Material Properties

Fuel Consumption of Conventional Reactor

A typical nuclear power plant has an electric-generating capacity of 1000 MWe. The heat source in the nuclear power plant is a nuclear reactor. As is typical in all conventional thermal power stations the heat is used to generate steam which drives a steam turbine connected to a generator which produces electricity. The turbines are heat engines and are subject to the efficiency limitations imposed by the second law of thermodynamics. In modern nuclear power plants the overall thermodynamic efficiency is about one-third (33%), so 3000 MWth of thermal power from the fission reaction is needed to generate 1000 MWe of electrical power.

This thermal power is generated in a reactor core, which contains especially the nuclear fuel (fuel assemblies), the moderator and the control rods. The core of the reactor contains all the nuclear fuel assemblies and generates most of the heat (fraction of the heat is generated outside the reactor – e.g. gamma rays energy). The assemblies are exactly placed in the reactor according to a fuel loading pattern.

A typical 1000 MWe (3000 MWth) nuclear core may contain 157 fuel assemblies composed of over 45,000 fuel rods and some 15 million fuel pellets. Generally, a common fuel assembly contain energy for approximately 4 years of operation at full power. Once loaded, fuel stays in the core for 4 years depending on the design of the operating cycle. During these 4 years the reactor core have to be refueled. During refueling, every 12 to 18 months, some of the fuel – usually one third or one quarter of the core – is removed to spent fuel pool, while the remainder is rearranged to a location in the core better suited to its remaining level of enrichment. The removed fuel (one third or one quarter of the core, i.e. 40 assemblies) has to be replaced by a fresh fuel assemblies. It follows, there are about 3-4 fuel batches which differ each other in the fuel burnup.

The total energy released in a reactor is about 210 MeV per ²³⁵U fission, distributed as shown in the table. In a reactor, the average recoverable energy per fission is about 200 MeV, being the total energy minus the energy of the energy of antineutrinos that are radiated away. This means that about 3.1⋅10¹⁰ fissions per second are required to produce a thermal power of 1 W. Since 1 gram of any fissile material contains about 2.5 x 10²¹ nuclei, the fissioning of 1 gram of fissile material yields about 1 megawatt-day (MWd) of heat energy.

Summary:

Consumption of a 3000MWth (~1000MWe) reactor (12-months fuel cycle)

It is an illustrative example, following data do not correspond to any reactor design.

• Typical reactor may contain about 165 tonnes of fuel (including structural material)
• Typical reactor may contain about 100 tonnes of enriched uranium (i.e. about 113 tonnes of uranium dioxide).
• This fuel is loaded within, for example, 157 fuel assemblies composed of over 45,000 fuel rods.
• A common fuel assembly contain energy for approximately 4 years of operation at full power.
• Therefore about one quarter of the core is yearly removed to spent fuel pool (i.e. about 40 fuel assemblies), while the remainder is rearranged to a location in the core better suited to its remaining level of enrichment (see Power Distribution).
• The removed fuel (spent nuclear fuel) still contains about 96% of reusable material (it must be removed due to decreasing k_inf of an assembly).

• Annual natural uranium consumption of this reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

• Annual enriched uranium consumption of this reactor is about 25 tonnes of enriched uranium.

• Annual fissile material consumption of this reactor is about 1 005 kg.

• Annual matter consumption of this reactor is about 1.051 kg.

• But it corresponds to about 3 200 000 tons of coal burned in coal-fired power plant per year.

Uranium 235 consumption in a nuclear reactor

A typical thermal reactor contains about 100 tons of uranium with an average enrichment of 2% (do not confuse it with the enrichment of the fresh fuel, that is about 4%). For the reactor of power of 3000MW_th determine the consumption of ²³⁵U that must undergo fission each day to provide this thermal power.

Solution:

This problem can be solved very simply. The average recoverable energy per ²³⁵U fission is about E_r = 200.7 MeV/fission. Since we know that each second we need 3000 MJ of energy, the required reaction rate can be determined directly as:

Since each atom of ²³⁵U has a mass of 235u x 1.66 x 10^-27 kg/u = 3.9 x 10^-25 kg, the daily consumption of a reactor is:

9.33 x 10¹⁹ fissions/sec x 3.9 x 10^-25 kg x 86400 sec/day = 3.14 kg/day

For comparison, a 1000 MWe coal-fired power plant burns about 10 000 tons (about 10 million kg) of coal per day.

Since a typical fuel cycle takes about 320 days (12 month fuel cycle), the annual fuel consumption is about:

3.14 kg/day x 320 days = 1 005 kg of ²³⁵U

Fissile material consumption in a nuclear reactor

All commercial light water reactors contains both fissile and fertile materials. For example, most PWRs use low enriched uranium fuel with enrichment of ²³⁵U up to 5%. Therefore more than 95% of content of fresh fuel is fertile isotope ²³⁸U.

