The presence of the steam generator, improves the performance of the
cycle, such that the thermal efficiency is higher.
- The net power developed is approximately 9.3403 MW
- The thermal efficiency of the cycle is approximately 35.9 %
Reasons:
State 1 - 2; Process through the turbine
From steam tables, at 8 MPa, 560°C, we have;
h₁ = 3545.3 kJ/kg
s₁ = s₂ = 6.9072 kJ/(kg·K)
The turbine exit pressure, p₂ = 8 kPa (given)
[tex]\displaystyle x_2 = \mathbf{\frac{s_2 - s_f}{s_g - s_f}}[/tex]
[tex]s_g[/tex] = 8.2287
[tex]s_f[/tex] = 0.5926
[tex]\displaystyle x_2 = \frac{6.9072 - 0.5926}{8.2287 - 0.5923} \approx 0.8269[/tex]
[tex]h_{2s} = h_f + x_2 \cdot h_{fg}[/tex] = 173.88 + 0.8269 × 2403.1 ≈ 2161.00339
[tex]h_{2s}[/tex] = 2161.00339 kJ/kg
State 3 is saturated liquid enters the pump, h₃ = 173.88 kJ/kg
p₃ = 8 kPa (given)
h₃ = [tex]h_{f3}[/tex] = 173.88 kJ/kg
State 4 to state 1; The boiler
p₄ = The pressure in the boiler = p₁ = 8 MPa
[tex]h_{4s}[/tex] = h₃ + v₃·(p₄ - p₃)
At 8 kPa, saturated liquid, v₃ = [tex]v_{f3}[/tex] = 1.0084 × 10⁻³ m³/kg
Therefore;
[tex]h_{4s}[/tex] = 173.88 kJ/kg + 1.0084 × 10⁻³ m³/kg × (8 MPa - 0.008 MPa) = 181.94 kJ/kg
[tex]\displaystyle Turbine \ efficiency, \ \eta_t = \mathbf{\frac{h_1 - h_2}{h_1 - h_{2s}}}[/tex]
Which gives;
[tex]\displaystyle0.88= \mathbf{\frac{3545.3 - h_2}{3545.3 - 2161.0339}}[/tex]
Which gives;
h₂ = 2327.15 kJ/kg
[tex]\displaystyle Pump \ efficiency, \ \eta_p = \frac{h_{4s} - h_3}{h_4 - h_3}[/tex]
Which gives;
[tex]\displaystyle 0.82= \frac{181.94 - 173.88}{h_4- 173.88}[/tex]
h₄ ≈ 183.71 kJ/kg
[tex]\displaystyle \frac{\dot Q_{in}}{\dot m} = \mathbf{ h_1 - h_4}[/tex]
[tex]\dot m \cdot ( h_1 - h_{4}) = \dot Q_{in}[/tex]
[tex]\displaystyle \dot m = \mathbf{ \frac{\dot Q_{in}}{ ( h_1 - h_{4})}}[/tex]
[tex]\dot Q_{in}[/tex] = 26 MW
Which gives;
[tex]\displaystyle \dot m = \frac{26,000}{ ( 3545.3- 183.71)} \approx 7.73[/tex]
The mass flowrate, [tex]\dot m[/tex] ≈ 7.73 kg/s
[tex]\displaystyle \mathbf{\frac{\dot W_t}{\dot m}} = h_1 - h_2[/tex]
[tex]\displaystyle \frac{\dot W_p}{\dot m} = \mathbf{h_4 - h_3}[/tex]
[tex]\dot W_{cycle} = \dot W_t - \dot W_p[/tex]
Therefore;
[tex]\dot W_{cycle} = \mathbf{\dot m \times ((h_1 - h_2) - (h_4 - h_3))}[/tex]
[tex]\dot W_{cycle}[/tex] ≈ 7.73 kg/s ×((3545.3 kJ/kg - 2327.15 kJ/kg) - (183.71 kJ/kg - 173.88 kJ/kg)) ≈ 9340.3 kW = 9.3403 MW
- The net power developed, [tex]\dot W_{cycle}[/tex] ≈ 9.3403 MW
The thermal efficiency, η, is given as follows;
[tex]\displaystyle \eta =\frac{\dot W_{cycle}}{\dot Q_{in}} = \mathbf{\frac{(h_1 - h_2) - (h_4- h_3)}{h_1 - h_4}}[/tex]
Therefore;
[tex]\displaystyle \eta = \frac{((3545.3 - 2327.15) - (183.71 - 173.88)) }{3545.3 - 183.71} \approx 0.359 = 35.9\%[/tex]
- The thermal efficiency, for the cycle, η ≈ 35.9%
Learn more about vapor power cycles here:
https://brainly.com/question/15311479
