Unformatted text preview: (3) Δ U = m · ( u 2u 1 ) change in internal energy of the steam (4) u 1 = 2709 . 9 [kJ / kg] initial speciﬁc internal energy (5) u 2 = 2659 . 6 [kJ / kg] ﬁnal speciﬁc internal energy (6) QW pwW piston = Δ U energy balance to determine W piston (7) $TabStops 0.2 3.0 in Solution Δ U =251 . 5 [kJ] m = 5 [kg] Q = 80 [kJ] u 1 = 2710 [kJ / kg] u 2 = 2660 [kJ / kg] W piston = 350 [kJ] W pw =18 . 5 [kJ]...
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 Spring '07
 MARTIN
 Thermodynamics, Energy, Heat, Heat Transfer, Joule

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