= Solution
For a <Burgers vortex>, integrating the axial <vorticity> gives the azimuthal <velocity>:
$$
u_\theta(r)=\frac\Gamma{2\pi r}(1-e^{-r^2/\delta^2}),\qquad \delta^2=\frac{4\nu}{\alpha}.
$$
The swirl contribution to the <viscous dissipation> per unit length and per unit density is
$$
\mathcal D_\theta=\nu\int_0^\infty
\left(\frac{du_\theta}{dr}-\frac{u_\theta}{r}\right)^2 2\pi r\,dr
=\nu\int_0^\infty\omega_z^2\,2\pi r\,dr.
$$
For the equality, expand the two squares using $\omega_z=u_\theta'+u_\theta/r$; their integrated difference is proportional to $[u_\theta^2]_0^\infty$, which vanishes. Substitution of the Gaussian <vorticity> profile gives
$$
\mathcal D_\theta
=\frac{2\nu\Gamma^2}{\pi\delta^4}\int_0^\infty r e^{-2r^2/\delta^2}\,dr
=\frac{\nu\Gamma^2}{2\pi\delta^2}
=\boxed{\frac{\alpha\Gamma^2}{8\pi}}.
$$
This <excess dissipation of a Burgers vortex> is independent of <kinematic viscosity> at fixed strain $\alpha$ and <circulation> $\Gamma$. Multiply by the <mass density> if a dimensional power per length is wanted.
The qualification "excess" matters. The imposed uniform strain itself has <viscous dissipation> density $3\nu\alpha^2$ per unit mass, so its integral over an infinite cross-section diverges. One must subtract that background or restrict to a finite cross-section. In a core-sized area, its contribution is of order $\nu^2\alpha$, smaller than the swirl contribution by order $(\nu/\Gamma)^2$ as $\Gamma/\nu\to\infty$.
Meanwhile $\delta\propto\nu^{1/2}$ shrinks and the central <vorticity> $\Gamma/(\pi\delta^2)$ grows. The finite dissipation becomes concentrated in a narrow tube: a useful local model of <internal intermittency> and the <turbulent dissipation anomaly>, though not a proof that an entire turbulent flow consists of such vortices.
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