Saturn’s Mean Temperature Equation Tmean.saturn is:

Tmean.saturn = [Φ (1-a) So (1/R²) (β*N*cp)¹∕ ⁴ /4σ]¹∕ ⁴

Saturn’s sidereal rotation period is10 h 33 min 38 sec, or 10,56 h

N = 24h/10,56h rotations/per day

R = 9,5826 AU, 1/R² = 1/9,5826² = 0,01089 times lesser is the solar irradiation on Saturn than that on Earth.

So = 1.362 W/m² is Solar constant

Saturn’s albedo, asaturn = 0,342 

Saturn is a gaseous planet, Saturn’s surface irradiation accepting factor Φsaturn = 1

(Saturn has not surface to reflect the incident sunlight. Accepted by a Gaseous Hemisphere with radius r sunlight is S*Φ*π*r²(1-a), where Φ = 1)

Cp.saturn = 3,1388 cal/gr oC , H₂ specific heat at 175 K, Saturn has not surface,  Saturn's atmosphere consists from 96,3 ± 2,4 % H₂

β = 150 days*gr*oC/rotation*cal – it is the Rotating Planet Surface Solar Irradiation Absorbing-Emitting Universal Law constant

σ = 5,67*10⁻⁸ W/m²K⁴, a Stefan-Boltzmann constant

So we have: Saturn’s mean temperature Tmean.saturn is:

Tmean.saturn = {1*(1-0,342)1.362*0,01089(W/m²) [150*(24h/10,56h)*3,1388]¹∕ ⁴ /4*5,67*10⁻⁸(W/m²K⁴) }¹∕ ⁴ =

Tmean.saturn = 125,25 K is the calculated.

And below is the measured by satellites

Tsat.mean.saturn = 134 K (at 1bar level)

Tsat.mean.saturn = 84 K (at 0,1 bar level).

Let's, for comparison reason, calculate Saturn’s Effective Temperature Equation Te.saturn.incompl is:

Te.saturn.incompl = [ (1-a) So (1/R²) /4σ]¹∕ ⁴

Te.saturn.incompl = [(1-0,342)1.362*0,01089(W/m²) /4*5,67*10⁻⁸(W/m²K⁴) ]¹∕ ⁴ =

Te.saturn.incompl = 80,99 K is the calculated.

And below is the measured by satellites

Tsat.mean.saturn = 134 K (at 1bar level)

Tsat.mean.saturn = 84 K (at 0,1 bar level).

And comparing with the Mean Temperature Equation Tmean.saturn = 125,25 K we may conclude here that the without-atmosphere Mean Temperature Equation, when applied to Gaseous Giant Saturn gives us much closer to the measured by satellites result.