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Practica No. 7 “Primera ley de la termodinámica en un ciclo que corresponde a una maquina frigorífica”

Enviado por   •  16 de Abril de 2018  •  1.138 Palabras (5 Páginas)  •  463 Visitas

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= [pic 90][pic 91]

= [pic 92][pic 93]

= [pic 94][pic 95]

= [pic 96][pic 97]

= [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:

= [pic 125][pic 126]

= [pic 127][pic 128]

= [pic 129][pic 130]

= [pic 131][pic 132]

= [pic 133][pic 134]

= [pic 135][pic 136]

= [pic 137][pic 138]

= [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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