Solar Loading on Aluminium Enclosure 4

[ArcEnergy]

Hello all,
I have already gone through the threads in this forum related to Solar gain, Temperature rise. I have got enough Information but some how stuck with unrealistic values through my calculations. I hope some you you can help me in this.

My question is ,to estimate the amount of temperature rise inside the enclosure due to solar loading.
I have a Switchgear enclosure (made of aluminuim) of LxWxH of 2.4mx1.15mx1.97m , Thickness would be around 0.01m
The internal power Dissipation is 900W .. it is because of a current of 12.5kA flowing in a conductor of resisrance 6x10^-6 Ohms
The ambient temperature (Tamb) is 40°C
I am considering the Solar Radiation of 1100W/m^2. The enclousure is painted with light Grey colour with Absorption coefficient of 0.5 and emittence coeff of 0.87 (correct me if iam wrong)
h for the Aluminium as 2 W/m^2K (not sure if this value make sense)

I would like to know how much would be the temperature rise inside the enclosure (Tenc).

i have been using the formule mentioned in other threads like

Sun Radiation * Absorption coeff = emittence coeff * 5.67x10^-8 * (Tenc^4 - Tamb^4) + h * (Tenc-Tamb)

By using this formulae i am getting the enclousure temp as 95°C .. that means a temp rise of 55°C .. which i feel is too high.

My end aim is to calculate the amount of flowing current that i should reduce when I place my switchgear in outdoors.

Since i am an electrical engineer this temperature stuff is Kind of new to me. I hope some one will throw some lighjt in this Topic.

Thanks in advance
Regards,
ArcEnergy

 

[racookpe1978]

Look very, very closely at the "assumptions" made in your radiation equilibrium formula!

Your formula is a valid approximation for a "black body" of emissivity = .87 (to correct it to a grey body assumption) ... with no other heat losses but long-wave thermal radiation.

First: Your solar absorbtion is just over one wall of 2.4 m x 1.15 M (worst case), and then you need to add the internal resistance heat gain.

But the thermal losses are going to occur from ALL 6 walls of the outside of the box using natural convection into the air AND conduction losses AND long wave thermal radiation back into the basic 40 degree atmosphere.

So, to simplify things, assume no solar radiation, and just the 900 watts/box heat load trying to get out of a closed Al box into 40 degree C air. Determine what the approximate temperatures will be for the 6 walls of the box: Top will be hottest (internal air rises up the sides, cools at the top, then falls down to the bottom, then rises again; bottom outside will be "cooled" by the outside air slightly differently than the top and side walls.) So, use three approximate convection equations that total into the 900 watts being dispursed into 40 deg C air.

From that approximation of bottom, sides, and top wall temperature, determine the radiation losses into 40 deg C (273 + 40) deg air. You can probably ignore the conduction losses down the supports of the box.

So, you have one internal heat input (900 watts) and 12 approximate heat losses (the radiation and convection losses from each of 6 different wall positions) into 40 degree air. Subtract the 6 radiant heat losses from the 900 original watts, and re-calculate the 6 wall temperatures. Re-caclculate the radiant heat losses (again) - The differences should quickly become near-zero for this coarse an approximation.

Then, calculate the single heat gain at each hour of the day from the sun - it will NOT be 1100 watts/m^2 except at noon on the highest day of the year (probably June 22) for your latitude! More likely, it will be substantially less every other day of the year and every other hour of the year. Give me your latitude - I can get you a spreadsheet that gives approximate (theoretical clear sky, direct radiation) solar gains per exposed sq meter for each hour, each day of the year at any latitude.

Add that extra radiation heat gain to the 900 watt internal load for the hours around noon. Then re-calculate the previous equilibrium wall temperatures to approximate the 9:00 AM, 10:00 Am, 11:00 AM, noon, 13:00 PM, and 14:00 PM and 15:00 PM air temperatures and solar heat gains. Note that actual outside air temperatures will maximize between 13:00 and 15:00 each (summer and winter) day, but will not be (on average) ever highest at noon on any given day.

Unless you are desert near-equator conditions, there will only be a few weeks of the year with high daily solar radiation and high air temperatures at the same time. You still have to design for worse case conditions of course, but the air will not be 40 deg C all day and all night. And the sun will not be shining at 1100 watts all day either.

