Answer:
a) Â 358.8 KJ/kg
b) Â 0.0977 Â KJ/K- kg
c) Â 83.28%
Explanation:
N2 at 300 k. ( use the properties of N2 at 300 k (T1) )
Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk , Â R = 0.1297 KJ/kgk , y = 1.4 ,
Given data:
T2 = 645 k
P1 = 1 bar , P2 = 10.5 bar
a)Determine the work input in KJ/Kg of N2 flowing
Winput = h2 - h1 Â = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg
b) Determine the rate of entropy in KJ/K- kg  of N2 flowing
Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1
                  = 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )
                  = 0.0977  KJ/K- kg
c) Determine isentropic compressor efficiency
Isentropic compressor efficiency = 83.28%
calculated using the relation below
( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )
where T'2 = 587.314
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