Answer:
The net work per unit mass [tex]W_{net}[/tex] = 692.14 kJ/kg
The thermal efficiency for the cycle = [tex]\mathbf{26.78 \ \%}}[/tex]
Explanation:
For an Ideal Rankine cycle:
The following properties were obtained from the steam tables at 1 MPa.
[tex]h_1 = h_g = 2777.1 \ kJ/kg[/tex]
[tex]s_1 = 6.585 \ kJ/kg.k[/tex]
The properties obtained at 10 kPa are:
[tex]h_f = 191.81 \ kJ/kg\\ \\ h_f_g = 2392.1 \ kJ/kg \\ \\ s_f= 0.6492 \ kJ/kg .k \\ \\ s_f_g = 7.4996 \ kJ/kg/k \\ \\ v_f = 0.001010 \ m^3 /kg[/tex]
For 1 - 2 isentropic expansion:
[tex]s_1 = s_2[/tex]
[tex]s_1 = (s_f+x_2 \times 7.4996)[/tex]
6.585 = 0.6492 + x₂ × 7.4996
6.585 - 0.6492 = 7.4996x₂
5.9358 = 7.4996x₂
x₂ = 5.9358/7.4996
x₂ = 0.791
[tex]h_2 = h_f + x_2 \ hfg[/tex]
[tex]h_2 = 191.81 + 0.791 \times 2392.1[/tex]
[tex]h_2 = 2083.96 \ kJ/kg[/tex]
At 10 kPa;
[tex]h_3= h_f = 191.81 \ kJ//kg[/tex]
The pump work for the process:
[tex]h_4 - h_3 = vf(\Delta \ P)[/tex]
[tex]h_4 - h_3 = 0.001010(1000 -10)[/tex]
[tex]h_4 - h_3 = 0.9999 kJ/kg[/tex]
[tex]h_4 - 191.81 \ kJ/kg = 0.9999 kJ/kg[/tex]
[tex]h_4 = 191.81 \ kJ/kg + 0.9999 kJ/kg[/tex]
[tex]h_4 = 192.8 \ kJ/kg[/tex]
However, the turbine work [tex]W_T[/tex] can be computed by using the formula:
[tex]W_T = h_1 -h_2[/tex]
[tex]W_T =[/tex] ( 2777.1 - 2083.96 ) kJ/kg
[tex]W_T =[/tex] 693.14 \ kJ/kg
Thus, the net work [tex]W_{net}[/tex] can be determined as:
[tex]W_{net}[/tex] = [tex]W_T - W_p[/tex]
[tex]W_{net}[/tex] = 693.14 - 0.9999
[tex]W_{net}[/tex] = 692.14 kJ/kg
Similarly, to determine the thermal efficiency of this cycle, we need to first know the heat addition [tex]Q_s= h_1 - h_4[/tex]
[tex]Q_s= (2777.1 -192.8 ) \ kJ/kg[/tex]
[tex]Q_s= 2584.3 \ kJ/kg[/tex]
Finally, the thermal efficiency can be calculated by using the formula:
[tex]n_{th} = \dfrac{W_{net}}{Q_s}[/tex]
[tex]n_{th} = \dfrac{692.14}{2584.3}[/tex]
[tex]n_{th} = 0.2678[/tex]
The thermal efficiency for the cycle = [tex]\mathbf{26.78 \ \%}}[/tex]
