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Let $M_{\ast }$ be the central mass, so ${\Omega }^2=G{M}_{\ast }/r^3$ and the specific orbital energy is $\epsilon =-G{M}_{\ast }/(2r)$. Define the inward accretion rate $\dot{M}>0$. The rate at which the gas acquires binding energy in moving from $b$ to $a$ isThe vertically integrated viscous heating per unit horizontal area, including the entire column, is $\bar{\nu }\Sigma (r{\Omega }^{\prime }{)}^2$. Since $r{\Omega }^{\prime }=-3\Omega /2$, the dissipated power in the annulus isHere the column integral already includes dissipation associated with both disk faces; adding another factor of two would double-count it. Performing the elementary integrals givesFor the whole disk, take $a=r_{\mathrm{in}}$ and $b\to \infty$. Both expressions giveIf $a\gg r_{\mathrm{in}}$, the bracket $1-\sqrt{r_{\mathrm{in}}/r}$ is uniformly close to one throughout the annulus. Comparing the positive integrals showswith the correction negative. Thus far from the inner edge the viscous dissipation is approximately three times the local binding-energy acquisition.

There is no violation of conservation of energy because the viscous torque in an accretion disk also transports mechanical energy. The outward viscous torque in an accretion disk isand its outward energy flux is $\Omega \mathcal{G}=G{M}_{\ast }\dot{M}[r^{-1}-\sqrt{r_{\mathrm{in}}}{r}^{-3/2}]$. The explicit expressions satisfyEnergy transported from the inner disk supplies the extra dissipation at larger radii. The torque power vanishes at the zero-torque inner boundary condition and tends to zero at infinity, leaving global balance. This is torque transport of energy in a steady accretion disk.