Download Science of Energy I by Metin C¸ akanyıldırım 1 Forms of Energy 2

Science of Energy I by Metin Çakanyıldırım
OPRE 6389 Lecture Notes
Compiled at 16:46 on Thursday 26th January, 2017
1 Forms of Energy
The laws of science imply that energy and mass are conserved, i.e., a closed system’s energy or mass cannot
be increased without an energy or a mass transfer from outside the system. The energy in a closed system,
without getting lost, changes from one form to another. Major forms of energy are potential, kinetic, thermal, chemical, nuclear, electric and magnetic. Elastic energy obtained by compressing or extending a string
is a another form but with limited use in energy industry, hence it is not discussed here.
Potential, kinetic and thermal energies are easier to observe and feel for human senses. For example,
potential energy and kinetic energy must have been observed by pre-historic humans who discovered the
wheel and rolled it down from a hill. Similarly, they felt cold winters and hot summers, and accordingly
designed rudimentary clothing. So we can group potential, kinetic and thermal energies and call them
traditional energy forms.
Chemical and nuclear energies are associated with a fuel as they involve some rearrangement of the
fuel’s chemical bonds or nucleus. Discovery of fire by our ancestors is basically turning chemical energy to
thermal energy by using a chemical reaction – technically known as burning or combustion. Although our
ancestors looked at nuclear reactions taking place in the Sun for centuries, they could not understand or
explain the source of sunlight or solar energy. So attempts to control and use nuclear energy started only
in the first half of the 20th century. Even today most of our energy comes from either chemical reactions
(such as burning wood, coal, oil, gas) or nuclear reactions (such as splitting uranium’s nucleus). Because
of their association with a fuel, we call chemical and nuclear energies as fuel-based energy forms.
Perhaps for lacking a visible fuel, understanding magnetic and electrical energies took a while for the
scientists. Our ancestors lived in Earth’s magnetic field, used magnets and compasses without understanding magnetism and the source of magnetic fields. They saw lightning during thunderstorms but could not
define it as a discharge of electrically charged clouds. The theory for electricity had been developed first,
then followed magnetism, and eventually came electromagnetism – the combination of electrical and magnetic theories together. Electromagnetism can be used to explain electromagnetic waves (such as sunlight
or radio waves) or wireless communication (electromagnetic energy transfer between antennas). Electrical
and magnetic energies can be classified as electromagnetic energy forms.
As the energy and mass are always conserved, we may use the term energy transformation instead of
energy generation. For example, gas powered power plant converts chemical energy to electric energy and
the Sun converts nuclear energy to electromagnetic energy.
2 Traditional Energy Forms
2.1 Potential Energy
Gravity of Earth pulls objects towards its surface. Earth exerts a gravitational force on every object proportional to its mass. The proportionality constant for Earth’s gravitational force, measured approximately at
the surface of Earth, is g = 9.8 metres per second2 . The units of metres per second2 measures the change in
the speed (metres per second) of an object in every second. The change in the speed is called acceleration,
Science of Energy I by Metin Çakanyıldırım
which has units of metres per second2 . Denoting the mass by m, the gravitational acceleration constant by
g, the weight (gravitational force) is
Weight = Gravitational Force = Mass × Gravitational acceleration constant = mg.
When the mass is in kilogram and gravitational acceleration is in metres per second2 , their multiplication
gives the gravitational force in Newtons. That is, 1 Newton (after Isaac Newton) is 1 kilogram metre per
second2 . Said differently, 1 Newton can increase the speed of 1 kilogram mass at the rate of 1 metre per
second2 .
For consistency and convenience, we use the Metric system: kilogram for mass measurement instead
of slug (=14.6 kg) of Imperial system and newton for force measurement instead of pound (=4.5 newton).
In the Imperial system, g=32.2 feet per second2 .
Example What is the weight (gravitational force applying to) of a person with 80 kilogram mass? What
is the same weight in pounds?
