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284             Renewable Energy Devices and Systems with Simulations in MATLAB  and ANSYS ®
                                                                                ®

                       100                           Averaged electrical power (kW)  100
                                                     Capacity factor (%)  =280.5 (kW) 80
                      Averaged electrical power (kW)  60  factor (C ) f  ae  P rated (C f  = 30%) C =30%  60  Capacity factor (%)
                        80
                                                  P =84.2 (kW) with P
                                                   ae
                                                                rated
                                     Averaged electrical
                                       power (P )
                                                       = 280.5 (kW)
                                  Capacity
                        40
                                                                             40
                                                            f
                        20


                                                                             0
                         0                                                   20
                          0      100      200     300     400      500     600
                                             Rated power (kW)

            FIGURE 11.17  Annual averaged electrical power output (P ae ) and capacity factor for different rated powers
            (assuming a conversion efficiency of the PTOS of 80%).

              Example Problem 11.3

              Given the information in Figures 11.15 and 11.16 and assuming a PTO conversion efficiency of 80%
              and a P rated  of 280.5 kW,

                 a.  Compute the P ae  for the given WEC system
                 b.  Calculate the device capacity factor
                                                         .
                 c.  Calculate the AEP, assuming an availability of η 2 =  095 (95%) and a transmission efficiency
                   of η 3 =  098 (98%)
                         .

              Solution

                 a.  The electrical power matrix for the two-body floating-point absorber can be obtained follow-
                   ing Equation 11.4. P ae  is then obtained by summing the product of the electrical power matrix
                   and the joint probability distribution for the reference site (Figure 11.15) and P ae  = 84.2 kW.
                 b.  The capacity factor = P ae /P rated  = 30%.
                 c.  AEP = 84.2*8766*98%*95% = 687 MWh.


            11.5.2  CEC Generator
            A CEC generator is usually designed very similarly to wind turbine generators, so these problems
            and exercises are derived in similar analogy to designing wind turbine generators. The available
            hydrokinetic power of a water turbine can be computed from the water flow and the turbine dimen-
            sion as
                                                       2
                                           P turb nei = 0.5ρπ R CV 3                  (11.7)
                                                         p
            where
              R is the radius of the swept area of the turbine
              ρ is the water density
              V is the water flow
              C  is the performance coefficient of the turbine
                p
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