Known

As shown in the schematic, a horizontal water jet exits a nozzle with a uniform velocity of \(V_1\), strikes a stationary vane, and is turned through an angle \(\alpha\). The water flow is assumed to be a steady one-dimensional flow with a constant cross-sectional area. Gravity and viscous effects are negligible.

Find

Determine the horizontal and vertical components of the force exerted by the vane on the water flow.

Analysis

Define a control volume shown by the dash lines.

Assume \(F_x\) is pointing left and \(F_z\) is pointing upward, as shown in the schematic.

For the steady one-dimensional water flow with a constant cross-sectional area, we have the speed \(V_1=V_2\) and the flow cross-sectional area \(A_1=A_2\) of the flow stream entering and leaving the control volume.

Apply the linear momentum equation to the defined control volume;

\[ \sum \mathbf{F} = \frac{\partial}{\partial t}\int_{CV}\mathbf{V}\rho\,d\mathcal{V} + \int_{CS}\mathbf{V}\rho\,\mathbf{V}\cdot\hat{\mathbf{n}}\,dA \]

In the \(x\) direction:

\[ -F_x=u_1\rho(\mathbf{V}_1\cdot\hat{\mathbf{n}}_1)A_1 +u_2\rho(\mathbf{V}_2\cdot\hat{\mathbf{n}}_2)A_2 \] \[ -F_x=V_1\rho(-V_1)A_1+V_2\cos\alpha\,\rho V_2A_2 \] \[ F_x=\rho A_1V_1^2-\rho A_2V_2^2\cos\alpha =\rho A_1V_1^2(1-\cos\alpha) \]

In the \(z\) direction:

\[ F_z=w_1\rho(\mathbf{V}_1\cdot\hat{\mathbf{n}}_1)A_1 +w_2\rho(\mathbf{V}_2\cdot\hat{\mathbf{n}}_2)A_2 \]

Since \(w_1=0\), we have

\[ F_z=w_2\rho(\mathbf{V}_2\cdot\hat{\mathbf{n}}_2)A_2 \] \[ F_z=V_2\sin\alpha\,\rho V_2A_2 =\rho A_2V_2^2\sin\alpha =\rho A_1V_1^2\sin\alpha \]

Input:

\(\rho\) of water: \(998.2~\mathrm{kg/m^3}\)
\(A_1: 0.01~\mathrm{m^2}\)
\(F_x\)
--
\(F_z\)
--
\(\dot{m}\)
--
\(\rho A_1V_1^2\)
--

\(F_x\) and \(F_z\) vs \(V_1\)

\(F_x\) and \(F_z\) vs \(\alpha\)

Scale of \(F_x\) and \(F_z\).

\(F_{\max}<45~\mathrm{N}\quad [0,50]\)

\(45~\mathrm{N}\le F_{\max}<180~\mathrm{N}\quad [0,200]~\mathrm{N}\)

\(F_{\max}\ge 180~\mathrm{N}\quad [0,600]~\mathrm{N}\)

keep the two plots having the same \(F_x\) & \(F_z\) scale.