Answer:
Total power generation will be 291.465 MW
Step-by-step explanation:
We have given that river is flowing at rate of V =
![500m^3/sec](https://img.qammunity.org/2020/formulas/physics/college/mvrdlhvic4spcabx6wixh0i7t9zfqczqcd.png)
Average speed v = 3 m/sec
Height h = 58 m
We know that total mechanical energy is given by
E = KE +PE
![E=(1)/(2)mv^2+mgh](https://img.qammunity.org/2020/formulas/physics/middle-school/1rgod8nd4qgjndtjdmvgrzm9lzgkd38v35.png)
Energy per unit mass
![(E)/(m)=(1)/(2)v^2+gh=(1)/(2)* 3^2+9.81* 58=583.29j/kg](https://img.qammunity.org/2020/formulas/physics/college/g0zguij9uhusvpbns0ax5u3a6q8an7xcf3.png)
Total power generation from the river is given by
![P=energy\ per\ unit\ mass* volume\ flow\ rate* density](https://img.qammunity.org/2020/formulas/physics/college/sob1i7w3at717efliedgteeb0j5srfzai3.png)
So
![P=583.29* 500* 1000=291.465MW](https://img.qammunity.org/2020/formulas/physics/college/gz9aqs03k5yik5kbax3aufwz4ghn7nw18t.png)