Question

A steam power plant operating on a simple ideal Rankine cycle maintains the boiler at 6000 kPa, the turbine inlet at 600 °C, and the condenser at 50 kPa. Compare the efficiency of this cycle when it is operated so that the liquid enters the pump as a saturated liquid against that when the liquid enters the pump 11.3 °C cooler than a saturated liquid at the condenser pressure.

Answer 1

Answer:

ηa=0.349

ηb=0.345

Explanation:

The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:

$h_{3}=3658.8kJ/kg\\ s_{3}=7.1693kJ/kg$

The quality at state 4 is determined from the condition  $s_{4} =s_{3}$ and the entropies of the components at the condenser pressure taken from table:

 $q_{4} =\frac{s_{4}-s_{liq50} }{s_{evap,50} } \\=\frac{7.1693-1.0912}{6.5019}=0.935$

The enthalpy at state 4 then is:  

$h_{4} =h_{liq50} +q_{4} h_{evap,50}\\ (340.54+0.935*2304.7)kJ/kg\\=2495.43kJ/kg$

Part A

In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:  

$h_{1}=340.54kJ/kg\\ \alpha _{1}=0.00103m^3/kg$

The enthalpy at state 2 is determined from an energy balance on the pump:

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

    =346.67 kJ/kg

The thermal efficiency is then determined from the heat input and output in the cycle:  

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.349$

Part B  

In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from  and the saturated liquid values:  

$h_{1}=293.7kJ/kg\\ \alpha _{1}=0.001023m^3/kg$

The enthalpy at state 2 is then determined from an energy balance on the pump:  

$h_{2} =h_{1} +\alpha _{1}( P_{2}-P_{1} )$

    =299.79 kJ/kg  

The thermal efficiency in this case then is:  

$=1-\frac{q_{out} }{q_{in} } \\=1-\frac{h_{4} -h_{1} }{h_{3} -h_{2}} \\=0.345$