Bernoulli equation for normal to streamline is
Pρ+∫V2rdn+gz=CWith ignoring the gravity, differentiate the equation.
\require{cancel} \frac{P}{\rho}+\int{\frac{V^2}{r}}dn+\cancelto{0}{gz}=C\\ \frac{dP}{\rho}+\frac{V^2}{r}dn=0\\ \therefore dP=-\frac{\rho V^2}{r}dnthe coordinate n points in a direction oppositethat of the radial coordinate so,
dn = -dr\\ \therefore dP=-\frac{\rho V^2}{r}dn=\frac{\rho V^2}{r}drThe velocity V(r) is defined as
V(r)=\left\{\begin{array}{lr} V_{max}\cdot\frac{r}{R_0} & \text{for } 0\leq r\leq R_0\\ V_{max}\cdot\frac{R_0}{r} & \text{for } r>R_0\end{array}\right\}\\The boundary condition for pressure is
\lim_{r \to \infty }P(r)=P_{atm}To get P(0)
\require{cancel} \lim_{a \to \infty}\int^{P(a)}_{P(0)}dp=\lim_{a \to \infty}\int^a_0{\frac{\rho V^2}{r}}dr\\ P_{atm}-P(0)=\int^{R_0}_0{\frac{\rho V_{max}^2r}{R_0^2}}dr+\lim_{a \to \infty}\int^a_{R_0}{\frac{\rho V_{max}^2R_0^2}{r^3}}dr\\ = \frac{\rho V_{max}^2}{2R_0^2}(R_0^2-\cancelto{0}{0^2})-\lim_{a \to \infty}\frac{\rho(V_{max}R_0)^2}{2}(\cancelto{0}{\frac{1}{a^2}}-\frac{1}{R_0^2})\\ =\frac{\rho V_{max}^2}{2}+\frac{\rho V_{max}^2}{2}=\rho V_{max}^2\\ \therefore P(0)=P_{atm}-\rho V_{max}^2