- #1

harsh_23

- 5

- 0

- Homework Statement
- We need to prove that fraction of total radiation from plate i falling on plate j is equal to 1-sin(α/2) where both plates are infinite long, share a common edge , have same width and are inclined at an angle α

- Relevant Equations
- in general fab represent fraction of total radiation from a falling on b

so sum of all fai =1

The figure is infinite out and in the page.

We need to find $$f_{ij}$$ means if $$I$$ is the total radiation plate i is emitting then $$f_{ij}$$ represent fraction of total radiation Falling on plate $$j$$ from plate $$i$$

$$So in general fab represent fraction of total radiation from a falling on b$$

$$so \ sum \ of \ all\ fai =1$$

So in the above image I want to prove why $$f_{ij} = 1-sin(α/2)$$

First thing I have observed in others proofs that they consider radiation going to the surrounding through the gap between the plates and they consider the gap as an imaginary surface let's call $$k$$ with width l

Now we can see that side l is simply $$2w×sin(α/2)$$

Also $$f_{ki}=f_{kj}=1/2$$

(by symmetry)(means 1/2 of $$k$$ radiation is going to plate $j$ and other to I)

Then by reciprocity theorem

(It's a general theorem in radiation)

(It states that $$A_a×f_{ab}=A_b×f_{ba}$$ (here $$A$$ represent area))

$$A_i×f_{ik}=A_k×f_{ki}$$ (now area of plates is proportional to width)

$$W×f_{ik}=2wsin(α/2)×1/2$$

$$f_{ik}=sin(α/2)$$

And $$f_{ik}+f_{ij}=1$$

So $$f_{ij}=1-sin(α/2)$$

Now my doubt is that radiation shouldn't only be going through the gap but also to the surrounding

So $$f_{ik}+f_{ij}≠1$$

If I call surrounding s then

$$f_{ik}+f_{ij}+f_{is}=1$$

And should be same for all

And even if I don't consider the radiation to surrounding then considering the gap be bit spherical (concave) then some radiation should go to itself so saying

$$f_{ki}=f_{kj}=1/2$$ should be false as some of the total radiation is going to itself.

So someone pls help Clearing My doubt...

#### Attachments

Last edited: