Question

A jet of water moving with a velocity of 20 m/s impinges on a curved vane, which is moving with a velocity of 10 m/s. The jet makes an angle of 20° with the direction of motion of vane at inlet and leaves at angle of 130° to the direction of motion of vane at outlet. Determine: i.) The angles of curved vane tips so that water enters and leaves without shock; ii.) The work done per N of water entering the vane.

Answer 1

Explanation:

v1= 20 m/s

u1=10 m/s

angle made by jet at..

A=20°

Answer 2

Answer:

The work done per N of water entering the vane=$62.41 \mathrm{Nm} / \mathrm{N}$

Explanation:

Given

A jet of water moving with a velocity of

Jet velocity $v_2=20 \mathrm{~m} / \mathrm{s}$

impinges on a curved vane which is moving with a velocity of

vane velocity, $4=10 \mathrm{~m} / \mathrm{s}$

The jet makes an angle of

$\alpha=20^{\circ}, \beta=180-130=50^{\circ}$ to the direction of motion of vane at outlet.

Step 1: Vane angle at inlet $\Theta$ and exit , $\phi$

from inlet velocity $\triangle A C D$

$\begin{aligned}& V_w=A c=v_1 \cos \alpha=30.31 \mathrm{~m} / \mathrm{s} \\& v s_1=C D=V_1 \text { sinalpha }\end{aligned}$

$\begin{aligned}& v_{r_1} B D=\sqrt{(B C)^2+(D)^2} \\& \sqrt{(10.31)^2+(17.8)^2} \\& 20.31 \mathrm{~m} / \mathrm{s} \\& \Theta=\tan ^{-1}\left(\frac{C D}{B C}\right) \\& \tan ^{-1}\left(\frac{17.5}{10.31}\right)=59.5\end{aligned}$

The angles of curved vane tips so that water enters and leaves without shock is 50.5

Step 2: Consider outlet velocity $\triangle E G H$ and neglecting blade friction i,e

$V r_2=v r_1=20.31 \mathrm{~m} / \mathrm{s}$

Applying sine rule to $\triangle E F G$

$\begin{aligned}& \frac{V r_2}{\sin (180-\beta)}=\frac{4}{\sin (B-\phi)} \\& \frac{20.31}{\sin (180-60)}=\frac{20}{\sin (60-\Phi)} \\& \sin (60-\phi)=0.8528 \\& (60-\phi)=58.52^{\circ} \\& \phi=1.48^{\circ} \\& \mathrm{FH}==V w_2=\mathrm{EH}-\mathrm{EF}\end{aligned}$

$\begin{aligned}& v r_2 \cos \phi-4 \\& 20.31 \cos 1.48-20 \\& =0.303 \mathrm{~m} / \mathrm{s}\end{aligned}$

Step 3: work done $\mathrm{N}$ weight of water, $\mathrm{W}$

$\begin{aligned}& w=\frac{1}{2}\left(V w_1+V w_2\right) u=\frac{1}{9.81} \times(30.31+0.303) \\& =62.41 \mathrm{Nm} / \mathrm{N} \text { weight of water }\end{aligned}$

Hence work done per N of water entering the vane is $62.41 \mathrm{Nm} / \mathrm{N}$.

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