7- 117.24(a) ΔΔΔ±±±HEEQWkps++=−; Δ±Ek, Δ±Ep, ±WHQs=⇒=0Δ²H4003278°=C, 1 atmkJ kgbg(Table B.7) ²H1002676°⇒=C, sat'd1 atmkJ kg(Table B.5) 22oo100 kg H O(v)/s 100 kg H O(v)/s100 C, saturated 400 C, 1 atm(kW)Q±±..Q=−=×100 kgkJ10 Jskg1 kJJs332782676 0602 107(b)ΔUEW−; ΔEk, ΔEp, WUQ=⇒=0Δ()3finalmˆˆˆTable B.5 100 C, 1 atm2507, 1.673 400 C, UVVP⇒°=°==°Interpolate in Table B.7 to find Pat which ˆV=1.673 at 400oC, and then interpolate again to find ˆUat 400oC and that pressure: 3ofinal3.11 1.6731.673 m /g 1.04.03.3 bar , (400 C, 3.3 bar) = 2966 kJ/kg3.11 0.617VPU−=+=−⎛⎞⎜⎟⎝⎠( )[ ]( )37ˆ29662507 kJ kg 10 J kJ4.59 10 JQUmU⇒=Δ=Δ=−=×The difference is the net energy needed to move the fluid through the system (flow work). (The energy change associated with the pressure change in Part (b) is insignificant.) 7.25²,.HlH OCkJ kg2ch20839(Table B.5) ².Hsteam, sat'd20 bars,2797 2=(Table B.6) o[kg H O(l)/h] [kg H O(v)/h]20 C 20 bar (sat'd)=0.65(813 kW)528 kWmmQ=±(a)
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.