Help needed with heat loss calculation from a steam pipe...please!by physicsidiott Tags: calculation, heat, loss, pipeplease, steam 

#1
Jan1810, 03:44 AM

P: 1

I’m an entrepreneur in South Africa with the rights to a type of insulating coating. It’s new on the market here and no real research and development has been done on the coating to determine how much energy it can save. I’m a marketer and I don’t have the educational background to answer this question that I am faced with almost every day. I have done test applications for some large factories with promising results that show substantial decreases in surface temperatures for only a 1mm thick application. I was hoping that the engineers at these factories would be able to calculate the possible energy savings that the coating would give them but it seems that they are as inadequate as I am. This is particularly frustrating as this is paramount to completing a feasibility study so that these factories can place the orders for which I have been working so hard to achieve. I’m hoping that someone on the world wide web would be willing to help me speed up this process.
I believe that I can use the formula for conductive heat loss for a pipe or cylinder which is: q = 2 π k (to  ti) / ln(ro/ri) It’s a 6” diameter steel steam pipe with an uncoated surface temperature of 130 °C After coating the pipe the temperature was reduced to 108 °C I think the thermal conductivity of the steel pipe is 68 W/mK (should K be converted to °C?...and how would I do that?) In South Africa we work in Watts and Kilowatts and °C and not Btu’s, °F or K. I need to see how this formula works in a stepbystep calculation so that a ‘dumb ass’ like myself can understand it, explain it and use it again in future. I’m also wondering how heat loss through radiation would come into the equation here? This formula has caused me even bigger headaches which is: P = ε б A (T⁴  T⁴c) It uses the Stefan Boltzmann constant which is: б = 5.6703 × 10⁻⁸ watt / m² K⁴ The emissivity of the steel pipe is 0.32 ...or is it 0.95? And what should the emissivity of the coating be...I would imagine, much closer to 0.1? This formula will also require a stepbystep calculation in order for me to understand it. If anybody on this forum is able to help and has the time to do so, the input will be greatly appreciated. 


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