Understanding the value of Q_RADI

[elangsz]

Hello everybody, im trying to understand the impact of setting a net heat flux of a certain value (10 kW/m2 on a 1 m2 horizontal surface) with convection coefficient set to 0, towards the value of Q_RADI. From what i read in the FDS user guide and technical reference, if the value of Q_RADI is positive it means that theres heat from a source that comes into the computational domain via radiation. As you can see in the .csv files, the value of Q_RADI is only about 2.2 kW instead of 10 kW, can anybody explain why? Is it lost to the boundary surface? because i also checked that the Q_CONV is also only about 2.2 kW, and was wondering about the rest of the radiative heat being emitted from that particular surface. I just wanted to be sure with my inputs for future simulations, thats all. Earlier in a different simulation where i was focusing on convective heat transfer, i turn off the radiation solver and set surface emissivity to zero, with convection coefficient set to default, and set heat flux of the 1m2 surface to 10 kw/m2, and i get the Q_COND equals to 10 kW as a result in which it is easier for me to understand.

Oh and i also set a couple of devices there, like the convective, net, radiative heat flux of the surface. Its also written that the surface radiative heat flux is 10 kW but the Q_RADI is still about 2.2 kW. There are also other slices like temperature, velocity vector, and radiation loss. Id be glad if someone takes the time to help me understand this, thankyou! (FDS file attached below)

case 9.txt

[drjfloyd]

If you had no participating media (i.e. no ambient CO2 or H2O) then you would be inputting 10 kW from the hot source and loosing 10 kW at the open boundaries. The net heat input into the gas phase domain of the MESH would be 0 KW (10 kW in - 10 kW out). This is not the case with your inputs as you have ambient CO2 and H2O. So the net heat input into the gas phase domain is now 10 kW in minus some amount less than 10 kW where that amount less is the radiation absorbed in the domain.

[elangsz]

Ah, okay, that makes sense. So the 2.2 kW of radiative heat was absorbed by the participating medium (i.e., CO₂ and H₂O, as you mentioned earlier), while the remaining 7.8 kW was not absorbed and was instead lost through the boundaries of the computational domain. Is there any device, whether gas phase, solid phase, or one of the spatially integrated output, that I can use to take into account the radiative heat loss? If so, what type should I use, where should I place it, and what should I measure? Thanks!

[drjfloyd]

See the 'RADIATION ....' quantitities in Table 22.4.