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 $R_2>R_1$ as shown in figure. The clearance between the cylinder and the bottom of the vessel is very small, the fluid density is $\rho$. Find the static gauge pressure of the fluid in the clearance as a function of the distance r from axis of the orifice (and the cylinder), if the height of the fluid is equal to h.

• A. $\rho gh(1-\frac{{R_1}^2}{r^2})$
• B. $\rho gh(1+\frac{{R_1}^2}{r^2})$
• C. $\rho gh(1-\frac{{R_2}^2}{r^2})$
• D. $\rho gh(1+\frac{{R_2}^2}{r^2})$

Answer 1

The correct option is A $\rho gh(1-\frac{{R_1}^2}{r^2})$

Consider three section 1 ,2 & 3 at radius $R_{1,}r\&R_2$ respectively. Water flows through the small clearance into the orifice. Let d be the clearance. Then from the equation of continuity `(2piR_1d)v_1=(2pird)v=(2piR_2d)v_2`
or `v_1R_1=vr=v_2R_2` ... (1)

where, `v_1`, $v$ and `v_2` are respectively the inward radial velocities of the fluid at section 1, 2 and 3.
$P_{1,}P\&P_2$ be the pressure at section 1, 2 & 3 respectively and $P_0$ be atmospheric pressure.
Large amount of water is flowing befoere section 3 having pressure $P_0+\rho gh$ and negligible velocity to section 3 having d , $P_2\&v_2$ as clearance, pressure and velocity respectively. Applying Bernoulli's theorem
`P_0+hrhog=P_2+1/2rhov_2^2` ...(2)
Applying the same theorem at section3, 2 and 1, we get
$P_2+\frac{\rho {v_2}^2}{2}=P+\frac{\rho v^2}{2}=P_0+\frac{\rho {v_1}^2}{2}...\left(3\right)$
(since the pressure in the orifice is atmospheric (`P_0`)

From Eqs. (2) and (3)
`v_1=sqrt(2gh)` ...(4)
and `P=P_0+1/2rhov_1^2(1-(v/v_1)^2)`
`=P_0+hrhog(1-(R_1/r)^2)` [Using (1) and (4)]
So, gauge pressure is $P-P_0$ which is `hrhog(1-(R_1/r)^2)`