Answer:
Explanation:
The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]
Calculate the inlet and exit velocity of water jet
[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]
The conservation of mass equation of steady flow
[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]
[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]
[tex]A_e\ \texttt {is the exit area of the jet}[/tex]
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
[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]
R is the radius
t is the thickness of the jet
D_j is the diameter of the inlet jet
[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]
(b)
[tex]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)][/tex]
[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]
[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]
The negative sign indicate that the direction of the force will be in opposite direction of our assumption
Therefore, the horizontal force is -7603N
