# Power

(Difference between revisions)
 Revision as of 22:39, 13 July 2010 (view source)← Older edit Revision as of 14:52, 25 July 2010 (view source)Newer edit → Line 3: Line 3:
$E=\frac{1}{2}m{V^2}\qquad \qquad(1)$
$E=\frac{1}{2}m{V^2}\qquad \qquad(1)$
- (2-2) When mass is expressed in kilograms and velocity in meters per second, kinetic energy will be in joules. When mass is expressed in kilograms and velocity in meters per second, kinetic energy will be in joules. - Example 2-1: Comets are time capsules that offer clues about the formation and evolution of our solar system, some 4.5 billion years ago. To better understand how the outer solar system was formed a team of NASA scientists designed a probe called “Deep Impact” which was to smash into a nearby comet (actually 83 million miles away!) at very high speeds and for the first time ever look into the comet. The 820-lb probe collided with the comet at 23,000 mph on July 4, 2005, making a hole the size of a football field. What was the kinetic energy of the probe at the time of impact? + Example: Comets are time capsules that offer clues about the formation and evolution of our solar system, some 4.5 billion years ago. To better understand how the outer solar system was formed a team of NASA scientists designed a probe called “Deep Impact” which was to smash into a nearby comet (actually 83 million miles away!) at very high speeds and for the first time ever look into the comet. The 820-lb probe collided with the comet at 23,000 mph on July 4, 2005, making a hole the size of a football field. What was the kinetic energy of the probe at the time of impact? Solution: Converting the data into SI units, the probe’s mass is $820/2.2 = 372 kg$; this mass collides with the comet at $(23,000x1.609)/3600 = 10.28 km/s = 10,280 m/s$. The kinetic energy of impact is: Solution: Converting the data into SI units, the probe’s mass is $820/2.2 = 372 kg$; this mass collides with the comet at $(23,000x1.609)/3600 = 10.28 km/s = 10,280 m/s$. The kinetic energy of impact is: Line 18: Line 17: ====Potential Energy==== ====Potential Energy==== + As its name implies, potential energy is energy at rest that is waiting, capable of changing in the shape or position of an object in a force field. Energy from a hydroelectric dam, from tides, and from a slingshot are all examples of potential energy. There are two types of potential energy: 1) ''Potential energy of form'' is the result of twisting, compressing, stretching, and bending matter out of its natural shape; 2) ''Potential energy of position'' is the energy of a mass caused by its higher elevation relative to its surroundings. The changes in potential energy due to deformation of bodies are not considered in this book. As its name implies, potential energy is energy at rest that is waiting, capable of changing in the shape or position of an object in a force field. Energy from a hydroelectric dam, from tides, and from a slingshot are all examples of potential energy. There are two types of potential energy: 1) ''Potential energy of form'' is the result of twisting, compressing, stretching, and bending matter out of its natural shape; 2) ''Potential energy of position'' is the energy of a mass caused by its higher elevation relative to its surroundings. The changes in potential energy due to deformation of bodies are not considered in this book. Line 23: Line 23:
${E_{pot}}=mgh\qquad \qquad(2)$
${E_{pot}}=mgh\qquad \qquad(2)$
- (2-3) In this equation, g is a constant signifying the gravitational acceleration between two objects (equal to 9.81 m/s2 on earth), m is mass in kilograms, h is height in meters, and Epot is potential energy in joules. In this equation, g is a constant signifying the gravitational acceleration between two objects (equal to 9.81 m/s2 on earth), m is mass in kilograms, h is height in meters, and Epot is potential energy in joules. Line 29: Line 28: Gravitational energy exists between any two objects separated by a distance. In the following example, the earth constitutes the other object. Gravitational energy exists between any two objects separated by a distance. In the following example, the earth constitutes the other object. - Example 2-2: Calculate the gravitational energy released as a result of the collapse of the 110-story World Trade Center in New York City on September 11, 2001. Each tower had a mass of about 0.5 million tons and a height of 415 m.[[#References|( 1 )]] + Example: Calculate the gravitational energy released as a result of the collapse of the 110-story World Trade Center in New York City on September 11, 2001. Each tower had a mass of about 0.5 million tons and a height of 415 m.