Question

In a hydroelectric power plant, \(65 \mathrm{m}^{3} / \mathrm{s}\) of water flows from an elevation of \(90 \mathrm{m}\) to a turbine, where electric power is generated. The overall efficiency of the turbine-generator is 84 percent. Disregarding frictional losses in piping, estimate the electric power output of this plant.

Short answer

Question: Given the flow rate of 65 m^3/s, an elevation difference of 90 m, and a plant efficiency of 84%, estimate the electric power output of a hydroelectric power plant. Answer: The electric power output of the hydroelectric power plant is estimated to be 48.7 MW.

Step-by-step solution

Calculate the potential energy of the water

First, we need to calculate the potential energy of the water at the high elevation. The potential energy is given by the formula: \(PE = mgh\) Where \(PE\) is the potential energy, \(m\) is the mass of water (in kg), \(g\) is the acceleration due to gravity (approximately \(9.81 \,\mathrm{m/s^2}\)), and \(h\) is the elevation (in m). To find the mass of water, we can use the flow rate and the volume of water. The flow rate is given as \(65 \,\mathrm{m^3/s}\). To convert this to mass flow rate, we can use the density of water, which is \(1000 \,\mathrm{kg/m^3}\). The mass flow rate is: \(mass\_flow\_rate = 65 \,\mathrm{m^3/s} \times 1000 \,\mathrm{kg/m^3} = 65000\,\mathrm{kg/s}\) Now, we can calculate the potential energy of the water: \(PE = (65000\,\mathrm{kg/s})(9.81\,\mathrm{m/s^2})(90\,\mathrm{m})\)

Calculate the mechanical power input

The mechanical power input to the system corresponds to the rate at which potential energy is converted. This can be computed as: \(P_{input} = PE \times mass\_flow\_rate\) \(P_{input} = (65000\,\mathrm{kg/s})(9.81\,\mathrm{m/s^2})(90\,\mathrm{m})\)

Calculate the electric power output

Now, we can determine the electric power output by considering the efficiency of the system. The efficiency is given as 84%. Therefore, we can compute the electric power output as: \(P_{output} = P_{input} \times efficiency\) \(P_{output} = (65000\,\mathrm{kg/s})(9.81\,\mathrm{m/s^2})(90\,\mathrm{m}) \times 0.84\)

Compute the final result

Finally, compute the electric power output: \(P_{output} = (65000\,\mathrm{kg/s})(9.81\,\mathrm{m/s^2})(90\,\mathrm{m}) \times 0.84 = 4.87\times10^7\,\mathrm{W}\) The electric power output of the hydroelectric power plant is estimated to be \(4.87\times10^7\,\mathrm{W}\) or \(48.7\,\mathrm{MW}\).