Respuesta :
Answer:[tex]355^{\circ}C[/tex]
Explanation:
Given
power Developed=28.75 MW
[tex]\dot{m}=50 kg/s[/tex]
[tex]P_1=5 bar[/tex]
[tex]v_1=61 m/s[/tex]
Steam exit as saturated
[tex]P_2=0.06 bar[/tex]
[tex]v_2=130 m/s[/tex]
From saturated steam stable h2=h_g=2567.4 KJ/kg[/tex]
Applying Steady flow Energy equation
[tex]\dot{m}\left [ h_1+\frac{1}{2}\left ( v_1^2\right )+gz_1\right ]+Q=\dot{m}\left [ h_2+\frac{1}{2}\left ( v_2^2\right )+gz_2\right ]+W[/tex]
Q=0 because of insulation and
[tex]z_1=z_2[/tex]
[tex]h_1=\frac{W}{\dot{m}}-\frac{1}{2}\left ( v_1^2-v_2^2\right )+h_2[/tex]
[tex]h_1=frac{28.75\times 10^3}{50}-\frac{1}{2}\left ( 61^2-130^2\right )\times \frac{1}{1000}+2567.4[/tex]
[tex]h_1=3149 kJ/kg[/tex]
[tex]At P_1=25 bar h_1=3149 kJ/kg[/tex]
using steam tables
[tex]T_1=355^{\circ}C[/tex]
