Question

Review: Figure P14.61 shows a valve separating a reservoir from a water tank. If this valve is opened, what is the maximum height above point B attained by the water stream coming out of the right side of the tank? Assume $\mathit{h}\mathbf{=}\mathbf{10}\mathbf{}\mathit{m}\mathbf{,}\mathit{L}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathit{m}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{\theta }\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$and assume the cross-sectional area at A is very large compared with that at B.

Short answer

The maximum height above point B attained by the water stream coming out of the right side of the tank is $y_{max}=2.25m$

Step-by-step solution

Given Data

$$\begin{gathered} h=10m \\ L=2.00m \\ \theta =30.0° \end{gathered}$$

Concept

The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume has the same value at all points along a streamline for an ideal fluid. This result is summarized in Bernoulli’s equation:

$P+\frac{1}{2}\rho v^2+\rho gy=\text{constant}$

Find the velocity

To find maximum height attain by the water first we compute the velocity of the output stream.

By using concept and the formula, we get

$$\begin{gathered} P_A+\rho gh+\frac{1}{2}\rho {v_A}^2=P_B+\rho gL\sin \theta +\frac{1}{2}\rho {v_B}^2=\text{constant} \\ As,P_A=P_B=P_0 \end{gathered}$$

$\rho gh=\rho gL\sin \theta +\frac{1}{2}\rho v_B^2$ $(v_a<<v_B)$

$$\begin{gathered} \frac{V_B^2}{2}=g(h-L\sin \theta ) \\ {V}_B^2=2g(h-L\sin \theta ) \\ {V}_B=\sqrt{2g(h-L\sin \theta }) \\ {V}_B=\sqrt{2\times 9.8(10-2\times \sin 30)} \\ {V}_B=\sqrt{2\times 9.8\left(10-2\times \frac{1}{2}\right)} \\ {V}_B=\sqrt{2\times 9.8(10-1)} \\ {V}_B=\sqrt{2\times 9.8\times 9} \\ {V}_B=13.3m/s \end{gathered}$$

Find the maximum height

To find the maximum height,

$$\begin{gathered} {v_y}^2={v_{iy}}^2-2g{y}_{\max } \\ {y}_{\max }=\frac{{v_{iy}}^2}{2g} \\ {y}_{\max }=\frac{{\left(v_B\sin \theta \right)}^2}{2g} \\ {y}_{\max }=\frac{13.3\times \sin 30}{2\times 9.8} \end{gathered}$$

$$\begin{gathered} y_{max}=\frac{13.3\times {\displaystyle \frac{1}{2}}}{2\times 10}......................(g=10m/s^2) \\ {y}_{max}=2.25m \end{gathered}$$

Hence, the maximum height above point B attained by the water stream coming out of the right side of the tank is $y_{max}=2.25m$