Question

In a steam power plant operating on a ideal Rankine cycle, superheated steam enters the turbine at $3$ MPa and ${350}^o$C. The condenser pressure is $75$kPa. The thermal efficiency of the cycle is .......
Given data: For saturated liquid at $p=75$ kPa,
$h_f=384.39$ $kJ/kg,$ $v_f=0.001037m^3/kg$,
$s_f=1.213kJ/kgK$
At $75kPa,h_{fg}=\mathrm{2278..6}kJ/kg$,
$s_{fg}=6.2434$ $kJK$
At $p=3Mpa$ and $T={350}^{o}C$ (superheated steam),
$h=3115.3kJ/kg,s=6.7428/,kJ/kgK$

• A. 26.01

Answer 1

The correct option is A 26.01

Calculating dryness fraction at point $2$
$s_1=s_2$
$6.7428=s_f+x_2{f}_{fg}$
$=1.213+x_2\times 6.2434$
or $x_2=\frac{6.7428-1.213}{6.2434}$
$=0.8857$
Calculating enthalpy at point $2$,
$h_2=h_f+x_2{h}_{fg}$
$=384.39+0.8857\times 2278.6$
$=2402.546kJ/kg$
Given: $h_1=3115.3kJ/kg$
$h_4=h_3+vdP$
$=384.39+0.001037\times [3000-75]$
$=387.423kJ/kg$
Pump work = vdP
$=0.001037\times [3000-75]$
$=3.033kJ/kg$
$W_{net}=W_T-W_P=(h_1-h_2)-3.033$
$=(3115.3-24-2.546)-3.033$
$=709.721kJ/kg$
Heat input = $h_1-h_4$
$=3115.3-387.423$
$2727.877kJ/kg$
Efficenciy of cycle$=\frac{Workdelivered}{Heatinput}$
$\frac{709.721}{2727.877}$
$0.2601=26.01$%