Question about heat loss from a surface

In summary, to calculate the heat loss from the surface area of a box into air, you would use the heat convection equation q = hA\Delta T, and to calculate the heat loss from a heated gas (into ambient air), you would use the Nusselt Number and the Prandtl Number.
  • #1
nuby
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How can I calculate the heat loss from the surface area of a box into air?

Also, how can I calculate the heat/energy loss from a heated gas (into ambient air)?.

Thanks in advance.
 
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  • #2
Consequently, if the box were in a forced convection flow region, it would probably be easier. If the air is stagnant, we have what's called Free Convection in that the air is moving due to the natural heating and cooling it experiences with proximity to the box.

Anyways, the general equation for heat convection is:
[tex]q = hA\Delta T[/tex]
A is the area, dT is the difference in temperature, and q is the heat transfer. The only problem is h, the heat convection coefficient.

We will use experimental results to "guess" the value of h. We can relate the heat transfer coefficient to a non-dimensional number called the Nusselt Number:
[tex]\bar{Nu_L} = \frac{\bar{h}L}{k}[/tex]
Where L is a characteristic length, k is the thermal conductivity, and h is the heat transfer coefficient that we want. We can get a relation for the Nusselt Number FOR A VERTICAL PLATE as:
[tex]\bar{Nu_L} = 0.68 + \frac{0.670 {Ra}^{1/4}_L}{\left[1+(0.492/Pr)^{9/16}\right]^{4/9}} [/tex]
The Prandtl Number is easy enough to search for and fine the relation, its easy. The Raylaeigh (sp?) number is then given by:
[tex]Ra_L = \frac{g\beta(T_s - T_{\infty}L^3}{\nu\alpha}[/tex]
Which has more easy material constants that you can look up.

For a horizontal plate, you can say that:
[tex]\bar{Nu_L}=0.54Ra^{1/4}_L \mbox{For 10^4<Ra_L<10^7}[/tex]
and
[tex]\bar{Nu_L}=0.15Ra^{1/3}_L \mbox{For 10^7<Ra_L<10^11}[/tex]

Then just add the effects all 5 sides together. Good luck,

Edit: Figured I'd give you some of those coefficients
Code:
Thermal Conductivity, k*10^3 (W / mK)
100K  9.34
300K 26.3
500K 40.7
Code:
Dynamic Viscosity, v (nu) *10^6
100K 2.00
300K 15.89
500K 38.79
Code:
Thermal Diffusivity, a (alpha)*10^6
100K 2.54
300K 22.5
500K 56.7
For ideal gasses, the volumetric thermal expansion coefficient, [tex]\beta[/tex] is simply equal to 1/T (where T is in absolute temperature).
 
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  • #3


To calculate the heat loss from the surface area of a box into air, you can use the formula Q = U x A x (T1-T2), where Q is the heat loss, U is the overall heat transfer coefficient, A is the surface area, and T1-T2 is the temperature difference between the box and the ambient air. The overall heat transfer coefficient takes into account the thermal conductivity of the materials and any factors such as convection or radiation that may affect the heat transfer.

To calculate the heat/energy loss from a heated gas into the ambient air, you can use the formula Q = m x Cp x (T1-T2), where Q is the heat loss, m is the mass of the gas, Cp is the specific heat capacity of the gas, and T1-T2 is the temperature difference between the heated gas and the ambient air. This formula assumes that the gas is at a constant pressure and that the heat loss is due to convection. If radiation is also a factor, you may need to use a more complex formula that takes into account the emissivity of the gas and the surrounding surfaces.

It is important to note that these calculations are simplified and may not account for all factors that can affect heat loss. Other variables such as humidity, air flow, and insulation can also play a role in heat loss and should be considered when making calculations. Additionally, these formulas assume steady-state conditions and may not accurately represent transient heat loss. It is always best to consult with a professional or use more advanced heat transfer analysis software for more accurate calculations.
 

FAQ: Question about heat loss from a surface

1) What is heat loss from a surface?

Heat loss from a surface refers to the transfer of thermal energy from a surface to its surrounding environment. This can occur through conduction, convection, or radiation.

2) What factors affect heat loss from a surface?

The rate of heat loss from a surface is influenced by several factors, including the temperature difference between the surface and its surroundings, the material and thickness of the surface, and the presence of insulation or other barriers.

3) How does heat loss from a surface impact energy efficiency?

The amount of heat loss from a surface can significantly impact the energy efficiency of a system or structure. Higher rates of heat loss can lead to increased energy consumption and higher heating or cooling costs.

4) How can heat loss from a surface be reduced?

There are various ways to reduce heat loss from a surface, such as using insulation to limit heat transfer, sealing any gaps or cracks in the surface, and implementing energy-efficient design strategies.

5) How is heat loss from a surface measured?

Heat loss from a surface can be measured using a variety of methods, including thermography, heat flux sensors, and thermal conductivity measurements. These techniques can provide valuable data for evaluating the effectiveness of insulation and other energy-saving measures.

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