In fact, during fuel burnup the fertile materials (conversion of ²³⁸U to fissile ²³⁹Pu known as fuel breeding) partially replace fissile ²³⁵U, thus in a nuclear reactor more fissile isotopes are involved in power generation. Since the ²³⁹Pu fission releases very similar amount of energy, this example can be generalized to:

Annual consumption is: 1 005 kg of all fissile material involved.

The fuel breeding permits power reactors to operate longer before the amount of fissile material decreases to the point where reactor criticality is no longer manageable.

The fuel breeding in the fuel cycle of all commercial light water reactors plays a significant role. In recent years, the commercial power industry has been emphasizing high-burnup fuels (up to 60 – 70 GWd/tU), which are typically enriched to higher percentages of ²³⁵U (up to 5%). As burnup increases, a higher percentage of the total power produced in a reactor is due to the fuel bred inside the reactor.

At a burnup of 30 GWd/tU (gigawatt-days per metric ton of uranium), about 30% of the total energy released comes from bred plutonium. At 40 GWd/tU, that percentage increases to about forty percent. This corresponds to a breeding ratio for these reactors of about 0.4 to 0.5. That means, about half of the fissile fuel in these reactors is bred there. This effect extends the cycle length for such fuels to sometimes nearly twice what it would be otherwise. MOX fuel has a smaller breeding effect than ²³⁵U fuel and is thus more challenging and slightly less economic to use due to a quicker drop off in reactivity through cycle life.

Natural uranium consumption in a nuclear reactor

Natural uranium refers to uranium with the same isotopic ratio as found in nature. It consists primarily of isotope ²³⁸U (99.28%), therefore the atomic mass of uranium element is close to the atomic mass of ²³⁸U isotope (238.03u).  Natural uranium also consists of two other isotopes: ²³⁵U (0.71%) and ²³⁴U (0.0054%). The abundance of  isotopes in the nature is caused by difference in the half-lifes. All three naturally-occurring isotopes of uranium (²³⁸U, ²³⁵U and ²³⁴U)  are unstable. On the other hand these isotopes (except ²³⁴U) belong to primordial nuclides, because their half-life is comparable to the age of the Earth (~4.5×10⁹ years for ²³⁸U).

Since natural uranium contains only 0.71% of fissile isotope ²³⁵U and most of current power reactors require enriched uranium, this natural uranium must be enriched. The level of enrichment required depends on specific reactor design (e.g.  PWRs and BWRs require 3% – 5% of 235U) and specific requirements of the nuclear power plant operator. Without required enrichment these reactors are not able to initiate and sustain a nuclear chain reaction for such a long period as 12 months (or more).

The enrichment process separates gaseous uranium hexafluoride into two streams, one being enriched to the required level and known as low-enriched uranium, the other stream is progressively depleted in U-235 and is called ‘tails’, or simply depleted uranium. Typically, to produce 1 kg of enriched uranium with 5% of ²³⁵U, about 10 kg of natural uranium is required with a byproduct of about 9 kg of depleted uranium. Therefore annual natural uranium consumption of 3000MWth reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

Matter consumption in a nuclear reactor

In general, the nuclear fission results in the release of enormous quantities of energy. This energy comes from the nuclear binding energy which holds nuclei together. The binding energy per nucleon is higher for fission fragments than for uranium nuclei. It was found the sum of the rest masses of products of fission is measurably smaller than the rest mass of uranium nucleus.

It is due to the fact during the nuclear fission some of the mass of the nucleus gets converted into these huge amounts of energy and thus this mass is removed from the total mass of the original particle, and the mass is missing in the resulting nuclei.

According to the Einstein relationship (E = mc²) this binding energy is proportional to this mass difference and it is known as the mass defect.

The annual mass defect of a typical 3000MW_th reactor  can be calculated directly from the Einstein relationship (E = mc²) as:

Δm = ΔE/c²

Δm = 3000×10⁶ (W = J/s) x 31.5×10⁶ (seconds in year) / (2.9979 x 10⁸)² = 1.051 kg

See also:

Nuclear Fuel

Uranium 235 consumption in a nuclear reactor

A typical thermal reactor contains about 100 tons of uranium with an average enrichment of 2% (do not confuse it with the enrichment of the fresh fuel, that is about 4%). For the reactor of power of 3000MW_th determine the consumption of ²³⁵U that must undergo fission each day to provide this thermal power.

Solution:

This problem can be solved very simply. The average recoverable energy per ²³⁵U fission is about E_r = 200.7 MeV/fission. Since we know that each second we need 3000 MJ of energy, the required reaction rate can be determined directly as:

Since each atom of ²³⁵U has a mass of 235u x 1.66 x 10^-27 kg/u = 3.9 x 10^-25 kg, the daily consumption of a reactor is:

9.33 x 10¹⁹ fissions/sec x 3.9 x 10^-25 kg x 86400 sec/day = 3.14 kg/day

For comparison, a 1000 MWe coal-fired power plant burns about 10 000 tons (about 10 million kg) of coal per day.

