Question

A 2-kg disk is constrained horizontally but is free to move vertically. The disk is struck from below by a vertical jet of water. The speed and diameter of the water jet are 10 m/s and 25 mm at the nozzle exit. Obtain a general expression for the speed of the water jet as a function of height, h. Find the height to which the disk will rise and remain stationary

Answer 1

Answer:

h = 4.281 m

Explanation

Given Data,

Weight of disk = 2kg

Speed of water jet (V) = 10 m/s

diameter of water jet = 25 mm

Cal. the velocity of the water jet as function height by applying Bernoulli's  eqtn of water surface to the jet

$\frac{P}{\rho } +\frac{V^{2}}{2} + gz$ = Constant

$\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g$

$V=\sqrt{V_{0}^{2}-2gh}$

Relation between  $V_{0}$ & V

$m=\rho V_{0}A_{0}$

$\rho VA=\rho V_{0}A_{0}$

$VA= V_{0}A_{0}$

Momentum

$F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA$

$-mg = w_{1}[-\rho VA] +w_{2}[\rho VA]$

$w_{1}$ = V

$w_{2}$ = 0

$mg = \rho V^{2}A$

$mg = \rho VV_{0} A$

$mg = \rho VV_{0} A_{0}$

$mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }$

Solving for h

$h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}]$

g is gravitational acc.

$= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ]$

$= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}]$

$= \frac{100-16.0078}{19.62}$

h = 4.281 m

h of disk on which it remains stationary.