A sphere is moving in water with a velocity of 1.6 m/s. Another s
A. 5 m/s
B. 10 m/s
C. 20 m/s
D. 40 m/s
Please scroll down to see the correct answer and solution guide.
Right Answer is: B
SOLUTION
Concept:
|
Serial No |
Number |
Equation |
Significance |
|
01 |
Reynolds No. |
\(\frac{{Fi}}{{Fv}} = \frac{{\rho VL}}{\mu }\) |
Flow in closed conduit. Resistance experienced by sub-marines, airplanes, fully immersed bodies. |
|
02 |
Froude No. |
\(\sqrt {\frac{{{F_i}}}{{{F_g}}}} = \frac{V}{{\sqrt {gL} }}\) |
Where a free surface is present and gravity force is predominant such as Spillway, weirs, Open Channels, waves in ocean. Flow of jet from an orifice or nozzle. Fluids of different densities flow over one another. |
|
03 |
Euler No. |
\(\sqrt {\frac{{{F_i}}}{{{F_P}}}} = \frac{V}{{\sqrt {p/\rho } }}\) |
In cavitation studies. Where pressure force is predominant. (Used in a closed pipe where turbulence is fully developed so that viscous forces are negligible.) |
|
04 |
Mach No. |
\(\sqrt {\frac{{{F_i}}}{{{F_e}}}} = \frac{V}{C}\) |
Where fluid compressibility is important. Launching of rockets, aeroplanes and projectile moving at supersonic speed. Underwater testing of torpedoes. Water-hammer problems. |
|
05 |
Weber No. |
\(\sqrt {\frac{{{F_i}}}{{{F_\sigma }}}} = \frac{V}{{\sqrt {\sigma /\rho L} }}\) |
In Capillary studies. Where Surface tension is predominant. Capillary rise in narrow passages. Capillary movement of water in soil. Capillary waves in channels. Flow over weirs for small heads.
|
Here the flow is fully immersed flow, so Reynold’s number for the prototype and model will be same.
Calculation:
\(\left( {\frac{{\rho VD}}{\mu }} \right)_p = {\left( {\frac{{\rho VD}}{\mu }} \right)_m}\)
Vp = 1.6 m/s, \(\frac{{{D_m}}}{{{D_p}}} = 2,\)
\(\frac{{{\rho _m}}}{{{\rho _p}}} = \frac{1}{{750}}\)
\(\frac{{{\mu _m}}}{{{\mu _p}}} = \frac{1}{{60}}\)
So, \(\frac{{{\rho _p}\;{V_p}{D_p}\;}}{{{\mu _p}}} = \frac{{{\rho _m}{V_m} \times {D_m}}}{{{\mu _m}}}\)
\( \Rightarrow \frac{{{\rho _m}}}{{{\rho _p}}} \times \frac{{{D_m}}}{{{D_p}}} \times \frac{{{\mu _p}}}{{{\mu _m}}} = \frac{{{V_p}}}{{{V_m}}}\)
\( \Rightarrow \frac{1}{{750}} \times 2 \times 60 = \frac{{1.6}}{{{V_m}}}\)
\(\Rightarrow {V_m} = \frac{{1.6 \times 750}}{{2 \times 60}} = 10\;m/s\)