A simple ideal Rankine cycle which uses water as the working fluid operates its condenser at 408C and its boiler at 3008C. Calculate the work produced by the turbine, the heat supplied in the boiler, and the thermal efficiency of this cycle when the steam enters the turbine without any superheating

871.8008 KJ/kg

2573.42085 KJ/kg

0.3743

Explanation:

Solution:-

- We are to analyze an ideal Rankine cycle, where the condenser and boiler operating temperatures are defined.

- We start of by evaluating the properties of water at each state before and after the process.

State 1: Condenser Exit / Pump Inlet

           T1 = 40°C   ---> P1,sat = 7.3851 KPa , v1 = vf = 0.001008 m^3/kg

            sat liquid         h1 = hf = 167.53 KJ/kg , s1 = sf = 0.5724 KJ/kg.K

            P1 = condenser pressure

State 2: Pump Exit / Boiler Inlet

           P2 = P3 = Psat,300°C = 8587.9 KPa

Process 1: " Isentropic Compression - constant volume "

The work done by pump in the compression process is:

          wp = v1* ( P2 - P1 )

          wp = ( 0.001008 ) * ( 8587.9 - 7.3851 )

          wp = 8.64915 KJ /kg

Determine the enthalpy at " State 2 " by energy balance on pump ( control  Volume) :

          h2 = h1 + wp

          h2 = 167.53 + 8.64915

          h2 = 176.17915 KJ/kg

State 3: Boiler Exit / Turbine Inlet

          T1 = 300°C   ---> P3,sat = 8587.9 KPa

           sat vapor         h3 = hg =  2749.6 KJ/kg , s3 = sg =  5.7059 KJ/kg.K

           P3 = Boiler pressure

Process 2: " Heat Addition - constant pressure "

The heat supplied in the boiler is:

          qb = h3 - h2

          qp = ( 2749.6 - 176.17915 )

          qb = 2573.42085 KJ /kg     Answer ( b )

State 4: Turbine Exit / Condenser Inlet ( Isentropic )

          P4 = P1 = 7.3851 KPa         sfg = 7.685 KJ/kg.K

          s4 = s3 = 5.7059 KJ/kg.K       sf = 0.5716994 KJ/kg.K

                                                           hf = 170.32524 KJ/kg

                                                          hfg = 2406.11 KJ/kg

Process 3: Isentropic Expansion - Determine the quality of liquid-vapor mixture phase ( x ) at state (4):

          x = (s4 - sf) / sfg

          x = (5.7059 - 0.5716994) / 7.685

          x = 0.66808

          h4 = hf + x*hfg

          h4 = 170.32524 + 0.66808*2406.11

          h4 = 1777.79920 KJ/kg

- The work-done by the turbine in the isentropic expansion process ( wt ) is:

          wt = h3 - h4

          wt = 2749.6 - 1777.7992

          wt = 971.8008 KJ/kg   ... Answer ( a )

- To determine the thermal efficiency ( nth ) of the rankine cycle. We need to determined the net work produced by the cycle ( wn ). The net work is the energy balance between the isentropic compression ( work done - pump ) and isentropic expansion ( work produced - turbine ):

          wn = wt - wp

          wn = 971.8008 - 8.64915

          wn = 963.15165 KJ/kg

- The thermal efficiency of a power cycle is the ratio of net work-produced ( wn ) and the heat supplied to the working fluid in the boiler ( qb ) as follows:

         nth = wn / qb

         nth = 963.15165 / 2573.42085

         nth = 0.3743     Answer ( c )

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explanation:

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