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Páginas: 2 (294 palabras) Publicado: 23 de abril de 2012
Gv1=385232 pie3/min
tG1= 25°C
tw1= 17°C
Z = 6.08 m
Kya= 1800 kg /m3hΔy
S= 4.32(8.08) =34.9056m2
tL2= 35 °C
2000 GAL/min
tG2 =?
tL1= 23°C
tG1= 25°C Y1= 0.013
tw1= 17°C

H1=0.24+0.45Y1×tG!+597.2Y1
H1=0.24+0.45*0.013×25+597.20.013=13.9098kcalkg
tL1=23
H1tL2=35
H2 máx.
H2 máx.= 30 Kcal/Kg

H2=CpL*L`Gas´×tL2 -tL1+H1
VH=128.82+Y1180.8205tG!+273.15PtVH=128.82+0.013180.820525+273.15585760=1.1257m3kg
Gv1=335232 ft3min ×1 m335.3145 ft3×60 min1 hr=569565.47m3hr
Gas=Gv1VH=569565.47m3hr1.1257m3kg=505965.6kghrGas´=GasS=505965.6kghr34.9056 m2=14 495.3558kghr m2
2000GALmin×1 m3854.1721 gal×60 min1 hr=454.2493m3hr
L=454.2493m3hr×1000 kg1 m3=454 249.3kghr
L´=LS=454249.3kghr34.9056 m2=13013.652kghm2

H2=113013.652kghm24 495.3558kghr m2×35-22+13.9098=25.5810kcalkg
i | tL | H | H* | 1H*-H | y |
0 | 22| 13.9098 | 15.2 | 0.775 (-) | 0.7751 |
1 | 24.2667 | 15.855 | 17.1 | 0.8032 (4) | 3.2128 |
2 | 26.334 | 17.8002 | 19.5 | 0.5883 (2)| 1.1766 |
3 | 28.5001 | 19.7454 | 22.0 | 0.4435 (4) | 1.774 |
4 | 30.6668 | 21.6906 | 24.5 | 0.3559 (2) | 0.7118 |
5 | 32.8335 |23.6358 | 27.3 | 0.2538 (4) | 1.0352 |
6 | 35 | 25.581 | 30.9 | 0.1880 (-) | 0.1880 |
| | | | | Σy= 8.8735 |

htL=35-226=2.1667hH=25.581-13.90986=1.9452
NUD=ht3Σy=2.166738.8735=6.4087
AUD=L´Ky∙a=13013.652kghm21800kgm3∆y=7.229
Z=NUDAUD=6.40877.229=46.3285m
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