SolHW16ME2320

SolHW16ME2320 - under steady – state conditions, the...

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5 – 54 GIVEN: Steam Adiabatic Turbine: P i = 10 MPa; T i = 500 o C; P e = 10 kPa; x e = 0.9; Ŵ = 5MW FIND: mass flow rate. SOLUTION : Assuming that the changes in kinetic energy are negligible and the process occurs under steady – state conditions, the energy eqn. can be written as e i i e h h W m h h m W - = - = - ) ( ; For P i = 10 MPa; T i = 500 o C b h i = 3375.1 kJ/kg For P e = 10 kPa; x e = 0.9 b h e = 2344.5 kJ/kg s kg m h h W m e i / 85 . 4 ; 5 . 2344 1 . 3375 5000 = - = - = 5 - 59
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5 – 60 GIVEN: Adiabatic Gas Turbine (air): P i = 1 MPa; T i = 500 o C; P e = 100 kPa; T e = 150 o C; A i = 0.2 m 2 ; V i = 40 m/s; A e = 1 m 2 FIND: mass flow rate and power produced. SOLUTION : Assuming air is an ideal gas and the process occurs
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Unformatted text preview: under steady – state conditions, the energy eqn. can be written as ) 2 1 2 1 ( 2 2 i i e e V h V h m W--+ =-& & The mass flow rate can be calculated from s kg m P RT A V v A V m i i i i i i i / 06 . 36 ) 273 500 )( 287 . ( ) 2 . )( 40 )( 1000 ( / = + = = = & & s m V A P RT m A v m V e i e e i i e / 7 . 43 ) 1 )( 100 ( ) 273 150 )( 287 . )( 06 . 36 ( = + = = = & & Substituting the mass flow rate into the energy eqn. MW W W V h V h m W i i e e 228 . 13 ) 2000 40 791 2000 78 . 43 424 ( 06 . 36 ) 2 1 2 1 ( 2 2 2 2 =--+ =---+ =-& & & &...
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This homework help was uploaded on 04/07/2008 for the course ME 232 taught by Professor Monefort during the Spring '08 term at Western Michigan.

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SolHW16ME2320 - under steady – state conditions, the...

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