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All (car) radiators are parallel, it is the only way to get the flow rate and reduce the fluid velocity in each tube sufficiently to allow the time for heat transfer. However, some radiators are series parallel where each section is in series and the section has the small tubes in parallel. Not seen that in a car though. If you can carefully « extract » ...

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If I get you right, there is a box having a small device that dissipates 30 Watts of heat continuously into the air. The air is blocked into a box, no way in or out. There are no fans/blowers trying to cool the device itself nor circulating the air inside the box. The size of the box is to be defined and the air temperature inside the box has to be figured ...

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This question can only be answered after many tests as to the convective properties of the box, it's geometry with respect to the geometry of the void, ignoring radiation. Because convection which is going to be the major heat transfer mechanism is basically a complex movement of air flowing up in a widening plume, and losing its temperature. Here is a link ...

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Any diameter wire, multiple parallel heated elements describes a toaster, an electric oven, any number of radiant heaters already in existence. Your problem is not whether you can do it with such a system, but more of a question if you can avoid melting the plastic bed on which the powder sits. It's unlikely that you can regulate the heat in such a ...

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If you are worried about the solar radiation (which is understandable), a much more affordable solution might be to put the whole thing in a shade if its possible, with enough clearance from the actual shading material. If you stick solar shields on the box, they are bound to get hot and, you won't be able to avoid conductive or convective heat transfer. ...

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You are missing several things. You need to know the emissivity of marshmallow. That will tell you how much of the 150W of radiation is absorbed, and how much is reflected. You need to know the latent heat of fusion, which tells you how much heat is needed to change the marshmallow from "solid" to "liquid". That may not be a very well ...

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Thermally radiated photons are emitted in whole piece of material, but only those photons which were born near the surface have a substantial possibility to avoid getting caught inside the piece. So this makes the radiated energy sound to be said as "per unit area". The stream of the emitted photons doesn't occur as pulses when our observation time ...

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Thermal equilibrium will be reached when heat in = heat lost. Find the U-value for each surface and calculate the heat loss rate in W/K (watts/kelvin) for each surface. Add them up and you've got the heat loss rate for the whole room. From this you can calculate the ΔT between inside and outside at equilibrium. For a give outside temperature you can now ...

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Greenhouses generally work by trapping radiation (infrared to be more specific). More specifically what they do is allow most of the sun radiation to come through and then trap the infrared radiation which is re-emitted from the objects within the greenhouse. types of heat transfer For the greenhouse equilibrium you need to consider all 3 types of heat ...

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Parallel, each tube just opens into the end-cap or top/bottom chamber ( On those I have seen, which is a bunch ,but they were all old copper/brass types).

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Briefest start. Written from a hosp[ital bed just post operation :-) - better could follow if useful. Solar loading will be peak in hours around noon. Angles shift across year. Find your site 0n Gaisma There are more relevant sites but that's an excellent start. Peak daily insolation is typically ABOUT 1 kW/m^2 BUT this is site specific and can be about 1....

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For anyone coming later: I ended up using a procedure described here: Kevin C. Gross, Phenomenological model for IR emission from high explosive detonation fireballs In one sentence: I ended up measuring the blackbody spectrum at increasing temperatures. Then fitting a linear relation between measured and theoretical value for emission intensity (per ...

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The equation for efficiency in thermodynamics is as follows: $$\eta = W_{out}/W_{in}$$ and to get this as a percentage you just multiply $\eta \;\mathrm{x}100=\%$ See the link below! https://en.wikipedia.org/wiki/Thermal_ef

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Imagine the system below at equilibrium. The energy balances are as follows: At the ball ... $$\epsilon_b T_b^4 =\epsilon_w T_w^4$$ At the wall ... $$\epsilon_w A_w \sigma T_w^4 = h_a A_w (T_a - T_w) + \epsilon_a A_w \sigma T_a^4$$ In these balances, $\epsilon_j$ are emissivities, $A_j$ are areas, $\sigma$ is the Stefan-Boltzmann constant, $T_j$ are ...

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