Heat transfer: All 3 modes together

AI Thread Summary
To determine the temperature rise in a steel plate exposed to sunlight, an energy balance approach is recommended, considering energy absorbed from solar radiation, emitted radiation, and convective heat loss. The energy absorbed can be calculated using the plate's absorptivity and the solar radiation power density, with a typical value of 750 watts/m² and an absorptivity of 0.5 for steel. The re-radiated energy is calculated using the emissivity of the plate and the Stefan-Boltzmann constant, factoring in the sky temperature, which is suggested to be around -5°C rather than 3 K. Newton's law of cooling is applied for convective heat loss, with the heat transfer coefficient obtained from natural convection correlations. The solution involves iterative calculations to balance the absorbed and lost energy, ultimately leading to the temperature rise of the plate.
chetanladha
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Hi.
How can i find out the temperature rise in a steel plate of known dimensions, when exposed to sunlight, losing heat to air flowing over it by convection, and heat getting transferred to another body by convection?
I hope to hear about the different approaches that could be taken.

Thanks in advance.
 
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Basically, you will perform an energy balance from radiation absorbed from the sun, radiant energy emitted from the plate, and convection energy loss as follows:

Energy absorbed by sun (Es) = absorptivity X solar radiation power density

For a steel plate, you can use absorptivity = 0.5. A good average value of energy from the sun at the surface of the Earth is 750 watt / m^2.

Now the plate will "re-radiate" energy back into space as follows

Er = emmissivity * Stefan Boltzmann constant X surface area of plate (plate surface temperature ^ 4 - sky temperature ^ 4)

you can use emmissivity = 0.21 for your steel plate and sky temperature (not ambient temperature) = - 5 C. Make sure your temperatures are in Kelvin and Rankine.

Finally, use Newton's law of cooling for the convection:

Econv = A * h * (T-surface - T-ambient)

And you can find h from a vertical plate natural convection correlation.

The solution is trial-and-error. Assume a value of T-surface until:

Es = Er + Econv
 
Many Thanks for your response.
Can u please explain why have u used sky temperature radiated energy, and not the ambient temperature.

Er = emmissivity * Stefan Boltzmann constant X surface area of plate (plate surface temperature ^ 4 - sky temperature ^ 4)

Also, is it appropriate to use sky temperature as -5, or should it be 3 Kelvin?
 
chetanladha said:
Many Thanks for your response.
Can u please explain why have u used sky temperature radiated energy, and not the ambient temperature.

Er = emmissivity * Stefan Boltzmann constant X surface area of plate (plate surface temperature ^ 4 - sky temperature ^ 4)

Also, is it appropriate to use sky temperature as -5, or should it be 3 Kelvin?

Atmospheric emmision is often estimated as blackbody, hence the "sky" temperature. You want to use an average temperature for the atmosphere so 3 K is too low (would suggest - 5C to -10C).
 
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