Question

Air enters an insulated turbine operating at steady state at 8 bar, 500K, and 150 m/s. At the exit the conditions are 1 bar, 320 K, and 10 m/s. There is no in elevation. Determine the work developed and the exergy destruction, each in kJ/kg of air flowing. Let To=300K and po=1bar significant change

Answer 1

given,

P₁ = 8 bar T₁ = 500 K V₁ = 150 m/s

P₂ = 1 bar T₂ = 320 K V₂ = 10 m/s

writing energy equation

h₁ + (KE)₁ + (PE)₁ + Q m = h₂ + (KE)₂ + (PE)₂ + W

`W = (h_1 - h_2 ) + (V_1^2-V_2^2)/(2000)`

ideal gas property of air

T₁ = 500 K h₁ = 503.02 KJ/kg S₁ = 2.21952 kJ/kgK

T₂ = 320 K h₂ = 320.29 KJ/kg S₂ = 1.7679 kJ/kgK

`W = (503.02-320.29) + (150^2-10^2)/(2000)`

W = 193.93 KJ/Kg

calculation of energy destruction

=
`T_0(S_2-S_1-Rln((P_2)/(P_1)))`

=
`T_0(S_2-S_1+Rln((P_1)/(P_2)))`

=
`300(1.7679-2.21952-(8.314)/(28.97)ln((8)/(1)))`

=
`300 * 0.145152`

=43.54 KJ/Kg