and equal to 2.15 m/sec .The guide blades make an angle of $10^{\circ}$ to the tangent of wheel and discharge at the outlet of the turbine is radical

Determine

i) The runner blade angles of inlet and outlet

ii) Head at inlet of the turbine

iii) Power developed

iv) Hydraulic efficiency

v) Draw the inlet and outlet velocity triangles

Inlet diameter or external diameter

$D_{1}$=1.1m

$D_{2}$=0.55m

N=210 rpm

$B_{1}=25 cm=0.25 m$

$vE_{1}=vf_{2}$=2.15 m

Guide blade angle d=$10^{\circ}$

Discharge at outlet is radical i.e $\beta=90^{\circ}$

$vw_{2}$=0

i) Runner blade angle at inlet $\Theta$ and at outlet $\Theta$:-

u=$\frac{\pi D_{1},N}{60}=\frac{\pi \times 11\times 210}{60}=12.1 m/s$

$u_{2}=\frac{\pi D_{2}N}{60}=\frac{\pi\times 0.55\times 210}{60}$

Q=$\pi D_{1}B_{1}V_{1}=\pi\times 1.1\times 0.25\times 2.15=1.8575m^{3}/s$

mass flow rate

m=pQ=1000$\times 1.8375$=18575 kg/s

from inlet velocity $\triangle$ACD we get

$Vm_{1}=AD=\frac{VA}{tan\alpha}=\frac{2+5}{tan10}$=12.193 m/s

$V_{1}=\frac{Vw_{1}}{cos\alpha}=\frac{12.193}{cos10}$=12.381 m/s

BD-$Vm_{1}-U_{1}$=12.193-12.1=0.093 m/s