Respuesta :
Answer:
(a) The percentage thermal efficiency is approximately 30.54%
(b) The temperature of the cooling water exiting the condenser is approximately 28 °C
Explanation:
In the question, we note that the boiler is the rector, therefore
T₁ = 620 °C = 893.15 K
P₁ = 100 bar
(a) From super heated steam tables at 100 bar, 600 °C
S₆₀₀ = 6.9045 kJ·kg⁻¹·K⁻¹
S₆₅₀ = 7.0409 kJ·kg⁻¹·K⁻¹
By interpolation, we have S₆₂₀ given by
[tex]S_{620} = S_{600} + (S_{650} - S_{600})\frac{620 - 600}{650 - 600}[/tex]
[tex]S_{620} = 6.9045 + (7.0409 - 6.9045 )\frac{620 - 600}{650 - 600}[/tex]
S₆₂₀ = 6.95906 kJ·kg⁻¹·K⁻¹
Therefore, S₁ = S₆₂₀ = 6.95906 kJ·kg⁻¹·K⁻¹
Similarly
h₆₀₀ = 3625.84 kJ·kg⁻¹
h₆₅₀ = 3748.32 kJ·kg⁻¹
[tex]h_{620} = 3625.84 + (3748.32 - 3625.84 )\frac{620 - 600}{650 - 600}[/tex]
h₆₂₀ = 3674.832 kJ·kg⁻¹
h₁ = h₆₂₀ = 3674.832 kJ·kg⁻¹
Therefore, T₂ is given by
T₂ = Temperature at 1 bar = 99.6059 °C = 372.7559 K
At S₁ = S₂ we have
S'₂ at 1 bar = 1.3026 kJ·kg⁻¹·K⁻¹
S''₂ at 1 bar = 7.3588 kJ·kg⁻¹·K⁻¹
Therefore, steam fraction x₂ is given by;
S₁ = S'₂ + (S''₂ - S'₂)×x₁
6.95906 =1.3026 + (7.3588 - 1.3026)×x₁
x₂ = 0.93399 ≈ 0.934
From which, h₂ = 417.436 + (2674.95 - 417.436)×0.934 = 2,524.94 kJ·kg⁻¹
Where saturated liquid exits the condenser, we have
h₃ = 417.436 kJ·kg⁻¹ it
[tex]v = v_f \ at 1 \ bar[/tex]
v = 0.00104315 = m³·kg⁻¹
Work done by pump = [tex]v_f \times (p_4 - p_3)[/tex]
∴ Work done by pump = 0.00104315 × (10000000 - 100000) = 10431.395685 J/kg = 10.431 kJ/kg
Therefore, h₄ - h₃ = 10.431 kJ/kg
The percent the thermal efficiency of the cycle is then
[tex]Cycle \ efficiency = \frac{(h_1-h_2)-(h_4-h_3)}{Gross\ heat \ supplied}[/tex]
[tex]Cycle \ efficiency = \frac{(h_1-h_2)-(h_4-h_3)}{(h_1-h_3)-(h_4-h_3)}[/tex]
Where the turbine efficiency = 87% and the condenser efficiency = 78% we have
[tex]Cycle \ efficiency = \frac{(h_1-h_2) \times 0.87-(h_4-h_3) \times 0.78}{(h_1-h_3)-(h_4-h_3) \times 0.78}[/tex]
[tex]Cycle \ efficiency = \frac{(3674.832 -2,524.94 )-10.431}{Gross\ heat \ supplied}[/tex]
[tex]Cycle \ efficiency = \frac{1149.892 \times 0.87-10.431 \times 0.78}{(3674.832 -417.436 )-10.431 \times 0.78} = 0.305383[/tex]
Therefore the cycle efficiency as a percentage = 0.305383 × 100 = 30.5383% ≈ 30.54%
(b) The temperature of the cooling water exiting the condenser is given by
Heat rejected in condenser = Heat absorbed by the cooling water
Heat rejected in condenser = -Q₂₃ = h₂ - h₃
∴ Heat rejected in condenser, -Q₂₃ = 2,524.94 kJ·kg⁻¹ - 417.436 kJ·kg⁻¹
∴ -Q₂₃ = 2107.504 kJ·kg⁻¹
Net cycle power = -∑ = W₁₂ - W₃₄ = 1149.892×0.87 - 10.431×0.78
Net cycle power = 1008.54222 kJ/s = 1.008542 MW
Where the developed net cycle power = 3 MW
The mass is given by 3/1.008542 = 2.975 kg
∴ Heat absorbed by the cooling water = 2107.504 kJ·kg⁻¹ × 2.975 kg
Heat absorbed by the cooling water = 6268.9612 kJ
Mass flow rate of cooling water = 114.79 kg/s
Therefore, from ΔQ = m·c·ΔT we have
Heat absorbed by 114.79 kg/s of cooling water = 6268.9612 kJ
Specific heat capacity of water = 4.2 kJ/kg
Therefore
6268.9612 kJ = 4.2×114.79×(T₂ - 15 °C)
∴T₂ = 15°C + 6268.9612 /(4.2×114.79) = 28.003 °C ≈ 28 °C.
