Power and Energy Calculations and Unit Conversions The Bay of Fundy is known for

Question

Power and Energy Calculations and Unit Conversions 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 power as water enters and leaves the bay is 1.9 GW. Calculate the amount of mass that would have to be held at 8.15 meters (half of 16.3 meters) above the ground to provide enough potential energy to supply this power over a 12 hour cycle where the basin drains. FYI - This would also be the energy entering the basin system over the next 12 hours as it refills -the average power available over a day is 1.9 GW. 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: Power = (rho g Z + (1/2) rho(v^2_f - v_i^2))(Q)(c) where rho = 1000kg/m^3 f or water, g = 0.8N/kg, Z = 160m total dam height, v_f and v_i are the inlet and outlet velocities which are negligible here, Q is the volumetric flow rate in m^3/s through the dam, and is the overall efficiency of conversion to electricity which can be estimated at 0.8.

Explanation / Answer

(a) Given , Height = 8.15m

Power = 1.9 GW = 1.9 * 109 W { 1GW = 109 W}

   Time = 12 hour = 12 * 60 * 60 sec = 43200 s

We know , Potential energy , P.E. = mgh

where m = mass

g = acceleration due to gravity = 9.8 m / s2

h = height

Also , Power , P = W / t

where W = work done

t = time

Work done can be calculated as difference between initial and final potential energies

(P.E.)initial = m*g*8.15 = (79.87*m) J {initial height = 8.15m} ; {J=joule (unit of energy)}

(P.E.)final = m*g*0 = 0 J {final height = 0 , ground level}

Hence , W = ( 79.87*m - 0 ) = 79.87*m J

Now , P = W / t

   1.9 * 109 = ( 79.87 * m ) / 43200

   m = 1.9 * 109 * 43200 / 79.87

   m = 1027.67 * 109 kg

  m = 102767 * 107 kg

Hence , required mass is 102767 * 107 Kg

(b) This can be solved by unitary method

Average energy usage per day of one house = 12.5 KWh = 12.5 KWh / 24h =0.52 KW

Average power of dam in one day = 1.9 GW = 1.9 * 109

0.52 KW of power = 1 house

1KW of power = 1 / 0.52 houses

1.9 * 109 KW of power = (1 * 1.9 * 109) / 0.52 = 3.654 * 109 = 3654 * 106 houses

Therefore , Dam can supply power to 3654000000 houses

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