[[#References|( 1 )]] + Solution: If it is assumed that the mass of the tower was uniformly distributed between floors, then the average height of the fall was $1/2(415) = 207.5$ m and the potential energy of the combined towers was $E = 2x5.{{10}^8}x9.81x207.5 = 2x{{10}^{12}}$ J. Solution: If it is assumed that the mass of the tower was uniformly distributed between floors, then the average height of the fall was $1/2(415) = 207.5$ m and the potential energy of the combined towers was $E = 2x5.{{10}^8}x9.81x207.5 = 2x{{10}^{12}}$ J. ====Torque==== ====Torque==== + + [[Image:Torque.jpg |thumb|400 px|alt= Torque | Figure 1: Torque]] + Torque is a measure of work (energy) required to rotate an object. Intuitively, torque can be interpreted as “force with a twist,” since it results in rotation of the object upon which the force is applied. To tighten a screw or to pedal a bicycle we must apply torque. Torque can be calculated by multiplying the force and perpendicular distance between the axis of rotation (or pivot) and the “line of action” of the force. This distance is called “lever arm” or the Torque is a measure of work (energy) required to rotate an object. Intuitively, torque can be interpreted as “force with a twist,” since it results in rotation of the object upon which the force is applied. To tighten a screw or to pedal a bicycle we must apply torque. Torque can be calculated by multiplying the force and perpendicular distance between the axis of rotation (or pivot) and the “line of action” of the force. This distance is called “lever arm” or the “arm of the force.” (See Figure 1). “arm of the force.” (See Figure 1).
${\tau}=F.d \qquad \qquad(3)$
${\tau}=F.d \qquad \qquad(3)$
- (2-4) In comparing this with equation 1 from [[Work]] $W=F.d$, we can see that torque has the same unit as work (N•m). In comparing this with equation 1 from [[Work]] $W=F.d$, we can see that torque has the same unit as work (N•m). - - [[Image:energy2_(4).jpg |thumb|400 px|alt= Torque | Figure 1: Torque  ]] - 2-1 [[Image:energy2_(5).jpg |thumb|400 px|alt= Work and Power FYI |  ]] [[Image:energy2_(5).jpg |thumb|400 px|alt= Work and Power FYI |  ]] Line 50: Line 49:
$P=\frac{W}{t}\qquad \qquad(4)$
$P=\frac{W}{t}\qquad \qquad(4)$
- 2-5 We know work is force times distance; therefore, power must be force times distance traveled per second. Since distance per second describes velocity, power equals force times velocity: We know work is force times distance; therefore, power must be force times distance traveled per second. Since distance per second describes velocity, power equals force times velocity:
$P=F.V\qquad \qquad(5)$
$P=F.V\qquad \qquad(5)$
- 2-6 Line 63: Line 60:
$P= \tau . \omega$
$P= \tau . \omega$
- 2-7 - In this equation $\omega =2 \pi N$ is the angular velocity and N is the shaft speed in revolutions per second. In this equation $\omega =2 \pi N$ is the angular velocity and N is the shaft speed in revolutions per second. - From the definition, we see that the SI unit of power is J/s, commonly called a watt (W) in honor of James Watt, the inventor of the steam engine. Because a watt is a relatively small unit of power, a kilowatt $(1 kW = 1,000 W)$ is often used instead. In the US, power is customarily expressed in horsepower, which is equivalent to 746 watts $(1 h.p. = 746 W)$. Table 2-1 compares the power output of an average man and horse with that of several machines. + From the definition, we see that the SI unit of power is J/s, commonly called a watt (W) in honor of James Watt, the inventor of the steam engine. Because a watt is a relatively small unit of power, a kilowatt $(1 kW = 1,000 W)$ is often used instead. In the US, power is customarily expressed in horsepower, which is equivalent to 746 watts $(1 h.p. = 746 W)$. + Table 2-1 compares the power output of an average man and horse with that of several machines. [[Image:energy2_(6).jpg |thumb|400 px|alt= Power Outputs of a Few Basic Machines | Table 2-1: Power Outputs of a Few Basic Machines  ]] [[Image:energy2_(6).jpg |thumb|400 px|alt= Power Outputs of a Few Basic Machines | Table 2-1: Power Outputs of a Few Basic Machines  ]] - [[Image:energy2_(7).jpg |thumb|400 px|alt= Watt and Horsepower FYI | Watt and HorsePower FYI ]] + [[Image:energy2_(7).jpg |thumb|400 px|alt= Watt and Horsepower FYI | Watt and HorsePower FYI]] - Example 2-3: How much energy does it take an 80 kg man to run up a flight of stairs 6-m high? What is the power expenditure if it takes 10 seconds to climb the stairs? Assume $g = 10 m/s2$. + Example: How much energy does it take an 80 kg man to run up a flight of stairs 6-m high? What is the power expenditure if it takes 10 seconds to climb the stairs? Assume $g = 10 m/s2$. Solution: The energy consumed is $E = mgh = 80x10x6 = 4800 J$. The power expenditure is $P = E/t = 4,800/10 = 480 W (0.63 h.p.)