Elementary Principles 248 - 7.5 (a) Mass flow rate: 42.0 m...

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7- 3 7.5 (a) Mass flow rate: ± . m == 42.0 m 0.07 m 10 L 273 K 130 kPa 1 mol 29 g s 4 1 m K 101.3 kPa 22.4 L STP mol gs 3 3 π bg 2 573 127 9 ± ± . E mu k ⋅⋅ = 2 2 2 14 2 0 113 127.9 g kg m 1 N 1 J 2 s 1000 g s 1 kg m / s N m Js 2 2 2 (b) ( ) 3 3 22 127.9 g 1 mol 673 K 101.3 kPa 22.4 L STP 1 m 4 49.32 m s s 29 g 273 K 130 kPa 1 mol 10 L (0.07) m = ± ± . . ±± ) ± ) E mu EE E k kk k = 2 2 2 9 3 2 1558 127.9 g kg m 1 N 1 J 2 s 1000 g s 1 kg m / s N m J/s = (400 C - (300 C = (155.8 -113) J / s = 42.8 J / s 43 J / s 2 2 2 Δ DD (c) Some of the heat added goes to raise T (and hence U ) of the air 7.6 (a) ΔΔ Em g z p =− 1 gal 1 ft 62.43 lb ft lb 7.4805 gal s 32.174 lb ft / s ft lb 3 m f 3 2 m 2 f 32174 10 1 834 . . (b) mu mg z u g z kp = ⇒ = = F H G I K J L N M O Q P = Δ 2 12 2 2 2 32174 10 254 b g .. ft s ft ft s 2 (c) False 7.7 (a) Δ ± E positive k When the pressure decreases, the volumetric flow rate increases, and hence the velocity increases. Δ ± E negative p The gas exits at a level below the entrance level.
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