If its speed is $200 rpm $ head 10 m and discharged at outlet is radial find

a) runner vane angles at inlet and outlet

b) work done/sec/kg of water

c) Hydraulic efficiency

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

Marks: 08

Years: DEC 2015

Given :

Internal diameter $D_1=0.5 m$

External diameter $D_2 =1.0 m$

Guide blade angle $\alpha = 15^\circ$

Velocity of flow $Vf_1=Vf_2=4 m/s$

speed $N= 200 rpm$

Head $H= 10 m$

Discharge at outlet = Radial

$\therefore Vw_2 = O \space \space Vf_2=V_2$

Tangential velocity of runner at inlet and outlet are :-

$U_1 = \dfrac {\pi D_1N}{60} =\dfrac {\pi\times 0.5\times 200}{60}=5.235 m/s \\ U_2 = \dfrac {\pi D_2N}{60} =\dfrac {\pi\times 1.0\times 200}{60}=10.471 m/s $

From the inlet velocity triangle $\tan \alpha = \dfrac {Vf_1}{Vw_1}\\ Vw_1 = \dfrac {Vf_1}{\tan \alpha}= \dfrac 4{\tan 15^\circ} = 14.928 m/s $

i) Runner vane angles at inlet and outlet are $\theta$ and $\phi$

$\tan\theta = \dfrac {Vf_1}{Vw_1-U_1} = \dfrac {4}{14.928-5.235}=0.4126 \\ \theta = \tan^{-1}(0.4126) = 22.42 \space or \space 22^\circ 25'$

From outlet velocity triangle, $\tan \phi = \dfrac {Vf_2}{U_2}\\ = \dfrac 4{10.471} = 0.3820 \\ \phi = \tan^{-1} (0.3820) = 20.90^\circ \space or \space 20^\circ 54'$

ii) Work done/sec/kg of water

$=\dfrac 1gVw_1U_1 \space \space \space [\because Vw_2=0] \\ =\dfrac {1}{9.81}\times 14.928\times 5.235 \\ =7.966 Nm/N$

iii) Hydraulic efficiency

$N_n = \dfrac {Vw_1U_1}{gH} = \dfrac {14.928\times 5.235}{9.81\times 10} =0.796=79.66\%$