SolHW15ME2320

SolHW15ME2320 - i =3231.7 kJ/kg; v i =0.09938 m 3 /kg...

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5 – 34 GIVEN: Diffuser with: P i = 100 kPa; T i = 20 o C; V i = 500 m/s; P e = 200 kPa; T e = 90 o C; FIND: V e . SOLUTION : Assume air is an ideal gas (Enthalpy only function of temperature) and the process is under steady–state The diffuser has only one inlet and one exit, therefore the mass flow rate is the same at the inlet and at the exit. From energy eqn. we have that ) 2 1 2 1 ( 2 2 i i e e V h V h m W Q - - + = - ; since there are not heat and work interactions we have that s m V V h h V e i e i e / 66 . 331 ) 500 ( 10 ) 363 293 ( 2 ) ( 2 2 3 2 = + × - = + - = 5 – 36 Given: Adiabatic Nozzle: Steam P i = 3 MPa; T i = 400 o C; V i = 40 m/s; P e = 2.5 MPa; V e = 300 m/s; FIND: T e , A i / A e SOLUTION : At the inlet: P i = 3 MPa; T i = 400 o C; V i = 40 m/s; Which is superheated steam with h
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Unformatted text preview: i =3231.7 kJ/kg; v i =0.09938 m 3 /kg Adiabatic system and no work is being produced which results in: kg kJ h h V V h h V h V h e e e i i e i i e e / 5 . 3187 10 ) 300 ( 2 1 10 ) 40 ( 2 1 7 . 3231 2 1 2 1 2 1 2 1 3 2 3 2 2 2 2 2 = ×-× + =-+ = →--+ =--Thus at the exit: P e = 2.5 MPa, h e =3187.5 kJ/kg since h e > h [email protected] the fluid at the exit is still superheated vapor with T e ≈ 377 o C and v e ≈ 0.1153 m 3 /kg 5 – 36 Cont. 464 . 6 ; 40 09938 . 1153 . 300 = = = → = → = e i e i i i e e e i e e e i i i e i A A A A V v v V A A v A V v A V m m & & 5 - 44...
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This note 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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SolHW15ME2320 - i =3231.7 kJ/kg; v i =0.09938 m 3 /kg...

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