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THERMODYNAMICS - CASE STUDY SOLUTION

   

A reheat process is considered for a simple Rankine cycle to reduce the moisture content at the turbine exit. The pressure at which the steam should be reheated and the net work output are to be determined.

Assumptions:

  • All the components in the cycle operate at steady state.
  • Kinetic and potential energy changes are negligible.
     


The Schematic and the T-s Diagram
of the Reheat Rankine Cycle

Saturated Water Temperature Table
Saturated Water Pressure Table
Superheated Steam Table

 

Model the cycle as an ideal reheat Rankine cycle. The schematic and the T-s diagram are shown on the left.     

(1) Determine the pressure at which the steam should be reheated

The reheat pressure can be determined from the requirement that the entropies at state 5 and state 6 be the same. State 6 is saturated mixture with a pressure of 10 kPa. The requirement of the moisture content at state 6 gives that the quality of steam at state 6 is greater than 0.9. Thus, assume the quality of steam at state 6 equals to 0.9. Then the entropy at state 6 can be determined as following:

      x6 = 0.9
      s6 = sf@10 kPa + x6sfg@10 kPa
          = 0.6493 + 0.9(7.5009) = 7.4001 kJ/(kg-K)

where sf@10 kPa and sfg@10 kPa can be obtained from the saturated water table.

Since the steam will be reheated to the inlet temperature of the high-pressure turbine, and the entropies at state 5 and state 6 are the same, the pressure at state 5 can be determined from superheated vapor table.

      T5 = 600oC (given)
      s5 = s6 = 7.4001
      P5 = 3.82 MPa
      h5 = 3675.1 kJ/kg

Therefore, steam should be reheated at a pressure of 3.82 MPa to prevent a moisture content at the low-pressure turbine above 10.0 percent.

(2) Determine the net work output after the addition of the reheat process

To determine the net work output of the cycle, the enthalpies at all other states need to be obtained first. They can be found from water tables.

State 1: Saturated liquid water
      P1 = 10 kPa (given)
      h1 = 191.83 kJ/kg
      v1 = 0.00101 m3/kg

State 3: Superheated vapor
      T3 = 600oC      P3 = 16 MPa (given)
      h3 = 3569.8 kJ/kg
      s3 = 6.6988 kJ/(kg-K)

State 4: Superheated vapor
      P4 = P5 = 3.82 MPa
      s4 = s3= 6.6988 kJ/(kg-K)
      h4 = 3151.6kJ/kg

 State 5: Superheated vapor
      T5 = 600oC (given)
      s5 = s6 = 7.4001
      P5 = 3.82 MPa
      h5 = 3675.1 kJ/kg   

State 6: Saturated mixture
      P6 = 10 kPa (given)
      s6 = s5 =7.4001 kJ/(kg-K)
      x6 = 0.9
      h6= hf@10 kPa + x6hfg@10 kPa
         = 191.83 + 0.9(2392.8) = 2345.4 kJ/kg

     
   

The net work output of the cycle equals the difference between the turbines work and the pump work.

      wpump,in = v1 (P2 - P1)
                   = 0.00101(16,000 - 10) = 16.1 kJ/kg

      wturb,out = w turb,high + w turb,low
                  = (h3 - h4) + (h5 - h6)
                  = (3569.8 -3151.6) + (3675.1 - 2345.4)
                  = 1747.9 kJ/kg

Hence, the net work output of the cycle is

      wnet,out = wturb,out - wpump,in
                = 1747.9 - 16.1 = 1,731.8 kJ/kg

Heat input to the cycle equals the heat added by the primary heating process plus heat added in the reheat process.

      qin = qprimary + qreheat
           = (h3 - h2) + (h5 - h4)
           = (3569.8 -207.93) + (3675.1 - 3151.6)
           = 3885.4 kJ/kg

where h2  is determined from the energy balance of the pump.

      h2 = h1 + wpump,in
          = 191.83 + 16.1 = 207.93 kJ/kg

Thus, the thermal efficiency of the cycle can be determined from

      ηth = wnet,out/qin = 1,731.8/3885.4 = 44.6%

     
   

Note: the net work output, heat input and thermal efficiency of the cycle before adding the reheat process can be determined in a similar way using the simple ideal Rankine cycle model.

      wnet,out = 1448.2 kJ/kg
      qin = 3361.87
       ηth = 43.1%
      xexit of the turbine = 0.81

This shows that the net work output, heat input, thermal efficiency are all increased after adding a reheat process. And the moisture content in the steam at the exit of the turbine is reduced from 19.0 percent to 10.0 percent.