$. Obviously, it takes the same person a bit longer than 20 seconds (maybe 25 s) to climb two flights of stairs. In this case, although the amount of energy spent is exactly twice as much (9600 J), the power reduces to $9,600/25 = 384 W (0.5 h.p.)$. Although humans can demonstrate impressive muscle power for a short time, even the best of athletes cannot sustain a power output of more than one-tenth of a horsepower. Solution: The energy consumed is $E = mgh = 80x10x6 = 4800 J$. The power expenditure is $P = E/t = 4,800/10 = 480 W (0.63 h.p.)$. Obviously, it takes the same person a bit longer than 20 seconds (maybe 25 s) to climb two flights of stairs. In this case, although the amount of energy spent is exactly twice as much (9600 J), the power reduces to $9,600/25 = 384 W (0.5 h.p.)$. Although humans can demonstrate impressive muscle power for a short time, even the best of athletes cannot sustain a power output of more than one-tenth of a horsepower. - Example 2-4: To accelerate an automobile from 0 to 100 km/hr in 10 seconds, how much total energy must be provided by the fuel? What is the power rating? Would it make a difference if the car were accelerated to 100 km/hr in 20 seconds instead? Assume that the automobile has a mass of 1200 kg. + Example: To accelerate an automobile from 0 to 100 km/hr in 10 seconds, how much total energy must be provided by the fuel? What is the power rating? Would it make a difference if the car were accelerated to 100 km/hr in 20 seconds instead? Assume that the automobile has a mass of 1200 kg. + Solution: The final velocity of the car is $v = 100,000/3600 = 27.78 m/s$. The kinetic energy is then equal to $0.5x1200x27.782 = 463,963 J$. Whether this acceleration is carried out in 10 s or 20 s is irrelevant. The power rating, however, depends on the rate at which energy is consumed. If the car accelerates in 10s, power will be $P = E/t = (463,963 J)/(10 s) = 46,396 J/s$ (46.4 kW or 62.2 h.p.). If the car’s acceleration takes place in 20 s, the power requirement would be reduced by half (23.2 kW or 31.1 h.p.). Solution: The final velocity of the car is $v = 100,000/3600 = 27.78 m/s$. The kinetic energy is then equal to $0.5x1200x27.782 = 463,963 J$. Whether this acceleration is carried out in 10 s or 20 s is irrelevant. The power rating, however, depends on the rate at which energy is consumed. If the car accelerates in 10s, power will be $P = E/t = (463,963 J)/(10 s) = 46,396 J/s$ (46.4 kW or 62.2 h.p.). If the car’s acceleration takes place in 20 s, the power requirement would be reduced by half (23.2 kW or 31.1 h.p.). - Example 2-5: It is expected that the world population will reach 10 billion by the year 2050. If, on the average, each person requires approximately 2 kW of electric power for his needs, how much total electric power must be produced in 2050? If 40% of all the electric power is produced by burning oil (as is done today), how much oil do we need annually to meet the demand? Each barrel of oil has an energy equivalent of 1,700 kWh. + Example: It is expected that the world population will reach 10 billion by the year 2050. If, on the average, each person requires approximately 2 kW of electric power for his needs, how much total electric power must be produced in 2050? If 40% of all the electric power is produced by burning oil (as is done today), how much oil do we need annually to meet the demand? Each barrel of oil has an energy equivalent of 1,700 kWh. + Solution: The total electric output must exceed $(10x109 persons) x (2x103 W) = 2x1013 W = 20,000 GW$ of electricity. As we will see later, power plants produce electricity with efficiencies of around 33%; we need to use three units of petroleum energy to produce one unit of electricity. If 40% of the electric power is produced by oil, we will need $(0.4)(60,000 GW) x (365x24 h) = 2.1x108 GWh = 2.1x1014 kWh$ of thermal energy. The amount of oil needed would be $2.1x1014/1,700 = 1.24x1011$ barrels $= 124$ billion barrels of oil (bbo). The current world total petroleum reserves are estimated at 3021 bbo (See Table 6-5). Solution: The total electric output must exceed $(10x109 persons) x (2x103 W) = 2x1013 W = 20,000 GW$ of electricity. As we will see later, power plants produce electricity with efficiencies of around 33%; we need to use three units of petroleum energy to produce one unit of electricity. If 40% of the electric power is produced by oil, we will need $(0.4)(60,000 GW) x (365x24 h) = 2.1x108 GWh = 2.1x1014 kWh$ of thermal energy. The amount of oil needed would be $2.1x1014/1,700 = 1.24x1011$ barrels $= 124$ billion barrels of oil (bbo). The current world total petroleum reserves are estimated at 3021 bbo (See Table 6-5). - Example 2-6: An electric motor runs at 1500 rpm and delivers 