×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Urgent Help - Heat flux received and View Factors

Urgent Help - Heat flux received and View Factors

Urgent Help - Heat flux received and View Factors

(OP)
Hi everyone,

I have been given a problem at work where I have been told to evaluate the amount of heat/radiation received by a worker at the bottom of a exchanger. If the exchanger is at a certain temperature then we can evaluate the heat flux emitted by the body using stephen boltzamn law (correct me if I am wrong). From some publications I have managed to get an equation for incident received flux which is I= t x VF x SEP. kW/m2
Where t is the atmospheric transmissivity, VF is the view factor and SEP is the surface emissive power. SEP we can calculate using the boltzman law, t is transmissivty which I think certain values can be assumed but lastly I am having problems calculating the view factor. The typical view factor equation avalilible on wikipedia seems a difficult task if I am to make it an excel version as I would need to evaluate the flux at different distances and locations from the exchanger, furthermore the person could be standing between 2 different bundles emitting heat. Once we evaluate the heat from one bindle, assmuming that the person is standing between the 2, can we just add the flux received by 2 sources? Can someone please advise on the problem?

Thanks

RE: Urgent Help - Heat flux received and View Factors

If the subject is able to receive energy from two sources, then they add.

Seems to me that the wki artical is pretty clear, particularly: http://en.wikipedia.org/wiki/View_factor#View_factors_of_differential_areas  You can basically take all viewable area and divide into small, differential areas, say ~1 sq. in., and crank the equation for each one of the squares.  You could treat the viewable area of the person from each differential area as a flat plate or a group of flat plates.  There are probably  programs for doing this sort of thing, though.

TTFN

FAQ731-376: Eng-Tips.com Forum Policies
Chinese prisoner wins Nobel Peace Prize

RE: Urgent Help - Heat flux received and View Factors

For 2 dimensional problems , the easiest menas of developing a view factor is the "hottel's crossed string method".

RE: Urgent Help - Heat flux received and View Factors

(OP)
Many thanks guys. I have been able to solve the problem by assuming a value which has given me reasonable answers.
I did look at all the links that everyone provided and it was all very useful

Thanks again.

Regards

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources