written 7.8 years ago by | • modified 7.8 years ago |
Mumbai university > Mechanical Engineering > Sem 3 > Thermodynamics
Marks: 12M
Year: May 2016
written 7.8 years ago by | • modified 7.8 years ago |
Mumbai university > Mechanical Engineering > Sem 3 > Thermodynamics
Marks: 12M
Year: May 2016
written 7.8 years ago by |
Pressure and condition of steam , at inlet to the turbine,
$p_1=6bar$,x=0.9
Exhaust pressure, $p_2=0.7bar$
From steam table at 6 bar: $h_1=h_f1+xh_gf1$=670.38+0.9×2085.8=2547.6KJ/Kg,
$s_1$=1.9308+.9×4.8785=6.32145 KJ/kgK
From steam table at 0.7 bar:
$h_f=377.04kJ/kg,h_fg=2282.5kJ/kg,v_f=0.001036 m^3⁄kg$
$s_f$=1.1929kj/kgk,s_fg=6.2853 kj/kgk
Pump work:
work=$(p_4-p_3 ) v_f=(6-.7)×0.001036×10^5$
=549.08 W
Turbine work:
$s_1=s_2=s_f2+x_2 s_fg2$
6.32145=1.1929+6.2853x
x=0.816
$h_2=h_f2+x_2 h_fg2=377.04+0.816×2282.5=2239.56KJ/Kg$
Therefore turbine work:
=m($h_1-h_2$ )
=2547.6-2239.56
=308.04 kW
The rankine efficiency:
$η_rankine=\frac{(h_1-h_2)}{(h_1-h_f2 )}$
=$\frac{308.04}{2547.6-377.04}$
=0.142
=14.2%