5.0 hp. How much torque does it deliver? + Example: An electric motor runs at 1500 rpm and delivers 5.0 hp. How much torque does it deliver? + Solution: The angular velocity of an electric motor and the power delivered are: Solution: The angular velocity of an electric motor and the power delivered are: $w = 2pN= (2p rad/rev)(1500 rev/min)(1 min/60s) = 157$ rad/s $w = 2pN= (2p rad/rev)(1500 rev/min)(1 min/60s) = 157$ rad/s Line 95: Line 94: Question: A cart filled with bricks is to be loaded onto a truck. Three ramps of different lengths are available (See figure). Question: A cart filled with bricks is to be loaded onto a truck. Three ramps of different lengths are available (See figure). + a. Which situation requires the least amount of force? a. Which situation requires the least amount of force? + b. Which situation requires the least amount of work? b. Which situation requires the least amount of work? + c. Which situation requires the least amount of power? c. Which situation requires the least amount of power? + Answer: Situation (a) requires the least amount of force, but the force must be applied over the longest distance. The product of force times distance, or work, is the same in all cases. The power expenditure depends on how long it takes to perform this work. If you pull faster, the time is shorter and more power is needed. Assuming speeds are the same in all cases, it takes longest for the cart to move up the longest ramp (a), and the least power is consumed. Can you verify these conclusions from equations (1) from [[Work]] through equation (2) of this article? Answer: Situation (a) requires the least amount of force, but the force must be applied over the longest distance. The product of force times distance, or work, is the same in all cases. The power expenditure depends on how long it takes to perform this work. If you pull faster, the time is shorter and more power is needed. Assuming speeds are the same in all cases, it takes longest for the cart to move up the longest ramp (a), and the least power is consumed. Can you verify these conclusions from equations (1) from [[Work]] through equation (2) of this article? Line 104: Line 107: Question: Two identical cars are leaving Los Angeles at the same time. The first car arrives in San Francisco in 5 hours, while it takes 10 hours for the second car to reach the same destination. Which car consumes more energy? Which car puts out more power? Question: Two identical cars are leaving Los Angeles at the same time. The first car arrives in San Francisco in 5 hours, while it takes 10 hours for the second car to reach the same destination. Which car consumes more energy? Which car puts out more power? Answer: Assuming that the cars’ efficiencies are equal at all speeds [in reality, cars are designed to give a better gas mileage at about 80-100 km/h (50-60 mph)], both cars use the same amount of fuel (energy), i.e. work the same amount. The first car, however, uses up fuel twice as fast, therefore putting out twice the power of the second car. Answer: Assuming that the cars’ efficiencies are equal at all speeds [in reality, cars are designed to give a better gas mileage at about 80-100 km/h (50-60 mph)], both cars use the same amount of fuel (energy), i.e. work the same amount. The first car, however, uses up fuel twice as fast, therefore putting out twice the power of the second car. + Question: Suppose a small passenger sedan with a power rating of 80 hp can accelerate from 0-60 mph in 12 seconds. How long would it take a sports car with a power rating of 240 hp to accelerate by the same amount? Which car performs more work? Question: Suppose a small passenger sedan with a power rating of 80 hp can accelerate from 0-60 mph in 12 seconds. How long would it take a sports car with a power rating of 240 hp to accelerate by the same amount? Which car performs more work? + Answer: Both cars accelerate to the same speed (and gain the same amount of kinetic energy), therefore they perform the same amount of work. The second car uses three times the power—which means it does the same work in one third of the time, or 4 seconds. Answer: Both cars accelerate to the same speed (and gain the same amount of kinetic energy), therefore they perform the same amount of work. The second car uses three times the power—which means it does the same work in one third of the time, or 4 seconds. ==References== ==References== + (1) See FEMA Evaluation Report at http://www.fema.gov/library/wtcstudy. (1) See FEMA Evaluation Report at http://www.fema.gov/library/wtcstudy. + + (2) Toossi Reza, "Energy and the Environment:Sources, technologies, and impacts", Verve Publishers, 2005 ==Further Reading== ==Further Reading== ==External Links== ==External Links==

