Question

Water, in a 100-mm-diameter jet with speed of 30 m/s to the right, is deflected by a cone that moves to the left at 14 m/s. Determine (a) the thickness of the jet sheet at a radius of 230 mm. and (b) the external horizontal force needed to m

Answer 1

Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same $V_i=V_e=V$

Calculate the inlet and exit velocity of water jet

$V=V_j+V_e\\\\V=30+14\\\\V=44m/s$

The conservation of mass equation of steady flow

$\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0$

$A_i\ \texttt {is the inlet area of the jet}$

$A_e\ \texttt {is the exit area of the jet}$

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

$A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}$

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

$t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm$

(b)

$R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)]$

$1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j$

$R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N$

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

Therefore, the horizontal force is -7603N