Calculo de la Tension Superficial con la correlacion de Hakim (ϒ Hakim) para cada valor de temperatura experimental

4.- Calculo de la Tension Superficial con la correlacion de Hakim (ϒ Hakim) para cada valor de temperatura experimental.

tfria= 16 ° C = 16 + 273.15 = 289.15 K

tambiente= 24°C = 24 + 273.15 = 297.15 K

tcaliente= 50° C = 50 + 273.15 = 323.15 K

ϒ Hakim = Pc 2/3 Tc 1/3 q (( 1-Texp/Tc)/0.4) M

ϒ Hakim a tfria = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(289.15 K)/(647.13 K))/0.4) 1.0033= 78.423 dina/cm

ϒ Hakim a tambiente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(297.15 K)/(647.13 K))/0.4) 1.0033= 76.665 dina/cm

ϒ Hakim a tcaliente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(323.15 K)/(647.13 K))/0.4) 1.0033= 70.952 dina/cm

5.-Tensión superficial Exaxta ( ϒ exacta )

ϒ exacta = 75.65 – 0.14(t exp) – 2.7 x10-4 (t exp)2

ϒ exacta a tfria = 75.65 – 0.14(16 °C) – 2.7 x10-4 (16 °C)2= 73.340 dina/cm

ϒ exacta a tambiente = 75.65 – 0.14(24 °C) – 2.7 x10-4 (24 °C)2= 72.134 dina/cm

ϒ exacta a tcaliente = 75.65 – 0.14(50 °C) – 2.7 x10-4 (50 °C)2= 67.975 dina/cm

4.- Calculo de la Tension Superficial con la correlacion de Hakim (ϒ Hakim) para cada valor de temperatura experimental.

tfria= 16 ° C = 16 + 273.15 = 289.15 K

tambiente= 24°C = 24 + 273.15 = 297.15 K

tcaliente= 50° C = 50 + 273.15 = 323.15 K

ϒ Hakim = Pc 2/3 Tc 1/3 q (( 1-Texp/Tc)/0.4) M

ϒ Hakim a tfria = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(289.15 K)/(647.13 K))/0.4) 1.0033= 78.423 dina/cm

ϒ Hakim a tambiente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(297.15 K)/(647.13 K))/0.4) 1.0033= 76.665 dina/cm

ϒ Hakim a tcaliente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(323.15 K)/(647.13 K))/0.4) 1.0033= 70.952 dina/cm

5.-Tensión superficial Exaxta ( ϒ exacta )

ϒ exacta = 75.65 – 0.14(t exp) – 2.7 x10-4 (t exp)2

ϒ exacta a tfria = 75.65 – 0.14(16 °C) – 2.7 x10-4 (16 °C)2= 73.340 dina/cm

ϒ exacta a tambiente = 75.65 – 0.14(24 °C) – 2.7 x10-4 (24 °C)2= 72.134 dina/cm

ϒ exacta a tcaliente = 75.65 – 0.14(50 °C) – 2.7 x10-4 (50 °C)2= 67.975 dina/cm

4.- Calculo de la Tension Superficial con la correlacion de Hakim (ϒ Hakim) para cada valor de temperatura experimental.

tfria= 16 ° C = 16 + 273.15 = 289.15 K

tambiente= 24°C = 24 + 273.15 = 297.15 K

tcaliente= 50° C = 50 + 273.15 = 323.15 K

ϒ Hakim = Pc 2/3 Tc 1/3 q (( 1-Texp/Tc)/0.4) M

ϒ Hakim a tfria = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(289.15 K)/(647.13 K))/0.4) 1.0033= 78.423 dina/cm

ϒ Hakim a tambiente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(297.15 K)/(647.13 K))/0.4) 1.0033= 76.665 dina/cm

ϒ Hakim a tcaliente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(323.15 K)/(647.13 K))/0.4) 1.0033= 70.952 dina/cm

5.-Tensión superficial Exaxta ( ϒ exacta )

ϒ exacta = 75.65 – 0.14(t exp) – 2.7 x10-4 (t exp)2

ϒ exacta a tfria = 75.65 – 0.14(16 °C) – 2.7 x10-4 (16 °C)2= 73.340 dina/cm

ϒ exacta a tambiente = 75.65 – 0.14(24 °C) – 2.7 x10-4 (24 °C)2= 72.134 dina/cm

ϒ exacta a tcaliente = 75.65 – 0.14(50 °C) – 2.7 x10-4 (50 °C)2= 67.975 dina/cm

4.- Calculo de la Tension Superficial con la correlacion de Hakim (ϒ Hakim) para cada valor de temperatura experimental.

tfria= 16 ° C = 16 + 273.15 = 289.15 K

tambiente= 24°C = 24 + 273.15 = 297.15 K

tcaliente= 50° C = 50 + 273.15 = 323.15 K

ϒ Hakim = Pc 2/3 Tc 1/3 q (( 1-Texp/Tc)/0.4) M

ϒ Hakim a tfria = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(289.15 K)/(647.13 K))/0.4) 1.0033= 78.423 dina/cm

ϒ Hakim a tambiente = (220.55 bar) 2/3 (647.13 K) 1/3 (0.180) (( 1-(297.15 K)/(647.13 K))/0.4) 1.0033= 76.665 dina/cm

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