## Contents

#### Kinetic Energy

Kinetic energy is the energy of motion. Since any displacement can be viewed as a combination of a linear motion and a rotation, there are two forms of kinetic energy-- linear (the energy of motion from one location to another) and rotational (the energy due to rotational motion about the center of mass). The energy from wind and rotating flywheels are two examples of kinetic energy. In this book we concern ourselves mainly with the kinetic energy of a body in linear motion, which for an object of mass (m) and velocity (v) is:

$E=\frac{1}{2}m{V^2}\qquad \qquad(1)$

When mass is expressed in kilograms and velocity in meters per second, kinetic energy will be in joules.

Example: Comets are time capsules that offer clues about the formation and evolution of our solar system, some 4.5 billion years ago. To better understand how the outer solar system was formed a team of NASA scientists designed a probe called “Deep Impact” which was to smash into a nearby comet (actually 83 million miles away!) at very high speeds and for the first time ever look into the comet. The 820-lb probe collided with the comet at 23,000 mph on July 4, 2005, making a hole the size of a football field. What was the kinetic energy of the probe at the time of impact?

Solution: Converting the data into SI units, the probe’s mass is 820 / 2.2 = 372kg; this mass collides with the comet at (23,000x1.609) / 3600 = 10.28km / s = 10,280m / s. The kinetic energy of impact is:

$E=\frac{1}{2}m{V^2}=\frac{(372).{{(10,280)}^2}}{2}=1.97x{{10}^{10}}J=19.7GJ$

Which is equivalent to 1.9x1010 / 4.2x106 = 4,680kg = 4.68 tons of TNT.

#### Potential Energy

As its name implies, potential energy is energy at rest that is waiting, capable of changing in the shape or position of an object in a force field. Energy from a hydroelectric dam, from tides, and from a slingshot are all examples of potential energy. There are two types of potential energy: 1) Potential energy of form is the result of twisting, compressing, stretching, and bending matter out of its natural shape; 2) Potential energy of position is the energy of a mass caused by its higher elevation relative to its surroundings. The changes in potential energy due to deformation of bodies are not considered in this book.