ANSWER The weight is 784=80(9.8) kilogram metres per second2 or 784 newtons. You can directly compute the equivalent weight as 174.2=784/4.5 pounds. Indirectly, 80 kilograms is 5.48=80/14.6 slugs. The
gravitational force applying on 5.48 slugs is 176.5=5.48(32.2) pounds. The difference between 174.2 and
176.5 pounds is due to rounding errors.
In daily life, 0.453 ≈ 4.5/9.8 is used as a multiplier to convert pounds to kilograms. As in this context
pound can be used as a mass measurement, then it should be called pound-mass and denoted by lbm
(=0.453 kg). Without a qualifier, pound often measures force and is denoted by lb. Sometimes pound measures the mass and other times the force. Similarly, ounce is used as both the measure of mass and volume.
These potentially cause confusion. ⋄
Potential energy of an object is proportional to the gravitational force applying on the object and the
height of the object. More specifically, the potential energy of an object is measured with respect to a
reference point by using the difference between the altitude of the object and the altitude of the reference
point. This difference is referred to as height and denoted by h.
Potential Energy = Gravitational Force × Height = Mass × Gravitational acceleration constant × Height
= mgh.
Continuing to measure the distances in metres, the units for potential energy is kg metre2 per second2 or it
is newton × metre. Since this energy unit is often used, it gets the special name joule (after James P. Joule).
1 Joule is equivalent to applying 1 newton force on an object to move it by 1 metre. 9.8 joule is the energy
required to raise a 1 kilogram mass by 1 metre approximately on the surface of the Earth.
Example What is the potential energy gained by 80 kg person walking stairs up for 10 metres? Would
your answer change if the person takes an elevator ?
ANSWER It is 80(9.8)10 kg metres2 per second2 or simply 78,400 joule. The answer does not change whether
the person walks up the stairs or takes an elevator. The potential energy does not depend the path an object
takes from one point to another but on the distance between points. ⋄
It is surprising that Earth can pull objects without touching them. The fact that the Earth can exert
a gravitational force on an object from a distance has been puzzling. To explain this interaction of apart
objects, a theory of field is developed. Every object has a gravitational field around it. Larger objects have
stronger fields and the field of an object becomes weaker further from the object. The gravitational field
multiplied by the mass yields the gravitational force, so the gravitational force is proportional to the gravitational field. There are other fields and forces associated with these other fields; the force applied by a
field is always proportional to the field.
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Science of Energy I by Metin Çakanyıldırım
2.2 Kinetic Energy
If an object is dropped from a height of h with an initial speed of zero, it gains a speed of g for every
second it falls. Setting t = 0 at the beginning of the drop, the speed at t ≥ 0 is gt. Let T be the time the
object completes its drop, we have the initial and ending speeds as v(t = 0) = 0 and v(t = T ) = gT. The
average speed during the fall is (v(0) + v( T ))/2 = gT/2. This average speed covers h in T seconds, so
h = ( gT/2) T = gT 2 /2. When the object drops from a height of h, its potential energy mgh becomes kinetic
energy captured by its final speed v( T ):
Potential Energy Lost = mgh = mg
gT 2
1
1
= m( gT )2 = m(v( T ))2 = Kinetic Energy Gained.
2
2
2
In general, kinetic energy of an object with mass m and speed v is
Kinetic Energy =
1 2
mv .
2
The units of kinetic energy is kg metre2 per second2 or joule.
Example An accounting book weighs 1 kg and is dropped from 1 metre, what is its kinetic energy at
the bottom of the drop?
