Temperature rise of object in the sun 1

[Dan68c]

Hi,
I have been trying for ages to work out the temperature rise of an object in the sun. Such as a roof of a house.
What i have done so far is used the constant on 1200W/m2 delivered from the sun.
From here i use the reflectivity value to reduce the amount absorbed and then the maximum temperature will occur during at steady state when the heat added is equal to the heat loss.
The heat loss i have done is from radiation and convection and as the temperature rises so does the heat loss.
However doing all this i get a huge temperature rise which doesnt seem real. I am using the correct constants and the temperature in kelvin.
Any help would be much appreciated.

 

[ione]

In the 1960 it has been calculated an average value for the specific solar thermal power on a surface perpendicular to the sun beams G=1353 W/m2 (instead of the 1200 W/m2 you have quoted) out of the atmosphere.
Thanks God the atmosphere is there.
The thermal radiation is shielded by the atmosphere and the atmospheric gases absorb part of the thermal energy. The sun specific radiation energy drops to 950 W/m2 in a very very shiny day.
Now consider that for your calculations you don’t have to take into account all the solar specific radiation quoted above, but just the part which hits the surface perpendicularly (plus the diffuse solar radiation). IMO the first bug in your calculations could be the assumption of 1200 W/m2 for the specific energy delivered by the sun.

 

[Dan68c]

Thanks, maybe i have over estimated that value. However looking at this say you have 10m2 area. This equates to 9500W received. Now Watts as power is one joule/sec. So if the object is exposed for say one hour so this value is multiplied by 3600. Now this is a total energy in joules. When i calcualted heat loss such as Q= mass * Cp * Temperaturechange, for natural convection this is also in joules. Now i think iv just got my head messed up with units but does this seem the correct way to go with multiplying it by the time exposed.

 

[ione]

You should consider 950 kW as an upper limit. (In the real world this value is reached in very few and particular occasions).

Moreover, part of the solar radiation is absorbed and part is reflected. Coefficients for absorption and emission are tabulated for different materials.

The net thermal power Qrad transmitted by radiation is

Qrad = alfa*G + epsilon *sigma*(Tsky^4 – To^4)

Where:

alfa = coefficient of absorption
G = Gp *cos(theta) + Gdiff [where Gp is the solar radiation, theta is the angle between the solar beams and the line perpendicular to the exposed surface, Gdiff is the diffused thermal radiation]
epsilon = coefficient of emissivity
sigma = 5.67 x 10^(-8) W/(m^2*K^4) Stefan-Boltzmann constant
Tsky = a value between 230 K and 285 K [atmosphere is assumed to be a radiating body]
To = temperature of the exposed surface [K]

You have then to evaluate heat losses (or gain if the temperature of the body is higher than surrounding ambient) from the “body” by convection (natural or forced), to compute Qtot.

The formula you have indicated and that I have reported below introducing Qtot:

Qtot = M*cp*deltaT

could be used to evaluate the temperature increase deltaT of your body:

deltaT = (M*cp)/Qtot

where:

M = mass of the body exposed to radiation [kg]
Cp = specific heat [J/(kg*K)]

 

[IRstuff]

Q= mass * Cp * Temperaturechange is for heat capacity, not heat loss

Q = area*htc*deltaT, where htc is in W/m^2-K is for convective heat loss, and results in watts, as is your solar load. If it gets sufficiently hot, you should also include radiated gains/losses.

I suggest that you get a copy of MIL-HDBK-310, which shows what a military 99-percentile hot day looks like. And the standard value is 1120 W/m^2.

What's also not accounted for is your object's emissivity, which determines the amount of solar load actually absorbed.



TTFN

FAQ731-376

 

[Dan68c]

Thanks for all the help i have search anywhere and everywhere and thats the best information yet.
Thanks

 

[Dan68c]

In reply to ione comments. The equation you have provided looked good but at a second look it needs to include area i believe. Do i just multiply G by this .
Thanks

 

[prex]

Just to stay with orders of magnitude.
The thermal exchange of an object to calm air close to room temperature can be evaluated to 10 W/m²°C (includes convection and radiation, roughly 50% each).
So with a heat input of 1 kW a black body would raise its temperature by 100°C.
This result should be corrected because of two main effects (besides of course the angle of exposure that reduces the effective area of heat input):
-an emissivity of 0.6 is much more common for a real world object with high absorptivity (or emissivity, it's the same)
-an object like a roof, exposed to the sun, will also loose heat by conduction to its supporting structure. An example of a nearly non conducting object is a car: that's why cars get really hot, even if they are white painted.
And of course air movement (wind) will also raise the exchange coefficient and proportionally reduce the temperature rise: with a sustained but non stormy wind (a breeze) the exchange coefficient will at least double.
Conclusion: the theory provides a good explanation of the temperatures anyone can experiment, but it is very difficult to calculate the temperature of an object in the sun with an approximation better than, say, 20°C.

prex

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[ione]

Dan68c,

Yes, that was a specific power (W/m2) so you have to enter the area to complete calculations.

