Question

A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 125 kPa in the condenser. Saturated steam enters the turbine. Irreversibilities in the turbine cause the steam quality at the outlet of the turbine to be 70 percent. Determine the isentropic efficiency of the turbine and the thermal efficiency of the cycle. Use steam tables. The isentropic efficiency of the turbine is %. The thermal efficiency of the cycle is

Answer 1

Answer:

The right solution is "28.45%".

Explanation:

The given values are:

$P_4=50\ kPa$

$h_4=0.7(2304.7)+340.5$

    $=1953.83 \ KJ/Kg$

and,

$P_3=15 \ mPa$

$h_3=hg$

    $=2610.8 \ KJ/Kg$

$s_3=sg$

    $=5.3108 \ KJ/Kgh$

At 45,

⇒ $x_{45} = \frac{5.3108-1.0912}{6.5019}$

          $=0.66$

At $P_4=50 \ Kpa$,

$h_f=340.54$

or,

$V_f=0.001030 \ m^3/Kg$

then,

⇒ $h_2=340.54+0.001030(15\times 10^{3}-50)$

        $=355.94 \ kJ/kg$

hence,

The isentropic efficiency of turbine will be:

⇒ $n_T=\frac{h_3-h_4}{h_3-h_{45}}$

         $=\frac{2610.8-1953.83}{2610.8-1836.26}$

         $=84.818$ (%)

The thermal efficiency of cycle will be:

⇒ $n_C=\frac{W_T-W_P}{2_{in}}$

         $=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}$

         $=28.45$ (%)