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Chapter 12: Problem 4

Each minute, \(73,800 \mathrm{~m}^{3}\) of water passes over a waterfall \(96.3 \mathrm{~m}\) high. Assuming that \(58.0 \%\) of the kinetic energy gained by the water in falling is converted to electrical energy by a hydroelectric generator, calculate the power output of the generator. (The density of water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).)

Short Answer

Expert verified
The power output of the hydroelectric generator is 679.35 GW.

Step by step solution

01

Calculating the mass of the water

The volume of water passing over the waterfall each minute is \(73,800 \mathrm{~m}^{3}\). Since the density of water is \(1,000 \mathrm{~kg/m}^{3}\), we can use this relationship to calculate the mass of the volume of water. The mass (m) is calculated by the formula \(m = \rho * V\) where \(\rho\) is the density and \(V\) is the volume. Substituting the given values, we get \(m = 1,000 \mathrm{~kg/m}^{3} * 73,800 \mathrm{~m}^{3} = 73,800,000 \mathrm{~kg}\)
02

Calculating the potential energy of the water

The potential energy (U) of an object of mass \(m\) at a height \(h\) is given by the formula \(U = m*g*h\), where \(g\) is the acceleration due to gravity. Here, the height is \(96.3 \mathrm{~m}\) and \(g\) is approximately \(9.8 \mathrm{~m/s}^{2}\). Substituting these values into the formula, we get \(U = 73,800,000 \mathrm{~kg} * 9.8 \mathrm{~m/s}^{2} * 96.3 \mathrm{~m} = 70,277,640,000,000 \mathrm{~J}\)
03

Calculating kinetic energy and electrical energy

Assuming all the potential energy is converted into kinetic energy, we can now find out how much of this is converted into electrical energy. It is mentioned that 58.0% of the kinetic energy is converted into electrical energy. This would be \(0.58 * 70,277,640,000,000 \mathrm{~J} = 40,761,031,200,000 \mathrm{~J}\)
04

Calculating power output

Finally, power output is the conversion of energy per unit time. Here, since we're considering energy conversion per minute, we need to divide the total electrical energy by 60 to convert from joules (J = watt seconds) to watts (power = energy/time). Thus, the power output (P) would be \(P = 40,761,031,200,000 \mathrm{~J} / 60 \mathrm{~s} = 679,350,520,000 \mathrm{~W} or 679.35 \mathrm{~GW}\)

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