Refrigeração (Aula 08)
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Expansão simples
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Separador 2
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• Um ciclo operando a expansão simples opera com R134a a -32 graus C no evaporador e 12 bar no condensador, o mesmo possui dois separadores, um para o evaporador e outro na pressão parcial do sistema, considere a temperatura de 40 na saida do condensador e uma vazão massica de 1 kg/s , Determine: A pressão parcial, a fração de refrigerante, a capacidade Frigorífica e a potência isetrópica do sistema e o COP.
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• h1= 229,14kJ/kg• s1= 0,9433kJ/kg K• h2= 275,95kJ/kg• h3= 115,76kJ/kg• h4= 115,76kJ/kg• h1’= 248,66kJ/kg• s1’= 0,9177kJ/kg K• h2’= 255,84kJ/kg• h3’= 106,19kJ/kg• h4’= 106,19kJ/kg
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• Pi=3,2 bar
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• g=1-(229,14-115,76)/(255,84-106,19)
• g= 0,242
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• Qe=(1(1-0,242))(229,14-106,19)
• Qe= 93,1kW
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• Wc=1((1-0,242)(255,84-229,14)+(275,95-248,66))
• Wc= 47,5 kW
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• COP=Qe/Wc
• COP= 1,96
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Ciclo em cascata
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TCI
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• h1= 110,81 btu/lb• s1= 0,2175 btu/lb F• h2s= 118,87 btu/lb• h2= 120,29 btu/lb• h3= 53,44 btu/lb• h4= 53,44 btu/lb• h1’= 98.81 btu/lb• s1’= 0,225 btu/lb F• h2s’= 114,83 btu/lb• h2’= 117,66 btu/lb• h3’= 32,53 btu/lb• h4’= 32,53 btu/lb
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• Qe=m1(h1’-h4’)• m1= Qe/(h1’-h4’)• m1= 360.000/(98.81- 32,53)
• m1=5.431,5 lb/h
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• m1(h2’-h3’)=m2(h1-h4)
• m2= m1((h2’-h3’)/(h1-h4))• m2=5.431,5((117,66’- 32,53)/ 110,81- 53,44))• m2= 8.059,7 lb/h
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• Wc=[m1(h2’-h1’)]+[m2(h2-h1)]
• Wc=[5.431,5(117,66-98.81)]+[8.059,7(120,29- 110,81)]
• Wc= 178.789,73 btu/h
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• COP=Qe/Wc
• COP=360.000/178.789,73
• COP=2