Question

The horizontal bottom of a wide vessel with an ideal fluid has a round orifice of radius $R_1$ over which a round closed cylinder is mounted, whose radius $R_2>R_1$ (figure shown above). The clearance between the cylinder and the bottom of the vessel is very small, the fluid density is $\rho$. Find the static pressure of the fluid in the clearance as a function of the distance $r$ from the axis of the orifice (and the cylinder), if the height of the fluid is equal to $h.$

Answer 1

Water flows through the small clearance into the orifice. Let $d$ be the clearance. Then from the equation of continuity
$\left(2\pi R_{1}d\right)v_1=\left(2\pi Rd\right)v=\left(2\pi R_{2}d\right)v_2$
or $v_1{R}_1=vR=v_2{R}_2$ (1)
where $v_1$, $v_2$ and $v$ are respectively the inward radial velocities of the fluid at 1, 2 and 3. Now by Bernoulli's theorem just before 2 and just after it in the clearance.
$p_0+h\rho g=p_2+\frac{1}{2}\rho {v_2}^2$ (2)
Applying the same theorem at 3 and 1 we find that this also equals
$p+\frac{1}{2}\rho v^2=p_0+\frac{1}{2}\rho {v_1}^2$ (3)
(since the pressure in the orifice is $p_0$)
From equations (2) and (3) we get,
$v_1=\sqrt{2gh}$ (4)
and $p=p_0+\frac{1}{2}\rho v_1^2(1-{\left(\frac{v}{v_1}\right)}^2)$
$=p_0+h\rho g(1-{\left(\frac{R_1}{r}\right)}^2)$ [Using (1) and (4)]