Since a typical fuel cycle takes about 320 days (12 month fuel cycle), the annual fuel consumption is about:

3.14 kg/day x 320 days = 1 005 kg of ²³⁵U 

Typically, when uranium 235 nucleus undergoes fission, the nucleus splits into two smaller nuclei (triple fission can also rarely occur), along with a few neutrons (the average is 2.43 neutrons per fission by thermal neutron) and release of energy in the form of heat and gamma rays. The average of the fragment atomic mass is about 118, but very few fragments near that average are found. It is much more probable to break up into unequal fragments, and the most probable fragment masses are around mass 95 (Krypton) and 137 (Barium). Most of the fission fragments are highly unstable(radioactive) nuclei and undergo further radioactive decays to stabilize itself.

Energy Release per Fission

In general, the nuclear fission results in the release of enormous quantities of energy. The amount of energy depends strongly on the nucleus to be fissioned and also depends strongly on the kinetic energy of an incident neutron. In order to calculate the power of a reactor, it is necessary to be able precisely identify the individual components of this energy. At first, it is important to distinguish between the total energy released and the energy that can be recovered in a reactor.

The total energy released in fission can be calculated from binding energies of initial target nucleus to be fissioned and binding energies of fission products. But not all the total energy can be recovered in a reactor. For example, about 10 MeV is released in the form of neutrinos (in fact antineutrinos). Since the neutrinos are weakly interacting (with extremely low cross-section of any interaction), they do not contribute to the energy that can be recovered in a reactor.

See also: Energy Release from Fission

Summary:

Consumption of a 3000MWth (~1000MWe) reactor (12-months fuel cycle)

It is an illustrative example, following data do not correspond to any reactor design.

• Typical reactor may contain about 165 tonnes of fuel (including structural material)
• Typical reactor may contain about 100 tonnes of enriched uranium (i.e. about 113 tonnes of uranium dioxide).
• This fuel is loaded within, for example, 157 fuel assemblies composed of over 45,000 fuel rods.
• A common fuel assembly contain energy for approximately 4 years of operation at full power.
• Therefore about one quarter of the core is yearly removed to spent fuel pool (i.e. about 40 fuel assemblies), while the remainder is rearranged to a location in the core better suited to its remaining level of enrichment (see Power Distribution).
• The removed fuel (spent nuclear fuel) still contains about 96% of reusable material (it must be removed due to decreasing k_inf of an assembly).

• Annual natural uranium consumption of this reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

• Annual enriched uranium consumption of this reactor is about 25 tonnes of enriched uranium.

• Annual fissile material consumption of this reactor is about 1 005 kg.

• Annual matter consumption of this reactor is about 1.051 kg.

• But it corresponds to about 3 200 000 tons of coal burned in coal-fired power plant per year.

See also:

Fuel Consumption

Natural uranium consumption in a nuclear reactor

Natural uranium refers to uranium with the same isotopic ratio as found in nature. It consists primarily of isotope ²³⁸U (99.28%), therefore the atomic mass of uranium element is close to the atomic mass of ²³⁸U isotope (238.03u).  Natural uranium also consists of two other isotopes: ²³⁵U (0.71%) and ²³⁴U (0.0054%). The abundance of  isotopes in the nature is caused by difference in the half-lifes. All three naturally-occurring isotopes of uranium (²³⁸U, ²³⁵U and ²³⁴U)  are unstable. On the other hand these isotopes (except ²³⁴U) belong to primordial nuclides, because their half-life is comparable to the age of the Earth (~4.5×10⁹ years for ²³⁸U).

Since natural uranium contains only 0.71% of fissile isotope ²³⁵U and most of current power reactors require enriched uranium, this natural uranium must be enriched. The level of enrichment required depends on specific reactor design (e.g.  PWRs and BWRs require 3% – 5% of 235U) and specific requirements of the nuclear power plant operator. Without required enrichment these reactors are not able to initiate and sustain a nuclear chain reaction for such a long period as 12 months (or more).

The enrichment process separates gaseous uranium hexafluoride into two streams, one being enriched to the required level and known as low-enriched uranium, the other stream is progressively depleted in U-235 and is called ‘tails’, or simply depleted uranium. Typically, to produce 1 kg of enriched uranium with 5% of ²³⁵U, about 10 kg of natural uranium is required with a byproduct of about 9 kg of depleted uranium. Therefore annual natural uranium consumption of 3000MWth reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

Summary:

Consumption of a 3000MWth (~1000MWe) reactor (12-months fuel cycle)

It is an illustrative example, following data do not correspond to any reactor design.

