The velocity of flow through the runner is constant and is equal to 1.8 m/s. The guide blades make an angle of 10^\circ to the tangent of the wheel and the discharge at the outlet of the turbine is radial. Draw the inlet and outlet velocity triangles and determine :-

i. The absolute velocity of water at inlet of runner

ii. The velocity of whirl at inlet

iii. The relative velocity at inlet

iv. The runner blade angle

v. Width of the runner at outlet

vi. Mass of water flowing through the runner per second

Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics

Marks: 08

Years: DEC 2014

Given

External Dia $= D_1 = 0.9m $

Internal Dia $= D_2= 0.45 $

Speed $N=200 rpm$

Width at inlet $ B_1 = 200mm =0.2 m$

Velocity of flow $vf_1 =vf_2 =1.8 m/s $

Guide blade angle $\alpha =10^\circ $

Discharge at outlet = Radial

$\therefore \space\space\space\space B=90^\circ $ and $Vw_2=0$

Tangential velocity of wheel at inlet and outlet are :-

$U_1 =\dfrac {\pi D_1N}{60}=\dfrac {\pi\times 0.9\times 200}{60}=9.424 m/s \\ U_2 =\dfrac {\pi D_2 N}{60} =\dfrac {\pi\times0.45\times 200}{60}= 4.712 m/s $

ii. Absolute velocity of water at inlet of the runner i.e. $V_1 $

From inlet velocity triangle

$V_1\sin \alpha = Vf_1 \\ V_1= \dfrac {Vf_1}{\sin\alpha} = \dfrac {1.8}{\sin 10^\circ}=10.365 m/s $

ii. Velocity of whirl at inlet i.e $Vw_1\\ Vw_1 = V_1\cos \alpha = 10.365 \times \cos 10^\circ \\ =10.207 m/s $

iii. Relative Velocity at inlet i.e $Vr_1$

$Vr_1 =\sqrt{Vr_1^2+ (Vw_1 - U_1)^2}\\ =\sqrt{1.8^2+(10.207-9.424)^2} \\ \sqrt{3.24+0.613}\\ =1.963m/s $

iv. The runner blade angle means the angle $\theta $ and $ \phi$

$\tan \theta = \dfrac {Vf_1}{(Vw_1-U_1)} =\dfrac {1.8}{(10.207-9.424)}=2.298 \\ \theta = \tan^{-1} 2.298 = 66.48^\circ \space or \space 66^\circ 29'$

From outlet velocity triangle we have

$\tan \theta = \dfrac {Vf_2}{U_2} =\dfrac {1.8}{4.712}= \tan 20.9^\circ \\ \phi = 20.9^\circ \space or \space 20^\circ 54.4'$

v. Width of runner at outlet i.e $B_2$

$\pi D_1 B_1 Vf_1 = \pi D_2B_2Vf_2 \space or \space D_1B_1=D_2B_2 \\ (\because \pi Vf_1 =\pi Vf_2 \space as \space Vf_1 =Vf_2) \\ B_2 =\dfrac {D_1B_1}{D_2} =\dfrac {0.90\times 0.20}{0.45} = 0.40 m=400 mm $

vi. Mass of water flowing through the runner per second

$Q=\pi D_1B_1Vf_1= \pi \times 0.8\times 0.20 \times 1.8 \\ =1.0178 m^3/s$