The Bay of Fundy is known for the largest tidal range in the world. A water basi
ID: 1334631 • Letter: T
Question
The Bay of Fundy is known for the largest tidal range in the world. A water basin within the bay, Minas Basin, has a location on it's shores that has a tidal range of up to 16.3 meters (53.5 ft) from low to high tide. We will model the situation as a conversion from potential energy to kinetic. The average rate of kinetic energy out of the bay on a 12.42 hour cycle is 1.9 GW. Calculate the amount of mass that would have to be held at 8.15 meters (half of Hi.3 meters) above the ground to provide enough potential energy to supply this kinetic energy over a 12.42 hour cycle where the basin drains. FYI - This would also be the energy entering the basin system over the next 12.42 hours also. At 12.5 kWh of energy usage per day per household, how many homes can the basin power? The Grand Coulee on the Columbia River in Washington is the largest hydroelectric power plant in the US providing 6480 MW of electricity. The available electrical power output of a dam can be expressed as: where rho = 1000kg/m3 forwater, g = 9.8N/kg, Z = 160mtotal dam height, v_f and v_i are the inlet and outlet velocities which are negligable here, Q is the volumetric flow rate in kg/m3 through the dam. and e is the overall efficiency of conversion to electricity which can be estimated at 0.8. For large impoundment dams like the Grand Coulee, the kinetic term can be neglected as the potential term is much larger. Calculate the volumetric flow rate of the dam. At 12.5 kWh of energy usage per day per household, how many homes can the dam power?Explanation / Answer
The energy produced in 12.42 hour is by rate 1.9GW
so we can say in 12.42 hour the total energy produced is
Energy = 1.9*109*12.42*3600 = 84952.8*109 J
Potential energy = mgh
m = 84952.8*109 / (9.81*8.15) = 1062.554*109 kg
Energy required per day by one household
12.5*103*3600 = 45000 kJ
Energy produced in one day by tide
= 24*1.9*109 /12.42 = 3.67*109 W
Energy in kJ = 3.67*109*24*3600 = 317217.39*109 J
No. of house = total energy /energy consumption by one house
= 317217.39*109 / (45000*103) = 7.05*106
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