 

[IRstuff]

attached is what the calculation in Mathcad looks like. You should be able to run the same problem in Excel, or Matlab, if you don't have Mathcad.

TTFN

FAQ731-376

 

[ione]

Tamb = 25 K

A bit cold!

 

[IRstuff]

That's added to T.b = 273.15K for the absolute temperature.
Convection is a delta temperature calculation, so absolute temperature is not needed.

I leave it in K just for convenience.

TTFN

FAQ731-376

 

[athomas236]

Can I just ask why the left hand side of the equation is sun times epsilon rather than just sun.

Regards,

athomas236

 

[ione]

Irstuff,

I can’t understand the way you can calculate the final temperature of the plate exposed to the sun, without taking into account its specific heat.

 

[IRstuff]

specific heat is a static property, which only shows up in transient calculations. In steady state, power in = power out, which is about heat flow. The specific heat is about changing the temperature of the object. If the object's temperature is constant, specific heat drops out.

The sun's energy is only partially absorbed. A mirror would have a steady state different temperature than a blackbody, because a mirror absorbs very little of the energy to start with. The emissivity as is often used is a crude approximation, at best. For radiometric accuracy, one has do the emissivity as a function of wavelength, but that's just too messy foe this type of estimation.

TTFN

FAQ731-376

 

[ione]

I considered (power in – power out) = net power (absorbed by an object or emitted by the object) and then from the net power (introducing specific heat and taking it as a constant ) evaluate the temperature rise of the object. Obviously this should be done in an iterative manner as net power changes as the temperature of the object changes.

 

[IRstuff]

Right, but the end result is that there is no longer a delta T to apply to the specific heat. This would be the steady state condition, which doesn't involve specific heat anymore, at which point the steady state equations are valid. And the final temperature of the object at that point minus the ambient temperature is the temperature "rise."

The key to whether specific heat is involved is when the question is couched in terms of time, i.e., how long does it take to reach steady state, or what is the temperature rise at t<infinity. In both cases, the change in temperature of the object itself affects the answer, so specific heat is relevant. Part of the discontinuity here is possibly the wording of the question. "Rise" connotes a transient phenomenon, but since time is specified, I assume that the problem reverts to steady state.

The specific OP question does technically require specific heat, because insolation is not constant, and the 1120 W/m^2 value only applies for about 2 hours a day. But, since an object could potentially stabilize in temperature within 20 minutes or so, a steady state problem formulation is plausible, albeit only valid for about 2 hours.

TTFN

FAQ731-376

 

[mrpi]

Isn't this problem also independent of surface area?

If you assume that all surfaces are exposed to the same amount of solar radiation and that that same surface areas possess the same convective heat transfer coefficient, the actual surface area is eliminated from the equilibrium temp equation.

If you sum the energy equations, at equilibrium the absorbed solar radiation must equal the convected and radiated energy from the surface (for the worst case where the body is insulated from conduction heat transfer).

q/A_sun = solar load (1120 W/m2)
a_sun = solar absorptivity (.95 ?)

a_lowtemp = surface absorptivity or emissivity at equlib temp (.95 ?)
sigma = Stephan-Boltzmann constant
T2 = equilibrium temp
T0 = ambient temp (325K ?)
h = convective heat transfer coeff. (10 W/m2-K)

So I have the following energy equations independent of surface area:

q/A_sun * a_sun = Solar loading

a_lowtemp * sigma * (T2^4 - T0^4) = Radiation loss from surface

h * (T2-T0) = Convective loss from surface

Balancing as:

q/A_sun * a_sun = a_lowtemp * sigma * (T2^4 - T0^4) + h * (T2-T0)


Which leaves you solving a 4th power polynomial of the form ax^4 + bx + c = 0

Actual absorptivity and emissivity numbers are the hard part, I would say.

I was getting calculated temp rises of ~55°C which seemed high to me. I have not attempted to experimentally measure this. The local weather station has a website that posts solar radiation levels in W/m2, so it wouldn't be too hard to actually try and measure it.

I think your calculations will be worst-case-type values, where solar load is at a peak, the surface is absorbing a lot of the radiation and emitting very little and the whole thing is in still air.

How did it work out for you?





Beat to fit, paint to match.

 

[prex]

Well, mrpi, it's independent of surface only for surfaces exposed at right angles to solar radiation. If there is an angle, the surface effective for absorbing power is the projection of actual surface on a plane normal to sun rays, whilst convection and radiation to the environment involve the actual full surface.

prex

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes and launchers for fun rides

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[mrpi]

Yeah, I just figured Dan's after the greatest expected temp rise.

Beat to fit, paint to match.