Answer:
cycle efficiency: 0.11
steam amount needed: 8.327 kg/s
Explanation:
Points 1,2,3 and 4 are depicted in first figure attached. h means enthalpy and s means entropy
.
Assumption [tex]s_2 = s_1 [/tex]
Interpolation to get [tex]h_1[/tex] (see second figure)
p (MPa) h (kJ/kg)
4 3445.3
5 [tex]h_1[/tex]
6 3422.2
(3445.3 - 3422.2)/(4 - 6) = (3445.3 - [tex]h_1[/tex])/(4 - 5)
[tex]h_1 = 3433.75 \; kJ/kg[/tex]
In a similar way [tex]s_1[/tex] is computed as 6.9852 kJ/(kg K)
At 12.35 kPa, i. e., 0.1235 bar (see third figure): [tex]s_f[/tex] = 0.6922; [tex]s_g[/tex] = 8.093; [tex]h_f[/tex] = 205.82 ; [tex]h_g[/tex] = 2590.575 (again from interpolation, units omitted)
Quality (x) = [tex]\frac{s_2 - s_f}{s_g - s_f} = 0.85[/tex]
Enthalpy = [tex]h_2 = (h_g - h_f)*x + h_f = 2232.86[/tex]
In a similar way [tex]h_3[/tex] is computed as 205.83 kJ/kg.
Point 4 is computed as saturated liquid at 5 MPa; then, [tex]h_4[/tex] = 1154.2 kJ/kg (see fourth figure).
Efficiency is calculated as
[tex]\eta = 1 - \frac{h_2 - h_3}{h_1 - h_4}[/tex]
[tex]\eta = 1 - \frac{2232.86 - 205.83}{3433.75 - 1154.2}[/tex]
[tex]\eta = 0.11[/tex]
Energy balance at turbine gives
[tex]\frac{\dot{W}}{\dot{m}} = h_1 - h_2[/tex]
[tex]\dot{m} = \frac{\dot{W}}{h_1 - h_2}[/tex]
[tex]\dot{m} = \frac{10,000 \; KW}{3433.75 \; kJ/kg - 2232.86 \; kJ/kg}[/tex]
[tex]\dot{m} = 8.327 \; kg/s [/tex]
