Projeto Do Módulo 2 Smacna
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1. 1) P=101,325 ∗( 1−2,2560∗10 −5 ∗ALT) 5,2560 P=101,325 ∗( 1−2,2560∗10 −5 ∗1172) 5,2560 =88,0137 kPA 2) P V ( TBU )=P V ( 18,6 °C )=2,1444 kPA ( INTERPOLAÇÃODA TABELA ) 3) ω ( UR;TBS)= 0,622 ∗ ( UR 100 ) ∗P V ( TBS) P− ( UR 100 ) ∗P V ( TBS ) ω ( 100 % ; 18,6 °C) = 0,622 ∗ ( 100 100 ) ∗2,1444 88,0137− ( 100 100 ) ∗2,1444 =0,01553 kg kg 4) ω ( TBS;TBU)= ( 2501 −2,3810∗TBU ) ∗ω ( TBU )−( TBS−TBU )∗1,0048 2501+1,8050∗TBS−4,1868∗TBU ω ( 30,2 °C; 18,6 °C) = ( 2501−2,3810∗18,6)∗0,01553−( 30,2−18,6 )∗1,0048 2501+1,8050∗30,2−4,1868∗18,6 ω ( 30,2 °C; 18,6 °C) =0,01069 kg kg ω INS =0,01069 kg kg
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Transcript of Projeto Do Módulo 2 Smacna
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