Heat transfer and heating up pipes/vessel question

[Barow]

Hello again guys:
I'm having some difficulty with remembering some heat transfer concepts and I hope it is not an imposition to ask for a little bit of advise.I have a vessel or exposed pipe.... exposed to sunlight and the concern is that it can heat up to unacceptable conditions. I want to find out how much time it will take to heat up to upset conditions...or to prove that it won't ever because of maybe insufficient energy from sun to meet conduction rate?

+++++++++++++++
- I have calculated the amount of heat energy at upset condition for the contents of the vessel, = UpsetH.E (say UpsetH.E = 2000 Joules). I have also used Q= m*Cp*(T2-T1) for the metal to calculate the temperature differentials in the metal vessel need to achieve the UpsetH.E

a) Assuming the sunlight is the source of energy that will heat up the metal to upset conditions and I calculate that net flux density from sun (SunQQ" (Watts)) from the sunlight is .....say = SunQQ" = 20 Joules/second = 20 Watts

b) I have used formula for metal heat conduction Rate = k•A•(T1 - T2)/w.t (knowing all these values, conductivity, Area,wall, thickness e.t.c ) to calculate my heat transfer rate in Watts to the metal. Lets say, I find that my heat transfer rate = SteelRate = say 100 Joules/sec = 100 Watts

I want to intepret these results:
(i) Can I conclude that my upset condition cannot be reached because I have calculated a heat transfer rate of SteelRate = 100 Watts and I have only SunQQ" =20 Watts from my heat source (Sunlight). That is that the Sun's 20 Watts is insufficient transfer rate to get the vessel to heat up to Upset conditions ?

OR

(ii ) Can I say , irrespective of the heat trasfer rate in the metal, the effect of sun is cummulative and heat transfer will occure at the SunQQ =20 Watts and that I will reach my upset condition within a certain time .......given by 2000/20 = 100 seconds?. This is irrespective of my conduction rate?

I would really appreciate your answers


Thanks very much

 

[IRstuff]

Not sure where you got the 20W solar load from; that's like 4 in the afternoon, hardly a peak sunlight condition, or an incredibly tiny surface area. The normally accepted peak solar load is at least 1 kW/m^2


Perhaps you should present some pictures, actual facts/data, etc. Your last statement is only valid if thermal conductivity is infinite.

TTFN
I can do absolutely anything. I'm an expert!
homework forum: //

http://www.engineering.com/AskForum/aff/32.aspx

faq731-376 forum1529

 

[racookpe1978]

What latitude are you going to install the piece at?
How will it be aligned with respect to the noonday sun?
What shape is the vessel? (Horizontal & round?, vertical and round? spheroidal? Box-like or building- like with rectangular sides?

 

[Barow]

>>Not sure where you got the 20W solar load from; that's like 4 in the afternoon, hardly a peak sunlight condition, or an incredibly tiny surface area. The normally accepted peak solar load is at least 1 kW/m^2
Perhaps you should present some pictures, actual facts/data, etc. Your last statement is only valid if thermal conductivity is infinite.<<

The numbers are not actual numbers and were just for illustrative purpose to describe the concept and yes, the actual solar load would be more like 787 W/m2 -1040W/m2. This is just a piece of exposed pipe that I want to see how long it takes to heat up to an unacceptable temperature/pressure starting from some assumed pressure. I calculated a delta temp rise of about 30 degrees in the gas (translates to about delta 7 degs) in the pipe .

The real numbers I calculated for heat energy in the gas =UpsetH.E is 14 million Joules for the Upset conditions .....and heat transfer rate SunQQ from sun in reality is about 11, 000 Wats after doing Qin - Qout from the surface area.

So it looks like it will take 14,000000/11,000 seconds (about 0.35 hours )to heat up the system?.

I'm trying to understand the concept a bit more clearly having calculated that the QSteelRate to see where it is useful otherwise it seems there would be no need for this heat transfer rate i get using metal heat conduction Rate = k•A•(T1 - T2)/w which is about 1 million watts for the pipe exceeds the 11,000watts from the sun.

a) So i'm trying to understand if this merely means that the pipe has the ability to transfer up to 1 million watts (QSteelRate) and will thus have the ability to tranfer the 11,000 watts from the sun into the system, meaning the 11,000 watts will have the effect of overheating the system within 14,000000/11, 000 secs.

Or

b) Does it mean that the pipe has the ability to absorb the 11,000 watts incoming from the sun -without having an adverse effect on the gas inside? Thats the concept I'm trying to remember?


Does it mean that if I had calculated the Qsteelrate to be less than the incoming net flux from the sun, then it means the opposite and that the sun will overheat they pipe and then the gas inside?

I'm having a hard time accepting the (a) answer because I can calculate the delta t in the pipe associated with the net QQsun when I set it as governing Rate = k•A•(T1 - T2)/wt for the pipe (about 0.06C) instead of 7C i got at upset.... and it is no way near what the delta T at the upset energy is. OR does it mean that the small delta t from the sun is cumulative and adds up every second and will eventually grow

 

[IRstuff]

You've neglected the thermal losses. You've short cut the equations. There governing relationship is the continuity of heat transfer, i.e., conservation of heat flow. The temperatures reflect what the heat flow does, not the other way around.

TTFN
I can do absolutely anything. I'm an expert!
homework forum: //

http://www.engineering.com/AskForum/aff/32.aspx

faq731-376 forum1529

 

[georgeverghese]

Agreed, solar radiation on pipe / vessel surfaces in the tropics / subtropics is some 900-1000w/m2, for which bare pipe surface temperatures (out in the open) range up to 50-55degC in most locations, and up to 70degC in the Middle East. This results in upper design temps being no lower than 50degC for uninsulated vessels and pipes out in the open.

This may be much lower at much higher / lower latitudes.

This incident heat from the sum would be dissipated out from the metal surface only at this temp through radiation and natural convection modes - must say I havent checked this out myself though, but that would be the theory.