Lösung von Aufgabe 17

1. p-V-Diagramm:
   
2. Zustandsgrößen:
   • Punkt 1
     • p₁ = 150 bar, T₁ = 313.15 K
     • V₁ = m R_i T / p = 0.5996 m³
     • U₁ = 0
     • H₁ = U₁ + p₁ V₁ = 8.994 MJ
   • Punkt 2
     • p₂ = 150 bar
     • Q₁₂ = m c_p (T₂ - T₁)
     •     ⇒ T₂ = Q₁₂/(m c_p) + T₁ = 811.2 K
     • V₂ = V₁ T₂ / T₁ = 1.553 m³
     • U₂ = m c_v (T₂ - T₁) + U₁
     •     = m (c_p - R_i) (T₂ - T₁) + U₁ = 35.70 MJ
     • H₂ = U₂ + p₂ V₂ = 58.99 MJ
   • Punkt 3
     • T₃ = T₁ = 313.15 K
     • Aus p^n-1 / Tⁿ = const. folgt:
     • p₃ = p₂ (T₃/T₂) ^{n/(n - 1)} = 0.4966 bar
     • V₃ = m R_i T₃ / p₃ = 181.1 m³
     • U₃ = U₁ = 0   (Isotherme!)
     • H₃ = U₃ + p₃ V₃ = 8.994 MJ
3. Nutzarbeit W_k = Σ W_Vi
   • W_V12 = p₁ (V₁ - V₂) = - 14.30 MJ
   • W_V23 = (m R_i)/(n - 1) (T₃ - T₂) = - 71.51 MJ
   • W_V31 = p₃ V₃ ln(p₁/p₃) = 51.36 MJ
   • W_k = - 34.46 MJ
4. thermischer Wirkungsgrad:
   • η = (|W_k|)/(Q₁₂ + Q₂₃)
   • Q₁₂ = 50 MJ
   • Q₂₃ = W_V23 (n - κ)/(κ - 1) = 35.82 MJ
   • ⇒    η = 0.4015
   • η_Carnot = 1 - T₃/T₂ = 0.6139