Potential energy of position (also called gravitational energy) can be calculated as the work needed to move an object from one location to a new location at a higher elevation. It is the product of weight (mg) and the height (h) to which the object was raised:

${E_{pot}}=mgh\qquad \qquad(2)$

In this equation, g is a constant signifying the gravitational acceleration between two objects (equal to 9.81 m/s2 on earth), m is mass in kilograms, h is height in meters, and Epot is potential energy in joules.

Gravitational energy exists between any two objects separated by a distance. In the following example, the earth constitutes the other object.

Example: Calculate the gravitational energy released as a result of the collapse of the 110-story World Trade Center in New York City on September 11, 2001. Each tower had a mass of about 0.5 million tons and a height of 415 m.( 1 )

Solution: If it is assumed that the mass of the tower was uniformly distributed between floors, then the average height of the fall was 1 / 2(415) = 207.5 m and the potential energy of the combined towers was E = 2x5.108x9.81x207.5 = 2x1012 J.

#### Torque

Figure 1: Torque

Torque is a measure of work (energy) required to rotate an object. Intuitively, torque can be interpreted as “force with a twist,” since it results in rotation of the object upon which the force is applied. To tighten a screw or to pedal a bicycle we must apply torque. Torque can be calculated by multiplying the force and perpendicular distance between the axis of rotation (or pivot) and the “line of action” of the force. This distance is called “lever arm” or the “arm of the force.” (See Figure 1).

${\tau}=F.d \qquad \qquad(3)$

In comparing this with equation 1 from Work W = F.d, we can see that torque has the same unit as work (N•m).

#### Power

A rock climber and a trail hiker of similar mass moving to the top of a cliff use the same amount of work. They each have to lift their own weight the entire height of the cliff. The hiker, however, can reach the top of the cliff much faster than the rock climber. The difference is the rate at which the rock climber and hiker are doing the work, or their rate of energy consumption or power. Power is defined as:

$P=\frac{W}{t}\qquad \qquad(4)$

We know work is force times distance; therefore, power must be force times distance traveled per second. Since distance per second describes velocity, power equals force times velocity:

$P=F.V\qquad \qquad(5)$

The rock climber performs the same task at a slower rate, using less power. For the same reason, you cannot carry a heavy load as quickly as a light load because your power output is too low.

Just as we showed that power for motion along a straight line is equal to force times velocity, we can show that rotational power is calculated as torque times angular velocity:

P = τ.ω

In this equation ω = 2πN is the angular velocity and N is the shaft speed in revolutions per second.

From the definition, we see that the SI unit of power is J/s, commonly called a watt (W) in honor of James Watt, the inventor of the steam engine. Because a watt is a relatively small unit of power, a kilowatt (1kW = 1,000W) is often used instead. In the US, power is customarily expressed in horsepower, which is equivalent to 746 watts (1h.p. = 746W).

Table 2-1 compares the power output of an average man and horse with that of several machines.

File:Energy2 (6).jpg
Table 2-1: Power Outputs of a Few Basic Machines

File:Energy2 (7).jpg
Watt and HorsePower FYI

Example: How much energy does it take an 80 kg man to run up a flight of stairs 6-m high? What is the power expenditure if it takes 10 seconds to climb the stairs? Assume g = 10m / s2. Solution: The energy consumed is E = mgh = 80x10x6 = 4800J. The power expenditure is P = E / t = 4,800 / 10 = 480W(0.63h.p.). Obviously, it takes the same person a bit longer than 20 seconds (maybe 25 s) to climb two flights of stairs. In this case, although the amount of energy spent is exactly twice as much (9600 J), the power reduces to 9,600 / 25 = 384W(0.5h.p.). Although humans can demonstrate impressive muscle power for a short time, even the best of athletes cannot sustain a power output of more than one-tenth of a horsepower.