ANSWER By using the conservation of energy, we can say the kinetic energy gained
is equal
√
√ to the potential energy lost, which is mgh = 1(9.8)1 = 9.8 joule. You
√ can also use T = 2h/g = 2/9.8 to first
√
√
compute the speed v( T = 2/9.8) = gT = 9.8 2/9.8 = 2(9.8) and then to obtain the kinetic energy as
√
2
(1/2)m(v( T ))2 = (1/2)1 2(9.8) = 9.8 joule. ⋄
Example Consider a horizontal cylinder which is subject to air flow from left to right. There is a propeller at the right-hand side of the cylinder. The air flow passes 75% of its kinetic energy to the propeller to
rotate it. Assuming that the energy conversion is perfect (no energy is lost when air flow’s kinetic energy
is passed to the propeller), how much does the air flow slows down after passing through the propeller?
ANSWER The kinetic energy of each incoming air particle is mv2i /2 at the entry and mv2i /2 − (0.75)mv2i /2 =
m(vi /2)2 /2 = m(vo )2 /2 at the exit for vo = vi /2. Hence, air particles lose half of their speed after transferring their energy to a propeller. ⋄
Example Suppose that the horizontal cylinder mentioned above has an area of A on its right-hand side
where the propeller rotates. Let ρ be the density of the air that reaches the propeller at the speed of v and
leaves it at the speed of v/2. What is the amount of kinetic energy transferred to the propeller over T units
of time?
ANSWER The volume of the air that passes through the propeller over T time is vTA, whose weight is
ρATv. The kinetic energy of the air reaching the propeller is ρATv3 /2. The propeller keeps the 75% of this
kinetic energy, which is 3ρATv3 /8 over T. Note that the amount of energy transferred to the propeller is
proportional to the cube of the speed of the air flow. ⋄
Power is the amount of energy gained or loss per time. Since energy is in joule, power is in joule per
second, which is called watt (after James Watt).
Example An accounting book weighs 1 kg and is raised for 1 metre in 2 seconds by spending the same
amount of power in every second. What is the power spent every second?
ANSWER The book gains 9.8 joule in 2 seconds or 4.9 joule per second. The power spent is 4.9 watt. ⋄
Example Suppose that wind with density ρ hits the propeller of a wind turbine at the speed of v and
leaves it at the speed of v/2. If the propeller rotates in an area of A, how much power does it generate?
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Science of Energy I by Metin Çakanyıldırım
Blade radius: r
Kinetic Energy
݉‫ ݒ‬ଶ Ȁʹ
Area covered by the
blade: ‫ ܣ‬ൌ ߨ‫ ݎ‬ଶ
Displacement of a single air particle in T time: v T.
Volume of air displaced in T time: A v T.
Mass of air displaced in T time: ߩ ‫ܶ ݒ ܣ‬.
Kinetic energy reaching the blades in T time: ߩ ‫ ݒ ܣ‬ଷ ܶ / 2.
Power reaching the blades: ߩ ‫ ݒ ܣ‬ଷ / 2.
Energy density per unit of area in T time: ߩ ‫ ݒ‬ଷ ܶ / 2.
Power density per unit of area: ߩ ‫ ݒ‬ଷ / 2.
Figure 1: Kinetic energy and power reaching a wind turbine.
ANSWER The energy kept by the propeller is 3ρATv3 /8 over T units of time. So the power is of the turbine
is 3ρAv3 /8, which is proportional to the cube of the wind speed. ⋄
Energy density is the amount of energy stored or contained in per unit of volume or area. Energy density
is not as commonly used as power, so it does not get special units. Its units can be joule per metre3 or joule
per m2 .
Example What is the energy density of wind per area over T time units when the wind has density ρ
and speed v?
ANSWER The energy brought by the wind is ρATv3 /2 over T units of time. Energy density per area is
ρTv3 /2. ⋄
Power density is the amount of power obtained per unit of volume or area in every unit of time.
Example What is the power density of wind per area when the wind has density ρ and speed v?
ANSWER Energy density per area is ρTv3 /2, so the power density per area is ρv3 /2. The amount of wind
power that can be obtained per area is proportional to the density of the air and the cube of the speed of
wind. ⋄
2.3 Thermal Energy
At the atomic level, the kinetic energy of an atom is proportional to its temperature measured in kelvin K.
