Problem Three ow (SSSF) adiabatic steam (H,O) turbine with singl produces a spec
ID: 2088631 • Letter: P
Question
Problem Three ow (SSSF) adiabatic steam (H,O) turbine with singl produces a specific work output, w +565.131 A steady-state, steady-flow (ssshea utfoww On the inflow T-500 °C (temperature oneed on the Vi- 1.0 met on the inflow). meter/sec (flow speed on the inflow) 0.0767875059 m2 (inflow cross-sectional area)- On the outflow: 2.0 MPa 2000.0 kPa (pressure on the outflow); de 2346.184 kJ/kg (specific internal energy on the outtlow Changes in kinctic and potential energs are negligibe Determine: (Part 1: 10 points; Part 2: 15 points; Part 3: 10 points)
Explanation / Answer
Applying 1st law of thermodynamics:
Q = ?u + W
=> Q = (ue – ui) + we
[where, Q = heat transfer; ue = specific internal energy out flow= 2346.184 kJ/kg; ui = specific internal energy in flow and we = specific work output =565.131]
=> 0 = (ue – ui) + we [since, the process is an adiabatic process, heat flow, Q = 0]
=> 0 = (2346.184 – ui) + 565.131
=> ui = 2911.315 kJ/kg
According to ideal gas law:
Pi × vi = R×Ti
[Pi = inflow pressure; vi = specific inflow volume = 1.0×0.0767875059 = 0.00767875059 m3/sec; R = characteristic gas constant of steam = 461.52 J/kg K and Ti = 500°C = 500 +273.16 = 773.16 K]
=> Pi × vi = 461.52×773.16 = 356828.8 J/kg
Steady state steady flow energy equation for turbine:
hi = he + we
[where, hi = specific enthalpy on inflow and he = specific enthalpy on outflow]
=> (ui + Pi × vi ) = he + we [since, hi = ui + Pi × vi]
=> (2911.315+ 356828.8) = he + 565.131
=> he = 359175.182 J/kg = 359.175 J/kg
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