Practica No. 7 “Primera ley de la termodinámica en un ciclo que corresponde a una maquina frigorífica”
Enviado por Jillian • 16 de Abril de 2018 • 1.138 Palabras (5 Páginas) • 546 Visitas
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= [pic 90][pic 91]
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= [pic 98][pic 99]
= [pic 100][pic 101]
= [pic 102][pic 103]
= [pic 104][pic 105]
= [pic 106][pic 107]
= [pic 108][pic 109]
Resultado:
= 0.695[pic 110]
= 19.805[pic 111]
= 19.11[pic 112]
= 18.415[pic 113]
= 17.72[pic 114]
= 17.02[pic 115]
= 60239.6826[pic 116]
=63108.7921[pic 117]
= 66228.6258[pic 118]
= 69631.848[pic 119]
= 73384.9651[pic 120]
Proceso 3 a 1
Volumen (L)
Presión (Pa)
= 20.5[pic 121]
=57593.37238[pic 122]
V=19.805
P=60239.6826
V=19.11
P=63108.7921
V=18.415
P=66228.6258
V=17.72
P=69631.848
V=17.02
P=73384.9651
= 16.33[pic 123]
=72287.1019[pic 124]
Formulas:
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= [pic 135][pic 136]
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= [pic 139][pic 140]
= [pic 141][pic 142]
= [pic 143][pic 144]
= [pic 145][pic 146]
Sustitución:
= [pic 147][pic 148]
= [pic 149][pic 150]
= [pic 151][pic 152]
= [pic 153][pic 154]
= [pic 155][pic 156]
= [pic 157][pic 158]
= [pic 159][pic 160]
= [pic 161][pic 162]
= [pic 163][pic 164]
= [pic 165][pic 166]
= [pic 167][pic 168]
Resultado:
= 0.695[pic 169]
= 19.805[pic 170]
= 19.11[pic 171]
= 18.415[pic 172]
= 17.72[pic 173]
= 17.02[pic 174]
= 59614.44755[pic 175]
=61782.5292[pic 176]
= 64114.2619[pic 177]
= 69369.2205[pic 178]
[pic 179][pic 180][pic 181]
10. Calcula la variación de energía interna, la variación de entalpia, el trabajo y el calor para cada proceso y para el ciclo.
Para todos los procesos R= 8.314 Gas diatómico = R = R[pic 182][pic 183][pic 184][pic 185][pic 186]
Proceso politropico (1-2)
Formula:
= n ([pic 187][pic 188][pic 189]
= ([pic 190][pic 191][pic 192]
W = - [pic 193]
Q = [pic 194]
Solución:
= (0.4787 mol) (20.785) ([pic 195][pic 196]
= ([pic 197][pic 198][pic 199]
W = - [pic 200]
Q = 209.1443 - 276.5545
Resultado:
= 209.1443 [pic 201]
= 292.8021[pic 202]
W= 276.5545
Q= - 67.4102
Proceso isocorico (2-3)
Formula:
= n ([pic 203][pic 204][pic 205]
= ([pic 206][pic 207][pic 208]
W= 0
Q = [pic 209]
Solución:
= (0.4787 mol) (20.785) ([pic 210][pic 211]
= ([pic 212][pic 213][pic 214]
W= 0
Q= -209.1443 - 0
Resultado:
= -209.1443[pic 215]
= -292.8021[pic 216]
W= 0
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