Question

Electric power is to be generated in a hydroelectric power plant that receives water at a rate of \(70 \mathrm{m}^{3} / \mathrm{s}\) from an elevation of \(65 \mathrm{m}\) using a turbine-generator with an efficiency of 85 percent. When frictional losses in piping are disregarded, the electric power output of this plant is \((a) 3.9 \mathrm{MW}\) \((b) 38 \mathrm{MW}\) \((c) 45 \mathrm{MW}\) \((d) 53 \mathrm{MW}\) \((e) 65 \mathrm{MW}\)

Short answer

Based on the given information, we can calculate the electric power output of the plant by first finding the potential energy (PE) of water using the formula \(PE = m \cdot g \cdot h\). After determining the mass flow rate and plugging in all the values, the potential energy is found to be 44,765,500 J/s. Next, we can determine the electric power output by multiplying the potential energy by the efficiency of the turbine-generator, which is 85%. This gives us a power output of 38,050,675 J/s. Lastly, we need to convert the power output from J/s to MW by dividing the value by 1,000,000, resulting in a final power output of 38.05 MW.

Step-by-step solution

1. Calculate potential energy

To find the potential energy (PE) of water, we'll use the formula: \(PE = m \cdot g \cdot h\) where: - \(m\) is the mass of water (in kg), - \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)), - \(h\) is the elevation or height (in meters). But first, we have to find out the mass flow rate (mass per unit time: kg/s). To do this, we can use the formula: \(MassFlowRate = WaterFlowRate \cdot WaterDensity\) Water density is approximately \(1000 \mathrm{kg/m^3}\) at room temperature. Using the given water flow rate, we have: \(MassFlowRate = 70 \mathrm{m^3/s} \cdot 1000 \mathrm{kg/m^3} = 70,000 \mathrm{kg/s}\) Now, let's plug this value into the potential energy formula with given height and gravity acceleration: \(PE = 70,000 \mathrm{kg/s} \cdot 9.81 \mathrm{m/s^2} \cdot 65 \mathrm{m} = 44,765,500 \mathrm{J/s}\)

2. Determine electric power output

The electric power output of the plant can be obtained by multiplying the potential energy by the efficiency of the turbine-generator. Since the efficiency is 85%, we have: \(PowerOutput = PE \cdot Efficiency\) \(PowerOutput = 44,765,500 \mathrm{J/s} \cdot 0.85 = 38,050,675 \mathrm{J/s}\)

3. Convert to megawatts

Lastly, convert the power output from J/s (joules per second) to MW (megawatts) by dividing the value by 1,000,000. \(PowerOutputMW = \dfrac{38,050,675 \mathrm{J/s}}{1,000,000 \mathrm{J/s/MW}} = 38.05 \mathrm{MW}\) So, the correct option is: \((b) 38 \mathrm{MW}\)