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

Answer

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

$W = (h_1 - h_2 ) + \dfrac{V_1^2-V_2^2}{2000}$

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

$W = (503.02-320.29) + \dfrac{150^2-10^2}{2000}$

W = 193.93 KJ/Kg

calculation of energy destruction

= $T_0(S_2-S_1-Rln(\dfrac{P_2}{P_1}))$

= $T_0(S_2-S_1+Rln(\dfrac{P_1}{P_2}))$

= $300(1.7679-2.21952-\dfrac{8.314}{28.97}ln(\dfrac{8}{1}))$

=$300 \times 0.145152$

=43.54 KJ/Kg