Steam enters a turbine at 12 MPa, 550 Solution Ti = 220 deg. C pi = 12 Mpa Vi =

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Explanation / Answer

Ti = 220 deg. C pi = 12 Mpa Vi = 60 m/s Po = 20 kPa Vo = 130 m/s moisture content, 1 - xo = 5% --> xo = 95% = 0.95 Q = - 150 kW W = 2.5 MW To = 25 deg, C (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. d. Possible increase, Energy balance, W' = m_dot*[(hi-ho) + (Vi^2-Vo^2)/2] W' = W - Q with, ho = h(po,xo) Exergy destruction Ed' = -W_max + m_dot(efi - efo), Q = 0 Ed' = m_dot(efi - efo) - W Exergetic efficiency, e' = W_max / (efi - efo) Ed' and e', should carefully counted as final state will be differ from actual condition. -- I still in doubt whether reversible process should count for heat transfer, as heat transfer is irreversible process. I assume turbine undergo internal reversible process and added Q to energy balance. My answer is only suggestion, if you have any other reasoning, then you should appy your own steps.

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