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Draw the triangles of velocities at inlet and outlet

A jet of water having a velocity of 35 m/s impinges on a series of vanes moving with a velocity of 20 m/s. The jet makes an angle of $30^0$ to the direction of motion of vanes when entering and leaves at an angle of $120^0$. Draw the triangles of velocities at inlet and outlet and find

i) The angles of vanes tips so that water enters and leaves without shock.

ii) The work done per unit weight of water entering the vanes and

iii) The efficiency.

Mumbai University > Civil Engineering > Sem 5 > Applied Hydraulics 1

Marks: 10M

Year: Dec 2014, May 2016

1 Answer
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Given:

velocity of jet $v_1$ = 35 m/s

Velocity of vane $u_1 = u_2$ = 20 m/s

Angle of jet at inlet$ \alpha = 30^0$

Angle made by the jet at outlet with the direction of motion of vanes = $120^0$

$\text{Angle B} = 180^0 - 120^0 = 60^0$

  1. Angle of vanes tips

    From inlet velocity triangle

    $V_{w1}= V_1 cos \alpha = 35 cos 30^0 = 30.31 m/s$

    $V_{f1}= V_1 sin \alpha = 35 sin 30^0 = 17.50 m/s$

    $u_1 = 11 m/s$

    $tan \theta = \frac{V_{f1}}{V_{w1}- u_1} = \frac{17.50}{30.31 - 20} = 1.697$

    $\therefore \theta = tan^{-1} (1.697) = 60^0$

    By sine rule,

    $\frac{V_{r1}}{sin 90^0} = \frac{V_{f1}}{sin \theta} or \frac{V_{r1}}{1} = \frac{17.50}{sin 60^0}$

    $V_{r1} = \frac{17.50}{0.866} = 20.25 m/s$

    $\text{Now} V_{r1} = V_{f1} = 20.25m/s$

    From outlet velocity triangle by sine rule

    $\frac{V_{r2}}{sin 120^0} = \frac{u_2}{sin (60^0 - \phi)} or \frac{20.25}{0.866} = \frac{20}{sin (60^0 - \phi)}$

    $sin(60^0 - \phi) = \frac{20 \times 0.866}{20.25} = 0.855$

    $(60^0 - \phi) = sin (0.855)$

    $60^0 - \phi = 58.75^0$

    $\phi = 60^0 - 58.75 = 1.25^0$

  2. Work done per unit weight of water entering $\frac{1}{g}(V_{w1} + V_{w2}) \times u_1$

    $V_{w2} = V_{r2} cos \phi - u_2 = 20.25 cos 1.25^0 - 20.0 = 0.24 m/s$

    $\therefore \text{work done/unit weight} = \frac{1}{9.81}[30.31 + 0.24] \times 20 = 62.28 Nm/N$

  3. $\text{Efficiency} = \frac{\text{work done per kg}}{\text{Energy supplied per kg}} = \frac{62.28}{\frac{V_1^2}{2g}}$

    $\frac{62.28 \times 2 \times 9.81}{35 \times 35} = 99.74 %$

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