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THERMODYNAMICS - CASE STUDY SOLUTION


 

A hydraulic turbine is used to produce power using for the apartments' lighting. The power generated needs to be determined.

Assumptions:

  • Heat transfer is negligible
  • Potential energy change is negligible
  • Kinetic energy change is negligible
     


Control Volume

 

The energy balance for one-inlet-one-exit system is:

      

Applying the basic assumptions for turbines, the energy balance can be rewritten as:

      

Water enters the turbine at 300 K and 750 kPa. At 300 K and 750 kPa, water is compressed liquid. The enthalpy of compressed liquid can be approximated by

      h = hf@T + vf (P - Psat.)

where
      hf@T = the enthalpy of saturated water at
                temperature given
      vf = the specific volume of saturated water at
            temperature given
      Psat. = the saturation pressure at temperature given

From the water table, the saturated properties of water at 300 K are:

      hf@300 k = 104.89 kJ/kg
      vf = 0.001003 m3/kg
      Psat. = 3.169 kPa

Hence, enthalpy at 300 K and 750 kPa is:

      h1 = 104.89 + 0.001003(750 - 3.169)
          = 105.64 kJ/kg

At 300 K and 170 kPa, water is compressed liquid also. Using the same approximation, the enthalpy of water at 300 K and 170 kPa is

      h2 = 104.89 + 0.001003(170 - 3.169)
          = 105.06 kJ/kg

Substituting these enthalpies and the given mass flow rate to the energy balance equation yields

      - = 10,000/3,600(105.06 - 105.64) = -1.6 kW
       = 1.6 kW

The power is generated by the system, so it is positive.

The power needs for lighting an apartment is 500 W, which is smaller than the power generated by the turbine.

       500 W < 1.6 kW = 1,600 W

Hence, the result shows the turbine can be used to generate power for the lighting system.