Known

As shown in the schematic, a horizontal water jet exits a nozzle with a uniform velocity of \(V_1\), strikes a vane, and is turned through an angle \(\alpha\). The vane is moving horizontally at a constant velocity \(V_0\). 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.

Apply the mass conservation to the defined moving control volume at a constant velocity involving a steady flow

\[ \dot{m}_1=\dot{m}_2 \]

Thus,

\[ \rho V_{\mathrm{rel},1}A_1=\rho V_{\mathrm{rel},2}A_2 \]

Since \(A_1=A_2\), we have \(V_{\mathrm{rel},1}=V_{\mathrm{rel},2}=V_1-V_0\).

Apply the linear momentum equation to the defined moving control volume at a constant velocity involving a steady flow

\[ \sum \mathbf{F} = \int_{CS}\mathbf{V}_{\mathrm{rel}}\rho (\mathbf{V}_{\mathrm{rel}}\cdot\hat{\mathbf{n}})\,dA \]

In the \(x\) direction:

\[ -F_x=V_{\mathrm{rel},1,x}\rho(\mathbf{V}_{\mathrm{rel},1}\cdot\hat{\mathbf{n}}_1)A_1 +V_{\mathrm{rel},2,x}\rho(\mathbf{V}_{\mathrm{rel},2}\cdot\hat{\mathbf{n}}_2)A_2 \] \[ -F_x=V_{\mathrm{rel},1}\rho(-V_{\mathrm{rel},1})A_1 +V_{\mathrm{rel},2}\cos\alpha\,\rho V_{\mathrm{rel},2}A_2 \] \[ -F_x=-\rho A_1V_{\mathrm{rel},1}^2 +\rho A_2V_{\mathrm{rel},2}^2\cos\alpha \] \[ F_x=\rho A_1V_{\mathrm{rel},1}^2-\rho A_2V_{\mathrm{rel},2}^2\cos\alpha =\rho A_1V_{\mathrm{rel},1}^2(1-\cos\alpha) \] \[ F_x=\rho A_1(V_1-V_0)^2(1-\cos\alpha) \]

In the \(z\) direction:

\[ F_z=V_{\mathrm{rel},1,z}\rho(\mathbf{V}_{\mathrm{rel},1}\cdot\hat{\mathbf{n}}_1)A_1 +V_{\mathrm{rel},2,z}\rho(\mathbf{V}_{\mathrm{rel},2}\cdot\hat{\mathbf{n}}_2)A_2 \]

Since \(V_{\mathrm{rel},1,z}=0\), we have

\[ F_z=V_{\mathrm{rel},2,z}\rho(\mathbf{V}_{\mathrm{rel},2}\cdot\hat{\mathbf{n}}_2)A_2 \] \[ F_z=V_{\mathrm{rel},2}\sin\alpha\,\rho V_{\mathrm{rel},2}A_2 =\rho A_2V_{\mathrm{rel},2}^2\sin\alpha =\rho A_1V_{\mathrm{rel},1}^2\sin\alpha \] \[ F_z=\rho A_1(V_1-V_0)^2\sin\alpha \]

Input:

\(\rho\) of water: \(998.2~\mathrm{kg/m^3}\)
\(A_1: 0.01~\mathrm{m^2}\)
\(F_x\)
--
\(F_z\)
--
\(V_{\mathrm{rel},1}=V_1-V_0\)
--
\(\dot{m}\)
--

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

\(F_x\) and \(F_z\) vs \(V_0\)

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

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

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

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

\(180~\mathrm{N}\le F_{\max}<550\quad [0,600]~\mathrm{N}\)

\(F_{\max}\ge 550\quad [0,1000]~\mathrm{N}\)

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