Question 5.7: A plate with frictionless surfaces is oriented inclined to a...
A plate with frictionless surfaces is oriented inclined to an incipient water jet, as shown in Fig. 5.9. Velocity is uniform over each flow section. The fluid flow may be approximated as incompressible and inviscid. The angle made by the plate with the vertical is \theta. Express Q_{1} and Q_{2} as functions of Q_{0} and \theta. Assume density of the water is constant. Neglect the change in height between the various points.
Choose a fixed CV as shown by the dotted line in Fig. 5.9(a).
From the conservation of mass for the control volume, we get
0=0+\rho\left(-Q_{0}\right)+\rho Q_{1}+\rho Q_{2}Or Q_{1}+Q_{2}=Q_{0} (5.18)
Or A_{1} V_{1}+A_{2} V_{2}=A_{0} V_{0}
where V_{0}, V_{1} and V_{2} are the uniform velocities at section 0, 1 and 2, respectively, and A_{0}, A_{1} and A_{2} are the corresponding cross-sectional areas.
0, A1 and A2 are the corresponding cross-sectional areas. From the conservation of linear momentum of the control volume following Eq. (5.17), one can write
\sum \vec{F}_{C V}=\frac{\partial}{\partial t} \int_{C V} \rho \vec{V} d \forall r+\int_{C S} \rho \vec{V}\left(\vec{V}_{r} . \hat{n}\right) d A (5.17)
F_{S} \hat{\varepsilon}_{S}+F_{n} \hat{\varepsilon}_{n}=0+\rho\left(V_{0} \sin \theta \hat{\varepsilon}_{s}-V_{0} \cos \theta \hat{\varepsilon}_{n}\right)\left(-Q_{0}\right)+\rho V_{1} \hat{\varepsilon}_{S} Q_{1}+\left(-\rho V_{2} \hat{\varepsilon}_{S}\right)\left(Q_{2}\right) (5.19)
Where \hat{\varepsilon}_{S} and \hat{ E }_{n} are unit vectors tangential and normal to the inclined plates, respectively. Further, F_{s} and F_{n} are the respective force components exerted by the plate on the water. Equating the force components along the surface of the plate we get
F_{s}=-\rho V_{0} \sin \theta+\rho V_{1} Q_{1}-\rho V_{2} Q_{2} (5.20)
Since the plate is frictionless, the net tangential force acting on the plate is zero. Therefore, Eq. (5.20) becomes
0=-\rho V_{0} \sin \theta+\rho V_{1} Q_{1}-\rho V_{2} Q_{2}Or Q_{2} V_{2}-Q_{1} V_{1}=-V_{0} Q_{0} \sin \theta (5.21)
Since the flow is steady, inviscid, and of constant density, Bernoulli’s equation may be applied along a streamline. Accordingly, applying Bernoulli’s equation along a streamline connecting the points and b, we get
\frac{p_{a}}{\rho}+\frac{V_{a}^{2}}{2}+g z_{a}=\frac{p_{b}}{\rho}+\frac{V_{b}^{2}}{2}+g z_{b}Since p_{a}=p_{b} and z_{a}=z_{b}, V_{a}=V_{b}, i.e, V_{0}=V_{1}
Thus, from Eqs (5.18) and (5.21), we get
Q_{1}=\frac{Q_{0}}{2}(1+\sin \theta) and \frac{Q_{0}}{2}(1-\sin \theta)
Note: From Eq. (5.19), equating the force components normal to the plate, we get
F_{n}=\rho V_{0} \cos \theta Q_{0}=\rho A_{0} V_{0}^{2} \cos \theta (5.22)
This is the force exerted by the plate on the water present in the control volume. Therefore, the force exerted by the water on the plate is -\rho A_{0} V_{0}^{2} \cos \theta
