Ch 10: Interactions and Potential Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 10: Interactions and Potential Energy

Problem 14b

Chapter 10, Problem 14b

In a hydroelectric dam, water falls 25 m and then spins a turbine to generate electricity. Suppose the dam is 80% efficient at converting the water's potential energy to electrical energy. How many kilograms of water must pass through the turbines each second to generate 50 MW of electricity? This is a typical value for a small hydroelectric dam.

Verified step by step guidance

Step 1: Start by identifying the relevant physical principles. The problem involves gravitational potential energy and energy conversion efficiency. The gravitational potential energy of water is given by the formula:

E = m g h

, where

m

is the mass of water,

g

is the acceleration due to gravity (9.8 m/s²), and

h

is the height (25 m in this case).

Step 2: Calculate the power output in terms of energy per second. The dam generates 50 MW of electrical power, but it is only 80% efficient. This means the actual energy input required is higher. Use the efficiency formula:

Efficiency = \frac{Useful}{Total}

. Rearrange to find the total energy input:

Total = \frac{Useful}{Efficiency}

Step 3: Relate the total energy input to the gravitational potential energy of the water. Since the energy input comes from the falling water, equate the total energy input to the gravitational potential energy per second:

Power = m g h

. Rearrange this equation to solve for the mass flow rate

m

(mass of water per second):

m = \frac{Power}{g}

Step 4: Substitute the known values into the equation. Use

Power = \frac{50}{0.8}

MW (accounting for efficiency),

g = 9.8

m/s², and

h = 25

m. Ensure the power is converted to watts (1 MW = 10⁶ W) before substituting.

Step 5: Perform the calculation to find the mass flow rate

m

. This will give the amount of water (in kilograms) that must pass through the turbines each second to generate the required electrical power.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Potential Energy

Potential energy is the energy stored in an object due to its position in a gravitational field. In the context of a hydroelectric dam, the potential energy of water is determined by its height above the turbine. The formula for gravitational potential energy is PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height. This energy is converted into kinetic energy as the water falls, which is then used to spin the turbine.

Efficiency

Efficiency in a system refers to the ratio of useful output energy to the input energy, often expressed as a percentage. In this case, the dam's efficiency of 80% means that 80% of the potential energy of the falling water is converted into electrical energy, while the remaining 20% is lost, typically as heat or sound. Understanding efficiency is crucial for calculating how much energy is actually available for conversion into electricity.

Power and Energy Conversion

Power is the rate at which energy is transferred or converted, measured in watts (W), where 1 watt equals 1 joule per second. In this scenario, the dam needs to generate 50 MW (megawatts) of electrical power, which is equivalent to 50 million watts. To find out how much water is needed, one must relate the power output to the energy derived from the falling water, taking into account the efficiency of the system.

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Textbook Question

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