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THERMODYNAMICS - CASE STUDY SOLUTION

 

The volumetric flow rates of the ocean surface water and deep ocean water need to be determined for a new Ocean Thermal Energy Conversion (OTEC) power plant.

Assumptions:

Assume the ocean water has the same properties as pure water and is modeled as incompressible fluid.

     


Schematic of an OTEC Power Plant

 

(1) Determine the heat transferred to the power plant from the surface water (H) and the heat transferred from the power plant to the deep ocean water ( L).

To operate an OTEC power plant involves both a heat source and a heat sink. Therefore, the hot ocean surface water serves as the heat source and the cold deep ocean water serves as a heat sink. The power plant generates net work. Hence, it is a heat engine.

The definition of thermal efficiency for a heat engine is

       

wherenet,out is the power generated by the power plant, which is 15,000 kW in this case.

If the ηth is given as 5%, heat transferred from the heat source to the power plant is

       

The energy balance of the power plant is

      

Hence the heat transferred to the sink from the power plant is

       = 300,000 - 15,000 = 285,000 kW
     


Evaporator: Heat Transferred to Propylene from the Ocean Surface Water


Condenser: Heat Transferred to the
Deep Ocean Water from Propylene

 

(2) Determine the volumetric flow rates of the ocean surface water and the deep ocean water

The ocean surface water is sent to the evaporator (a heat exchanger) where the working fluid propylene vaporizes. Consider the ocean water in the heat exchanger as a system, the heat transfer between the ocean water and the propylene can be determined as:

      

where ΔT is the temperature difference of the ocean water between the inlet and exit, which is

      ΔT = 27 - 22 = 5 oC

Hence the volumetric flow rate () of the surface water is

      

The deep ocean water is sent to the condenser (a heat exchanger) where the propylene vapor condenses to liquid. The flow rate of the deep ocean water can be determined using a similar way. That is,

      

At 1,000 meter depth of the sea, the temperature of the water is

      Tdeep = -0.022 h + Tsurface
              = -0.022(1,000) + 27 = 5oC

The water leaves the condenser at 10 oC. Hence, the temperature increase of the deep ocean water is

      ΔT = 10 - 5 = 5 oC

In heat exchangers, properties are determined using the average temperature at the inlet and exit. For the evaporator, the average temperature of the water is 24.5 oC.The density and the specific heat of water at that temperature are:

      ρsurface = 997 kg/m3 and cP,surface = 4.18 kJ/kg-oC

Also, the average temperature for the condenser is 7.5 oC, the density and the specific heat of water are:

      ρdeep = 997.8 kg/m3 and cP,deep = 4.22 kJ/kg-oC

Substituting all these numbers to the flow rate equations, the flow rates can be determined. They are:

      surface = 300,000/((997)(4.18)(5)) = 14.4 m3/s

      deep = 285,000/((997.8)(4.22)(5)) = 13.5 m3/s