Thermal Engineering

Polyurethane Foam

Polyurethane foam (PUR) is a closed cell thermoset polymer. Polyurethane polymers are traditionally and most commonly formed by reacting a di- or poly-isocyanate with a polyol. Polyurethane foam insulation is available in closed-cell and open-cell formulas. Polyurethane foam can be used as cavity wall insulation or as roof insulation, floor insulation, pipe insulation, insulation of industrial installations. Insulating panels made from PUR can be applied to all elements of the building envelope. Another important aspect is that PUR can also be injected into existing cavity walls, by using the existing openings and some extra holes.

Thermal Conductivity of Polyurethane Foam

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. Typical thermal conductivity values for polyurethane foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. polyurethane foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Polyurethane Foam Insulation

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of bricks with the thermal conductivity of k₁ = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use polyurethane foam insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m²K

The heat flux can be then calculated simply as:

q = 3.53 [W/m²K] x 30 [K] = 105.9 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 105.9 [W/m²] x 30 [m²] = 3177W

2. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m²K

The heat flux can be then calculated simply as:

q = 0.259 [W/m²K] x 30 [K] = 7.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 7.78 [W/m²] x 30 [m²] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

See also:

Insulation Materials

Thermal Conductivity of Polyurethane Foam

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. Typical thermal conductivity values for polyurethane foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. polyurethane foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

See also:

Insulation Materials

Example – Polyurethane Foam Insulation

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of bricks with the thermal conductivity of k₁ = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use polyurethane foam insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.028 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m²K

The heat flux can be then calculated simply as:

q = 3.53 [W/m²K] x 30 [K] = 105.9 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 105.9 [W/m²] x 30 [m²] = 3177W

2. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.028 + 1/30) = 0.259 W/m²K

The heat flux can be then calculated simply as:

q = 0.259 [W/m²K] x 30 [K] = 7.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 7.78 [W/m²] x 30 [m²] = 233 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

See also:

Insulation Materials

Polyisocyanurate Foam

Polyisocyanurate foam (PIR), also referred to as PIR, polyiso, or ISO, it is very similar as polyurethane foam. It is also a closed cell thermoset polymer formed by reacting a di- or poly-isocyanate with a polyol. The thermal conductivity can be lower than of polyurethane foam. Polyisocyanurate foam is typically used for metal faced panels, roof boards, cavity wall boards and pipe insulation. PIR foam panels laminated with pure embossed aluminium foil are used for fabrication of pre-insulated duct that is used for heating, ventilation and air conditioning systems. On the other hand, PIR cannot be used to insulate existing cavity walls, since there is no PIR foam that can be injected into existing cavity walls. It offers even better thermal stability and flammability resistance than polyurethane foam.

Thermal Conductivity of Polyisocyanurate Foam

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. Typical thermal conductivity values for polyisocyanurate foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. polyisocyanurate foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Polyisocyanurate Foam Insulation

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of bricks with the thermal conductivity of k₁ = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use polyisocyanurate foam insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.022 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m²K

The heat flux can be then calculated simply as:

q = 3.53 [W/m²K] x 30 [K] = 105.9 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 105.9 [W/m²] x 30 [m²] = 3177W

2. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.022 + 1/30) = 0.207 W/m²K

The heat flux can be then calculated simply as:

q = 0.207 [W/m²K] x 30 [K] = 6.21 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.21 [W/m²] x 30 [m²] = 186 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

See also:

Insulation Materials

Thermal Conductivity of Polyisocyanurate Foam

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. Typical thermal conductivity values for polyisocyanurate foams are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. polyisocyanurate foam) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

See also:

Insulation Materials

Example – Polyisocyanurate Foam Insulation

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of bricks with the thermal conductivity of k₁ = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use polyisocyanurate foam insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.022 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m²K

The heat flux can be then calculated simply as:

q = 3.53 [W/m²K] x 30 [K] = 105.9 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 105.9 [W/m²] x 30 [m²] = 3177W

2. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.022 + 1/30) = 0.207 W/m²K

The heat flux can be then calculated simply as:

q = 0.207 [W/m²K] x 30 [K] = 6.21 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 6.21 [W/m²] x 30 [m²] = 186 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

See also:

Insulation Materials

Cellulose Insulation

Cellulose insulation is made from recycled paper products, primarily newspapers and has a very high recycled material content. The obtained cellulose fibres have a wool like structure (therefore paper wool). In order to make the cellulose fibres moisture and flame retardant, boric acid or ammonium sulfate are added. Cellulose insulation is used in wall and roof cavities to insulate, draught proof and reduce free noise. Cellulose insulation is used in both new and existing homes, usually as loose-fill in open attic installations and dense packed in building cavities. Cellulose and the other loose-fill materials can be blown into attics, finished wall cavities, and hard-to-reach areas.

