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Chapter 6: Problem 61

Water flows from a reservoir at the rate of \(3000 \mathrm{~kg} / \mathrm{min}\), to a turbine \(120 \mathrm{~m}\) below. If the efficiency of the turbine is 80 percent, compute the power output of the turbine. Neglect friction in the pipe and the small KE of the water leaving the turbine. Don't forget that it's only 80 percent efficient.

Short Answer

Expert verified
The turbine outputs 47060.8 Watts.

Step by step solution

01

Find the Potential Energy Flow per Minute

Calculate the potential energy that flows from the reservoir to the turbine per minute using the formula: \(PE = m \cdot g \cdot h\), where \(m\) is the mass flow rate (in kg/min), \(g\) is the acceleration due to gravity (approximately \(9.81 \ m/s^2\)), and \(h\) is the height difference (120 m). So, \(PE = 3000 \times 9.81 \times 120\).
02

Calculate the Total Potential Energy per Second

To find the potential energy per second (i.e., power), divide the total potential energy flow per minute by 60, since there are 60 seconds in a minute. Compute \(\frac{PE}{60}\) to convert it to Joules per second (or Watts).
03

Apply the Turbine Efficiency

The power output of the turbine is not 100% efficient. Given the efficiency is 80%, calculate the actual power output by multiplying the potential power with the efficiency (as a decimal). Compute: \(Power_{output} = Power_{potential} \times 0.80\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion Efficiency
Efficiency is a way to measure how well a machine or process converts energy input into useful energy output. In this case, we're dealing with a turbine that converts potential energy from water into electrical energy. The efficiency of this conversion tells us how much of the input energy becomes useful output.
  • Efficiency is expressed as a percentage or a decimal.
  • The formula is: Efficiency = (Useful Energy Output / Total Energy Input) 脳 100.
If a turbine has an 80% efficiency, it means that 80% of the energy input is converted into power output. The rest is lost, typically as heat, sound, or other forms of non-electrical power. This concept is crucial in calculating the power output of a system, as real-life systems are never perfectly efficient.
Potential Energy
Potential energy is the stored energy of an object due to its position. Here, the water at the top of a reservoir has potential energy because it is elevated. The potential energy can be calculated using the formula:
  • \( PE = m \cdot g \cdot h \)
where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \, m/s^2\)), and \(h\) is the height.
The higher and heavier the object, the more potential energy it possesses. In the case of the turbine exercise, we calculate the potential energy using the mass flow rate of the water, the height of the fall, and the constant gravity to find out how much energy is stored due to its elevated position.
Mass Flow Rate
The mass flow rate is a measure of the amount of mass passing through a given point or system per unit time. It's crucial in hydraulic calculations because it tells us how much water is being used to generate energy.
  • In our exercise, the mass flow rate of the water is given as \(3000 \, kg/min\).
  • This measurement helps us understand and calculate the flow of potential energy through the system.
By knowing the mass flow rate, we can determine how much potential energy is available each minute, which is vital for calculating the total energy that the turbine can convert into power. Understanding the mass flow rate ensures that we accurately calculate how much energy is input into the system at any given time.
Turbine Efficiency
Turbine efficiency is a measure of how well a turbine converts potential energy into mechanical or electrical energy. It鈥檚 a critical factor in determining the actual power output a turbine can deliver.
  • Efficiency helps us understand the real-world constraints of energy conversion.
  • A turbine efficiency of 80% means only 80% of the potential energy gets converted into useful power.
To apply this, we take the total potential power, calculated based on the mass flow rate and the height of the waterfall, and multiply it by the efficiency percentage (expressed as a decimal: 0.80). This gives us the actual power output. Knowing the efficiency is essential for realistic power assessments, ensuring the system鈥檚 output expectations are aligned with its operational capacity.

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Most popular questions from this chapter

A narrowing ladder \(3.0 \mathrm{~m}\) long weighing \(200 \mathrm{~N}\) has its center of gravity \(120 \mathrm{~cm}\) from the bottom. At its top end is a \(50-\mathrm{N}\) can of paint. Compute the work required to raise the ladder from being horizontal, lying on the ground, to being vertical with its legs resting on the ground. In other words, how much work must be done to rotate the ladder into an upright vertical position, thereby raising both its center of gravity and the can of paint? The work done (against gravity) consists of two parts: the work to raise the center of gravity \(1.20 \mathrm{~m}\) and the work to raise the load at the end through \(3.0 \mathrm{~m}\). Therefore, Work done \(=(200 \mathrm{~N})(1.20 \mathrm{~m})+(50 \mathrm{~N})(3.0 \mathrm{~m})=0.39 \mathrm{~kJ}\)

How much work is done against gravity in lifting a \(3.0\) -kg object through a vertical distance of \(40 \mathrm{~cm}\) ? An external force is needed to lift an object. If the object is raised at constant speed, the lifting force must equal the weight of the object. The work done by the lifting force is referred to as work done against gravity. Because the lifting force is \(m g\), where \(m\) is the mass of the object, Work \(=(m g)(h)(\cos \theta)=(3.0 \mathrm{~kg} \times 9.81 \mathrm{~N})(0.40 \mathrm{~m})(1)=12 \mathrm{~J}\) In general, the work done against gravity in lifting an object of mass \(m\) through a vertical distance \(h\) is \(m g h\).

A proton \(\left(m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) that has a speed of \(5.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\) passes through a metal film of thickness \(0.010 \mathrm{~mm}\) and emerges with a speed of \(2.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). How large an average force opposed its motion through the film?

Compute the power output of a machine that lifts a 500 -kg crate through a height of \(20.0 \mathrm{~m}\) in a time of \(60.0 \mathrm{~s}\).

An advertisement claims that a certain 1200 -kg car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in a time of \(8.0 \mathrm{~s}\). What average power must the motor develop to produce this acceleration? Give your answer in both watts and horsepower. Ignore friction losses. The work done in accelerating the car is $$ \text { Work done }=\text { Change in } \mathrm{KE}=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{1}{2} m v_{f}^{2} $$ The time taken for this work to be performed is \(8.0 \mathrm{~s}\). Therefore, to two significant figures, $$ \text { Power }=\frac{\text { Work }}{\text { Time }}=\frac{\frac{1}{2}(1200 \mathrm{~kg})(25 \mathrm{~m} / \mathrm{s})^{2}}{8.0 \mathrm{~s}}=46875 \mathrm{~W}=47 \mathrm{~kW} $$ Converting from watts to horsepower, we have $$ \text { Power }=(46875 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=63 \mathrm{hp} $$
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