Correlating Indian measured sky luminance distribution and Indian Design clear sky model with five CIE Standard clear sky models

Abstract

This paper aims to find out representative CIE (International Commission on Illumination) Standard Clear Sky model(s) for three different seasons - winter solstice, equinox, and summer solstice applicable for prevailing clear sky climatic conditions in India [Roorkee]. Indian measured sky luminance distribution database is available only for Roorkee [29°51^′ N; 77°53^′ E]. To find out the best match between Indian measured sky luminance distribution and each of five CIE Standard Clear sky models, sky component of spatial illuminance distribution over the working plane of a room was simulated by MATLAB for three different seasons. The effectiveness of using Indian design clear sky model proposed in IS: 2440:1976 was also evaluated side by side. Daylight Coefficient method has been applied for the simulation using Indian sky luminance database. The simulation has been done for the room with eight different window orientations ranging from 0° to 315° with an interval of 45° to generate data for the entire sky vault. To find out the best fit CIE model(s), statistical test of significance (95% confidence level) has been carried out for the spatial illuminance distributions taking results obtained with Indian measured sky luminance distribution data as reference. Analysis revealed that CIE Standard General Clear sky type 15 described as “White-blue sky, turbid with a wide solar corona effect” is the best-fit clear sky model for both summer and equinox seasons and sky type 11 described as “White blue sky with a clear solar corona” is the best-fit clear sky model for winter season at Roorkee. Hence these sky types can be used for daylight prediction instead of Indian Design Clear Sky Model which shows large deviations.

Annexure I

Derivation of daylight coefficient [D_γα] for horizontal illuminance computation

Considering an unobstructed horizontal surface, the expression for horizontal illuminance [ΔE_h] due to sky element specified by two angles γ and α is given as

$$ \Delta {{\text{E}}_{\text{h}}} = \frac{{{{\text{I}}_{\gamma \alpha }}\cos (\frac{\pi }{2} - \gamma )}}{{{{\text{R}}^2}}} $$

(I.1)

where, I_γα – directional luminous intensity due to sky element along the station point

$(\frac{\pi }{2} - \gamma )$ – angle of incidence of sky light

R – radius of sky vault.

Now, as the area element is perpendicular to radius,

$$ {{\text{I}}_{\gamma \alpha }} = {{\text{L}}_{\gamma \alpha }} * {\text{dA}} $$

where, dA – area of sky element and

L_{γα -} luminance of that sky element

Hence above Eq. I.1 becomes

$$ \Delta {{\text{E}}_{\text{h}}} = \frac{{{\text{L}}{}_{\gamma \alpha } * {\text{dA}} * \cos (\frac{\pi }{2} - \gamma )}}{{{{\text{R}}^2}}} = {{\text{L}}_{\gamma \alpha }} * \sin \gamma * \frac{\text{dA}}{{{{\text{R}}^2}}} $$

Now $\frac{\text{dA}}{{{{\text{R}}^2}}} = \Delta {\omega_{\gamma \alpha }}$, the solid angle subtended by sky element at station point,

So,

$$ \Delta {{\text{E}}_{\text{h}}} = {{\text{L}}_{\gamma \alpha }} * \sin \gamma * \Delta {\omega_{\gamma \alpha }} $$

(I.2)

Now from Eq. 10

$$ \Delta {\text{E}} = {{\text{L}}_{\gamma \alpha }} * {{\text{D}}_{\gamma \alpha }} * \Delta {\omega_{\gamma \alpha }} $$

(I.3)

Comparing Eqs. I.2 and I.3

$$ {{\text{D}}_{\gamma \alpha }} = {\text{sin}}\gamma $$