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

Marks: 10M

Year: May 2015

Given:

velocity of jet $v_1$ = 22 m/s

Velocity of vane $u_1$ = 11 m/s

Angle made by the leaving jet, with the direction of motion = $130^0$

$B = 180^0 - 130^0 = 50^0$

In this problem $u_1 = u_2 = 11m/s$

$V_{r1}=V_{r2}$

1. Vane Angle means angle made by the relative velocities at inlet and outlet i.e $\theta$ and $\phi$

   $\triangle ABD$ we have $tan \theta = \frac{BD}{CD}$

   $\frac{V_{f1}}{AD - AC} = \frac{V_{f1}}{V_{w1}-u_1}$

   Where $V_{f1} = V_1 sin x = 22 sin 21^0 = 7.88 m/s$

   $V_{w1}= V_1 cos \alpha = 22 cos 21^0 = 20.53 m/s$

   $u_1 = 11 m/s$

   $tan \theta = \frac{7.88}{20.53 - 11} = 0.8268$

   $\theta = 39.58^0$

   $\theta = 39^035^1$

   From $\triangle ABC$

   $sin \theta = \frac{V_{f1}}{V_{r1}} or V_{r1} = \frac{V_{f1}}{sin \theta}$

   $\hspace{3cm} = \frac{7.88}{sin(39.58^0)} = 12.36$

   $\therefore V_{r2} = V_{r1} = 12.36 m/s$

   From $\triangle ABC$ applying sine rule we have

   $\frac{V_{r2}}{sin(180^0 - \beta)} = \frac{u_2}{sin (\beta - \phi)}$

   $\frac{12.36}{sin \beta} = \frac{11}{sin (\beta - \phi)}$

   $\frac{12.36}{sin 50^0}= \frac{11}{sin(50^0 - \phi)}$

   $sin(50^0 - \phi) = \frac{11 \times sin 50^0}{12.36} = 0.6817$

   $50^0 - \phi = 42.97$

   $\phi = 50^0 - 42.97^0$

   $\phi = 7.03^0$

2. Work done per second per unit weight of the water striking the vane per second is given by equation.

   $\frac{1}{g}[V_{w1}+V_{w2}] \times u Nm/N$ (+ve sign is taken as B is an accute angle)

   $V_{w1} = 20.53 m/s$

   $V_{w2} = GH - GF = V_{r2} cos \phi - u_2 \\ \hspace{2cm} = 12.36 cos 7.03 - 11 \\ \hspace{2cm} = 1.26 m/s$

   $u = u_1 = u_2 = 11m/s$

   Work done per unit weight of water

   $\frac{1}{9.81} [20.53 + 1.26] \times 11 = 24.43 Nm/m$