‘ = πππ‘πππ‘ βπππ‘ ππππ, πππ‘ππ£ππ‘π¦ πππππππππ‘ πΆπΏπΉ = πππππππ ππππ ππππ‘ππ 17 http://www.engproguides.com π΅π‘π’ βπ π΅π‘π’ βπ π΅π‘π’ βπ ∗ ππΉ ∗ ππ΄πΉ ∗ ππΉ π = π ∗ πππ‘π‘π ∗ π€ππ‘π‘π π€βππ π = ππ’ππππ ππ πππβπ‘ π‘π¦ππ. ππΉ = π’π πππ ππππ‘ππ ππ΄πΉ = π ππππππ πππππ€ππππ ππππ‘ππ ππΉ = π ππππ πππππ‘πππ 3.412 Lighting π (π»π) π΅π‘π’β ∗ π»π πΈπππππππππ¦ πππ‘ππ π»πππ‘ πΏππ π = πππ‘ππ π»πππ‘ ∗ (1 − πΈπππππππππ¦) πΈππ’ππππππ‘ π»πππ‘ πΏππ π = πππ‘ππ π»πππ‘ ∗ (πΈπππππππππ¦) πππ‘ππ π»πππ‘ = π = 2545 = πππ‘ππ π»πππ‘ ∗ (1 − π )∗πΉ ∗πΉ = πππ‘ππ π»πππ‘ ∗ (π )∗πΉ ∗πΉ π = βππππ πππ€ππ ππ πππ‘ππ = ππππππππππ¦ ππ πππ‘ππ π πΉ = π’π πππ ππππ‘ππ ππ π‘βπ πππ‘ππ πΉ = ππππ ππππ‘ππ ππ π‘βπ πππ‘ππ π π Miscellaneous Equipment ∗πΉ ∗πΉ π΅π‘π’ = ππππ’π‘ π‘π π‘βπ πππ’ππππππ‘ π β πΉ = πππππ‘πππ ππ π‘βπ π‘ππ‘ππ βπππ‘ π‘βππ‘ ππ ππππππ‘ππ π‘π π‘βπ π ππππ πΉ = π’π πππ ππππ‘ππ π=π π΅π‘π’ = 4.5 ∗ πΆπΉπ ∗ ββ [ ] ππ = 1.08 ∗ πΆπΉπ ∗ (π −π = 4,840 ∗ πΆπΉπ ∗ (π −π π = βπ’πππππ‘π¦ πππ‘ππ [πππ /πππ π Infiltration π π 18 http://www.engproguides.com ) ) ] Section 9.0 – Equipment & Components Term Simplified Sensible Heat Equation Affinity laws (For when given a system curve and diameter change is less than 5%) Contact Factor Equation for Coils Equation π Description π΅π‘π’ = 1.08 ∗ πΆπΉπ ∗ βπ[β] β π· π = ; ππ π ππππ ππ βπππ ππππ π‘πππ‘, π π· π€βπππ π = ππππ€ πππ π· = ππππππ‘ππ π π = ; ππ ππππππ‘ππ ππ βπππ ππππ π‘πππ‘ π π π· π» = ; ππ π ππππ ππ βπππ ππππ π‘πππ‘, π» π· π€βπππ π» = ππππ π π’ππ, π· = ππππππ‘ππ π» π = ; ππ ππππππ‘ππ ππ βπππ ππππ π‘πππ‘ π» π π· π = ; ππ π ππππ ππ βπππ ππππ π‘πππ‘ π π· π€βπππ π = πππ€ππ πππ π· = ππππππ‘ππ π π = ; ππ ππππππ‘ππ ππ βπππ ππππ π‘πππ‘ π π −β β πΆπππ‘πππ‘ πΉπππ‘ππ = β −β Contact Factor Equation for Coils πΆπππ‘πππ‘ πΉπππ‘ππ = Contact Factor Equation for Coils πΆπππ‘πππ‘ πΉπππ‘ππ = Bypass Factor ∗ πππ ππππππ‘ππππ ππ‘ 70β πππ 1 ππ‘π. −π π π where h is equal to the enthalpy where T is equal to the dry bulb temperature −π π π −π −π π΅π¦πππ π πΉπππ‘ππ = 1 − πΆπππ‘πππ‘ πΉπππ‘ππ π» = 60 ∗ π ∗ π ∗ (π −π ) π = π‘βπ βπ’πππππ‘π¦ πππ‘ππ πππ‘πππππ ππ ππππ£πππ π‘βπ π