Figure 16.11(a) shows a jet of water with a velocity v1 striking a vane that is moving with a velocity v0. Determine the forces exerted by the vane on the water if v1 = 20 m/s and v0 = 8 m/s. The jet is 50 mm in diameter.

Question

Figure 16.11(a) shows a jet of water with a velocity [latex] v_1[/latex] striking a vane that is moving with a velocity [latex] v_0[/latex]. Determine the forces exerted by the vane on the water if [latex] v_1 = 20 m/s[/latex] and [latex]v_0 = 8 m/s[/latex]. The jet is 50 mm in diameter.

Accepted Answer

The system with a moving vane can be converted into an equivalent stationary system as shown in Fig. 16.11(b) by defining an effective velocity [latex]v_e [/latex] and an effective volume flow rate [latex] Q_e[/latex]. We then have

➭ Effective Velocity and Volume Flow Rate

[latex] v_e = v_1 - v_0[/latex] (16–11)

[latex] Q_e = A_1 v_e[/latex] (16–12)

where [latex] A_1[/latex] is the area of the jet as it enters the vane. It is only the difference between the jet velocity and the vane velocity that is effective in creating a force on the vane. The force equations can be written in terms of [latex] v_e[/latex] and [latex] Q_e[/latex]. In the x direction,

[latex] R_x = ho Q_e v_e \cos heta - \left(- ho Q_e v_e ight)[/latex]

[latex] = ho Q_e v_e \left(1+ \cos heta ight) [/latex] (16–13)

In the y direction,

[latex] R_y = ho Q_e v_e \sin heta -0[/latex] (16–14)

We know that

[latex] v_e = v_1 - v_o = \left(20-8 ight) m/s = 12m/s[/latex]

[latex] Q_e = A_1 v_e = \left(1.964 imes 10^{-3}m^2 ight) \left(12m/s ight) =0.0236m^3/s[/latex]

Then the reactions are calculated from Eqs. (16–13) and (16–14):

[latex] R_x = \left(1000 ight) \left(0.0236 ight) \left(12 ight) \left(1+\cos45° ight) =483N[/latex]

[latex] R_y = \left(1000 ight) \left(0.0236 ight) \left(12 ight) \left(\sin45° ight) =200N[/latex]

Question 16.5: Figure 16.11(a) shows a jet of water with a velocity v1 stri...

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