Question

The variation of the spectral transmissivity of a 0.6cm thick glass window is 0.92 in the wavelength range from 0.3-3micrometer. Determine the average transmissvity of this window for solar radiation at T=5800K. Also, determine the amount of solar radiation transmitted through the window for incident solar radiation of 650W/m².

Answer 1

Given: Variation of spectral transmissivity of glass window, =0.92

Wavelength Range, ₁=0.3 m and ₂= 3 m

Source temperature, T=5800 K

Incident solar radiation, G_incident = 650 W/m²

Obtain the value of dimensionless quantity called the black body radiation function,   as follows.

From the Black body Radiation Functions Table, obtain the value of corresponding to ₁T=1740 mK.

For T=1600  mK, =0.019718 and for T=1800  mK, =0.039341. Interpolate the values.

Thus, =0.0334

From the Black body Radiation Functions Table, obtain the value of corresponding to ₂T=17400 mK.

For T=16000  mK, = 0.973814 and for T=18000  mK, =0.980860. Interpolate the values.

Thus, =0.9787

The average transmissivity of this window at T=5800 K is

where is the Transmissivity

₁ is the Variation of Transmissivity

and   is the Black body radiation function for the corresponding wavelength and temperature.

The amount of solar radiation transmitted through the window, G is obtained as follows

Therefore the average transmissivity for solar radiation at 5800 K is =0.87 and the amount of solar radiation transmitted through the window is G=565.5 W/m².