Question 18.25: A jet of water having a velocity of 25 m/s impinges on a cur......

Given data:

Velocity of jet                            V_1=25 \mathrm{~m} / \mathrm{s}

Velocity of vane                        u =10 m/s

Angle made by jet with the direction of motion of vane at entry \alpha_1=20^{\circ}

Angle made by leaving jet with the direction of motion of vane = 90°

\therefore                   \alpha_2=180^{\circ}-90^{\circ}=90^{\circ}

The inlet and outlet velocity triangles are shown in Fig. 18.20.

Here        u_1=u_2=u=10 \mathrm{~m} / \mathrm{s}

V_{r 1}=V_{r 2}                       (\because \text { friction neglected })

From the inlet velocity triangle, we have

V_{w 1}=V_1 \cos \alpha_1=25 \times \cos 20^{\circ}=23.49 \mathrm{~m} / \mathrm{s}

V_{f 1}=V_1 \sin \alpha_1=25 \times \sin 20^{\circ}=8.55 \mathrm{~m} / \mathrm{s}

\tan \beta_1=\frac{V_{f 1}}{V_{w 1}-u_1}

or                      \tan \beta_1=\frac{8.55}{23.49-10}=0.6338

or                    \beta_1=\tan ^{-1}(0.6338)=39.33^{\circ}

Again from the inlet velocity triangle, we get

\sin \beta_1=\frac{V_{f 1}}{V_{r 1}}

or                              V_{r 1}=\frac{V_{f 1}}{\sin \beta_1}=\frac{8.55}{\sin 39.33^{\circ}}=13.49 \mathrm{~m} / \mathrm{s}

From the outlet velocity triangle, we obtain

  \cos \beta_2=\frac{u_2}{V_{r 2}}=\frac{10}{13.49}=0.7413

or                  \beta_2=\cos ^{-1}(0.7413)=42.16^{\circ}

The work done per second per unit weight of water striking the vane is given by Eq. (18.27) as

={\frac{(V_{{w1}}\pm V_{{w2}})\times u}{g}}        (18.27)

=\frac{V_{w 1} \times u}{g}

=\frac{23.49 \times 10}{9.81}=23.94 \mathrm{~N}-\mathrm{m} / \mathrm{N}