Propiedades Del Aire Atmosférico Saltillo

ll
PROPIEDADES DEL AIRE ATMOSFÉRICO (h = 1600 m)

TBS = 28 ° C

Para Saltillo, Coahuila, México

TBH = 15° C



Altura de Saltillo sobre el nivel del mar.h = 1600 m = 5248 ft

Pb = a + bh (ft) = 29.42 + (-0.0009)(5248) = 24.6968 inHg

Pb = 24.6968 inHg = 627.29 mmHg = 83.631 KPa



W = 0.622 Pw . W =Ws*hfg - Cp (TBS - TBH)

Pb – Pw hg – hf*



Ws* = 0.622 Pws* .

Pb – Pws*

[pic]TABLA DE VAPOR

TBH = 15° C

hfg* = hg* - hf* = 2465.91 KJ/Kg

hg* = 2458.9 KJ/ Kg

hf * = 62.99 KJ/Kg

Pws* = 0.01705 bar (100 KPa) = 1.705 KPa



TBS = 28 ° C

hg = 2552.6 KJ/KgWs* = 0.622 Pws* . = 0.622 1.705 KPa . = 0.0129 Kgw/Kga

Pb – Pws* 83.631 – 1.705



W = Ws*hfg - Cp (TBS - TBH)hg – hf*

W = [(0.0129kgw/Kga)(2465.91KJ/Kg)] – [1KJ/Kg°C(28-15)°C]

2552.6 KJ/Kg – 62.99 KJ/Kg

| |
|W =0.00755 Kgw/Kga |


Φ = [Pw / Pws ] * 100 W = 0.622 Pw .

Pb– Pw

Si Pb = 83.631 KPa

0.622Pw = 0.00755(83.631) – 0.00755Pw

0.622Pw + 0.00755Pw = 0.6314

Pw = 0.010029 bar = 1.0029 KPa



Φ = [Pw/Pws] * 100 = [0.010029/0.03782] * 100

Pwsobtenido de la tabla para TBS = 28 ° C

|Φ = 26.51 % |


h = CpT + Whg = [1 KJ/Kg°C][28°C] + [0.00755 Kgw/kga][2552.6KJ/Kg]

| |
|h = 47.27 KJ/Kg|






Con Pw se obtendrá la temperatura de rocío.

TPR = ? Pw = 0.010029 bar



8 ° C – 0.01072 bar

X °C – 0.010029 bar X= 7.114 °C

10...