太陽能小屋的設(shè)計(jì)及電池鋪設(shè)畢業(yè)論文設(shè)計(jì)(編輯修改稿)
2025-07-21 20:44 本頁面
【文章內(nèi)容簡(jiǎn)介】 0的光伏電池, 選取逆變器為4個(gè)SN6和1個(gè)SN5還有1個(gè)SN1. 每20個(gè)光伏電池A3并聯(lián)一個(gè)逆變器SN6, 還有10個(gè)A3并聯(lián)在逆變器SN5, 剩下的27個(gè)C10光伏電池并聯(lián)在逆變器SN1. . , , . 靜態(tài)投資回收期, 其回收的年限為：即總成本除以每年的收益. . , , . . 動(dòng)態(tài)投資回收期, 設(shè)貼現(xiàn)率為, 設(shè)動(dòng)態(tài)投資回收期為, 則可以得到：當(dāng) 時(shí), 當(dāng) 時(shí), 當(dāng) 時(shí), 由此可以計(jì)算出動(dòng)態(tài)投資回收年限. 電池和逆變器的總成本為76380元. , , , （程序見附件7）. 6 模型評(píng)價(jià)模型的優(yōu)點(diǎn)：在求解問題時(shí), 從單一吸收光能最強(qiáng)、效率最好的入手,引入太陽輻射強(qiáng)度最大房屋朝向的計(jì)算模型, 再引入逆變器的計(jì)算公式, 目標(biāo)明確, 將復(fù)雜問題簡(jiǎn)單化, 減少了計(jì)算量, 可得出較高工作效率、經(jīng)濟(jì)型的光伏電池與逆變器的相應(yīng)配從而縮小了回收年限. 模型的缺點(diǎn)：所設(shè)計(jì)的房屋追求太陽能輻射量最大, 所以設(shè)計(jì)的方位并不是實(shí)際生活最美觀. 7 參考文獻(xiàn)[1] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽B題 太陽能小屋的設(shè)計(jì).[2] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽B題 太陽能小屋的設(shè)計(jì). [3] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽 B題 張富順、安明梅、.[4] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽B題 太陽能小屋的設(shè)計(jì). [5] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽 B 題 基于貪心算法的光伏電池最優(yōu)鋪設(shè)方案. [6] 2012高教社杯全國大學(xué)生數(shù)學(xué)建模競(jìng)賽B題 趙志成、蔡玉漢、. [7] 中國期刊全文數(shù)據(jù)庫：（遠(yuǎn)程訪問) , 2013321[8] 韓明, 王家寶, （MATLAB版）. 同濟(jì)大學(xué)出版社, 2009.[9] 韓明, 張積林, 李林, 林杰, , 2012.8 英文摘要The designing of solar house and battery laying. Caiyuhan [Abstract] Promoting the use of solar house has great significance for solving the global energy crisis and protecting environment. Computational model of solar radiation intensity is established in this paper, and then according to the algorithm to calculate the house orientation and the best angle of photovoltaic cells when the sun radiation intensity is the largest. Thus the best solar house is designed. The laying of solar energy equipment requires the solar photovoltaic battery power of the outer surface of the house as large as possible at the same time the cost of unit capacity as little as possible. So, we build two objective functions. In order to convenient calculation, we bringing in the discount factor, and establishes the payback period of investment calculation model. This paper use Matlab to solve the problem according to the known data given in the question. The solar house is meters long, 5 meters wide. Its south wall is meters high, and the North support a flat plate on the roof, which on the same side of the south wall, and the projection in the horizontal plane of the other side is on the same side of the north wall. We determine the angle plate and the horizontal plane is degrees. When the discount rate is 0, we get that the static payback period of investment given the cabin is and the discount rate is , we can draw the conclusion that the dynamic investment recovery period of solar house is years. [Key words] design solar energy solar house photovoltaic cell laying payback period of investment9 附件附件1第一個(gè)模型尋找角度的程序// : 定義控制臺(tái)應(yīng)用程序的入口點(diǎn). // include”” include”iostream”include””using namespace std。typedef struct sum{ float mounts[8760]。//水平面散射輻射強(qiáng)度 float mountz[8760]。//法向直射輻射強(qiáng)度 float x1[8760]。//入射角公式的第一個(gè)參數(shù) sin float x2[8760]。//入射角公式的第一個(gè)參數(shù) cos float pangel[8760]。pangel為壁面太陽方位角, 計(jì)算方法為太陽方位角壁面方位角}。class Bestangel { sun s。 public: Bestangel()。 Void cal()。//計(jì)算最佳角度}；FILE*fp。Bestangel:: Bestangel(){ fp=fopen(“”,”rt”)。 for(int i=0。I8760。i++) fscanf(fp,”%f%f%f%f%f”,amp。[i],amp。s,x2[i],amp。[i],amp。[i],amp。[i])。}void Bestangel::cal(){ float book[600]={0}。 float bx[600]={0}。int bb=0。float ia=0。//傾斜角float ib=0。//壁面方位角 float max=0。float maxia=0。float maxib=0。/*************ia,ib為搜索變量, 根據(jù)各種情況修改即可*************//*壁面方位角ib從正北方向?yàn)?, 順時(shí)針0360度傾斜角ia根據(jù)實(shí)際情況設(shè)置范圍為090度*/for(ib=90。ib=270。ib=ib+)//******壁面方位角的選定范圍{ for(ia=0。ia=90。ia=ia+)//******傾斜角的選定范圍 { float sumz=0
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