Thermal Conductivity of Cellulose Insulation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer, and hence the greater the insulation’s effectiveness. Typical thermal conductivity values for cellulose insulation are between 0.022 and 0.035W/m∙K.

In general, thermal insulation is primarily based on the very low thermal conductivity of gases. Gases possess poor thermal conduction properties compared to liquids and solids, and thus makes a good insulation material if they can be trapped (e.g. in a foam-like structure). Air and other gases are generally good insulators. But the main benefit is in the absence of convection. Therefore, many insulating materials (e.g. cellulose insulation) function simply by having a large number of gas-filled pockets which prevent large-scale convection.

Alternation of gas pocket and solid material causes that the heat must be transferred through many interfaces causing rapid decrease in heat transfer coefficient.

Example – Cellulose Insulation

A major source of heat loss from a house is through walls. Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of bricks with the thermal conductivity of k₁ = 1.0 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

1. Calculate the heat flux (heat loss) through this non-insulated wall.
2. Now assume thermal insulation on the outer side of this wall. Use cellulose insulation 10 cm thick (L₂) with the thermal conductivity of k₂ = 0.04 W/m.K and calculate the heat flux (heat loss) through this composite wall.

Solution:

As was written, many of the heat transfer processes involve composite systems and even involve a combination of both conduction and convection. With these composite systems, it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

The overall heat transfer coefficient is related to the total thermal resistance and depends on the geometry of the problem.

1. bare wall

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 1/30) = 3.53 W/m²K

The heat flux can be then calculated simply as:

q = 3.53 [W/m²K] x 30 [K] = 105.9 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 105.9 [W/m²] x 30 [m²] = 3177W

2. composite wall with thermal insulation

Assuming one-dimensional heat transfer through the plane composite wall, no thermal contact resistance and disregarding radiation, the overall heat transfer coefficient can be calculated as:

The overall heat transfer coefficient is then:

U = 1 / (1/10 + 0.15/1 + 0.1/0.040 + 1/30) = 0.359 W/m²K

The heat flux can be then calculated simply as:

q = 0.359 [W/m²K] x 30 [K] = 10.78 W/m²

The total heat loss through this wall will be:

q_loss = q . A = 10.78 [W/m²] x 30 [m²] = 323 W

As can be seen, an addition of thermal insulator causes significant decrease in heat losses. It must be added, an addition of next layer of thermal insulator does not cause such high savings. This can be better seen from the thermal resistance method, which can be used to calculate the heat transfer through composite walls. The rate of steady heat transfer between two surfaces is equal to the temperature difference divided by the total thermal resistance between those two surfaces.

See also:

Insulation Materials

Regenerative Heat Exchanger

In general, the heat exchangers used in regeneration may be classified as either regenerators or recuperators.

• Regenerator (Regenerative Heat Exchanger) is a type of heat exchanger where heat from the hot fluid is intermittently stored in a thermal storage medium before it is transferred to the cold fluid. It has a single flow path in which the hot and cold fluids alternately pass through.

• Recuperator is a type of heat exchanger has separate flow paths for each fluid along their own passages and heat is transferred through the separating walls. Recuperators (e.g. economisers) are often used in power engineering, to increase the overall efficiency of thermodynamic cycles. For example, in a gas turbine engine. The recuperator transfers some of the waste heat in the exhaust to the compressed air, thus preheating it before entering the combustion chamber. Many recuperators are designed as counter-flow heat exchangers.

Heat Regeneration

In steam turbines theory,  significant increases in the thermal efficiency of steam turbine can be achieved through reducing the amount of fuel that must be added in the boiler. This can be done by transferring heat (e.g. partially expanded steam) from certain sections of the steam turbine, which is normally well above the ambient temperature, to the feedwater. This process is known as heat regeneration and a variety of heat regenerators can be used for this purpose. Sometimes engineers use the term economiser that are heat exchangers intended to reduce energy consumption, especially in case of preheating of a fluid.

As can be seen in the article “Steam Generator”, the feedwater (secondary circuit) at the inlet of the steam generator may have about ~230°C (446°F)and then is heated to the boiling point of that fluid (280°C; 536°F; 6,5MPa)and evaporated. But the condensate at the condenser outlet may have about 40°C, so the heat regeneration in typical PWR is significant and very important:

• Heat regeneration increases the thermal efficiency, since more of the heat flow into the cycle occurs at higher temperature.
• Heat regeneration causes a decrease in the mass flow rate through low-pressure stage of the steam turbine, thus increases LP Isentropic Turbine Efficiency. Note that at the last stage of expansion the steam has very high specific volume.
• Heat regeneration causes an increases in working steam quality, since the drains are situated at the periphery of turbine casing, where is higher concentration of water droplets.