 

[IRstuff]

MIL-HDBK-310 has tables for solar loading; my recollection is that 1100W/m^2 is basically the max for about 3 hrs during the hot day. Note that the ambient temperature is usually modeled to lag the solar peak.

You should be able to put racook's approach into Excel and configure an energy balance equation that you can goal seek, and just have Excel do the iterating.

TTFN
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[dbill74]

First you appear to be forgetting that the sun moves across the sky and will impart solar radiation at different angles at different times of day. This also means that 4 sides of the enclosure will receive solar radiation (up to three at one time) throughout the day. Also local weather conditions will change ambient temperature throughout the day as well. 95°C appears to be a reasonable temperature to reach inside the enclosure you described.

Then I get to your 'end aim'; and I have to ask why?
The switchgear is providing power to something, if you reduce the power going to ...whatever... it is not going to be able to do what it needs to do. If you want to keep the temperature in the enclosure under a certain point, you need to know net energy going into the enclosure, solar and internal so you can have an HVAC engineer calculate how much ventilation to provide.

 

[racookpe1978]

Well, it would be (at maximum) only two sides that could be exposed to direct radiation, not three.

But you do have a good idea to reduce maximum heat load.

1. Get rid of the gray paint - keep ALL of the Al walls shiny. This will reflect more incoming solar energy, because not only will the sun-lit side be reflecting more (absorbing less) but the secondary radiation from the ground and skies on the back sides be reflecting more.

2. Rotate the box so only the EDGE of the southern side (if northern hemisphere!) actually faces south. True, when the sun is at 9:00 Am or 15:00 PM, that one side will face more directly towards the sun, but the edge will face due south at noon (when sunshine is highest) and the two "flat" faces will be at 45 degree angles from the sun. It will be interesting to determine if the combined two faces - both angled AWAY from the sun - collect more radiation than a single flat face aimed due south.

3. Even simpler, why not mount a single flat 1/8 (or 1/4 inch) Al plate about 2 inches from the top and south wall as a sun shade? Use the north wall (if northern hemisphere) for the maintenance access opening/door. There will be some (not much!) secondary radiation from the exposed sunscreen, through the sunscreen plate, to the shaded electronics box, but much less. 2 inches of air space are big enough to serve as insulation, and is large enough for natural convection air flow to be unimpeded.

 

[IRstuff]

I'm not convinced that #2 with the tilted faces is going to make that much difference, since the projected area of the two tilted faces will still equal that of a single, untilted face.

TTFN
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[dbill74]

Max number of sides getting solar radiation is three sides. Top, south and east/west.
Assuming one side does face south (northern hemisphere), there is a limited amount of time during which only the top and the one side will see the sun, however the time is limited. The other sides will receive sufficient solar radiation as to not be negligible.

Shading is a good idea, however 900W of heat generated in the enclosure is significant and ventilation is still required.

 

[ArcEnergy]

Thanks alot for the responses.

I want to explain a little more clear about my calculations.

The switchgear is tested with 12.5kA of rated current at 40°C of ambient temperature. according to standards we are allowed to have a 65K of rise inside the switchgear cubicle. this value is mainly because the current carrying conductor has only tested to withstand 105°C. as you see the allowed temperature rise is 105°C-40°C = 65K

I have then calculated the effect different ambient temperatures on conductor current by keeping the conductor temperature of 105°C constant using the formula

Inew/Irated = sqrt[(105°C - Tamb new)/Temp rise allowed]

Eg: If the Tamb new is 50°C, then Inew = Irated x sqrt [55/65) gives us the new Current of 11.5kA .. That means when the out side temp is 50°, you are only allowed to pass 11.5kA in the conductor.

Now considering the situation of placing the switchgear in outdoors. Now the additional effect is solar radiation. According to the standard "C37.24-2003 - IEEE Guide for Evaluating the Effect of Solar Radiation on Outdoor Metal-Enclosed Switchgear"

for grey painted enclosures, the absorption is 0.50 which increases a temperature rise of 7.7K when the solar radiation is 1044 W/m2... So in my new calculation, i have added this 7.7K to my allowed 65K temp rise and calculated the Inew for the amb temp of 40°C. I get the value of 11.64 as allowed current in the conductor when placed my switchgear in outdoors at 40°C.

