Saved Bookmarks
| 1. |
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m^2 s^(-1). If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 m s^(-2)?). |
|
Answer» Solution :Here HEIGHT of water HEAD h = 300 m, VOLUME of water flow `V = 100 m^(3)s^(-1)` `therefore `Mass of water flow m = V`rho = 100 xx 10^(3) = 10^(5) kg m^(-1)` [`therefore` Density of water `rho = 10^(3) kg m^(-3)`] `therefore` Input power = Loss of potential ENERGY PER unit time = mgh `=10^(3) xx 9.8 xx 300 = 2.94 xx 10^(8)` As efficiency of turbine generator `eta = 60% = 60/100 = 0.6` `therefore` Power output `= eta xx ` input power `=0.6 xx 2.94 xx 10^(5) W = 176` MW. |
|