Steam enters a turbine at 12 MPa, 550 Solution i am just writting down the formu
ID: 2993108 • Letter: S
Question
Steam enters a turbine at 12 MPa, 550Explanation / Answer
i am just writting down the formulas. I know you engineer can easily solve simple mechanical sums. moisture content, 1 - xo = 5% --> xo = 95% = 0.95 a) Reversible power output, si = so Energy balance for actual condition, 0 = Q - W + m_dot*[(hi-ho) + (Vi^2-Vo^2)/2] m_dot = W - Q / [(hi-ho) + (Vi^2-Vo^2)/2] hi = h(Ti,pi) ho = h(xo,po) Energy balance for reversible turbine at steady state, dEcv/dt = 0 = Q - Wr + m_dot*[(hi-ho) + (Vi^2-Vo^2)/2] Wr = Q + m_dot*[(hi-ho) + (Vi^2-Vo^2)/2] hi = h(Ti,pi) ho = h(po,si) (b) The exergy destroyed, Ed Exergy balance for turbine at steady state, dExcv/dt = 0 = (1 - To/Tb)Q - W + m_dot(efi - efo) - Ed Ed = (1 - To/Tb)Q - W + m_dot(efi - efo) with exergy accompany by flow, efi - efo = (hi-ho) - To(si-so) + Vi^2-Ve^2/2 This can be solved if we may assume average outer surface temperature, Tb. c. Exergetic efiiciency e = W / [ (efi - efo) - (1-To/Tb)Q ], exergy accompany by heat transfer must be positive. placing values will give you the required answers.