3.31 - 3 3.31 25C, 3.0 106 Pa = 3 MPa 125 kg s (a) mass...

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3.31 25 ° C, 30 10 3 6 . ×= Pa MPa 125 kg s (a) mass balance (steady-state) 0 125 12 =+ ⇒= −= ±± MM kg s Energy balance (neglecting PE terms) 0 22 11 1 2 2 2 F H G I K J ++ F H G I K J ± ² ± ² MH v v ; ± Mv Am n v A mass density, n = molar density, == ρ = v = velocity, A = pipe area, m = molecular weight. ± . . . . .. M m P RT vA v v mv = × ×× ⋅ ×× × = × ×⋅ = 125 16 30 009 2283 2 16 21 4170 10 417 6 2 2 3 kg s kg kmol Pa 298.15 K 8.314 10 Pa m ms m kg kmol m s kg m Ns J kmol kJ kmol 33 2 2 af a f π Back to energy balance, now on a molar basis HH mv mv CT T p 2 2 1 2 −=−= − As a first guess, neglect kinetic energy terms . . . T T p C 1 2 02 −=⇒== ° 5 Now check this assumption v nv n Pv P v 2 2 2 6 1 6 20 34 24 === × × = . . . m s Recalculate including the kinetic energy terms m vv p Jkmo l 1 2 2 2 2 16 2 34 24 5209 = ch c h TT T 1 5209 368 1000 014 =− × =− ° l Jmo l mo lkmo l C . . Thus the kinetic energy term makes such a small contribution, we can safely ignore it. (b) Mass balance on compressor (steady-state) 0 NN 3.0 × 10 6 Pa T 2 = ? compressor
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3.31 - 3 3.31 25C, 3.0 106 Pa = 3 MPa 125 kg s (a) mass...

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