• Typical reactor may contain about 165 tonnes of fuel (including structural material)
• Typical reactor may contain about 100 tonnes of enriched uranium (i.e. about 113 tonnes of uranium dioxide).
• This fuel is loaded within, for example, 157 fuel assemblies composed of over 45,000 fuel rods.
• A common fuel assembly contain energy for approximately 4 years of operation at full power.
• Therefore about one quarter of the core is yearly removed to spent fuel pool (i.e. about 40 fuel assemblies), while the remainder is rearranged to a location in the core better suited to its remaining level of enrichment (see Power Distribution).
• The removed fuel (spent nuclear fuel) still contains about 96% of reusable material (it must be removed due to decreasing k_inf of an assembly).

• Annual natural uranium consumption of this reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

• Annual enriched uranium consumption of this reactor is about 25 tonnes of enriched uranium.

• Annual fissile material consumption of this reactor is about 1 005 kg.

• Annual matter consumption of this reactor is about 1.051 kg.

• But it corresponds to about 3 200 000 tons of coal burned in coal-fired power plant per year.

See also:

Fuel Consumption

Matter consumption in a nuclear reactor

In general, the nuclear fission results in the release of enormous quantities of energy. This energy comes from the nuclear binding energy which holds nuclei together. The binding energy per nucleon is higher for fission fragments than for uranium nuclei. It was found the sum of the rest masses of products of fission is measurably smaller than the rest mass of uranium nucleus.

It is due to the fact during the nuclear fission some of the mass of the nucleus gets converted into these huge amounts of energy and thus this mass is removed from the total mass of the original particle, and the mass is missing in the resulting nuclei.

According to the Einstein relationship (E = mc²) this binding energy is proportional to this mass difference and it is known as the mass defect.

The annual mass defect of a typical 3000MW_th reactor  can be calculated directly from the Einstein relationship (E = mc²) as:

Δm = ΔE/c²

Δm = 3000×10⁶ (W = J/s) x 31.5×10⁶ (seconds in year) / (2.9979 x 10⁸)² = 1.051 kg

Summary:

Consumption of a 3000MWth (~1000MWe) reactor (12-months fuel cycle)

It is an illustrative example, following data do not correspond to any reactor design.

• Typical reactor may contain about 165 tonnes of fuel (including structural material)
• Typical reactor may contain about 100 tonnes of enriched uranium (i.e. about 113 tonnes of uranium dioxide).
• This fuel is loaded within, for example, 157 fuel assemblies composed of over 45,000 fuel rods.
• A common fuel assembly contain energy for approximately 4 years of operation at full power.
• Therefore about one quarter of the core is yearly removed to spent fuel pool (i.e. about 40 fuel assemblies), while the remainder is rearranged to a location in the core better suited to its remaining level of enrichment (see Power Distribution).
• The removed fuel (spent nuclear fuel) still contains about 96% of reusable material (it must be removed due to decreasing k_inf of an assembly).

• Annual natural uranium consumption of this reactor is about 250 tonnes of natural uranium (to produce of about 25 tonnes of enriched uranium).

• Annual enriched uranium consumption of this reactor is about 25 tonnes of enriched uranium.

• Annual fissile material consumption of this reactor is about 1 005 kg.

• Annual matter consumption of this reactor is about 1.051 kg.

• But it corresponds to about 3 200 000 tons of coal burned in coal-fired power plant per year.

See also:

Fuel Consumption

Law of Conservation of Momentum

In general, the law of conservation of momentum or principle of momentum conservation states that the momentum of an isolated system is a constant. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by Newton’s laws. This principle is a direct consequence of Newton’s third law.

Let us assume the one dimensional elastic collision of two billiard balls,, the ball A and the ball B. We assume the net external force on this system of two balls is zero—that is, the only significant forces during the collision are the forces that each ball exerts on the other. According to Newton’s third law, the two forces are always equal in magnitude and opposite in direction. Hence, the impulses that act on the two balls are equal and opposite, and the changes in momentum of the two balls are equal and opposite.

These two balls are moving with velocities v_A and v_B along the x axis before the collision. After the collision, their velocities are v’_A and v’_B. The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision.

The law of conservation of linear momentum in words:

If no net external force acts on a system of particles, the total linear momentum of the system cannot change.

Conservation of Momentum and Energy in Collisions

The use of the conservation laws for momentum and energy is very important also in particle collisions. This is a very powerful rule because it can allow us to determine the results of a collision without knowing the details of the collision. The law of conservation of momentum states that in the collision of two objects such as billiard balls, the total momentum is conserved. The assumption of conservation of momentum as well as the conservation of kinetic energy makes possible the calculation of the final velocities in two-body collisions. At this point we have to distinguish between two types of collisions:

• Elastic collisions
• Inelastic collisions

Elastic Collisions

A perfectly elastic collision is defined as one in which there is no net conversion of kinetic energy into other forms (such as heat or noise). For the brief moment during which the two objects are in contact, some (or all) of the energy is stored momentarily in the form of elastic potential energy. But if we compare the total kinetic energy just before the collision with the total kinetic energy just after the collision, and they are found to be the same, then we say that the total kinetic energy is conserved.

