- 2069 Bhadra · 8 marks
Steam turbine: entropy generation per kg
Short answer
The entropy generation per kg of steam is 0.373 kJ/kg·K. Heat loss of 69.95 kJ/kg at 500 K contributes an entropy transfer of –0.1399 kJ/kg·K, while the steam's entropy increases by 0.2327 kJ/kg·K.
The conditions of steam at entrance and exit of a turbine are: h1 = 3456.5 kJ/kg, S1 = 7.2338 kJ/kgK, V1 = 150 m/s; and h2 = 2792.8 kJ/kg, S2 = 7.4665 kJ/kgK, V2 = 100 m/s respectively. The work output per kg of steam flow is 600 kJ. Heat transfer between of 500K. Determine the entropy generation per kg steam flow.
Reading of the problem. The statement "Heat transfer between of 500K" is read as: the turbine casing loses heat to surroundings at a boundary temperature T_b = 500 K.
Given: h₁ = 3456.5 kJ/kg, s₁ = 7.2338 kJ/kg·K, V₁ = 150 m/s; h₂ = 2792.8 kJ/kg, s₂ = 7.4665 kJ/kg·K, V₂ = 100 m/s; work w = 600 kJ/kg.
Step 1: Heat transfer per kg (steady-flow energy equation)
q − w = (h₂ − h₁) + (V₂² − V₁²)/2
(V₂² − V₁²)/2 = (10000 − 22500)/2 = −6250 J/kg = −6.25 kJ/kg h₂ − h₁ = 2792.8 − 3456.5 = −663.7 kJ/kg
q = 600 + (−663.7) + (−6.25) = −69.95 kJ/kg
The negative sign means heat is lost by the steam to the surroundings.
Step 2: Entropy balance for the steady-flow device (per kg)
s_gen = (s₂ − s₁) − q/T_b
s₂ − s₁ = 7.4665 − 7.2338 = 0.2327 kJ/kg·K q/T_b = −69.95/500 = −0.1399 kJ/kg·K
s_gen = 0.2327 + 0.1399
Answer: s_gen = 0.3726 kJ/kg·K ≈ 0.373 kJ/kg·K (positive, so the process is irreversible and possible).
✓ Answer checked by Claude Opus 5.5 (high effort)
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