Petroleo 404 DEBER
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EJERCICIO RESUELTO
![ln(−∆Tc) = −0.82868 − 0.08997(130.34) 0.45 + 2.4567(1.3325) 0.45
ln(∆TC) = −0.828368 − 0.80516 + 2.7955
ln(∆TC) = 1.16666
∆TC = 3.21
∆Tc = 130.34 − (3.21) = 127.13
∆Tc = 1.8(127.123) + 32 =260.834ºF
ln(−∆Tme) = −1.53181 − 0.0128(130.34) 0.6667 + 364.064(1.3325) 0.333
ln(−∆Tme) = −1.53181 − 0.32910 + 4.01779
ln(−∆Tme) = 2.15088
∆Tme = 8.59
Tb=130.34-(8.59)=121.75ºC
Tb=251.15ºF
PROMEDIO DE EBULLICION
Tw=272.606ºF
Tc=260.834ºF
Tm=241.826ºF
Tb=251.15ºF
SG =
1.41.5
131.5+62
SG =
1.41.5
193.5
= 0.73126
Tb= 251.15ºF + 460= 711.15 ºR
M = 20.486(711.15) 1.26027
− (0.73126) 4.9308 е(0.0001165(218.15)).7.78712(0.73)+(0.0011582)(9.1195)(0.73826)
M=80376.28.(0.73126) 4.9303 (6.6758*10−3)
M=11.807
CH = 3.4707[ 𝑒𝑥𝑝(1.48 ∗ 10−2)(394075) + 16.94(0.73126) − 1.299 ∗
10−2(394.75)(0.73126)]*394.75−2.725(0.73126) −6.798
CH=3.4707[228889.985]7.0641*10−7](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Petroleo 404 DEBER
- 1. VABP= 77.8+107.8+126.7+155+184.4 5 VABP= 651.7 5 = 130.34ºC VABP= 130.34 +273.15 =403.49ºK SL= 184.4−77.8 580 SL=1.3325 ln(−∆Tw) = −33.64991 − 0.02706(130.34) 0.6667 + 5.163875(1.3325) 0.25 ln(−∆Tw) = 1.202427 ∆Tw = −3.33 Tw = 130.34ºC − (−3.33) = 133.67ºC Tw = 272.606ºF ln(−∆Tm) = −1.15158 − 0.018(130.34) 0.6667 + 3.70612(1.3325) 0.333 ln(∆Tm) = −1.15158 − 2.30364 + 4.0787 ln(∆Tm) = 2.62265 ∆Tm = 13.77 ∆Tm = 130.34 − (13.77) = 116.57 ∆TC = 241.826ºF
- 2. ln(−∆Tc) = −0.82868 − 0.08997(130.34) 0.45 + 2.4567(1.3325) 0.45 ln(∆TC) = −0.828368 − 0.80516 + 2.7955 ln(∆TC) = 1.16666 ∆TC = 3.21 ∆Tc = 130.34 − (3.21) = 127.13 ∆Tc = 1.8(127.123) + 32 =260.834ºF ln(−∆Tme) = −1.53181 − 0.0128(130.34) 0.6667 + 364.064(1.3325) 0.333 ln(−∆Tme) = −1.53181 − 0.32910 + 4.01779 ln(−∆Tme) = 2.15088 ∆Tme = 8.59 Tb=130.34-(8.59)=121.75ºC Tb=251.15ºF PROMEDIO DE EBULLICION Tw=272.606ºF Tc=260.834ºF Tm=241.826ºF Tb=251.15ºF SG = 1.41.5 131.5+62 SG = 1.41.5 193.5 = 0.73126 Tb= 251.15ºF + 460= 711.15 ºR M = 20.486(711.15) 1.26027 − (0.73126) 4.9308 е(0.0001165(218.15)).7.78712(0.73)+(0.0011582)(9.1195)(0.73826) M=80376.28.(0.73126) 4.9303 (6.6758*10−3) M=11.807 CH = 3.4707[ 𝑒𝑥𝑝(1.48 ∗ 10−2)(394075) + 16.94(0.73126) − 1.299 ∗ 10−2(394.75)(0.73126)]*394.75−2.725(0.73126) −6.798 CH=3.4707[228889.985]7.0641*10−7
- 3. CH=5.611 𝑙𝑏𝐶 𝑙𝑏𝐻 %C= 5.611 6.611 X100 =84.4% %H= 1 6.11 𝑋100 = 15.1% Punto De Anilina AP=-183.3+0.27(62)(298.75) 1/3+0.317(394.75) AP=-183.3+122.79+125.13 AP=64.625ºC AP=173.91 Calor Latente De Vaporizacion ∆𝑯 𝒏𝒃𝒑 𝒗𝒂𝒑 = a⍬ 𝒃 .⍬ 𝒄 ∆𝑯 𝒏𝒃𝒑 𝒗𝒂𝒑 =37.32315(394.9ºK 𝟏.𝟏𝟗𝟎𝟔 ) ( 𝟎. 𝟕𝟑𝟏𝟐𝟔) 𝟗.𝟕𝟕𝟎𝟖𝟗𝒙𝟏𝟎 −𝟑 ∆𝑯 𝒏𝒃𝒑 𝒗𝒂𝒑 =37.38315(983.93)=34139.10J/mol-g 34109.62J/mol-gX 𝟒𝟓𝟒𝒎𝒐𝒍−𝒈 𝟏𝒎𝒐𝒍−𝒍𝒃 𝑿 𝟏𝒎𝒐𝒍𝒍𝒃 𝟔𝟓𝟔𝟏𝑱 𝒙 𝟏𝑩𝑻𝑼 𝟏𝟎𝟓𝟓𝒋 = 𝟏𝟎𝟏. 𝟒𝟕 𝒃𝒕𝒖 𝒎𝒐𝒍−𝒍𝒃 Calor De Combustion HHV=17.887 +57.5(62) HHV=3582.887 LHV=HHV-91.23 (%) LHV=3582.887 -91.23 (15.1) LHV=2205.314