THE WHOLE ABOVE CALCULATIONS ARE FINE TO ME IF I DONT GO DEEP INTO THE SUBJ OF SOLAR RADIATION.

But in the other thrds that posted here about solar radiation, i have read the explanations by IRSTUFF and IONE and realised this is not as easy as i thought. i have to consider convection radiation concepts which i am not aware of.

so I have posted my question in terms of Watts
according to racooks explanation i have to perform

Q1 x area of wall 1 = htc x area of wall 1 (Tenc1-Tamb) for 6 sides ..
But for every case i have two unknowns, Q and Tenc .. but as he said sum should be 900W, I will add Q1, Q2 ...Q6 = 900W.
So My equation would be

900W = 2x[5.52 (Tenc1-313) + 9.46 (Tenc2-313) + 4.54 (Tenc3-313)] ------> 2.76m2, 4.73m2 and 2.27m2 are the areas of 2x3 walls and htc of Al 2W/m2K.
Here for simplicity i have considered Tenc1 =Tenc2 =Tenc3 as 338K

the results didnt make sense to me. I am convecting more than the 900W . ..I am lost here even before considering the radiation effect..

I am pretty much sure i messed it up somewhere. I was kind of hoping a regular formula to measure the temp rise inside the enclouse based on the solar radiation. didnt see this coming. I would be really thankfull if anyone walk me through this stuff.

Regards
Arcenergy

 

[IRstuff]

I think the issue is that you've not iterated on what is supposed to be a system of equations that are solved simultaneously. The basic premise is conservation of energy flow, i.e., gizintas = gizoutas. When it's done correctly, you will get a self-consistent set of temperatures. Your guesses are fine, but they are not cast in concrete. You must iterate on the answer until everything converges, or use a math program that can do that for you.

TTFN
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[kjoiner]

You may also want to check out some of the enclosure manufacturers such as Hoffman or Haewa. They have some guidelines for determining heat loads. Below is a link to a Hoffman brochure.

http://www.pentairprotect.com/wcsst...ads/Literature Requests/Content_Cat-00048.pdf

Kyle

 

[denniskb]

ArcEnergy,

I got your email so will try to help out.

Wouldn’t it be nice if things were so simple that we had a formula to solve everything? The situation you describe is in reality complex and so is the solution unless we make some significant assumptions to simplify the solution without losing rigour.

I guess that a box of this size is supported on a concrete slab with an air gap on the underside of 75-100mm?

I also guess that the box is ventilated to allow air to enter near the bottom and exit near the top?

Given the equipment type I guess that even short periods above 105°C may not be acceptable?

I also guess that the unit will have a low thermal mass so the temperature will reflect the short term midday maximum conditions?

The indicated solar radiation level of 1100 W/m2 is extremely high. The sun radiates to the upper atmosphere at around 1300 W/m2. The atmosphere reduces this intensity to a maximum possible of ~1065 W/m2 direct summer sun on a clear (i.e. not hazy) cloudless day at noon and probably on top of a mountain. More typical are 850 W/m2 – direct summer sun clear sky maximum, 660 W/m2 - Australian desert direct summer sun, clear sky maximum, 400 W/m2 - Australian summer mean. As noted in the responses the maximum solar radiation intensity varies with your location on the planet, the time of day and atmospheric conditions. My guess is that you could safely design for a worst case of 850 W/m2 and perhaps lower?

The angle of the surface to the sun directly affects the solar intensity and the angle of the box to the sun directly affects the area exposed. I believe you can assume direct solar radiation on any one face only so you should assume the largest face which is the top of the box which also fits with the maximum mid-summer, midday conditions. This assumption does not imply that there is then no radiant transfer from the other surfaces of the box; simply that this is not direct solar radiant heating. There will still be radiant transfer with the surroundings.

For solar radiant heat transfer (and you must remember that this includes two way heat transfer and always between two surfaces) the wavelength is short and largely penetrates the atmosphere therefore the associated heat rejection is to the sky and not the local air which is at 40°C. When you fly in a plane the airline often displays the external air temperature which is often around -35°C. Radiant transfer heat rejection from the surface will on a clear day be to an assumed surface of between -43°C and -35°C. Yes on not clear days this radiation may be to a much higher temperature however under these conditions the incoming solar intensity will also be much lower. If you are using a design solar radiant intensity of 850 W/m2 then you should assume a sky temperature of 35°C.

