written 8.0 years ago by | • modified 8.0 years ago |
Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics
Marks: 08
Years: MAY 2016
written 8.0 years ago by | • modified 8.0 years ago |
Mumbai University > Electronics and telecommunication > Sem 7 > Applied Hydraulics
Marks: 08
Years: MAY 2016
written 8.0 years ago by |
Head on turbine H1=20m
Speed N1=200r.p.m
Discharge Q1=9m3/s
Overall efficiency No=90% or 0.90
Performance of the turbine under a head H2=20m means to find the speed discharge and power developed by the turbine when working under the head of 20 m
Let for the head H2=20m
Speed =N2
discharge Q2
Power P2
Using the relation
No=PW.P=P1p×g×Q1×H11000P1=No×p×g×Q1×H11000=0.90×1000×9.81×9×201000=1589.22KwN1√H1=N2√H2N2=N1√H2√H1=200×√20√20=200r.p.mAlsoQ1√H1=Q2√H2Q2=Q1×√H2√H1=9×√20√20=9m3/s
And P1H3/21=P2H3/22P2=P1H3/22H3/21=P1(H2H1)3/2=1589.22(2020)3/2=1589.22Kw