摘要
首次用解析的方法完整地研究了任意坡向的坡面相对于天文辐射日总量的最热坡度问题,建立了任意坡向的坡面相对于天文辐射日总量的最热坡度解析模式.研究表明,当坡面日出日没时角的配置关系和水平面或非水平面日出日没时角相同时,存在相对于坡面天文辐射日总量的最热坡度,而其余时角配置关系则不存在相对于坡面天文辐射日总量的最热坡度.当存在相对于坡面天文辐射日总量的最热坡度时,最热坡度可解析地求得.南、北坡的最热坡度计算公式是任意坡向最热坡度计算公式的特例.
This paper studies the problem of the warmest slope of daily extra_terrestrial solar radiation in random slope azimuth. According to the laws of the matching relations of sunrise and sunset hour angles and their changes, the matching relations of sunrise and sunset hour angles on slope vary with slope azimuth angle β, slope angle α, latitude φ and solar declination δ. For different slope azimuth scope, we can obtain different matching relations and their changes with solar declination. If the single order derivative of the daily solar radiation to slope is equal to zero, and the double order derivative is below zero,there is the warmest slope.When the matching relation is ω_1=-ω_0,ω_2=ω_0,ω_1 =ω_(s_1),ω_2=ω_(s_2)or ω_1=ω_(s_2),ω_2=ω_(s_1),there are the warmest slopes .But in other matching relation, there is not the warmest slope.When ω_1=-ω_0,ω_2=ω_0,the daily solar radiation can be expressed as:Qs=I_0Tπρ~2= Where, U=sinφcosα-cosφsinαcosβ, V=cosφcosα+sinφsinαcosβ. So, the warmest slope α_0 can be calculated by the following formula: tanα_0=sinφsinω_0-ω_0cosφtanδcosφsinω_0+ω_0sinφtanδcos βWhen ω_1=ω_(s_1),ω_2=ω_(s_2)or ω_1=ω_(s_2),ω_2=ω_(s_1), the daily solar radiation can be expressed as:Qs=I_0Tπρ~2(ω_xsin xsinδ+cos xcosδcosω_x)ω_x?x are two parameters and can be expressed as: sinx= sinφcosα-cosφsinαcosβ, ω_x=arccos(-tanxtanδ).So, the warmest slope can be calculated by following formula:sinα_(x_1)=-cosφsinx_1cosβ±sinφcos^2x_1-sin^2βcos^2φ1-sin^2βcos^2φWhere, x_1 is one of the boots of equation ω_xtanδ-tanxsinω_x=0.
出处
《南京大学学报（自然科学版）》
CAS
CSCD
北大核心
2003年第6期781-787,共7页
Journal of Nanjing University（Natural Science）
