Pressure diagram in rotating closed tank

Assumption: the tank is completely full and closed, and the pressure at the centre of the top cover (on the axis) remains atmospheric (0 gauge). Then the parabolic pressure variation starts from zero at the axis on the top.

Data: $R=0.5$ m, height 2 m, $N=100$ rpm so $\omega=\dfrac{2\pi\times100}{60}=10.472$ rad/s, $\rho=1000$ kg/m³.

Pressure at a point at radius $r$ and depth $z$ below the top:

$$p=\tfrac{1}{2}\rho\omega^{2}r^{2}+\rho g z$$

Along AC (top cover, $z=0$)

$$p=\tfrac{1}{2}(1000)(10.472)^{2}r^{2}=54\,831\,r^{2}\ \text{Pa}$$

The diagram is a parabola, zero at the centre and 13.71 kPa at A and at C (symmetrical).

Along AB (vertical wall, $r=0.5$ m)

$$p=13.71+9.81\,z\ \text{kPa}$$

The diagram is a straight line (trapezium) from 13.71 kPa at A to 33.33 kPa at B.

  A |->13.71 kPa          C
    |  \     parabola    /
    |   \_______________/  (0 at centre)
    |->23.52 (1 m)
    |->33.33 kPa at B

Answer: at A and C, 13.71 kPa; at B and D, 33.33 kPa; AC is parabolic (0 at the centre) and AB linear.