• Some large-scale interactions like the slingshot type gravitational interactions (also known as a planetary swing-by or a gravity-assist manoeuvre) between satellites and planets are perfectly elastic.
• Collisions between very hard spheres may be nearly elastic, so it is useful to calculate the limiting case of an elastic collision.
• Collisions in ideal gases approach perfectly elastic collisions, as do scattering interactions of sub-atomic particles which are deflected by the electromagnetic force.
• Rutherford scattering is the elastic scattering of charged particles also by the electromagnetic force.
• A neutron-nucleus scattering reaction may be also elastic, but in this case the neutron is deflected by the strong nuclear force.

Inelastic Collisions

An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. For example, in collisions of common bodies, such as two cars, some energy is always transferred from kinetic energy to other forms of energy, such as thermal energy or energy of sound. The inelastic collision of two bodies always involves a loss in the kinetic energy of the system. The greatest loss occurs if the bodies stick together, in which case the collision is called a completely inelastic collision. Thus, the kinetic energy of the system is not conserved, while the total energy is conserved as required by the general principle of conservation of energy. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy.

In nuclear physics, an inelastic collision is one in which the incoming particle causes the nucleus it strikes to become excited or to break up. Deep inelastic scattering is a method of probing the structure of subatomic particles in much the same way as Rutherford probed the inside of the atom (see Rutherford scattering).

In nuclear reactors, inelastic collisions are of importance in neutron moderation process. An inelastic scattering plays an important role in slowing down neutrons especially at high energies and by heavy nuclei. Inelastic scattering occurs above a threshold energy. This threshold energy is higher than the energy the first excited state of target nucleus (due to the laws of conservation) and it is given by following formula:

E_t = ((A+1)/A)* ε₁

where E_t is known as the inelastic threshold energy and ε₁ is the energy of the first excited state. Therefore especially scattering data of ²³⁸U, which is a major component of nuclear fuel in commercial power reactors, are one of the most important data in the neutron transport calculations in the reactor core.

Conservation Laws in Nuclear Reactions

A nuclear reaction is considered to be the process in which two nuclear particles (two nuclei or a nucleus and a nucleon) interact to produce two or more nuclear particles or ˠ-rays (gamma rays). Thus, a nuclear reaction must cause a transformation of at least one nuclide to another. Sometimes if a nucleus interacts with another nucleus or particle without changing the nature of any nuclide, the process is referred to a nuclear scattering, rather than a nuclear reaction.

In analyzing nuclear reactions, we apply the many conservation laws. Nuclear reactions are subject to classical conservation laws for charge, momentum, angular momentum, and energy (including rest energies).  Additional conservation laws, not anticipated by classical physics, are are electric charge, lepton number and baryon number. Certain of these laws are obeyed under all circumstances, others are not. We have accepted conservation of energy and momentum. In all the examples given we assume that the number of protons and the number of neutrons is separately conserved. We shall find circumstances and conditions in which  this rule is not true. Where we are considering non-relativistic nuclear reactions, it is essentially true. However, where we are considering relativistic nuclear energies or those involving the weak interactions, we shall find that these principles must be extended.

Some conservation principles have arisen from theoretical considerations, others are just empirical relationships. Notwithstanding, any reaction not expressly forbidden by the conservation laws will generally occur, if perhaps at a slow rate. This expectation is based on quantum mechanics. Unless the barrier between the initial and final states is infinitely high, there is always a non-zero probability that a system will make the transition between them.

For purposes of this article it is sufficient to note four of the fundamental laws governing these reactions.

1. Conservation of nucleons. The total number of nucleons before and after a reaction are the same.
2. Conservation of charge. The sum of the charges on all the particles before and after a reaction are the same
3. Conservation of momentum. The total momentum of the interacting particles before and after a reaction are the same.
4. Conservation of energy. Energy, including rest mass energy, is conserved in nuclear reactions.

Conservation of Momentum in Nuclear Reactions – Neutron Moderation

See also: Neutron Elastic Scattering

See also: Neutron Inelastic Scattering

See also: Neutron Moderators

It is known the fission neutrons are of importance in any chain-reacting system. All neutrons produced by fission are born as fast neutrons with high kinetic energy. Before such neutrons can efficiently cause additional fissions, they must be slowed down by collisions with nuclei in the moderator of the reactor. The probability of the fission U-235 becomes very large at the thermal energies of slow neutrons. This fact implies increase of multiplication factor of the reactor (i.e. lower fuel enrichment is needed to sustain chain reaction).