The values you suggest for the grey paint absorption and emittance are credible but note that you should have two absorption values – one for Solar (short wavelength) and one for Ambient (longer wavelength). I would suggest you use 0.5 for both. Alternative paint colours may help while everything is new and shiny but with time and a layer of dust the improvement will not last. Using bare Aluminium is not a good idea as its radiant properties are typically worse than paint.

For the sides radiant heat transfer will occur with the surroundings so what is located near to the box will affect this. It is typically reasonable to assume that the surroundings are at ambient temperature and located nearby (e.g the ground) and with radiant properties similar to soil. I would suggest you might ignore the bottom surface of the box as the wall temperature will likely be close to the concrete temperature.
All real world heat transfer problems involve i) conduction, ii) convection and iii) radiant heat transfer. It is only in the text books that you can solve just for one. Often one or two are dominant and in this case conduction can likely be assumed to be not significant. However care is needed because the dominant mode can vary as conditions approach equilibrium. I manage this by solving for all three from which the dominant ones become obvious.

Convective heat transfer with the surroundings has its own set of complexities largely due to the huge number of variations in film coefficients associated with various flow and heating/cooling conditions. I manage this by solving a large number of film coefficients from which I can choose while noting others with similar values.

Convective heat transfer through ventilation of the box is almost a different topic again and one that I do not have much experience with (particularly if it is natural rather than forced) so for now I plan to ignore this and keep is as a safety factor for the design. Others may know how much benefit is available from ventilation?

OK that was the background so how solve your problem?

Given the assumed conditions above fix the internal temperature at 40°C and calculate the heat transfer rate per m2 for solar and convective modes through the top surface and the walls. Multiply the rate per m2 by each surface area to calculate the overall heat inflow/outflow rate in kW.

Assuming a nett heat inflow fix the internal temperature at 60°C and do the same.

When the heat inflow equals or is close to the heat outflow you have found the equilibrium temperature for the empty box.

When the net heat outflow matches the 900W heat load from the equipment you have found the equilibrium temperature for the enclosure. I would not be at all surprised to see a 95°C result.

Note that at this point you have not found the actual temperature of the device but the temperature of the air in the box. However, I am guessing that this is the intended limit for the switchgear?

I will try to run some cases and provide the heat transfer rates for your surfaces and post these shortly.

TTFN suggestion 3 is however a very effective way to improve the outcome but I would suggest one improvement. On the underside of the shield attach an insulating layer (e.g. a glued on closed cell foam rubber mat – make sure the glue is suitable) of around 5-10 mm thickness. Without the insulation the box will experience radiant heat transfer direct with the shield at elevated temperature. With the insulation the rubber surface temperature will be much lower. Unfortunately this solution makes the heat transfer calculations even more difficult as you must now determine the natural draft convective heat transfer in this gap. As I said at the beginning “if only things sere so simple …”


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[racookpe1978]

Concur with the recommended sun shield, vents, and correcting his assumed (too high!) 1100 watt/m^2 solar insolation loading.

It is not clear why louvers would be rejected/not used - though sand infiltration could be a factor.

Rather than "gluing" rubber on the back of the sun screen, I'd recommend a wood-backing: A plain sheet of 3/8 plywood on the back of Al sunscreen would reduce the re-radiation without any threat of glue or rubber deterioration over time. Simpler to bolt on, rigid, low emiissivity for LW radiation, and an adequate through not perfect insulator so the hot Al sheet on the hot side has a lower thermal radiation on the cooler, shaded side.

But he DOES have to go through the math sequence of iterated equations for each of the six walls of the enclosure. No single formula works.

 

[ArcEnergy]

Hello Dennis,

Thanks alot for acknowledging my email and for your clear explanation.

Most of your assumptions are correct. The switch gear will be installed on a concrete slab. doesnt know the exact gap though but your assumption seems reasonable.

Your assumption of boy being ventilated is not true. It is completely closed with out any ventilation. i guess it comes under protection degree IP55 . But i thing you have suggested to ignore the effect of ventilation which fits to the situation.
and yes even the shorter periods above 105°C would be critical. That is why i am considering a worst case scenario of very high radiation of 1100W/m2 . even the standards use around 1044W/m2 i guess.