−273 celsius (after Anders Celsius) is 0 kelvin (after William T. Kelvin). 27 celsius (about room temperature)
is 300 kelvin. To compute kelvin temperature, add 273 to celsius temperature.
Kinetic Energy of an Atom/Molecule =
3
BK,
2
where B is Boltzman (after Ludwig Boltzman) constant 1.38 × 10−23 joule per kelvin. Interestingly, the
kinetic energy depends only on the temperature but not on the substance itself, the formula is the same for
different substances (say water molecules and methane gas). A single water molecule at 300 kelvin has little
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Science of Energy I by Metin Çakanyıldırım
kinetic energy, in the order of 10−23 . However, there are many molecules in 1 mole of water. The number
of atoms/molecules in 1 mole of substances is given by Avagadro (after Amedeo Avagadro) number 6.02
× 1023 .
Example What is the kinetic energy of 1 mole of water at 300 kelvin? How much does the kinetic
energy increase by when the water is heated to 350 kelvin?
ANSWER Inserting the constant into the kinetic energy formula,
3
3
1.38 × 10−23 × 300 × 6.02 × 1023 = 1.38 × 300 × 6.02
2
2
= 3738.42 joule
Kinetic Energy of 1 mole of water at 300 kelvin =
3
3
1.38 × 10−23 × 350 × 6.02 × 1023 = 1.38 × 350 × 6.02
2
2
= 4361.49 joule
Kinetic Energy of 1 mole of water at 350 kelvin =
Kinetic energy increases by 623.07 joule (4361.49-3738.42) by heating 1 mole of water from 300 kelvin to 350
kelvin. ⋄
Example The weight of a water molecule is 18 grams. 1 coffee cup takes 180 grams of water. What is
the total kinetic energy of 1 cup of water molecules at 300 kelvin?
ANSWER 180 grams of water has 10 moles of water molecules. Using the previous exercise, 10 moles of
water at 300 kelvin has 37,384.2 joule of kinetic energy. ⋄
The atomic perspective links thermal energy captured by the temperature of a substance to its kinetic
energy. Despite this nice linkage, atomic perspective has little use for industry level operations. At the
industrial level, thermal energy of an object can be computed similar to the potential energy:
Thermal Energy = Mass × Heat capacity constant × Temperature
= mCK.
Similar to the height in potential energy, temperature in thermal energy also indicates difference between
two temperatures measured in celsius or kelvin. The change in temperature is the same regardless of
whether temperature is in celsius or kelvin because kelvin and celsius temperature measures always differ
by constant 273. Similar to the gravitational acceleration constant in potential energy, heat capacity constant
C measures the amount of increase in energy as a function of increase in temperature difference. Unlike
gravitational acceleration constant, heat capacity constant is specific to the substance under consideration.
Sometimes, this constant is called specific heat capacity. When the energy is measured in joule, the mass is
in kilograms, and the temperature is in kelvin, the units of specific heat capacity is joule/(kg × kelvin). The
specific heat capacity of a substance takes the same value regardless of whether temperature is measured
in kelvin or celsius.
Example Specific heat capacity for water is 4200 joule/(kg × kelvin). To increase the temperature of
0.33 kilogram (330 millilitre, can soda size) of water by 50 celsius, how much thermal energy is needed?
Answer the same question for the same amount of sand whose specific heat capacity is 835 joule/(kg ×
kelvin).
