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Solar–wind hybrid renewable energy system                         231















           Figure 12.5  Optimum sizing criteria/methodology of solar wind HRES. HRES, hybrid
           renewable energy systems.

           4  Optimization techniques for hybrid solar–wind system

           Optimum sizing method is necessary to efficiently and economically utilize the RES.
           The optimum sizing of the components of HRES can ensure the lowest investment
           with the maximum utilisation of the system components so that the HRES can work at
           the optimum conditions in terms of investment and system power reliability require-
           ment. An optimum combination of the hybrid system is the one that can make the best
           compromise between power reliability and system cost. Hence, to select an optimum
           combination of components for a hybrid system, the evaluation must be carried out
           based on power reliability and system life-cycle cost. Fig. 12.5 present implicit crite-
           ria/methodology generally employed for optimum sizing of HRES [15].


           4.1  Criteria for hybrid solar–wind system optimization
           4.1.1  Power reliability analysis
           Solar radiation and wind speed characteristics highly influence the energy produc-
           tion from the hybrid system. Due to their intermittent characteristics, power reliability
           analysis is usually considered as an important step in optimum system design. Several
           methods have been studied to calculate the reliability of the hybrid systems. Among
           them, the loss of power supply probability (LPSP) is the most popular one.
              The LPSP is the probability that an insufficient power supply results when the
           hybrid system (PV, wind power and energy storage) is not able to satisfy the load
           demand [64–66]. LPSP can be represented by the (Eq. (12.5)).

                         T
                        ∑ DE()
                              t
                         =
                LPSP =  T  t 1
                      ∑  P ()∆  t                                          (12.5)                                       LPSP=∑t=1TDEt∑t=1TPloadt∆t
                             t
                          load
                       t 1=
           where DE(t) represents the deficit energy at hour t.
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