I have a report to get out gotta quit playing games and I wanta wait for Dave to come along and give his take:D:D
Is the drier clogged? Am I further out in left field?
No the dryer was Ok nothing at all wrong with the HP it was operating normal
It is in reverse operation. Normally the dryer would be cold
Look where the dryer is located by the MFG its between the outlet of the compressor and before the reversing valve that dryer is always going to be hot reguardless of the mode set on the thermostat normal cool or normal heat
PS I can see I need to give a class on the refrigerant cycle of a heat pump
I would certainly appreciate it as I spent about 35 minutes trying to solve this puzzle. Through all my research not once did I come across a terminology called a “dryer” in a heat pump…?
Its a dryer filer. It aDsorbs oil, moisture, acid and contaminants.
Juan
He said the unit was working normally in the hp phase, and the title is Understanding IR. Me think he just Scre*ing with us. He manipulated the picture
Hey guys I am not trying to belittle anyone here but the overall purpose of this thread was to indicate the difference between shooting images of a wet wall or a leaking shower pan and or mechanical equipment.
When shooting IR of electrical and mechanical equipment it is very important to understand the operation of the equipment you are imaging. Temp are very cruical and you have to have them correct on the IR side but you need to understand why they are correct when you are performing for pay.
Under normal conditions I would not have been concerned about the temp on the short copper discharge line on the HP I would not have given it a second glance I am just using it to prove a point
No John I am not screwing with anyone not my way I did raise the emissivity a bit in the software normal for copper would be 78 and I raised it to 90 on the image did not have much effect on the overall temp.
Just for you Jim I will do it this weekend when I have more time:D
The IR thermography takeaway from this is that you cannot point your thermal imager at a shiny piece of copper and be able to determine with any degree of accuracy what the actual temperature of that piece of copper is.
Polished copper has a long wave emissivity of .02. That means that 98% of the incident radiation that you are seeing is being reflected from other sources, which may be substantially hotter or colder than your target. Don’t attempt to do this.
If you need to use IR to measure the temperature of a shiny metal object, you must alter the emissivity of it in some way (usually by applying a coating or target coupon of known emissivity, such as 3M electrical tape).
So , instead of dissipation it’s reflection.
So to Chuck and John how do you reflect from a 200+ degree object and come up with a 65 degree temp between two hot objects;-)
Difference between painted steel and uncoated copper Final answer
Thanks, is it filer or filter?
This seems like an important or semi-important piece of knowledge I was not aware of.
You don’t know what is being reflected off of that piece of round shiny copper. On a clear day that sky behind you might read -40F or less, Out of range for your imager.
How do you point your imager at a material that is .02 emissivity and decide your imager is giving you an accurate reading, especially with your emissivity set at .9? That’s not in any reference or standard that I know of.
Let’s give the folks IR information that is of practical use to them rather than confound them.
Partially correct but not the total answer you guys are not considering what is happening inside the unit when its in full heat mode. The exterior condenser surroundering the compressor what is happening to temps in the heat mode. One has to consider every source of heat and or lack of it., Have you guys never placed your hand above the fan outside with the unit in the heat mode what is happening;-)
Chuck in this particular instance I was more concerned with the temp of the evap between 38 and 42 degrees and the fan drawing it across that 6 inch section of horizional discharge pipe than I was emissivity. Just proving my point that one needs to understand what he has his imager pointed at. You seem to be only conserned with emssivity which is important but not the only factor to be considered
If a 98% error factor doesn’t matter, then I guess emissivity doesn’t matter either.
But it doesn’t make sense to try to compare temps of two objects with radically different emissivity values in a single image without taking it into account.