T-134a—ENG Technical Information

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T-134a—ENG Technical Information

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  T-134a—ENG TechnicalInformation ThermodynamicProperties of  HFC- 134 a (  1,1,1,2 -tetrafluoroethane) The DuPont Oval Logo, The miracles of science™,and Suva ® , are trademarks or registered trademarks of E.I. du Pont de Nemours and Company. DuPont ™  Suva ® refrigerants DuPont Product Names:DuPont™ Suva  ®   134a RefrigerantDuPont™ Formacel  ®   Z-4 Blowing AgentDuPont™ Dymel  ®   134a Aerosol PropellantDuPont™ Dymel  ®   134a/P Aerosol Propellant(Pharmaceutical Grade)  New tables of the thermodynamic properties of HFC-134ahave been developed and are presented here. These tablesare based on experimental data from the database at theNational Institute of Standards and Technology (NIST).Equations have been developed, based on the ModifiedBenedict-Webb-Rubin (MBWR) equation of state, whichrepresent the data with accuracy and consistency through-out the entire range of temperature, pressure, and density. Physical Properties Chemical FormulaCH 2 FCF 3 Molecular Weight102.03Boiling Point atOne Atmosphere–14.9°F(–26.06°C)Critical Temperature213.9°F(101.08°C)673.6°R(374.23 K)Critical Pressure588.9 psia(4060.3 kPa [abs])Critical Density32.17 lb/ft 3 (515.3 kg/m 3 )Critical Volume0.031 ft 3  /lb(0.00194 m 3  /kg) Units and Factors t=temperature in °FT=temperature in °R = °F + 459.67P=pressure in lb/in 2  absolute (psia)v f  =volume of saturated liquid in ft 3  /lbv g =volume of saturated vapor in ft 3  /lbV=volume of superheated vapor in ft 3  /lbd f  =1/v f   = density of saturated liquid in lb/ft 3 d g =1/v g  = density of saturated vapor in lb/ft 3 h f  =enthalpy of saturated liquid in Btu/lbh fg =enthalpy of vaporization in Btu/lbh g =enthalpy of saturated vapor in Btu/lbH=enthalpy of superheated vapor in Btu/lbs f  =entropy of saturated liquid in Btu/(lb) (°R)s g =entropy of saturated vapor in Btu/(lb) (°R)S=entropy of superheated vapor in Btu/(lb) (°R)C p =heat capacity at constant pressure in Btu/(lb) (°F)C v =heat capacity at constant volume in Btu/(lb) (°F)v s =velocity of sound in ft/secThe gas constant, R = 10.732 (psia) (ft 3 )/(°R) (lb–mole)for HFC-134a, R = 0.1052 (psia) (ft 3 )/lb •  °ROne atmosphere = 14.696 psiaConversion factor from Work Units to Heat Units:J=0.185053Btu/lb=[(psia •  ft 3 )/lb] JReference point for enthalpy and entropy:h f  =0.0 Btu/lb at –40°Fs f  =0.0 Btu/lb •  °R at –40°F Thermodynamic Properties of HFC-134a Refrigerant(1,1,1,2-tetrafluoroethane) Engineering (I/P) Units Equations The Modified Benedict-Webb-Rubin (MBWR) equation of state was used to calculate the tables of thermodynamicproperties. It was chosen as the preferred equation of statebecause it provided the most accurate fit of the thermo-dynamic data over the entire range of temperatures andpressures presented in these tables. The data fit andcalculation of constants for HFC-134a were performed forDu Pont at the National Institute of Standards and Tech-nology (NIST) under the supervision of Dr. Mark O.McLinden.The constants were calculated in SI units. For conversionof thermodynamic properties to Engineering (I/P) units,properties must be calculated in SI units and converted toI/P units. Conversion factors are provided for each prop-erty derived from the MBWR equation of state. 1.Equation of State (MBWR)= ΣΣΣΣΣ  a n /V n  + exp (–V c2 /V 2 ) ΣΣΣΣΣ  a n /V 2n–17 where the temperature dependence of the coefficients isgiven by:a 1 = RTa 2 = b 1 T + b 2 T 0.5  + b 3  + b 4  /T + b 5  /T 2 a 3 = b 6 T + b 7  + b 8  /T + b 9  /T 2 a 4 = b 10 T + b 11  + b 12  /Ta 5 = b 13 a 6 = b 14  /T + b 15  /T 2 a 7 = b 16  /Ta 8 = b 17  /T + b 18  /T 2 a 9 = b 19  /T 2 a 10 = b 20  /T 2  + b 21  /T 3 a 11 = b 22  /T 2  + b 23  /T 4 a 12 = b 24  /T 2  + b 25  /T 3 a 13 = b 26  /T 2  + b 27  /T 4 a 14 = b 28  /T 2  + b 29  /T 3 a 15 = b 30  /T 2  + b 31  /T 3 + b 32  /T 4 where T is in K = °C + 273.15, V is in liters/mole(= m 3  /kg ∞  MW), V c  = 0.199334 liters/mole, P is in kPa,and R = 0.08314471 bar (absolute) ∞  liters/mole ∞  K. P100 9n=115n=10 1  MBWR coefficients for HFC-134a: b 1 =–6.5455235227E–02b 2 =5.8893751817E+00b 3 =–1.3761788409E+02b 4 =2.2693168845E+04b 5 =–2.9262613296E+06b 6 =–1.1923776190E–04b 7 =–2.7214194543E+00b 8 =1.6295253680E+03b 9 =7.2942203182E+05b 10 =–1.1724519115E–04b 11 =8.6864510013E–01b 12 =–3.0660168246E+02b 13 =–2.5664047742E–02b 14 =–2.4381835971E+00b 15 =–3.1603163961E+02b 16 =3.4321651521E–01b 17 =–1.0154368796E–02b 18 =1.1734233787E+00b 19 =–2.7301766113E–02b 20 =–6.6338502898E+05b 21 =–6.4754799101E+07b 22 =–3.7295219382E+04b 23 =1.2614735899E+09b 24 =–6.4742200070E+02b 25 =1.2362450399E+05b 26 =–1.5699196293E+00b 27 =–5.1848932204E+05b 28 =–8.1396321392E–02b 29 =3.0325168842E+01b 30 =1.3399042297E–04b 31 =–1.5856192849E–01b 32 =9.0679583743E+00 Ideal Gas Heat Capacity Equation (at constantpressure):C p  (J/mole •  K) = cp1 + cp2 T + cp3 T 2 cp1=1.94006 E+01cp3=–1.29665 E–04cp2=2.58531 E–01R=8.314471 J/mole •  KMW=102.03Properties calculated in SI units from the equation andconstants listed above can be converted to I/P unitsusing the conversion factors shown below. Please notethat in converting enthalpy and entropy from SI to I/Punits, a change in reference states must be included(from H = 200 and S = 1 at 0°C for SI units to H = 0and S = 0 at –40°C for I/P units). In the conversionequation below, H (ref) and S (ref) are the saturatedliquid enthalpy and entropy at –40°C. For HFC-134a,H (ref) = 148.4 kJ/kg and S (ref) = 0.7967 kJ/kg •  K.P (psia)=P (kPa) •  0.14504T (°F)=(T[°C] •  1.8) + 32D (lb/ft 3 )=D (kg/m 3 ) •  0.062428V (ft 3  /lb)=V (m 3  /kg) •  16.018H (Btu/lb)=[H (kJ/kg) – H (ref)] •  0.43021S (Btu/lb •  °R)=[S (kJ/kg •  K) – S (ref)] •  0.23901C p  (Btu/lb •  °F)=C p  (kJ/kg •  K) •  0.23901C v  (Btu/lb •  °F)=C v  (kJ/kg •  K) •  0.23901v s  (ft/sec)=v s  (m/sec) •  3.2808 2.Martin-Hou Equation of State (fit from MBWR data)As previously stated, the thermodynamic propertiespresented in these tables are based on the MBWRequation of state. Coefficients for the Martin-Houequation of state are presented below for the conve-nience of those who may have existing computerprograms based on this equation of state. While not asaccurate as the data from the MBWR equation of state,particularly in the superheated region, data calculatedusing these Martin-Hou coefficients should be suffi-cient for most engineering calculations. P = RT/(V–b) + ΣΣΣΣΣ  (A i  + B i T + C i  exp (–kT/T c ))/(V–b)  i For SI units T and T c  are in K = °C + 273.15, V is in m 3  /kg, andP is in kPaR = 0.0815 kJ/kg •  Kb, A i , B i , C i , k are constants:A 2 =–8.909485 E–02A 4 =1.778071 E–05B 2 =4.408654 E–05B 4 =–4.016976 E–08C 2 =–2.074834 E+00C 4 =–2.977911 E–04A 3 =–1.016882 E–03A 5 =–7.481440 E–08B 3 =2.574527 E–06B 5 =1.670285 E–10C 3 =2.142829 E–02C 5 =1.255922 E–06b=3.755677 E–04k=4.599967 E+00 O 5i=2 2  For I/P units T and T c  are in °R = °F + 459.67, V is in ft 3  /lb, andP is in psiaR = 0.1052 (psia)(ft 3 )/lb •  °Rb, A i , B i , C i , k are constants:A 2 =–3.315708 E+00A 4 =1.697907 E–01B 2 =9.115011 E–04B 4 =–2.131040 E–04C 2 =–7.721597 E+01C 4 =–2.843653 E+00A 3 =–6.061984 E–01A 5 =–1.144381 E–02B 3 =8.526469 E–04B 5 =1.419396 E–05C 3 =1.277414 E+01C 5 =1.921091 E–01b=6.016014 E–03k=4.599967 E+00 Ideal Gas Heat Capacity   (at constant volume):C v  = a + bT + cT 2  + dT 3  + f/T 2 For SI units C v  = kJ/kg •  KT is in K = °C + 273.15a, b, c, d, f are constants:a=3.154856 E+00d=–3.754497 E–08b=–1.656054 E–02f=–3.023189 E+04c=4.353378 E–05 For I/P units C v  = Btu/lb •  °RT is in °R = °F + 459.67a, b, c, d, f are constants:a=7.540287 E–01d=–1.538660 E–09b=–2.198925 E–03f=–2.341093 E+04c=3.211365 E–06 3.Vapor Pressurelog 10  P sat  = A + B/T + C log 10  T + D T +E ([F–T]/T) log 10  (F–T)For SI units T is in K = °C + 273.15 and P is in kPaA, B, C, D, E, F are constants:A=4.069889 E+01D=7.616005 E–03B=–2.362540 E+03E=2.342564 E–01C=–1.306883 E+01F=3.761111 E+02 For I/P units T is in °R = °F + 459.67 and P is in psiaA, B, C, D, E, F are constants:A=4.325629 E+01D=4.231114 E–03B=–4.293056 E+03E=2.342564 E–01C=–1.306883 E+01F=6.770000 E+02 4.Density of the Saturated Liquidd f   = A f   + B f   (1–T r ) (1/3)  + C f   (1–T r ) (2/3)  + D f   (1–T r )+ E f   (1–T r ) (4/3) For SI units T r  = T/T c , both in K = °C + 273.15 and d f   is in kg/m 3 A f  , B f  , C f  , D f  , E f   are constants:A f  =5.281464 E+02D f  =–9.491172 E+02B f  =7.551834 E+02E f  =5.935660 E+02C f  =1.028676 E+03 For I/P units T r  = T/T c , both in °R = °F + 459.67 and d f   isin lb/ft 3 A f  , B f  , C f  , D f  , E f   are constants:A f  =3.297110 E+01D f  =–5.925145 E+01B f  =4.714456 E+01E f  =3.705512 E+01C f  =6.421816 E+01 o o o 3
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