Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations KootK on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Pyranometer

Status
Not open for further replies.

Sinexus

Mechanical
Jan 13, 2006
7
I'm using a pyranometer that on it's side says to have an output unit of 4.55 V/(W*m^-2). I'm unsure how to interpret this unit, so if anyone has any experience with this I would appreciate it.
 
Replies continue below

Recommended for you

You hook up the output leads to a voltage datalogger or an oscilloscope, and then expose the meter to a source of light. The voltage signal you measure is proportional to the intensity of light at the spot where you have placed the meter. W/m2 is watts per square meter, or power of light received per square meter of exposed area.

An example is that on the surface of the earth the maximum intensity of solar radiation is about 1000 W/m2. Your meter would read 1000 x 4.55 or 4500V. That seems rather high to me. Check the label again to see if it doesn't say 4.55 mV/W m2.
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor