Radiative view factors in glass-melt solar storage

Radiative view factors between a cylinder and its endcaps. Diagram from Isidoro Martinez, "Radiative View Factors."

In radiative heat transfer, a view factor F_A→B is the proportion of the radiation which leaves surface A and (directly) strikes surface B. In some cases there may also be an adiabatic blackbody surface C which, at thermal equilibrium, must re-radiate all of the photons it receives. In this case we can define an effective view factor  F*_A→B that includes the re-radiated photons that reach B indirectly:

F*_A→B = F_A→B + (F_A→C)(F_C→B)

When all the surfaces are blackbodies (an assumption that greatly simplifies the math, and is approximately true in practice) the net radiative flux from A to B, averaged over  S_A, the area of A, is:

Φ_A = F*_A→B * σ(T_A⁴ - T_B⁴),

where σ is the Stefan–Boltzmann constant, σ = 5.7 E−8 W m−2 K−4.

The corresponding net flux at surface B, Φ_B, must be Φ_A multiplied by the ratio of the two areas:

Φ_B = Φ_A * (S_A/S_B) = F*_A→B * σ(T_A⁴ - T_B⁴) * (S_A/S_B).

When solar energy is stored in a glass melt, we can consider the free surface of the melt to be surface A (the lower cap of a cylinder,) the water wall tubes of the boiler to be surface B (the walls of the cylinder,) and the roof of the furnace to be the adiabatic surface C (the upper cap of the cylinder.)

Water wall tubes. Image quoted from http://tubeweld.com.

In the view factor diagram above, a reduced radius 'r' is related to the cylinder's radius, R, and its height, H, by:

r = R/H,

and a parameter ρ is defined to be:

ρ = (√(4 * r2 + 1) -1) / r.

The correspondences between our lettered surfaces and the diagram's numbered surfaces are:

A = 3, B = 1, C = 4.

From the diagram,

F_A→B = F_3→1 = ρ/2r

F_A→C = F_3→4 = 1 - ρ/2r

F_C→B = F_4→1 = F_3→1 = ρ/2r

So,

F*_A→B = F_A→B + (F_A→C)(F_C→B) =  ρ/2r + (1 - ρ/2r)(ρ/2r) = (ρ/2r) (2 - ρ/2r),

and the area ratio, S_A/S_B, is

S_A/S_B = πR² / 2πRH = R/2H = r/2.

The average flux on the water wall, Φ_B, is

Φ_B = F*_A→B * σ(T_A⁴ - T_B⁴) * (S_A/S_B) = (ρ/2r) (2 - ρ/2r) (r/2) * σ(T_A⁴ - T_B⁴)

Φ_B = (ρ/4) (2 - ρ/2r) * σ(T_A⁴ - T_B⁴)

With R = 51 m, H = 87 m, r = 0.58, ρ = 0.93, T_A = 1610 °K (1340 °C) , T_B = 730 °K (460 °C), as calculated in an earlier post,

Φ_B = (ρ/4) (2 - ρ/2r) * σ(T_A⁴ - T_B⁴) = 0.26 * 367 kw/m2 = 96 kw / m2

which is not nearly the 250 kW/m2 we need to see on a water tube wall.

We need to increase radiant transfer by upping the "empty" temperature of the glass melt. This will also incidentally decrease storage density, resulting in an increase in R and a decrease in H (as calculated on the basis of 250 kW/m2.)

A spreadsheet exploring radiant transfer in glass-melt thermal storage.

Exploring these relations in a Numbers spreadsheet shows that Tempty = 1570 °C gives a consistent solution with R = 63 m, H = 71, and the flux on the water wall tubes = 250 kw/m2. The glass temperature range from empty to full is just 230 °C.