Question

In a Rankine cycle power plant, the steam supply is at 16 bar and dry and saturated. The condenser pressure is 0.3 bar. The following data can be used:
At 16 bar: $T_{\text{sat}}={198.3}^{\circ }C,h_g=2789.9\text{kJ/kg},s_g=6.4406\text{kJ/kgK}$
At 0.3 bar: $T_{\text{sat}}={65.9}^{\circ }C,h_f=217.7\text{kJ/kg},h_{fg}=2319.2\text{kJ/kg},s_f=1.0261\text{kJ/kgK},s_{fg}=6.6448\text{kJ/kgK}$
The second law of efficiency of the cycle is ______ (Correct to two decimal place)

• A. 0.943

Answer 1

The correct option is A 0.943
$T_1={198.3}^{\circ }C=471.3\text{K}$
$T_2={55.9}^{\circ }C=338.9\text{K}$
${\eta }_{\text{Cannot}}=1-\frac{T_2}{T_1}=1-\frac{338.9}{471.3}=0.280$
${\eta }_{\text{Rankies}}=\frac{\text{Adiabatic heat drop}}{\text{heat supplied}}=\frac{h_1-h_2}{h_1-h_{f_2}}$
$h_2=h_{f_2}+x_2{h}_{f{g}_2}$
$=217.7+x_2\times 2319.2$
Now $s_1=s_2$
$6.4406=s_{f_2}+x_{2}s{f}_{g_2}=1.0261+x_2\times 6.6448$
$x_2=0.815$
$\therefore h_2=217.7+0.815\times 2319.2=2107.8\text{kJ/kg}$
$\therefore {\eta }_{\text{Rankine}}=\frac{2789.9-2107.8}{2789.9-217.7}=0.265$
${\eta }_{l1}=\frac{{\eta }_{\text{Rankine}}}{{\eta }_{\text{Carnot}}}=\frac{0.265}{0.280}=0.943$