Question 6.SP.3: A free jet of water with an initial diameter of 2 in she tri......

Take as a free body diagram the element of fluid (control volume, C V ) in contact with the blade. Assume the forces acting on the element are as shown in the sketch. The forces F_x and F_y represent the components (directions assumed) of the net force of the blade on the water (C V) in the x and y directions. This net force includes shear stresses tangential to the blade and pressure forces normal to the blade.

Applying Eq. (6.7a) along the x  axis and noting that A=\pi(2 / 12)^2 / 4=0.0218 \mathrm{ft}^2

\begin{aligned}-F_x=\rho Q\left(V_{2 x}-V_{1 x}\right) & =1.94(0.0218 \times 100)\left(95 \cos 30^{\circ}-100\right) \\& =4.23(-17.7)=-75.0 \mathrm{lb}\end{aligned}

So :           F_x=+75.0 \mathrm{lb}=75.0 \mathrm{lb} \leftarrow

The plus sign indicates that the assumed direction of  F_x  was correct. Applying Eq. (6.7b) along the  y  axis,

+F_y=\rho Q\left(V_{2 y}-V_{1 y}\right)=4.23\left(95 \sin 30^{\circ}-0\right)=+201 \mathrm{lb}=201 \mathrm{lb} \uparrow

The resultant force of the blade on the control volume is the sum of these two components. The force of the fluid on the blade is equal and opposite to this. The resultant force on the blade is 214 lb  at an angle of 69.5°  \sphericalangle

Note that if we neglected friction (i.e., V_2=V_1=100 \mathrm{fps} ), we would have calculated the forces as F_x=56.7 \mathrm{lb} and F_y=212 \mathrm{lb}. So, when the angle of deflection \theta from the initial direction of the jet is less than 90^{\circ}, we find that friction increases the value of F_x over the value it would have if there were no friction. But when \theta is greater than 90^{\circ}, friction decreases the value of F_x. On the other hand, friction decreases the value of F_y for any value of angle \theta.

If the flow had been in a vertical plane, we would have to consider the effect on V_2 of the higher elevation at exit from the blade, and we would have to estimate the weight of the liquid on the blade and add it to \rho Q\left(\Delta V_z\right) to get the total value of F_z.

\sum F_x=\dot{m}\left(\Delta V_x\right)=\rho Q\left(\Delta V_x\right)=\rho Q\left(V_{2 x}-V_{1 x}\right)          (6.7a)

\sum F_y=\dot{m}\left(\Delta V_y\right)=\rho Q\left(\Delta V_y\right)=\rho Q\left(V_{2 y}-V_{1 y}\right)          (6.7b)