I dont know exactly what a thermal mass means. but based on the explanation from google, the low thermal mass scenario would fit the application.

With the angle of sun, Since i am considering the worst case situation, I would say your assumption is good that the largest face area(4.68m2 )is directly exposed to the sun light.

about the solar radiant heat transfer, it is not 100% clear to me. did you mean that the radiation from the enclosure is of two types? one is into the surrounding atmosphere which is 40°C and other is to sky which is at -35°C ? Am i correct?

Now coming to the solving,
Only Box. no internal heat dissipation

You suggested me to assume the internal temperature of the enclosure is 40°C and told me to calculate heat transfer from solar and convective modes.

Heat transfer from solar to internal enclosure (direct sun so largest face) 1100W/m2 x 0.5 x 4.68 = 2574W is the amount of heat energy resulted through sun light into the enclosure.

Now the convection through the walls to the outside atmoshphere using
q = htc (Tinside-Toutside) x each face . But since our assumed internal temp and outside temp is 40°C, isnt that the heat convection is 0?

Solar radiation increases the heat inside to 60°C.
now using convection formulas q= htc (Tinside-Toutside) x each face ..
= 2 (333-313) x 4.68x2 +
x 2.76x2 +
x 2.25x2 ... The total Equals to 774W ..
radiation emission from all sides = 0.5x 5.67x10^-8 (333^4-313^4)[4.68x2 + 2.76x2 + 2.25x2] = 76.5 x 19.38 = 1482W

Total emission is 2256W which is not exactly equal to 2574 but i guess we could assume at 62°C internal temperature the solar heat inflow would be equal to heat out flow through convection and radiation.
This gives us a rough idea that due to solar radiation of 1100W/m2 shining directly on the top of the enclosure a temperature rise of 22K (62°C-40°C) is possible.

Please let me know if these above calculations make sense to you are not. And you are right i am mostly interested about the temperature of the air inside the box.

Regards,
ArcEnergy

 

[Tunalover]

ArcEnergy,
I have a well-documented Excel spreadsheet that solves your exact problem. How can I get it to you??
TL

Tunalover

 

[denniskb]

ArcEnergy,
“if only things sere so simple …”
You will most likely not be able to make this work with 1100 W/m2 and should change to a more realistic maximum of 850 W/m2

A. Your simple calculations have not accounted for all heat transfer paths nor included for absorptivity and emissivity.
Thermal radiation is emitted by all bodies that are at a temperature above absolute zero with an ideal intensity of H = B x Tabs^4 - where B is the Stefan Boltzman Constant of 5.67 x 10^-8 W/m2.K.
The emission levels are directly affected by the emissivity of the source body E1
The level received by another body in its path are affected by the absorptivity of the receiver body A2.
Therefore the transmitted energy from body 1 to body 2 is H = A2 x E1 x B x (T1 + 273)^4
However surface absorptivity is dependent on wavelength so you need two A2 values A2s and A2b for solar and background
With two bodies this can lead to five values E1, A1, E2, A2s, A2b but if you assume body 1 is a black body then E1 and A1 are = 1.00
So for the enclosure (body e) consider only Aes, Aeb, Ee
Note that while you have these values for your grey paint these will be for the new clean condition but will degrade as the paint oxidises and gets a coating of dust. In my experience you would be better using values of 0.75, 0.75, 0.85 respectively for Aes, Aeb, Ee.
Three radiant heat transfer paths can affect the enclosure;
 Absorption from the sun for which the intensity (Is) is known - Hs = Aes x Is
 Absorption from the background - Hb = Aeb x 1.00 x B x (Tb + 273)^4
 Emission from the enclosure – He = Ee x 1.00 x B x (Te + 273)^4
You must account for all three of these for the enclosure;
For the top surface;
 Solar absorption
 Absorption from the sky at -35°C – [Yes absorption NOT emission]
 Emission from the enclosure surface
For the four sides;
 Assume solar load is nil
 Absorption from the background at ambient surface temperature (say ambient + 10-15°C)
 Emission from the enclosure surface

