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(OP)
Hey ,

I have question I'm calculating the sun loading for Power generator installation in 20 ft hi cube container. The power generator itself after insulation the pipe produces 5kw of heat. Im thinking to do 2" thickness R7.5 insulation inside the container. S how do i calculate the sun loading and also to make sure is this required insulation. Because the insulation was recommended by a vendor who does insulation inside container. COuld anyone help me to solve this

You would normally assume some emissivity for the exposed surfaces and some level of insolation based on your location and that would establish the solar input. Being on the surface, though, means that a lot of the solar load will be radiated or convected back into the environment. This is all caveated by the ambient temperature of your installation. If you're in Furnace Creek, that's one extreme, and if you're in International Falls, that's a different extreme.

You can start off with crude resistor network model to do initial feasibility analysis.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

(OP)
Ok, but basically what all data do i need to find sun loading on a shipping container. Lets say my machine after insulation emits 5 kw of heat (internally). This container is going to be outdoor installation. So Kindly tell me what all values do i need to calculate the sun loading so that i can figure whether i need insulation or not.

Let's get some applicable details.

You have a typical steel shipping container except it's double high? 20 feet? How long?

You have a internal combustion engine driven generator built with-in it?

You're considering insulation needs of this system?

Why are you concerned about solar heat gain in this container?
During operation?
During not operation?

Does the container have any ventilation?

If so, does it all close up when not operating?

Are there batteries in this container?

What geographic area is this system going to be applied in?

Will it have power supplied to it when not generating?

Keith Cress
kcress - http://www.flaminsystems.com

(OP)
its a typical shipping container 20 ft hi cube.

yes it a waste heat recovery system. so there will input heat coming into system. So when its operating with insulation covering the heat exchanger's pipes the sytem itself emits 5 kw of heat
yes I'm planning to put ventilation
yes it does have batteries in electrical section
so i wanted to know i f have insulate or not.
if my stem itself creates more heat than outside temperature. I would think to go with low R-value for insulation
I'm concerned about heat load because it will be installed outside in Connecticut-United states.
Also i can size ventilation accordingly.
Also if i calculate the sun load i can figure if i need 2" thickness of insulation
Because during winter the temperature drops upto 14 F.
This is first time I'm calculating sunloading. That's why i wanted to what factors or data i should consider.

Okay. Thanks for the details.
Some rough numbers.

Heat Flux(Watts or BTU/hr) = U(insulation factor) x A(meters2 or ft2) X dt(delta temperature)

A horizontal metal surface has a U factor of 6.8W/m2

The sun hitting a flat metal surface can raise the temp 40C.

6.8W/m2 x 40C = 272W/m2 with a 40C temp difference.

For your 2.4m x 5.9m(20 foot) roof you have 14.2m2 x 272W/m2 = 3.8kW from just the roof.

However a flat metal roof with 2" of fiberglass insulation has a U valve of 0.91W/m2. This would reduce your heat flux to
0.91W/m2 x 40C = 36W/m2 with a 40C temp difference and hence a total load of 14.2m2 x 36W/2 = 516W of gain.

Keith Cress
kcress - http://www.flaminsystems.com

(OP)
how did you get the u factor for metal surface?

(OP)
could you explain how did you get this ? 3.8kw you got without insulation right?

Your problem reminds me of a test question on a PE exam taken back in the 1970's. So I have two methods that you can study. The first method deals with comfort cooling described in my Kent's ME Handbook-Power, 12th edition, pg12-78, table 3 detailing solar heat rate gain (Btu/hr). Obviously, latitude of your location as well as any interference from trees or building shadows and time of the day will alter the values in the reference listed. I have also included an attachment that calculates a theoretical value of the earth surface temperature from solar radiation. You can modify that attachment to suit your problem and compare both values with regard to solar loading. Now if you are looking for surface temperature in you problem consider no heat loss to ambient from container and no heat generated within container. Good luck.

One more thing is about the reference that I mentioned. If you do not have a copy, libraries in your area colleges or universities may have current or older copies; well stacked public libraries are also alternate sites for Kent ME Hdbk's.

@Chicopee,

I seem to be getting a discrepancy. Your calculated solar constant seems a bit high; the value typically cited is around 1366 W/m^2, although NASA says 1361 W/m^2, your comes out to be more like 1516 W/m^2.

btw for the engineering paper; reminds me of college.

That's at the mean value of Earth's orbit, so still need to factor in the atmosphere. The value used in MIL-HDBK-310 is 1120 W/m^2, but that's way more variable due to atmospheric transmission coupled with sun angle.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

@IRstuff--It could be that NASA has better data than I had?? Also being a heat transfer problem I am about 11.3% higher than NASA which might have taken into consideration other factors such as for example the atmosphere. I'll never forget what my heat transfer professor said once in my class and that was if your results are within 50% of actual values in essence you are not doing too badly. Disparity in my result could also be in the accuracy of my values.

Of couse in WA or anywhere near the equator at normal elevations, bare metal temperatures in the sun exceed 80 C, with insulated enclosure of a heat source, you need a professional engineer to design your faciliitiy. Best of luck with your project,

I believe the ASHRAE handbooks have calculations just like this, along with reasonable and detailed "sun loading" for the regions of the US of A.

The "Fundamentals" volume TOC lists these tasty sounding topics -
15. Fenestration (TC 4.5, Fenestration)
16. Ventilation and Infiltration (TC 4.3, Ventilation Requirements and Infiltration)
17. Residential Cooling and Heating Load Calculations (TC 4.1, Load Calculation Data and Procedures)
18. Nonresidential Cooling and Heating Load Calculations (TC 4.1)
19. Energy Estimating and Modeling Methods (TC 4.7, Energy Calculations)

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