jwaban
4
Po P =¿ M=¿ M=¿ M=¿ M=¿ M=¿ M=¿ T 1 T =1+ 1 2 ( k− 1) M 2 T= T 1 1+ 1 2 ( k−1 ) M 2 T= 378 1+ 1 2 ( 1,4−1 ) 1,435 2 T= 378 1+ 1 2 ( 0,4 ) 2,06 T= 378 1,412 = 267,7 K Aliran adiabatis steady, T adalah konstan T 1= T 2 = 378 K ¿ T =1 + 1 2 ( k −1) M 2 M=¿ M=¿ M=¿ Po P =¿ P=220 ¿ P=220 ¿ P=220 ( 4,84 ) P=1064,8 KPa P 0 P 1 =¿ P 0 P 1 =¿ P 0 P 1 =¿ P 0 P 1 =¿ P 0 =1,6( 800)=1280 KPa T 0 T 1 =1+ 1 2 ( k− 1) M 2 T 0 T 1 =1+ 1 2 ( 1,4−1 ) 0,6 2 T 0 T 1 =1+ ( 0 , 2 )( 0,36) =1,072 To=1,072 ( 378 )=405,2 K
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Transcript of jwaban
Aliran adiabatis steady, T adalah konstan T1= T2 = 378 K
=0,52 Kpa
Diketahui :
FLOW
P1 = 800 kPa (abs) T1 = 20oC V1 = 300 m/sec
FLOW
Jawab :1)
= 343,11
2)
= 457,1 Kpa
3)
4) Gas ideal