The neutrons released during fission with an average energy of 2 MeV in a reactor on average undergo a number of collisions (elastic or inelastic) before they are absorbed. During the scattering reaction, a fraction of the neutron’s kinetic energy is transferred to the nucleus. Using the laws of conservation of momentum and energy and the analogy of collisions of billiard balls for elastic scattering, it is possible to derive the following equation for the mass of target or moderator nucleus (M), energy of incident neutron (E_i) and the energy of scattered neutron (E_s).

where A is the atomic mass number.In case of the hydrogen (A = 1) as the target nucleus, the incident neutron can be completely stopped. But this works when the direction of the neutron is completely reversed (i.e. scattered at 180°). In reality, the direction of scattering ranges from 0 to 180 ° and the energy transferred also ranges from 0% to maximum. Therefore, the average energy of scattered neutron is taken as the average of energies with scattering angle 0 and 180°.

Moreover, it is useful to work with logarithmic quantities and therefore one defines the logarithmic energy decrement per collision (ξ) as a key material constant describing energy transfers during a neutron slowing down. ξ is not dependent on energy, only on A and is defined as follows:For heavy target nuclei, ξ may be approximated by following formula:From these equations it is easy to determine the number of collisions required to slow down a neutron from, for example from 2 MeV to 1 eV.

Example: Determine the number of collisions required for thermalization for the 2 MeV neutron in the carbon.

ξ_CARBON = 0.158

N(2MeV → 1eV) = ln 2⋅10⁶/ξ =14.5/0.158 = 92

For a mixture of isotopes:

Momentum of a Photon

A photon, the quantum of electromagnetic radiation,  is an elementary particle, which is the force carrier of the electromagnetic force. The modern photon concept was developed (1905) by Albert Einstein to explain of the photoelectric effect, in which he proposed the existence of discrete energy packets during the transmission of light.

In 1916, Einstein extended his concept of light quanta (photons) by proposing that a quantum of light has linear momentum. Although a photon is massless, it has momentum, which is related to its energy E, frequency f, and wavelength by:

Thus, when a photon interacts with another object, energy and momentum are transferred, as if there were a collision between the photon and matter in the classical sense.

Momentum of a Photon – Compton Scattering

The Compton formula was published in 1923 in the Physical Review. Compton explained that the X-ray shift is caused by particle-like momentum of photons. Compton scattering formula is the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays. In the case of Compton scattering the photon of frequency f collides with an electron at rest. Upon collision, the photon bounces off electron, giving up some of its initial energy (given by Planck’s formula E=hf), While the electron gains momentum (mass x velocity), the photon cannot lower its velocity. As a result of momentum conservation law, the photon must lower its momentum given by:

So the decrease in photon’s momentum must be translated into decrease in frequency (increase in wavelength Δλ = λ’ – λ). The shift of the wavelength increased with scattering angle according to the Compton formula:

where λ is the initial wavelength of photon λ’ is the wavelength after scattering, h is the Planck constant = 6.626 x 10^-34 J.s, m_e is the electron rest mass (0.511 MeV)c is the speed of light Θ is the scattering angle. The minimum change in wavelength (λ′ − λ) for the photon occurs when Θ = 0° (cos(Θ)=1) and is at least zero. The maximum change in wavelength (λ′ − λ) for the photon occurs when Θ = 180° (cos(Θ)=-1). In this case the photon transfers to the electron as much momentum as possible. The maximum change in wavelength can be derived from Compton formula:

The quantity h/m_ec is known as the Compton wavelength of the electron and is equal to 2.43×10⁻¹² m.

Conservation of Momentum in Fluid Dynamics

In general, the law of conservation of momentum or principle of momentum conservation states that the momentum of an isolated system is a constant. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by Newton’s laws.

In fluid dynamics, the analysis of motion is performed in the same way as in solid mechanics – by use of Newton’s laws of motion. As can be seen moving fluids are exerting forces. The lift force on an aircraft is exerted by the air moving over the wing.

A jet of water from a hose exerts a force on whatever it hits. But here it is not clear what mass of moving fluid we should use and therefore it is necessary to use a different form of the equation.

Newton’s 2nd Law can be written:

As can be seen from the picture the control volume method can be used to analyze the law of conservation of momentum in fluid. Control volume is an imaginary surface enclosing a volume of interest. The control volume can be fixed or moving, and it can be rigid or deformable. In order to determine all forces acting on the surfaces of the control volume we have to solve the conservation laws in this control volume.

The first conservation equation we have to consider in the control volume is the continuity equation (the law of conservation of matter). In the simplest form it is represented by following equation:

∑ṁ_in = ∑ṁ_out

Sum of mass flow rates entering per unit time = Sum of mass flow rates leaving per unit time

The second conservation equation we have to consider in the control volume is the momentum formula.