π¦π π‘ππ Moisture Transfer Equation ππ π = ππππ ππ‘π¦ ππ πππ [ ] ππ‘ ππ‘ π = πππ ππππ€ πππ‘π [ ] πππ ππ π» = ππππ π‘π’ππ π‘ππππ ππππππ [ ] βπ π ππππ = π Cooling Tower π΄ππππππβ = π [β] − π , , πΈπππππ‘ππ£ππππ π = [β] − π [β] , , [β] π ππππ π ππππ + π΄ππππππβ 19 http://www.engproguides.com [ππ ππ π€ππ‘ππ] [ππ ππ πππ¦ πππ] π[π΅π‘π’β] = 500 ∗ ππππ€ πππ‘π πππ ∗ π πππ πππ = ππ£ππππππ‘πππ πππ‘π πππ . 000943 ∗ πππππππ π‘ππ€ππ ππππ€ πππ‘π Cooling Tower Evaporation Rate Fans πππ ∗ (π πππ , −π , πππ‘ππ π»π = π΅π»π ∗ πππ π∗π£ = 2∗π ππ‘ ππ‘ − πππ π = 32.2 πππ − π ππ‘ π£ = ; π§ = ππ‘; ππ = ππ/ππ‘ π ππ = Shortcut Equation πΉππ [ππ. π€π] πΉππ π£ππππππ‘π¦ ππππ π π’ππ = 4005 π€βπππ πΉππ = π£ππππππ‘π¦ ππππ‘ πππ ππππ’π‘π π π π» π» π π π π π π π π π π π» π» π π Affinity Laws for all other cases (fans and pumps) N = speed, Q = flow rate, H = head or pressure, p = density, D = diameter, P = power ACFM vs. SCFM π π −π ) , 1 ) π΅ππππ βπππ ππππ€ππ = ππ»π ∗ ( πππ ππππππππππ¦ πΉππ πππβππππππ βπππ ππππ€ππ πΆπΉπ ∗ πππ[ππ. π€π] = 6,356 Fans , π ( ) π π π π· = ( ) π· π π π π π· = ( ) π· π π π» . π . π· = π· π» π π· π» . π . = ( ) π· π» π π· π» . π . = π· π» π π π· = ( ) π· π π π π· = ( ) π· π π π π π· = ( ) π· π π π 519 ππΆπΉπ = π΄πΆπΉπ ∗ ∗( ) π 14.7 = πππ‘π’ππ ππππ π π’ππ ππ πππ πππ’π‘π π‘ππππ (ππ ππ); = πππ‘π’ππ π‘πππππππ‘π’ππ ππ π ππππππ (°π ) = π· π· 20 http://www.engproguides.com 1 πππ‘ππ ππππππππππ¦ π π‘ 10 → π‘βππ π€πππππ ππππ π π’ππ π£ππ π ππ ππ π π’πππ‘πππ ππ πππ ππ¦πππππππππ π‘βππ π€πππππ ππππ π π’ππ π£ππ π πππ → π = π‘ ππ πππ π πβππππππ π‘βππ π€πππππ ππππ π π’ππ π£ππ π πππ → π = 2π π = π π‘πππ π (ππ π); π = ππππ π π’ππ (ππ π); π = πππππ’π (ππ) π‘ = π‘βππππππ π (ππ) Pressure vessel πΊππ ππππ π π’ππ ππππππ π£πππ£π π ππ§πππ π ∗ √ππ → π΄= πΆ ∗ πΎ ∗ π ∗ πΎ ∗ √π Pressure relief βπ ππΊ π=πΆ Control valve liquid π = 59.64 ∗ πΆ Control valve gas βπ ∗π 520 ππΊ ∗ π π 520 π =πΆ ∗π ∗ ππΊ ∗ π Control valve critical point π Boiler fuel energy = π ∗ π»π»π π΄ = π£πππ£π ππππππ‘ππ£π πππππππ ππππ (ππ ); πΆ = πππππππππππ‘ ππ πππ (315 πππ πππ); πΎ = πππππππππππ‘ ππ πππ πβππππ (π‘π¦πππππππ¦ 0.975); πΎ = πππππππ‘πππ ππππ‘ππ πππ ππππ ππππ π π’ππ π = ππππππ’πππ π€πππβπ‘ ππ πππ (π ππ ππππππππ π‘ππππ); π = ππππππ£πππ ππππ π π’ππ (ππ ππ); π = πππ πππ’π‘π π‘πππππππ‘π’ππ (°π ); πππ ; π = ππππππ£πππ πππππππ‘π¦ βπ π = πππππππ π ππππππ‘π¦ ππππ‘ππ π = π£πππ’πππ‘πππ ππππ€ πππ‘π (πππ); πΆ = π£πππ£π ππππππππππππ‘; βπ = ππππ π π’ππ ππππ (ππ π) ππΊ = π πππππππ ππππ£ππ‘π¦ ππ πππ’ππ π = π£πππ’πππ‘πππ ππππ€ πππ‘π (π ππβ − π π‘ππππππ ππ’πππ ππππ‘ πππ βππ’π); πΆ = π£πππ£π ππππππππππππ‘; βπ = ππππ π π’ππ ππππ (ππ π); π = πππ πππ’π‘π π‘πππππππ‘π’ππ (°π ) ππΊ = π πππππππ ππππ£ππ‘π¦ ππ πππ (πππ = 1.0); π = ππππππ‘πππ ππππ π π’ππ (ππ π) π€βπππ πΆ = πππ π£πππ£π πππππππππππ‘ π΅π‘π’ π»π»π = βππβππ βπππ‘πππ π£πππ’π ππ ππ’ππ [ ] ππ ππ π = πππ π ππππ€ πππ‘π ππ ππ’ππ βπ π Boiler efficiency Boiler energy for water temperature increase = (π ) ∗ (β π −β ∗ π»π»π , ∗ βπ = π ∗ π ∗ βπ = πππ /βπ π΅π‘π’ π = βπππ‘ πππππππ‘π¦ ππ π€ππ‘ππ = 1.00 πππ ∗ β βπ = πβππππ ππ π‘πππππππ‘π’ππ ππ πππ [β] π = 500 ∗ πΊππ π 21 http://www.engproguides.com ∗β π=π π΅π‘π’ πππ = πππ π ππππ€ πππ‘π ππ πππ /βπ π 1 ππ‘ 62.4 ππ 60 ππππ’π‘π ππππππ ππ π€ππ‘ππ ∗ ∗ ∗ 1 7.48 ππππππ ππ‘ βππ’π ππππ’π‘π β Boiler energy for vaporization of water = βπππ‘ ππ π£ππππππ§ππ‘πππ ππ π€ππ‘ππ π = 500 ∗ πΊππ ∗β =π Desuperheating Region ∗π , ∗ (π 1π π‘ ππ‘ππ: π = π Feedwater heater ∗β −π Condensing Region ∗β 2ππ ππ‘ππ: π = π Subcooling Region ∗π , ∗ π 3ππ ππ‘ππ: π = π −π 22 http://www.engproguides.com ) Section 10.0 – Systems & Components Term Rectangular Duct Oval Duct Equation π· = 1.30 ∗ (π ∗ π) . (π + π) . Description π€βπππ a and b are the width[ft] and height[ft] of the duct (π΄ ∗ π΄πππ) . (πππππππ‘ππ) . πππππππ‘ππ = π ∗ π + 2 ∗ (π΄ − π) π· = 1.55 ∗ π€βπππ A is the major axis and a is the minor axis Velocity Pressure as a Function of Air Velocity ππ = Friction loss due to length of duct πΉ ππ. π€π [ππ. π€π] = πΏ[ππ‘] ∗ π[ ] 100 ππ‘ Friction loss due to duct fitting πΉ [ππ. π€π] = πΆ ∗ ππ Compressed air piping loss 0.1025πΏπ ππ . Where, π = ππππ π π’ππ ππππ (ππ π) πΏ = ππππ πππππ‘β (ππ‘) π = πππππππ€ (πππ) πππ πβππππ ππππ π π’ππ @ ππππ πππ‘πππππ π = πππ‘ππ πππππππππ π πππ ππππ πππ ππππ π π’ππ π = πππ‘πππππ ππππ πππππ‘ππ (ππ) πΉππ 4005 πΉππ = πππ π£ππππππ‘π¦ ππ ππππ‘ πππ ππππ’π‘π ππ = π£ππππππ‘π¦ ππππ π π’ππ [ππ. π€π] [ππ. π€π] π= Darcy Weisbach Pressure h= fLv 2Dg 1 psi is equal to 2.31 feet of head (water) NPSH = Suction Head NPSHA = (P Net positive suction head ππππ»π΄ = π ππππ»π΄ = π Velocity pressure (Pumps) h = ft of head; f = Darcy friction factor; ft , v = velocity sec D = inner diameter [ft], ft g = gravity [32.2 ] sec π [ππ‘ ππ βπππ]; 2π P − Vapor Pressure − P ) − VP π −π or +π −π −π π£ππππππ‘π¦ ππ 23 http://www.engproguides.com ππ‘ ππ‘ ; ππππ£ππ‘π¦ = 32.2 π ππ sec π π , π π Air to air energy recovery devices π π = 4.5 ∗ πΆπΉπ = 4.5 ∗ πΆπΉπ , , , , , ∗ (β ∗ (β −β −β ) ) = 1.08 ∗ πΆπΉπ = 1.08 ∗ πΆπΉπ ∗ (π ∗ (π −π −π ) ) = 4,770 ∗ πΆπΉπ = 4,770 ∗ πΆπΉπ ∗ (π ∗ (π −π −π ) ) π π π π = π π = π , = π π , , , , , 24 http://www.engproguides.com Section 11.0 – Supportive Knowledge Term Fresh air flow based on ASHRAE 62.1 Equation πΉπππ β πππ ππππ€ πππ‘π = π ∗ (# ππ ππππππ) + π ∗ (π΄πππ) Refrigeration Room Ventilation Rate Vibration Control: Natural Frequency of Spring Vibration Control: Transmissibility and Vibration Isolation Efficiency Combining the Sound Levels of Multiple Sources Description π[πΆπΉπ] = 100ππΊ . , where G = lbs of refrigerant. π (π»π§) = 3.13 ∗ π= 1 π π‘ππ‘ππ πππππππ‘πππ (πππβππ ) πππππ πππ π ππππππ‘π¦% = 100% − ππππππ‘πππ πΌπ ππππ‘πππ πΈπππππππππ¦% 1 π π −1 πΏ = 10 ∗ log (10 Sound Level at a Distance from a Point Source (Spherical Propagation) + 10 + 10 + 10 + 10 +. . ) − 20 ∗ log π₯ − 1 πΏ =πΏ πΏ = πππ’ππ πππ£ππ ππ‘ π πππ π‘ππππ ππ π₯ [π·π΅] = πππ’ππ πππ£ππ ππ πππ’ππππππ‘ [π·π΅] πΏ π₯ = πππ π‘ππππ ππππ πππ’ππππππ‘ [ππ‘ ] Sound Level at a Distance from a Point Source (Half-Spherical Propagation) πΏ =πΏ − 20 ∗ log π₯+2 Sound Level at a Distance from a Point Source (Quarter-Spherical Propagation) πΏ =πΏ − 20 ∗ log π₯+5 Sound Level at a Distance from a Point Source (Eighth-Spherical Propagation) πΏ =πΏ − 20 ∗ log π₯+8 25 http://www.engproguides.com