Respuesta :
Answer:
[tex]\eta = 94.64[/tex]
entropy generation = 0.45908 kw/k
Explanation:
Given data:
pressure of steam is 6MPa
temperature of steam is 600 degree Celcius
For temperature 600 degree Celcius
latent heat h_1 = 3658.8 kJ/kg
specific heat s_1 = 7.1693 kJ/kg K
Power produced
[tex]P = m (h_1 -h_2)[/tex]
[tex]2626 = 2 (3658.8 -h_2)[/tex]
solving for h_2
[tex]h_2 = 2345.8 kJ/kg[/tex]
At 10 kPa
[tex]h_f = 191.81 kJ/kg[/tex]
hfg = 2392.1 kJ/kg
sf = 0.6492 kJ/kg K
sfg = 7.499 kJ/kg
h = hf + x hfg
[tex]2345.8 = 191.81 + x \times 2392.1[/tex]
x = 0.9
[tex]s_2 = sf + x\times sfg[/tex]
    [tex]= 0.6492 + 0.9 \times 7.4996[/tex]
[tex]s_2 = 7.39884 kJ/kg K[/tex]
Now considering [tex]s_1 = s_2[/tex] to determine enthalalpy
[tex]s_1 = s_2 = 7.1693 Â kJ/kg K[/tex]
s = sf + x sfg
[tex]7.1693 = 0.6492 + x \times 7.4996[/tex]
x2s = 0.869
[tex]h2s = hf + x2s hfg[/tex]
[tex]h2s = 191.81 + 0.869 \times 2392.1[/tex]
x = 2271.4778 kJ/kg
a) isotropic efficiency
[tex]\eta = \frac{h_1 - h_2}{h_1 -h_{2s}} \times 100[/tex]
[tex]\eta = \frac{3658.8 - 2345.8}{3658.8 -2271.4778} \times 100[/tex]
[tex]\eta = 94.64[/tex]
b) entropy generation
[tex] = m(s_2 -s_1)[/tex]
[tex]= 2\times (7.39884 - 7.1693)[/tex]
= 0.45908 kw/k
