Question combined mode radiation and convection ?

[manal950]

Here we have question with answer
Could please explain to me the answer of this question ?

and can explain to me why we don't use area 2pi . r^2 because we have pipe ?

and how we can know this combined modes ? I mean conduction and radiation ?

In addtion , As I know s = the stephan-Boltzmann constant 5.6697 x 10-8

why here we used that value for emissivy ?

 

[gneill]

Here we have question with answer
Could please explain to me the answer of this question ?

The solution is adding the heat losses from each mode (convection and radiation).

and can explain to me why we don't use area 2pi . r^2 because we have pipe ?

You want the surface area of the pipe, not the cross-sectional area.

and how we can know this combined modes ? I mean conduction and radiation ?

There are three possible modes of heat transfer: conduction, convection, and radiation. Heat transfer by conduction doesn't work so well for a gas, it's main method of transport is convection.

 

[manal950]

thanks but:

1 ) this value 5.67 X 10 ^ -8 for what ? becuse some time we used this for emissivy and some stme for stefan boltoman constant

2 ) Is alwyes in this formula we used surface area of the pipe ?

3 ) what is surface area for plate and cycle ?

4 ) How we know here in this question there is radiation of heat transfer ?

 

[gneill]

thanks but:

1 ) this value 5.67 X 10 ^ -8 for what ? becuse some time we used this for emissivy and some stme for stefan boltoman constant

It is Boltzmann's constant. See: Stefan-Boltzmann Law

2 ) Is alwyes in this formula we used surface area of the pipe ?

The surface area of a cylinder is given by the circumference multiplied by the length. The circumference is given by $2 \pi r$ or $\pi D$, where D is the diameter of the cylinder.

3 ) what is surface area for plate and cycle ?

(Circle?)
Surface for a plate is length x width. That gives the area of one side of the plate.
A circle is the usual $\pi r^2$

4 ) How we know here in this question there is radiation of heat transfer ?

The obvious hint is that they gave you a value for the emissivity, ε.

 

[manal950]

thanks so much but you said that is for Boltzmann's constant but here in this quetions we used 5.67 X 10 ^ -8 fro
emissivy ? why ?

 

[manal950]

yes now I got it ..

thanks so so muck gneill