written 7.1 years ago by
teamques10
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modified 7.1 years ago
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Given
Jet veloicty $v_{2}$=35m/s
vane velocity, 4=20m/s
$\alpha=30^{\circ},\beta=180-120=60^{\circ}$
1] Vane angle at inlet $\Theta$ and exit , $\phi$
from inlet velocity $\triangle ACD$
$V_{w}=Ac=v_{1}cos\alpha$=30.31 m/s
$vs_{1}=CD=V_{1}sin alpha$
35 sin 30=17.50m/s
BC=$Vw_{1}-\mu=30.31$-20=10.31m/s

$v_{r_{1}}BD=\sqrt{(BC)^{2}+(D)^{2}}$
$\sqrt{(10.31)^{2}+(17.8)^{2}}$
20.31 m/s
$\Theta=tan^{-1}(\frac{CD}{BC})$
$\tan^{-1}(\frac{17.5}{10.31})$=59.5
Consider outlet velocity $\triangle EGH$ and neglecting blade friction i,e
$Vr_{2}=vr_{1}$=20.31 m/s
Applying …
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