Momentum Formula

In the simplest form the momentum formula can be represented by following equation:

–

Choosing a Control Volume

A control volume can be selected as any arbitrary volume through which fluid flows. This volume can be static, moving, and even deforming during flow. In order to solve any problem we have to solve basic conservation laws in this volume. It is very important to know all relative flow velocities to the control surface and therefore it is very important to define exactly the boundaries of the control volume during an analysis.

See also:

Laws of Conservation

What is the Law of Conservation of Momentum

In general, the law of conservation of momentum or principle of momentum conservation states that the momentum of an isolated system is a constant. The vector sum of the momenta (momentum is equal to the mass of an object multiplied by its velocity) of all the objects of a system cannot be changed by interactions within the system. In classical mechanics, this law is implied by Newton’s laws. This principle is a direct consequence of Newton’s third law.

Let us assume the one dimensional elastic collision of two billiard balls,, the ball A and the ball B. We assume the net external force on this system of two balls is zero—that is, the only significant forces during the collision are the forces that each ball exerts on the other. According to Newton’s third law, the two forces are always equal in magnitude and opposite in direction. Hence, the impulses that act on the two balls are equal and opposite, and the changes in momentum of the two balls are equal and opposite.

These two balls are moving with velocities v_A and v_B along the x axis before the collision. After the collision, their velocities are v’_A and v’_B. The conservation of the total momentum demands that the total momentum before the collision is the same as the total momentum after the collision.

The law of conservation of linear momentum in words:

If no net external force acts on a system of particles, the total linear momentum of the system cannot change.

See also:

Laws of Conservation

Momentum of Photon

A photon, the quantum of electromagnetic radiation,  is an elementary particle, which is the force carrier of the electromagnetic force. The modern photon concept was developed (1905) by Albert Einstein to explain of the photoelectric effect, in which he proposed the existence of discrete energy packets during the transmission of light.

In 1916, Einstein extended his concept of light quanta (photons) by proposing that a quantum of light has linear momentum. Although a photon is massless, it has momentum, which is related to its energy E, frequency f, and wavelength by:

Thus, when a photon interacts with another object, energy and momentum are transferred, as if there were a collision between the photon and matter in the classical sense.

Momentum of a Photon – Compton Scattering

The Compton formula was published in 1923 in the Physical Review. Compton explained that the X-ray shift is caused by particle-like momentum of photons. Compton scattering formula is the mathematical relationship between the shift in wavelength and the scattering angle of the X-rays. In the case of Compton scattering the photon of frequency f collides with an electron at rest. Upon collision, the photon bounces off electron, giving up some of its initial energy (given by Planck’s formula E=hf), While the electron gains momentum (mass x velocity), the photon cannot lower its velocity. As a result of momentum conservation law, the photon must lower its momentum given by:

So the decrease in photon’s momentum must be translated into decrease in frequency (increase in wavelength Δλ = λ’ – λ). The shift of the wavelength increased with scattering angle according to the Compton formula:

where λ is the initial wavelength of photon λ’ is the wavelength after scattering, h is the Planck constant = 6.626 x 10^-34 J.s, m_e is the electron rest mass (0.511 MeV)c is the speed of light Θ is the scattering angle. The minimum change in wavelength (λ′ − λ) for the photon occurs when Θ = 0° (cos(Θ)=1) and is at least zero. The maximum change in wavelength (λ′ − λ) for the photon occurs when Θ = 180° (cos(Θ)=-1). In this case the photon transfers to the electron as much momentum as possible. The maximum change in wavelength can be derived from Compton formula:

The quantity h/m_ec is known as the Compton wavelength of the electron and is equal to 2.43×10⁻¹² m.

See also:

Photon

Law of Conservation of Angular Momentum

In physics, the linear momentum, p = mv, has a rotational analog. It is called angular momentum  (rarely, moment of momentum or rotational momentum), L, and for an object rotating about a fixed axis with angular velocity ω, it is defined as:

L = Iω

It is an important quantity in physics because it is a conserved quantity:

The angular momentum of an isolated system remains constant in both magnitude and direction.

When there is zero net torque acting on an object, and the object is rotating about a fixed axis or about an axis through its center of mass whose direction doesn’t change, we can write:

I₁ω₁ = I₂ω₂ = constant

Like conservation of energy and of linear momentum, this principle is a universal conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies. The angular momentum is a vector quantity and therefore it can be generalized as:

The vector sum of the angular momenta of the parts of an isolated system is constant.

This puts a strong constraint on the types of rotational motions which can occur in an isolated system. If one part of the system is given an angular momentum in a given direction, then some other part or parts of the system must simultaneously be given exactly the same angular momentum in the opposite direction.

The law of conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs closer to the vertical axis of rotation. By bringing part of the mass of her body closer to the axis she decreases her body’s moment of inertia. Because angular momentum is the product of moment of inertia (I) and angular velocity (ω), and the angular momentum remains constant according to this law, the angular velocity of the skater must increase.

