= Solution
The steady <Euler equations for an inviscid fluid> with no body force are $(\mathbf u\cdot\nabla)\mathbf u=-\nabla p/\rho$. Use the <convective acceleration identity>
$$
(\mathbf u\cdot\nabla)\mathbf u=\nabla(\tfrac12|\mathbf u|^2)-\mathbf u\times(\nabla\times\mathbf u)
$$
to obtain
$$
\boxed{\mathbf u\times\boldsymbol\omega=\nabla\left(\frac p\rho+\tfrac12|\mathbf u|^2\right).}
$$
For the question's <streamfunction> convention, $\mathbf u=(\psi_y,-\psi_x,0)$, so $\mathbf u\times\boldsymbol\omega=-\omega\nabla\psi$. Constant $\omega$ therefore gives the <Bernoulli function for planar constant-vorticity flow>
$$
\boxed{\tfrac12|\mathbf u|^2+\omega\psi+p/\rho=C}
$$
on the connected fluid region.
For circular <streamlines>, write $\mathbf u=w(r)\widehat{\boldsymbol\phi}$. The cylindrical <curl> gives $(rw)'/r=\omega$, so $w=\omega r/2+C_1/r$. The outer wall condition gives $C_1=-2\omega a^2$. Choose the positive azimuthal orientation so that $w(a)=V$, as implicit in the requested sign for $\omega$. Then
$$
\boxed{\omega=-\frac{2V}{3a},\qquad \mathbf u=\frac V{3a}\left(\frac{4a^2}{r}-r\right)\widehat{\boldsymbol\phi}.}
$$
Reversing the circulation reverses $\omega$ but leaves the speed and pressure difference unchanged. For this <uniform-vorticity circular flow in an annulus>, one convenient streamfunction is $\psi=V(r^2/2-4a^2\log(r/a))/(3a)+C_2$.
The radial component of the <Euler equations for an inviscid fluid> is $p_r=\rho w^2/r$. Thus pressure increases outwards, and
$$
\begin{aligned}
p(2a)-p(a)&=\frac{\rho V^2}{9a^2}\int_a^{2a}\left(\frac{16a^4}{r^3}-\frac{8a^2}{r}+r\right)\,dr\\
&=\frac{\rho V^2}{9a^2}\left[\frac{15a^2}{2}-8a^2\log2\right].
\end{aligned}
$$
Therefore
$$
\boxed{\Delta p=p(2a)-p(a)=\frac{15-16\log2}{18}\rho V^2.}
$$
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