Example: To accelerate an automobile from 0 to 100 km/hr in 10 seconds, how much total energy must be provided by the fuel? What is the power rating? Would it make a difference if the car were accelerated to 100 km/hr in 20 seconds instead? Assume that the automobile has a mass of 1200 kg.

Solution: The final velocity of the car is v = 100,000 / 3600 = 27.78m / s. The kinetic energy is then equal to 0.5x1200x27.782 = 463,963J. Whether this acceleration is carried out in 10 s or 20 s is irrelevant. The power rating, however, depends on the rate at which energy is consumed. If the car accelerates in 10s, power will be P = E / t = (463,963J) / (10s) = 46,396J / s (46.4 kW or 62.2 h.p.). If the car’s acceleration takes place in 20 s, the power requirement would be reduced by half (23.2 kW or 31.1 h.p.).

Example: It is expected that the world population will reach 10 billion by the year 2050. If, on the average, each person requires approximately 2 kW of electric power for his needs, how much total electric power must be produced in 2050? If 40% of all the electric power is produced by burning oil (as is done today), how much oil do we need annually to meet the demand? Each barrel of oil has an energy equivalent of 1,700 kWh.

Solution: The total electric output must exceed (10x109persons)x(2x103W) = 2x1013W = 20,000GW of electricity. As we will see later, power plants produce electricity with efficiencies of around 33%; we need to use three units of petroleum energy to produce one unit of electricity. If 40% of the electric power is produced by oil, we will need (0.4)(60,000GW)x(365x24h) = 2.1x108GWh = 2.1x1014kWh of thermal energy. The amount of oil needed would be 2.1x1014 / 1,700 = 1.24x1011 barrels = 124 billion barrels of oil (bbo). The current world total petroleum reserves are estimated at 3021 bbo (See Table 6-5).

Example: An electric motor runs at 1500 rpm and delivers 5.0 hp. How much torque does it deliver?

Solution: The angular velocity of an electric motor and the power delivered are: w = 2pN = (2prad / rev)(1500rev / min)(1min / 60s) = 157 rad/s P = (5.0hp)(746W / hp) = 3730W Torque is calculated using equation 2-7, t = P / w = 3730W / 157rad / s = 23.76N.m

File:Energy2 (8).jpg
Power of Words FYI

Question: A cart filled with bricks is to be loaded onto a truck. Three ramps of different lengths are available (See figure).

a. Which situation requires the least amount of force?

b. Which situation requires the least amount of work?

c. Which situation requires the least amount of power?

Answer: Situation (a) requires the least amount of force, but the force must be applied over the longest distance. The product of force times distance, or work, is the same in all cases. The power expenditure depends on how long it takes to perform this work. If you pull faster, the time is shorter and more power is needed. Assuming speeds are the same in all cases, it takes longest for the cart to move up the longest ramp (a), and the least power is consumed. Can you verify these conclusions from equations (1) from Work through equation (2) of this article?

File:Energy2 (9).jpg
Cart and Ramp

Question: Two identical cars are leaving Los Angeles at the same time. The first car arrives in San Francisco in 5 hours, while it takes 10 hours for the second car to reach the same destination. Which car consumes more energy? Which car puts out more power? Answer: Assuming that the cars’ efficiencies are equal at all speeds [in reality, cars are designed to give a better gas mileage at about 80-100 km/h (50-60 mph)], both cars use the same amount of fuel (energy), i.e. work the same amount. The first car, however, uses up fuel twice as fast, therefore putting out twice the power of the second car.

Question: Suppose a small passenger sedan with a power rating of 80 hp can accelerate from 0-60 mph in 12 seconds. How long would it take a sports car with a power rating of 240 hp to accelerate by the same amount? Which car performs more work?

Answer: Both cars accelerate to the same speed (and gain the same amount of kinetic energy), therefore they perform the same amount of work. The second car uses three times the power—which means it does the same work in one third of the time, or 4 seconds.

## References

(1) See FEMA Evaluation Report at http://www.fema.gov/library/wtcstudy.

(2) Toossi Reza, "Energy and the Environment:Sources, technologies, and impacts", Verve Publishers, 2005