Question

Water is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 12 MPa, and the condenser pressure is 10 kPa. The mass flow rate of steam entering the turbine is 150 kg/s. If the work input to the pump is 12 kJ/kg, what is the rate of heat transfer in the boiler in MW?

Answer 1

Answer:

 Qh = 372.24 MW

Explanation:

Solution:-

- Determine the state properties for working fluid water ( Rankine Cycle ):

- We are to assume ideal rankine cycle.

- The mass flow ( m^ ) = 150 kg/s

  Turbine Inlet - ( Saturated Vapor ) -

   P2 = P3 = 12 MPa, Tsat = 324.68°C

   h3 = hg = 2685.4 KJ/kg , s3 = sg = 5.4939 KJ/kg.K

   Condenser Inlet/Turbine exit - ( Liquid-vapor )

    P4 = P1 = 10 KPa , Tsat = 45.8056°C , s4 = s3 = 5.4939 KJ/kg.K

    sf = 0.64919 KJ / kg.K      ,    sfg = 7.49961 KJ / kg.K

    hf = 191.81 KJ/kg    ,      hfg = 2392.09 KJ/kg

    h4 = 1737.0762 KJ/kg

- Evaluate the quality "x" of liquid vapor mixture of H2O at condenser inlet:

       

              x = ( s4 - sf ) / sfg

              x = ( 5.4939 - 0.64919 ) / 7.49961

              x = 0.64599 ;

- The enthalpy at this state:

              h4 = hf + x*hfg

              h4 = 191.81 + 0.64599*2392.09

              h4 = 1737.0762 KJ/kg

- The work input by the pump, win,p = 12 KJ/kg

 Pump Inlet - ( Saturated liquid ) -

   P1 = 10 KPa, Tsat = 45.8056°C

   h1 = hf = 191.81 KJ/kg , s1 = sf =  0.64919 KJ / kg.K

Boiler Inlet / Pump exit

  P2 = 12 MPa, Tsat = 324.68°C , s2 = s1 = 0.64919 KJ / kg.K

  h2 ?

- The isentropic compression process by the pump can be modeled by applying the first law of thermodynamic law for the work done by pump. Assuming steady state conditions of flow and ideal operation of pump at 100% efficiency:

          Win,p = mass flow ( m^ ) * [ h2 - h1 ]

          mass flow ( m^ ) * win,p = mass flow ( m^ ) * [ h2 - h1 ]

          h2 = win,p + h1

          h2 = 12 + 191.81

          h2 = 203.81 KJ/kg

- The rate of heat transferred from boiler to the working fluid (Qh) is determined by applying steady state first law of thermodynamic on the boiler and water system:

         

           Qh = mass flow ( m^ ) * [ h3 - h2 ]

           Qh = 150 * [ 2685.4 - 203.81 ]

           Qh = 372238.5 MW / 1000

           Qh = 372.24 MW

   

Answer 2

Answer:

$\dot Q_{in} = 372.239\,MW$

Explanation:

The water enters to the pump as saturated liquid and equation is modelled after the First Law of Thermodynamics:

$w_{in} + h_{in}- h_{out} = 0$

$h_{out} = w_{in}+h_{in}$

$h_{out} = 12\,\frac{kJ}{kg} + 191.81\,\frac{kJ}{kg}$

$h_{out} = 203.81\,\frac{kJ}{kg}$

The boiler heats the water to the state of saturated vapor, whose specific enthalpy is:

$h_{out} = 2685.4\,\frac{kJ}{kg}$

The rate of heat transfer in the boiler is:

$\dot Q_{in} = \left(150\,\frac{kg}{s}\right)\cdot \left(2685.4\,\frac{kJ}{kg}-203.81\,\frac{kJ}{kg} \right)\cdot \left(\frac{1\,MW}{1000\,kW} \right)$

$\dot Q_{in} = 372.239\,MW$