= Standard thin-disk dissipation flux
Consider a steady <Keplerian disk> around a central <mass> $M$, with positive inward <accretion rate> $\dot M$ and zero <viscous torque in an accretion disk> at $R_{\rm in}$. <Angular momentum conservation> gives the outward internal <torque>
$$
\mathcal T(R)=\dot M[\ell(R)-\ell(R_{\rm in})],\qquad
\ell(R)=\sqrt{GMR}.
$$
Differential <rotation> dissipates <energy> at the rate $\mathcal T(-d\Omega/dR)dR$ in an annulus. Dividing by its projected <area> $2\pi R\,dR$ and using $\Omega=\sqrt{GM/R^3}$ gives the flux summed over both faces:
$$
F_{\rm diss}(R)=\frac{3GM\dot M}{4\pi R^3}
\left[1-\left(\frac{R_{\rm in}}R\right)^{1/2}\right].
$$
The <energy flux> through one face is $F_{\rm diss}/2$. In a locally radiating, optically <thick disk>, one therefore sets $\sigma T_{\rm eff}^4=F_{\rm diss}/2$, rather than equating a single-face blackbody flux to the two-face expression.<Dynamics of Astrophysical Discs, Lecture 5: Boundary Conditions and Steady Accretion by Gordon I. Ogilvie>[Section 5.2, printed p. 3, steady torque and surface-temperature equations]{ref}
The bracket enforces the zero-torque inner boundary. Far outside it, $F_{\rm diss}\propto R^{-3}$. Internal <torque> transports <energy> between annuli as well as transporting <angular momentum>, so local dissipation is not simply the local loss of orbital binding <energy>. A nonzero inner <torque>, strong advective transport or a <mass> outflow changes this model.
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