HW6 Solutions

HW6 Solutions - Solutions Chapter 24 j Since...

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Unformatted text preview: Solutions Chapter 24 j Since Pfi~I'V,noI'and liHfi~' are in the tables, you have to get a liH.noI' and Vrm.,vaporgiven an assumed Pfi",' (8) . that satisfY the equation. liHfi",' wilJ be composed of saturated liquid and vapor. liHL(8-m) + liH,(m) = liHr...,. m = Ib vapor. (7c) (7b) 2 ft' = Vl(8-m)+ Vv(m). VL,V.,liHL,liH" andp finalare all relatedin the steamtables. * Vof saturated vapor = 350.8 fe lIb so that 1.87 ft' represents 0.0053 Ib, a negligible amount relative to 2 lb. 24.2 (I) (2) The system is the mixing point of the 10OOF"air" and water at 70F Open system (3) and (4) Air at 10000F Products (100 Ib) H20@ 400F, 'Ii = 1201.2 Btullb air@ 400F, H = 2576 Btullb mol A H = 6984 Btullb mol The enthalpies of water come ITomthe steam tables. The properties of air come fi"omthe tables in Appendix D7 (with interpolation). The reference temperature for the enthalpy is 32F. (5) liE =[liU +liPE+liKE],,,,,,,,, = Q+ W-liH -liPE -liKE . 0 because steadv state Solutions Chapter 24 . Q is assumed to be 0 because time is short in transit...
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HW6 Solutions - Solutions Chapter 24 j Since...

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