wheel diameter=.0.12

Determine

i) The required discharge

ii) Wheel diameter

iii) Diameter and number of jets

iv) specific speed

$H=145m, \ \ \ N=225rpm \ \ \ \ p_{3}=8500kw$

$nmech=0.75 \ \ \ hh=0.88 \ \ \ Cx=0.98$

Speed ratio kw=$\frac{u}{V1}$=0.48

$\frac{jet \ diameter \,d}{Wheel \ diameter \ D}$=0.12

i) Required discharge Q

Input power=shaft power $\times \frac{1}{h_{h}}\times \frac{1}{n_{mech}}$

$pg QH=P_{s}\times \frac{1}{h_{h}}\times \frac{1}{h_{m}}$

$1000\times 9.81\times Q\times 145=(8500\times 10^{3})\times \frac{1}{0.88}\times \frac{1}{0.75}$

Q=9.054$m^{3}/s$

ii) Wheel diameter D

Velocity of jet V=Cr$\sqrt{2gH}=0.98\sqrt{2\times 9.81\times 145}$=52.27m/s

Ku=0.48=$\frac{wheel \ velocity \ u}{jet \ velocity \ V1}$

u=$0.48\times 52.27=25.09m/s$

$u=\frac{\pi DN}{60}$

25.096=$\frac{\pi\times D\times 225}{60}$

D=2.18m

iii)Jet diameter d and number of jets n

Given $\frac{d}{D}$=0.12;

d=$0.12\times 2.13$

=0.2556m

Discharge jet, q$A\times V_{1}$

=$\frac{\pi}{4}\times d^{2}\times V_{1}$

$\frac{\pi}{4}\times (0.2556)^{2}\times 52.27$

=$2.682m^{3}/s$

Number of jets,

m=$\frac{Total \ discharge \ of \ Q}{Discharge \ let \ q}=\frac{9.054}{2.682}$

=3.376 i.e 4 jets

iv) Sepcific Speed Ns:

Ns$\frac{N\sqrt{P_{3}}}{H^{5/4}}=\frac{225\sqrt{8500}}{(145^{5/4})}$

=41.227