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# How to calculate jacket loss?

Member Posts: 17
Has anyone ever calculated actual jacket loss? Rather than just using the estimated percentage, I want to know the true jacket loss. Does anyone know of any formulas or can help with this?

• Member Posts: 7,356
Jacket loss on what? A storage tank? Your piping?
• Member Posts: 17
A boiler. Can it be done by measuring the heat of a lot of points on the jacket and calculated?
• Member Posts: 7,356
A thermal imaging camera would get you close. A handful of temperature probes and a datalogger would allow you to correct that to a pretty decent number.
• Member Posts: 17
Is there a calculation to do once you have that data?
• Member Posts: 7,356
A couple of ways, best to do both and compare results.

Figure out how many square feet of boiler HX surface (exterior) there is and find the ΔT between that HX surface (at the inside of the insulation, averaged over several points) and the metal enclosure (again, over several points, though the TIC would help there once you calibrate it for emmissivity using a surface probe.) If it's a natural draft appliance, you would need to do the same thing with the underside of the HX and the concrete underneath it.

Another option would be to clock the gas going in to the burner, measure the heat lost to the chimney and how much comes out as hot water or steam, then see what's missing.
• Member Posts: 17
Thanks for the information, SWEI. Always a big help.
• Member Posts: 7,356
Happy to help, and interested to see what you learn.

Got a college nearby with a mechanical engineering department? This would be a great project for some aspiring ME.
• Member Posts: 17
I spoke with an engineer I know and was given a calculation for it although there seems to be some discrepancies. Once I clarify these things and get a solid number, I will post my findings and the calculation.
• Member Posts: 17
Thanks Hatterasguy, I will use both methods.
• Member Posts: 17
Just to follow up, the equation I was given is below:
T1=internal temperature + 460 to be in Kelvin
T2=external temperature + 460
Sigma= 0.173 x 10^-8
Emissivity=0.55

Q=Sigma x emissivity x square feet of panel x (T1^4 - T2^4)

This seems to be an ok way to calculate the btus that are being lost from each side. We took measurements of points all around the boiler and averaged them out for T2. We did this for the top, bottom, front, back, and sides.

If anyone has any input or suggestions, I would love to hear those as well.