A Kaplan Turbine Runner Has a Diameter of 5 M$80 \%$

Question

Examlex · Accepted Answer

The Answer is (a) The area of the runner is

( 3.14 ) imes ( 2.5 \mathrm {~m} ) ^ { 2 } = 19.6 \mathrm {~m} ^ { 2 }

The energy associated with the water flow is

\left( 1 \mathrm { MW } _ { \mathrm { e } } ight) / ( 0.8 ) = 1.25 \mathrm { MW }

Then from equation (11.7) we find

P = \frac { 1 } { 2 } A ho v ^ { 3 }

Solving for velocity gives

v = \frac { 2 P } { ho A } = \frac { 2 imes \left( 1.26 imes 10 ^ { 6 } \mathrm {~J} / \mathrm { s } ight) } { \left( 1000 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes \left( 19.6 \mathrm {~m} ^ { 2 } ight) } = 5.03 \mathrm {~m} / \mathrm { s }

(b) For an area of

19.6 \mathrm {~m} ^ { 2 }

the flow rate will be

\left( 19.6 \mathrm {~m} ^ { 2 } ight) imes ( 5.03 \mathrm {~m} / \mathrm { s } ) imes \left( 1000 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) = 9.85 imes 10 ^ { 4 } \mathrm {~kg} / \mathrm { s }

A Kaplan Turbine Runner Has a Diameter of 5 M $80 \%$