Question

A pipe of diameter 300 mm and length 3500 m is used for
the transmission of power by water. The total head at the
inlet of the pipe is 500 m. Find the maximum power
available at the outlet of the pipe, if the value of f = 0.006.​

Answer 1

Answer:

Explanation:

d=300mm=0.3m

l=3500m

h=500m

f=0.006

for maximum power transmission

hf=H/3=500/3=166.7m

now ,

hf=4flv^2/d*2g=14.27v^2

v=3.147m/s

q=v*area=0.2145m^3/s

heat available at the end of the pipe

H-hf=333.3m

therefore max power available

=pg*Q*(head available at the end of the pipe)/1000

=1000*9.81*0.2145*333.3/1000(kw)

=789.7kw

hope it helps u buddy

pls mrk me as brainliest : )..

Answer 2

Answer:

maximum power is =789.18