ANSWER The energy needed for water is 0.33(4200)50=69,300 joules. The energy needed for sand is
0.33(835)50=13,777.5 joules. The same amount of water can absorb significantly more thermal energy than
sand. ⋄
In the energy industry, thermal energy transfer is important, say from a gas fired combustion chamber
to a turbine. The material used in energy transfer must be able to absorb a large amount of energy per its
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Science of Energy I by Metin Çakanyıldırım
weight. From the example above, water is a better medium for energy transfer than sand. Water has very
high specific heat capacity not only with respect to sand but also 920 joule/(kg × kelvin) of aluminum and
790 joule/(kg × kelvin) of granite. Molten salt (such as Potassium Nitrate KNO3 and Sodim Nitrite NaNO2
mixture) also has relatively high specific heat capacity of 1,560 joule/(kg × kelvin) and so it is a reasonable
alternative for energy transfer and storage. Molten salts are in solid phase at slightly above 100 celsius,
where they do not flow. Inventing molten salts with lower melting temperatures such as 60-70 celsius is an
on-going research topic.
Since the water has C = 4, 200 joule/(kg × celsius), it takes 4.2 joule to increase the temperature of 1
gram of water by 1 celsius. Calorie is another measure of thermal energy and increases the temperature of
1 gram of water by 1 celsius. Then
4.2 joule = 1 calorie, and both increase the temperature of 1 gram of water by 1 celsius.
We can also write C = 1 calorie/(g × celsius) for water.
Example According to http://www.utdallas.edu/studentwellness/salads, a Taco salad has about 860
calories and it can be served with 2 tablespoons of sliced almonds (100 calories) toping and blue cheese
(140 calories) dressing. How many joules of energy do you obtain by eating this salad for lunch? How
many meters of stairs an 80 kg person can climb with this energy?
ANSWER The total calorie intake is 1000 calories, which is 4,200 joule. If an 80 kg person uses 4,200 joules,
s/he can climb 4, 200/(80 × 9.8) = 5.35 metres. At this point, you must be greatly surprised: if you use up
your energy from lunch by just climbing a few meters, say two floors of a building, how would your body
maintain its temperature and engage in other activities until dinner? It appears that there must be something wrong with the calculations. What is unfortunate is the way nutritionists quote the energy content
of food, because they use calories instead of kilocalories by awkwardly dropping kilo. That is, the salad
actually has 860 kilocalories, almonds have 100 kilocalories and blue cheese dressing has 140 kilocalories.
By eating this salad, you obtain 1000 kilocalories, which is 1,000,000 calories. To burn the calories from this
salad, you must climb a mountain of 5,350 meters. ⋄
We end this section by listing some notewhorty numbers/constants and putting them in Table 1.
• Gravitational accelaration constant g=9.8 metres per second2 .
• Newton is a unit of measure for force and it is 1 kilogram metre per second2 .
• Joule is a unit of measure for energy and it is 1 newton metre.
• Celsius is a unit of measure for temperature. x celsius is equivalent to 32 + 1.8x fahrenheit.
• Kelvin is a unit of measure for temperature and x kelvin is x − 273 celsius.
• Avagadro number 6.02 ×16023 is the number of atoms (or molecules) in 1 mole of atoms (molecules).
• Water has specific heat capacity 4,200 joules/(kilogram × celsius)=4.2 joules/(gram × celsius).
• Calorie is a unit of measure for energy and 1 calorie = 4.2 joules: 1 calorie increases the temperature
of 1 gram of water by 1 celsius.
• British Thermal Unit (BTU) is a unit of measure for energy and 1 BTU = 1055 joules = 251 calories. 1
BTU increases the temperature of 1 pound (≈ 452 gram=251*1.8) water by 1 fahrenheit. Specific heat
capacity of water is 1 BTU/(pound × fahrenheit).
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Science of Energy I by Metin Çakanyıldırım
Table 1: Metric vs. Imperial Units.
To measure
Distance
Volume
Time
Mass
Force
Temperature
Energy
Metric units
1 metre
1 cubicmetre
1 second
1 kilogram
1 newton
x celsius
1 joule
equivalent in imperial units
3.28084 feet
35.31467 cubicfeet
1 second
0.06852 slugs
0.22481 pounds
32 + 1.8x fahrenheit
0.00095 British thermal unit
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other
1000 litre = 264 gallon
1/60 minutes
2.20462 pounds
x + 273 kelvins
0.23809 calories