Question

A turbine in a steam power plant operating under steady state conditions receives superheated steam at 3MPa and 350ºC at the rate of 1kg/s and with a velocity of 50m/s at an elevation of 2m above the ground level. The steam leaves the turbine at 10kPa with a quality of 0.95 at an elevation of 5m above the ground level. The exit velocity of the steam is 120m./s. The energy losses as heat from the turbine are estimated at 5kJ/s. Estimate the power output of the turbine. How much error will be introduced, if the kinetic energy and the potential energy terms are ignored? 

Answer 1

Given that; the turbine is running under steady state condition.At inlet: P₁ = 3MPa; T₁ = 350°C; m_f = 1 kg/sec; V₁ = 50 m/s; Z₁=2m At exit: P₂ = 10kPa; x = 0.95; Z₂ = 5 m; V₂ = 120 m/s Heat exchanged during expansion = Q = – 5 kJ/sec From superheat steam table, at 3MPa and 350ºC h₁ = 3115.25 kJ/kgh₂ = h_f2 + xh_fg2 From saturated steam table; at 10kPa h₂ = 191.81 + 0.95(2392.82) = 2464.99 kJ/kg

From steady flow energy equation:

Q – W_s = m_f [(h₂ – h₁)+ 1/2(V²₂ – V¹₂) + g(z₂– z₁)]

– 5 – W_S = 1x [ (2464.99 – 3115.25) + { (120)² – (50)2}/2 × 1000

 + 9.8 (5 – 2 )/1000] – W_S = – 639.28 kJ/sec

W_S = 639.28 kJ/sec

If the changes in potential and kinetic energies are neglected; then SFEE as; Q – W_s = m_f (h₂ – h₁) – 5 – 1W₂ = 1 × (2464.99 – 3115.25) 1W₂ 

= 645.26 KJ/sec

% Error introduced if the kinetic energy and potential energy terms are ignored:

% Error = [(W_s – 1W₂)/W_s ] × 100

= [(639.28 – 645.26)/639.28] x 100

= – 0.935%

So Error is = 0.935%