Question

In a power generating plant, steam enters the turbine at 150 bar, ${600}^{\circ }C$ and exits at 0.5 bar. Isentropic efficiency of the turbine is $85\%$. Neglecting the pump work, the power generated by the plant is
Use the following data:

• A. 1073.55 kJ/kg
• B. 1263.00 kJ/kg
• C. 1134.56 kJ/kg
• D. 1346.23 kJ/kg

Answer 1

The correct option is A 1073.55 kJ/kg

$\begin{array}{cccc}\text{P(bar)}& l{(}^{\circ }C)& h\left(kJ/kg\right)& s\left(kJ/kgK\right)\\ 150& 600& 3583& 6.6797\\ 0.5& 81.32& \begin{array}{c}{h}_f=340\\ h_g=2645\end{array}& \begin{array}{c}{s}_f=1.091\\ s_g=7.593\end{array}\end{array}$

At state 1 $h_1=3583\text{kJ/kg}$
$s_1=6.6797\text{kJ/kgK}$
$s_1=s_{2s}=6.6797$
$6.6797=s_f+x(s_g-s_f)$
$x=\frac{6.6797-1.091}{7.593-1.091}$
$x=0.859$
$h_{2s}=h_f+x(h_g-h_f)$
$h_{2s}=340+0.859(2645-340)$
$h_{2s}=2319.995\text{kJ/kg}$
Actual work $={\eta }_{isen}\times \text{Isentropic work}$
$=0.85\times (3583-2319.995)$
Actual work $=1073.55\text{kJ/kg}$