B. The convective heat transfer is primarily governed by the inside and outside film coefficients and not the conductivity of the plate.
The solution requires algorithms to calculate the Nusselt Number (Nu) and requires various fluid properties, geometric data for the enclosure and fluid flow conditions. Further these must be determined at both inside and outside faces of the enclosure. Trust me this is no easy task. The heat transfer is directly related to Nu which can easily vary by a factor of 1 to 10. Therefore if you get this wrong your answer can be way off.
For example Nu will vary significantly with flow velocity (i.e. Reynolds Number) which for natural draft is unknown (I have assumed 0.15 m/sec). This is particularly important inside the enclosure because low Nu will lead to the enclosure wall being at a lower temperature and so reject less heat to outside. One option is to install fans inside to blow the air along the walls? Another is to weld ribs on the inside of the enclosure to increase the surface area and so again to increase the wall temperature.

C. Some of the heat transfer paths can have very high H values (as you have found) but some are absorption (+ve) and some are emission (-ve) and the balance, being the sum of all three, can be quite small (particularly near the equilibrium temperature). Since you need to reject an additional 900 W heat load using just the small balance you can get widely varying results from just minor variations.
For example solar absorption 638, background absorption 137, emission -689, convection -100 the balance is just -14. A change in the emission calculated of just 2% will either delete or double the heat balance!
What then becomes important is to able to test the sensitivity of the outcomes to possible changes in the assumptions and so to get a “feel” for the reliability of the result. For me I can do this because I have automated the simultaneous calculations for the three radiation paths and the convective algorithms and calculations and can re-run cases in less than a second. The problem then is that this opens up all of the issues noted above which in reality are much more important than the relatively “simple” calculations.
I have run some calculations with the following results (Rad is the sum of Solar + Backg + Emiss, Nett is Conv + Rad, Load is the total heat load)
Paint Weathered + Dust (assumed 0.75, 0.75, 0.85)
Natural Convection inside and out (assumed 0.15 m/sec)
Top Surface - 4.7 m2, Four sides – 10.1 m2
Inside Skin Conv Solar Backg Emiss Rad Nett Load
Top Surface °C °C W/m2 W/m2 W/m2 W/m2 W/m2 W/m2 W
40 69.4 -87.1 637.5 136.8 -662.4 111.9 24.8 117
50 70.1 -89.7 637.5 136.8 -667.8 106.5 16.8 79
60 70.8 -92.4 637.5 136.8 -673.2 101.1 8.7 41
70 71.5 -95 637.5 136.8 -678.4 95.9 0.9 4
80 72.1 -97.5 637.5 136.8 -683.6 90.7 -6.8 -32
90 72.8 -100 637.5 136.8 -688.7 85.6 -14.4 -68
100 73.4 -102.5 637.5 136.8 -693.7 80.6 -21.9 -104

Side Walls °C °C W/m2 W/m2 W/m2 W/m2 W/m2 W/m2 W
40 40.2 -0.1 0 463.7 -463.5 0.2 0.1 1
50 41.2 -1.6 0 463.7 -469.7 -6 -7.6 -76
60 42.9 -3.4 0 463.7 -475.6 -11.9 -15.3 -154
70 43.1 -5.3 0 463.7 -481.3 -17.6 -22.9 -230
80 44 -7.3 0 463.7 -486.8 -23.1 -30.4 -306
90 44.9 -9.4 0 463.7 -492.1 -28.4 -37.8 -380
100 45.7 -11.4 0 463.7 -497.4 -33.7 -45.1 -453

You will see that at 100°C the total heat rejection load is -104 + -453 = -557 W so has not yet achieved the desired load of 900 W.
As noted above doubling the inside velocity with fans or doubling the inside surface area would significantly increase the total heat rejection and likely achieve the desired outcome.

Even with the tools I have developed this is no simple task and needs expertise and time. Amazing considering this is just a simple box out in the sun but as I keep repeating “if only …..”


Dennis Kirk Engineering

http://www.ozemail.com.au/~denniskb

 

[ArcEnergy]

Tunalover,

Can you kindly forward that sheet to my email address

somebodyinnobody@ gmail .com.

Thanks in advance mate

regards
ArcEnergy

 

[chicopee]

Since solar radiation appears to be a problem regardless of the numbers that you crunch, put a canopy over your switch gear and be done with this matter.