Products

## Holooly Rewards

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

## Holooly Tables

All the data tables that you may search for.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Sources

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 12.4

Use the emission approximation to find the energy flux emerging from an isothermal gas layer at $T_g$ with an integrated mean absorption coefficient weighted by the blackbody spectrum at $T_g$ (the “Planck mean” in Section 12.6.1) of $κ_P = 0.010 cm^{−1}$ and thickness D = 1.5 cm, if the layer is bounded by transparent nonradiating walls and cold surroundings (Figure 12.1a).

## Verified Solution

If I(θ) is the total intensity emerging from the layer in direction θ, the emerging flux is $q=2\pi \int_{\theta =0}^{\pi /2}{I(\theta )}\cos \theta \sin \theta d\theta .$ The layer is isothermal with constant temperature $T_g$ so, for the inner integral of Equation 12.16, the quantities are independent of S* and the Planck mean of

$I(S)=\int_{S^*=0}^{S}{\left[\int_{\lambda =0}^{\infty }{k_\lambda } (S^*)I_{\lambda b}(S^*)d\lambda \right] }dS^*$                        (12.16)

$k_\lambda(T_g)$ gives $\pi \int_{\lambda =0}^{\infty }{\kappa _{\lambda }(T_{g})I_{\lambda b}(T_{g})d\lambda } =\kappa _{p}(T_{g})\sigma T^{4}_{g}$.Then Equation 12.16 can be integrated over any path $S = D/cos θ$ through the layer to yield $I (θ) =(1/\pi )\kappa _P(T_g)\sigma T_g^4D/\cos \theta .$ This gives the radiative flux from each boundary as

$q=2\int_{\theta =0}^{\pi /2}{\kappa _P(T_g)\sigma T_g^4D\sin \theta d\theta} =2\kappa_P(T_g)\sigma T_g^4D$                      (12.17)

which gives $q=0.03\sigma T_g^4$ for the specified numerical values. This is not a precise result, even though the layer thickness is optically thin: $κ_PD = 0.015 \ll 1.$ This is because the radiation reaching the layer boundary along each path has passed along a distance D/cosθ. For θ approaching π/2 the optical path length becomes very large, so the emission approximation cannot hold.

A more accurate solution including effects of the proper path lengths gives $q = 1.8κ_P(T_g)\sigma T_g^4D$ (see Section 10.8.4), which is a 10% decrease compared with Equation 12.17.