I₁ω₁ = I₂ω₂

Example: Neutron Star Rotation

A neutron star is the collapsed core of a large star (usually of a red giant). Neutron stars are the smallest and densest stars known to exist, but they are rotating extremely rapidly. This rapid rotation is a direct consequence of the law of conservation of angular momentum. As the star’s core collapses, its rotation rate must increase, because of conservation of angular momentum, hence newly formed neutron stars must rotate at up to several hundred times per second. Some neutron stars emit beams of electromagnetic radiation that make them detectable as pulsars.

For example:

Assume a neutron star of a radius of 7 x 10⁵ km, which collapses under its own gravitation to a radius of 10 km. This star is rotating at a frequency of 1.0 revolution every 30 days. Assume that the star is a homogenous sphere at all times, and loses no mass.

From the law of conservation of angular momentum:

I₁ω₁ = I₂ω₂

where the subscripts 1 and 2 refer to initial star and neutron star, respectively. The moment of inertia of a sphere about its central axis is:

I = ⅖ m₁r₁²

therefore

Conservation of Angular Momentum in Fluid Mechanics

Euler’s turbomachine equation, or sometimes called Euler’s pump equation, plays a central role in turbomachinery as it connects the specific work Y and the geometry and velocities in the impeller. The equation is based on the concepts of conservation of angular momentum and conservation of energy.

The Euler’s turbomachine equations are:

Shaft torque:                         T_shaft    =                                     ρQ(r₂V_t2 – r₁V_t1)

Water horsepower:             P_w         =     ω . T_shaft        =      ρQ(u₂V_t2 – u₁V_t1)

Pump head:                            H           =     P_w / ρgQ      =      (u₂V_t2 – u₁V_t1)/g

where

• r₁ and r₂ are the diameters of the impeller at the inlet and outlet respectively.
• u₁ and u₂ are the absolute velocities of the impeller (u₁ = r₁ . ω) at the inlet and outlet respectively.
• V_t1 and V_t2 are the tangential velocities of the flow at the inlet and outlet respectively.

Euler’s turbomachine equations can be used to predict the impact of changing the impeller geometry on the head. It does not matter when we deal with a pump or with a turbine. If torque and angular velocity are of like sign, work is being done on the fluid (a pump or compressor). If torque and angular velocity are of opposite sign work is being extracted from the fluid (a turbine). Thus for the design aspect of turbines and pumps, the Euler equations are extremely useful.

See also:

Laws of Conservation

Whar is Angular Momentum

In physics, the linear momentum, p = mv, has a rotational analog. It is called angular momentum  (rarely, moment of momentum or rotational momentum), L, and for an object rotating about a fixed axis with angular velocity ω, it is defined as:

L = Iω

It is an important quantity in physics because it is a conserved quantity:

The angular momentum of an isolated system remains constant in both magnitude and direction.

When there is zero net torque acting on an object, and the object is rotating about a fixed axis or about an axis through its center of mass whose direction doesn’t change, we can write:

I₁ω₁ = I₂ω₂ = constant

Like conservation of energy and of linear momentum, this principle is a universal conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies. The angular momentum is a vector quantity and therefore it can be generalized as:

The vector sum of the angular momenta of the parts of an isolated system is constant.

This puts a strong constraint on the types of rotational motions which can occur in an isolated system. If one part of the system is given an angular momentum in a given direction, then some other part or parts of the system must simultaneously be given exactly the same angular momentum in the opposite direction.

The law of conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs closer to the vertical axis of rotation. By bringing part of the mass of her body closer to the axis she decreases her body’s moment of inertia. Because angular momentum is the product of moment of inertia (I) and angular velocity (ω), and the angular momentum remains constant according to this law, the angular velocity of the skater must increase.

I₁ω₁ = I₂ω₂

See also:

Angular Momentum

Example: Neutron Star Rotation

A neutron star is the collapsed core of a large star (usually of a red giant). Neutron stars are the smallest and densest stars known to exist, but they are rotating extremely rapidly. This rapid rotation is a direct consequence of the law of conservation of angular momentum. As the star’s core collapses, its rotation rate must increase, because of conservation of angular momentum, hence newly formed neutron stars must rotate at up to several hundred times per second. Some neutron stars emit beams of electromagnetic radiation that make them detectable as pulsars.

For example:

Assume a neutron star of a radius of 7 x 10⁵ km, which collapses under its own gravitation to a radius of 10 km. This star is rotating at a frequency of 1.0 revolution every 30 days. Assume that the star is a homogenous sphere at all times, and loses no mass.

From the law of conservation of angular momentum:

I₁ω₁ = I₂ω₂

where the subscripts 1 and 2 refer to initial star and neutron star, respectively. The moment of inertia of a sphere about its central axis is:

I = ⅖ m₁r₁²

therefore

See also:

Angular Momentum