The spectral transmissivity of a glass cover used in a solar collector is given as Solar radiation is incident at a rate of \(950 \mathrm{~W} / \mathrm{m}^{2}\), and the absorber plate, which can be considered to be black, is maintained at \(340 \mathrm{~K}\) by the cooling water. Determine \((a)\) the solar flux incident on the absorber plate, \((b)\) the transmissivity of the glass cover for radiation emitted by the absorber plate, and (c) the rate of heat transfer to the cooling water if the glass cover temperature is also \(340 \mathrm{~K}\).

First, we need to determine the solar flux incident on the absorber plate. We know that the solar radiation is incident at a rate of \(950 \mathrm{~W} / \mathrm{m}^{2}\). Since there is no information given about losses, we can assume that there is no loss in the absorber plate, and thus the total incident radiation is equal to the flux on the absorber plate. So the solar flux incident on the absorber plate is \(950 \mathrm{~W} / \mathrm{m}^{2}\).

To find the transmissivity of the glass cover for radiation emitted by the absorber plate, we will use Planck's law. The law states that the spectral distribution of blackbody radiation is given by: \(B_{\lambda}(T) = \frac{2 \pi h c^{2}}{\lambda^{5}} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}\) where \(B_{\lambda}(T)\) is the spectral radiance, \(h\) is the Planck's constant, \(c\) is the speed of light, \(\lambda\) is the wavelength, \(k\) is the Boltzmann constant, and \(T\) is the temperature. However, before we use Planck's law, we must determine the average transmissivity \((\bar{\tau})\) of the glass cover for the wavelengths of interest. Using the given spectral transmissivity formula, we can integrate over the wavelengths of interest: \(\bar{\tau} = \frac{\int \tau_{\lambda} B_{\lambda}(T) d\lambda}{\int B_{\lambda}(T) d\lambda}\) Given that the absorber plate is black and maintained at \(340 \mathrm{~K}\), we can find the transmissivity using this equation. After calculating the average transmissivity, we can apply it to the radiative heat transfer between the absorber plate and glass cover.

With the average transmissivity from the previous step, we can now calculate the rate of heat transfer to the cooling water. Since the glass cover temperature is also \(340 \mathrm{~K}\), we can say that the heat transfer is due only to the radiation. We can calculate the net radiative heat transfer rate between the black absorber plate and the glass cover as follows: \(\dot{Q} = A (\sigma T^{4} - \sigma \bar{\tau} T_{glass}^{4})\) where \(\dot{Q}\) is the net radiative heat transfer rate, \(A\) is the area of the cover, \(\sigma\) is the Stefan-Boltzmann constant, \(T\) is the temperature of the absorber plate, and \(T_{glass}\) is the temperature of the glass cover. After we find the net radiative heat transfer rate, we can assume that this heat is transferred to the cooling water, so the rate of heat transfer to the cooling water is equal to \(\dot{Q}\).