Elementary Principles 268 - 7.43 (cont'd) (a) T2 = T ( P =...

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7-23 7.43 (cont’d) (a) TT P 2 70 == ( . bar, sat'd steam) =165.0 C o ± (( ) , ± ) , ( Hv 3 2 2954 2760 H O P = 7.0 bar, T = 250 C) kJ kg (Table B.7) H O P = 7.0 bar, sat'd) kJ kg Table B.6) 2 o 2 = = ΔΔ Δ EQW E ps k HH H H H H ,, , . ± . ± . ± . ± . ± (. ) Energy balance kg(2954 kJ / kg) -1.0 kg(2760 kJ / kg) bar, T kJ / kg T C 11 = ⇒= 0 02 9 6 1 9 6 1 0 1 9 6 2 9 6 10 0 3053 300 31 2 1 1 D (b) The estimate is too low . If heat is being lost the entering steam temperature would have to be higher for the exiting steam to be at the given temperature. 7.44 (a) P VP l v 1 30 = = ± ± bar, sat'd.) =133.5 C bar, sat'd.) = 0.001074 m / kg bar, sat'd.) = 0.606 m / kg 3 3 D V V m l space v = 0 001074 177 2 200 0 22.8 L 1 0606 0 0376 . . . . . m 1000 L 165 kg kg m L L -177.2 L = 22.8 L m 1 kg 1000 L m kg 3 3 3 3 m=165.0 kg P=3 bar V=200.0 L P max =20 bar Vapor Liquid (b) PP m total = + = max .. . . 20 0 1650 0 0376 16504 bar; kg P lv 1 20 0 20 0 20 0 ± ± bar, sat'd.) = 212.4 C bar, sat'd.) = 0.001177 m / kg; bar, sat'd.) = 0.0995 m / kg 33 D Vm V m V m V mm V t o t a ll v l l t o t a v =+ ⇒+ = ±± ± () ± . / ) (. / L m L kg m kg) + (165.04 - kg m kg) kg; kg 3 200 0 1 1000
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