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200_Lectures_11_21_post

m va v dmcv dt min mout 6 2 8 2 0 1 a

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Unformatted text preview: (1 %* % '+ A 99 F6 ! A 9#<< +! ! % % ' (188.4 + 273.15) K = 3.47 TR ,in = 133K 108.2kPa pR ,in = = 0.03 3.77 x103 kPa TR ,in > 2.0 and pR ,in < 0.1 (180 + 273.15) K = 3.41 TR ,out = 133K 101.4kPa pR ,out = = 0.03 3.77 x103 kPa TR ,out > 2.0 and pR ,out < 0.1 @ #9 % 0 ! *! , 0! % %% #$ ! %0 ! ' .0 = 6 D 1? . %* = 6 D 1? . 5 * :; % ;6 * 6D 1 E % 6 = ' ( $A 1 6&A: ; 6 A= 6+AD 1 ' !' ( % 0AE CV A Z in = Z out ≈ 1 % (7% ) ' ( ' Ain = Vout ρ out V Aout = out Vin ρin Vin Vout Vin Tin Tout 150 m s Ain = 73.2 m s Ain = 0.137 m %0 (7 dmCV = min − mout dt ( ρVA)in = ( ρVA )out Ain = ' 2 min = mout B pout Aout RTout - ' %. m= ' ( +! pout Aout pin 101.4kPa 108.2kPA ! C % RTin pin %* 0 VA v - = 6+ A D # 9 1. v ( 80o F ,50 psia ) ≈ v f ( 80o F ) = 0.01607 ft (188.4 + 273.15) K (180 + 273.15) K 0.07 m π 4 ft s 2 mwater = 4 2in 1 ft 12in 0.01607 ft mwater = 5.42 lbm s 3 lbm 2 3 lbm 2 % * % #9 ! 8. ' ( +! * %* R T M RT v= Mp pv = m= VA VA ( Mp ) = v RT 4 2 π 1 ft 2in 4 12in ft s m= (1545 ( 28.01 ft − lb f...
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