University:
Massachusetts Institute of Technology
Course:
16.01 | Unified Engineering I, II, III, & IV
Academic year:

2019
Views:

46

Pages:

2
Author:

Nomes67i

T8 SOLUTIONS (WAITZ) Problem a): Given: P₁ = 1x10⁵ Pa T₁ = 200 K C_p = 50 J/kgK P₂ = 5x10⁵ Pa Process: P.V^k = constant. C_v = ? Calculations: P₁V₁ = RT₁ => V₁ = 0.052 m³/kg P₂V₂ = RT₂ => V₂ = 0.012 m³/kg S.F.E.E (Steady Flow Energy Equation): h₁ + (C₁²/2) = h₂ + (C₂²/2) + w or w = h₁ - h₂ + (C₁²/2) - (C₂²/2) w = C_p(T₁ - T₂) + (C₁²/2) - (C₂²/2) P₂V₂^k = P₁V₁^k => T₂ = 230.8 K w = -78.2 kJ/kg Ẇ = ṁw = -7.8 MW Calculation for W_f (Flow Work): W_f = 260 (230.8 - 200) = 8 kJ/kg Ẇ_f = ṁw_f = 0.8 MW Calculation for W_s (Shaft Work): W_s = [C_p(T₁ - T₂) + (C₁²/2) - (C₂²/2)] - W_f W_s = -86.2 kJ/kg or Ẇ_s = -8.62 MW Section (b) C₂ = 50m/s, T₂ = 230.8 K, P₂ = 5x10⁵ Pa C₃ = 100m/s, T₃ = ?, P₃ = 5x10⁵ Pa Q - Ws = Cp (T₃ - T₂) + C₃²/2 - C₂²/2 120x10³ = 2800 (T₃ - 230.8) + 100²/2 - 50²/2 T₃ = 693.7 K Ws = 0 W₂ = R(T₃ - T₁) = 260(693.7 - 230.8) = 120 kJ or W₂ = 1.2 MW Section (c) Ws(TURBINE) = -Ws(PUMP) = 86.2 kJ W₂ = 86.6 MW Q - Ws = Cp (T₄ - T₃) + C₂²/2 - C₁²/2 -86.2x10³ = 2800(T₄ - 693.7) + 120²/2 - 100²/2 T₄ = 662 K Ws = R(T₄ - T₃) = 260(662 - 693.7) = -82 kJ/kg or W₂ = -0.82 MW

T8 Solutions

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