A cylindrical vessel with closed bottom and open top is 0.9 m in
A. 650 rpm
B. 600 rpm
C. 580 rpm
D. 96 rpm
Please scroll down to see the correct answer and solution guide.
Right Answer is: D
SOLUTION
The equation of parabolic profile is given by:
\(Z = \frac{{{\omega ^2}{r^2}}}{{2g}}\)
For point A: (0.2, y); \(y = \frac{{{{\rm{\omega }}^2}{{\left( {0.2} \right)}^2}}}{{2{\rm{g}}}}\)
For point B: (0.45, d + y + 0.5); \(d + y + 0.5 = \frac{{{{\rm{\omega }}^2}{{\left( {0.45} \right)}^2}}}{{2{\rm{g}}}}\)
Or
\(d + y = \frac{{{{\rm{\omega }}^2}{{\left( {0.45} \right)}^2}}}{{2{\rm{g}}}} - 0.5\)
By Volume Conservation:
Volume of fluid before rotation = Volume of fluid after rotation
i.e. V1 = V2
V1 = π (0.45)2 d
V2 = π (0.45)2(d + 0.5) - 1/2 {π(0.45)2(d + y + 0.5) - π (0.2)2y}
Now, V1 = V2
On solving we get,
d + y = 0.5 + 16y/81
From above equation 1 and 2, substitute the values of ‘d + y’ and ‘y’ in equation no 3.
\(\frac{{{{\rm{\omega }}^2}{{\left( {0.45} \right)}^2}}}{{2{\rm{g}}}} - 0.5 = 0.5 + \frac{{16}}{{81}}\left( {\frac{{{{\rm{\omega }}^2}{{\left( {0.2} \right)}^2}}}{{2{\rm{g}}}}} \right)\)
We get ω = 10 rad/sec
We know that
\(\omega = \frac{{2\pi N}}{{60}}\;or\;N = \frac{{60\omega }}{{2\pi }}\)
Or
\(N = \frac{{60 \times 10}}{{2\pi }}\) = 96 rpm
