3.3 - 3 3.3 (a) System: contents of the piston and cylinder...

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3.3 (a) System: contents of the piston and cylinder (closed isobaric = constant pressure) M.B.: MM M MMM 21 2 1 0 −= =⇒ = = E.B.: MU 22 11 ±± −= MH ± di 0 + + Q W s 0 z PdV M UU Q P d VQP d VV = = − − z z ch a f U Q PMV V c h QM UU M P VP V MU P V UP V H =− + − = + + ± ± ± ± ± ± 2 1 2 2 1 1 c h c h c h P =≈ 1013 01 .. bar MPa ± V ± U ± H T = 100 1.6958 2506.7 2676.2 T = 150 1.9364 2582.8 2776.4 Linear interpolation T 125 C 1.8161 2544.8 2726.3 Initial state Final state P = . MPa , m ± . V 2 36322 = 3 /kg T 500 C 3.565 3488.1 T 600 C 4.028 3704.7 Linear interpolation 36322 3565 4028 3565 500 600 500 5145 2 2 . = T T C 500 600 500 34881 3704 7 34881 35195 2 2 . ± . ± . = = H H Q = () = 1 27263 7932 kg kJ kg kJ . WP d V V V × × × × ⋅⋅ × × z 1 1 36322 18161 1 100 000 18161 1816 bar bar m kg bar Pa bar kg ms P a J msk g mk g kJ kg 3 2 3 a f ,. . (b) System is closed and constant volume M.B.: MM M == E.B.: ± 0 + + Q W s 0 PdV z 0 UU Here final state is P 0 1 3 0 2 . ~ . bar MPa ; . 18161 g 3 (since piston-cylinder volume is fixed)
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P = 02 . MP a ; ± . V 2 18161 = T ( ° C) ± V ± U 500 1.7814 3130.8 600 2.013 3301.4 18161 17814 2 013 17814 500 600 500 00347 02316 01498 015 .. . . .~ . = == T T = ° 515 C ± . . ± . U U 2 2 31308 33014 31308 01498 31564 kJ kg Q ( ) = 1 4 25448 6116 kJ kg 3156 kJ kg . (c) Steam as an ideal gas—constant pressure N PV RT RT RT =⇒ =
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3.3 - 3 3.3 (a) System: contents of the piston and cylinder...

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