Better read your charts again polished copper is .05 tarnished copper 0.78 that copper was not polished. I say again the cold air blowing across that copper had more effect than the e value or do you not understand the operation of a heat pump I will be holding a class this weekend:p:D
Make up your mind Charlie. In post 3 you said it was shined.
I’m using the correct value for long wave, although degrees of polish may vary. Besides, do you think there is a meaningful difference in your accuracy between .02 and .05 emissivity?
The point is none of the people here should be attempting to compare targets of radically different emissivity in a thermal image with their emissivity value set to .9, even if they are only doing qualitative thermography.
Let’s not teach bad habits.
I’m getting a headeache…1 1/2 hrs and it’s not making any logical sense to me other than two different types of material.
If it was a infrared thermometer would the temps be the same?
When realistic figures are changed or manipulated, every thing past that point is hypothetical.
No.
In visible light terms, think of it as a mirror. It’s very difficult to focus on the actual glass of a clean mirror, because most of what you are seeing is what is being reflected.
The same is true of infrared radiation. The properties involved are emissivity and reflectivity, which are assigned values. They are complimentary and combined always equal 1.00 for objects that are opaque to IR (we will leave IR transparency out of the discussion for now).
The higher the emissivity the greater the percentage of incident radiation being detected by the imager is actually coming from the target. The lower the emissivity, the greater the percentage of radiation is being reflected by the target from other sources. The greater the temperature difference between your target and what is being reflected, the more inaccurate the temperature reading is. So if your target has an emissivity value of .02, 98% of what you are seeing is coming from something else and has nothing to do with the actual target temperature. If there is a couple hundred degrees difference between the target and what is being reflected, imagine how inaccurate your imager will be. With a curved profile, like a pipe, it’s even worse, because you often can’t even tell what is being reflected.
Within reasonable E levels, you can compensate and calculate with relative accuracy, provided you know the actual emissivity and reflected apparent temperatures.
Keep in mind that many things that are very reflective to visible light may not be reflective to infrared or vice versa and emissivity is often quite different between the short and long range IR wavelengths and can change with temperature.
The easiest and most effective way to compensate for low emissivity targets is to apply a coating to alter their emissivity using something of a known emissivity (e.g., paint, Arid deodorant, 3M electrical tape, etc.).
I change my E valvue in the software not in the camera where are in the real world are you going to see or use.02 on copper you are trying to blow it out of portion I was not trying to compare one temp to the other I was trying to get understanding why the 65 degree temp on a line that had hot gas going thru the inside of the tube
Ok here is the same image with the E valve set at .78 as it would have been if this was for real, the original image had 0.90 E valve and the difference in temp is less than one degree F The cold air has more effect on the copper than the E value setting
Is the RAT really 68F? How do you know what is being reflected? How do you know that the E value for the copper pipe is really .78 or what the E value is of the target that you are comparing it to?
This quiz about reversing valves in HPs really isn’t helpful to folks in helping them “Understanding IR”
If you were to put black tape on it then would you have the correct E value?
You are talking above most of the heads of the viewers on this board I don’t know if you are trying to impress someone or not. I try to keep it simple. Its not up to you to decide what is helpful and what is not. My original intent was just trying to get folks to understand that images are not always as they appear and one needs to understand what they have the camera pointed at. The HP was in operation the aluminum fan blades rotating was reflecting temp from every where nothing was reality and here you are jumping on E value
Assuming you knew the E value of the tape. The convention is to use 3M electrical tape with an known E value of .95. Use the tape as the target.
BTW: you can use this method to calculate the E value of the object that you placed the tape on. However, you always want to use the highest E value available for best accuracy.
You still need to be concerned with the E value of the painted metal items that you are comparing to, but most paint is typically pretty high .9+
You should state that up front. It would be helpful for people to know that nothing in your example is reality as pertains to IR.
I didn’t come up with this self-promotion.
So if I use the tape (e=.95) then place it on the pipe for a period of time, the heat from the pipe will transfer to the tape giving me the correct temperature?
Exactly. Set your imager E to .95 and take your reading from the tape.
For greatest accuracy you would also check and set values for reflected temperature, distance and humidity, but that is well beyond what you would need to do for most applications at shorter distances.
Forgot about this discussion. My original opinion was based upon the low coefficient of emissivity of the copper. There would have been a strong probability that it was reflecting some of the heat that was collecting near the inside top of the exterior case. If the ambiemt temp was around 50 deg F and the heat pump was running for a short time, this could have been the case.
Jeff
Here’s a quick picture/thread on the IR thermometer readingand why you shouldn’t trust it unless you know how to use the equipment.
Jeff
Did I miss the picture/thread?
I missed putting in the link. It’s there now.
Jeff
Compressors normally heats up like that when they switch on and off too often. Could be low refrigerant!
Compressor is compressing gas, gas passes through venturi tube and is cooled to become liquid Both compressor and coil leading to venturi would be hot. My Guess