Analysis of Heat Exchangers

Heat exchangers are commonly used in industry, and proper design of a heat exchanger depends on many variables. In the analysis of heat exchangers,  it is often convenient to work with an overall heat transfer coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling. Moreover, engineers also use the logarithmic mean temperature difference (LMTD) to determine the temperature driving force for heat transfer in heat exchangers.

Special Reference: John R. Thome, Engineering Data Book III. Wolverine Tube Inc. 2004.

 

 

See also:

Heat Exchangers

Heat Generation in Nuclear Reactors

As was written, a nuclear power plant (nuclear power station) looks like a standard thermal power station with one exception. 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. But in nuclear power plants reactors produce enormous amount of heat (energy) in a small volume. The density of the energy generation is very large and this puts demands on its heat transfer system (reactor coolant system).

For a reactor to operate in a steady state, all of the heat released in the system must be removed as fast as it is produced. This is accomplished by passing a liquid or gaseous coolant through the core and through other regions where heat is generated. The heat transfer must be equal to or greater than the heat generation rate or overheating and possible damage to the fuel may occur. The nature and operation of this coolant system is one of the most important considerations in the design of a nuclear reactor.

It should be noted that from a strictly nuclear standpoint, there is theoretically no upper limit to the reactor thermal power, which can be attained by any critical reactor having sufficient excess of reactivity to overcome its negative temperature feedbacks. In each nuclear reactor, there is a direct proportionality between the neutron flux and the reactor thermal power. The term thermal power is usually used, because it means the rate at which heat is produced in the reactor core as the result of fissions in the fuel. Moreover, for a short period, a critical reactor does not need to have high excess of reactivity as in case of rapid reactivity excursions.

In short, almost any reactor is able to exceed the ability of heat removal of its coolant system. Beyond this point, the fuel would heat up and can reach very high temperatures. This situation must be avoided by reactor operator and by reactor safety systems. It is essential, that the heat generation – heat removal rate balance must be maintained to prevent these temperatures that might result in the failure of fuel or other structural materials. In reactor engineering, the thermal-hydraulics of nuclear reactors describe the effort involving the coupling of heat transfer and fluid dynamics to accomplish the desired heat removal rate from the core under both normal operation and accident conditions.

Heat Production in Fuel Elements

In nuclear reactors, there is a direct proportionality between the neutron flux and the reactor thermal power. This proporcionality is determined by the fission reaction rate per unit volume (RR = Ф . Σ). The fission reaction rate within a nuclear reactor is controlled by several factors. For simplicity let assume that the fissionable material is uniformly distributed in the reactor. In this case, the macroscopic cross-sections are independent of position. Multiplying the fission reaction rate per unit volume (RR = Ф . Σ) by the total volume of the core (V) gives us the total number of reactions occurring in the reactor core per unit time. But we also know the amount of energy released per one fission reaction to be about 200 MeV/fission. Now, it is possible to determine the rate of energy release (power) due to the fission reaction. It is given by following equation:

P = RR . E_r . V = Ф . Σ_f . E_r . V = Ф . N_U235 . σ_f²³⁵ . E_r . V

where:

P – reactor power (MeV.s^-1)

Ф – neutron flux (neutrons.cm^-2.s^-1)

σ – microscopic cross section (cm²)

N – atomic number density (atoms.cm^-3)

Er – the average recoverable energy per fission (MeV / fission)

V – total volume of the core (m³)

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.

As can be seen from the table, 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 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.

As can be seen from the description of the individual components of the total energy energy released during the fission reaction, there is significant amount of energy generated outside the nuclear fuel (outside fuel rods). Especially the kinetic energy of prompt neutrons is largely generated in the coolant (moderator). This phenomena needs to be included in the nuclear calculations.

For LWR, it is generally accepted that about 2.5% of total energy is recovered in the moderator. This fraction of energy depends on the materials, their arrangement within the reactor, and thus on the reactor type.

It must be also added, also the other reactor internals must be cooled sufficiently in order to prevent overheating of thir construction materials. One of most exposed components is the neutron reflector, especially the heavy reflector. While acting as a neutron shield, the heavy reflector is heated due to absorption of the gamma radiation. In order to avoid overheating, the heat in the reflector is removed by water flowing through cooling channels drilled through the reflector.

See also: Energy release per fission

See also: Residual Heat

Power Distribution in Conventional Reactor Cores

It should be noted the flux shape derived from the diffusion theory is only a theoretical case in a uniform homogeneous cylindrical reactor at low power levels (at “zero power criticality”). We have implicitly assumed that the core consisting of thousands of fuel and control elements, coolant, and structure can be represented by some effective homogeneous mixture. This is a very strong assumption, because it does not take into account the heterogeneity of a core.

See also: Diffusion Equation – Finite Cylindrical Reactor

In commercial reactor cores the flux distribution is significantly influenced by many factors. One of the most important aspects is the heterogeneity of fuel-moderator assembly. This problem is very complex and is described separately in:

See also: Power Distribution

Thermal Limits

The temperature in an operating reactor varies from point to point within the system. As a consequence, there is always one fuel rod and one local volume, that are hotter than all the rest. In order to limit these hot places the peak power limits must be introduced. The peak power limits are associated with such phenomena as the departure from nucleate boiling and with the conditions which could cause fuel pellet melt.

Therefore power distribution within the core must be properly limited. These limitations are usually divided into two basic categories:

• Limitation of global power distribution
  • Axial flux difference
  • Power tilt
• Limitation of local power distribution
  • Local Heat Flux
  • Enthalpy Rise in Hot Channel

Reactor Coolant delta T – Energy Balance

Another very useful relation is that the thermal power produced by a reactor is directly related to the mass flow rate of the reactor coolant and the temperature difference across the core.

On a straight thermodynamic basis, this heat generation is also related to the fluid temperature difference across the core and the mass flow rate of the fluid passing through the core. Thus, the size of the reactor core is dependent upon and limited by low much liquid can be passed through the core to remove the generated thermal energy. Note that, in PWRs, the core outlet temperature is limited. In a typical pressurized water reactor, the hot primary coolant (water 330°C; 626°F) is pumped into the steam generator through primary inlet. This requires maintaining of very high pressures to keep the water in the liquid state. In order to prevent boiling of the primary coolant and to provide a subcooling margin (the difference between the pressurizer temperature and the highest temperature in the reactor core), pressures around 16 MPa are typical for PWRs. The reactor pressure vessel is the key component, which limits the thermal efficiency of each nuclear power plant, since the reactor vessel must withstand high pressures. Many other factors affect the amount of heat generated within a reactor core, but its limiting generation rate is based upon how much energy can safely be carried away by the coolant.

Temperature Profile – Nuclear Fuel

See also: Cladding Surface Temperature

Most of PWRs use the uranium fuel, which is in the form of uranium dioxide. Uranium dioxide is a black semiconducting solid with very low thermal conductivity. On the other hand the uranium dioxide has very high melting point and has well known behavior. The UO₂ is pressed into cylindrical pellets, these pellets are then sintered into the solid.

These cylindrical pellets are then loaded and encapsulated within a fuel rod (or fuel pin), which is made of zirconium alloys due to its very low absorption cross-section (unlike the stainless steel). The surface of the tube, which covers the pellets, is called fuel cladding.

See also: Thermal Conduction of Uranium Dioxide

Thermal and mechanical behavior of fuel pellets and fuel rods constitute one of three key core design disciplines. Nuclear fuel is operated under very inhospitable conditions (thermal, radiation, mechanical) and must withstand more than normal conditions operation. For example temperatures in the centre of fuel pellets reach more than 1000°C (1832°F) accompanied by fission-gas releases. Therefore detailed knowledge of temperature distribution within a single fuel rod is essential for safe operation of nuclear fuel. In this section we will study heat conduction equation in cylindrical coordinates using Dirichlet boundary condition with given surface temperature (i.e. using Dirichlet boundary condition). Comprehensive analysis of fuel rod temperature profile will be studied in separate section.

Temperature in the centerline of a fuel pellet

Consider the fuel pellet of radius r_U = 0.40 cm, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. Instead of volumetric heat rate q_V [W/m³], engineers often use the linear heat rate, q_L [W/m], which represents the heat rate of one meter of fuel rod. The linear heat rate can be calculated from the volumetric heat rate by:

The centreline is taken as the origin for r-coordinate. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

Calculate the temperature distribution, T(r), in this fuel pellet, if:

• the temperatures at the surface of the fuel pellet is T_U = 420°C
• the fuel pellet radius r_U = 4 mm.
• the averaged material’s conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the linear heat rate is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

In this case, the surface is maintained at given temperatures T_U. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (r = 0) temperature (maximum) in this cylindrical fuel pellet at these specific boundary conditions will be:

The radial heat flux at any radius, q_r [W.m^-1], in the cylinder may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of r.

The following figure shows the temperature distribution in the fuel pellet at various power levels.

______

The temperature in an operating reactor varies from point to point within the system. As a consequence, there is always one fuel rod and one local volume, that are hotter than all the rest. In order to limit these hot places the peak power limits must be introduced. The peak power limits are associated with a boiling crisis and with the conditions which could cause fuel pellet melt. However, metallurgical considerations place an upper limits on the temperature of the fuel cladding and the fuel pellet. Above these temperatures there is a danger that the fuel may be damaged. One of the major objectives in the design of a nuclear reactors is to provide for the removal of the heat produced at the desired power level, while assuring that the maximum fuel temperature and the maximum cladding temperature are always below these predetermined values.

See also:

Heat Transfer

Heat Production in Fuel Elements

In nuclear reactors, there is a direct proportionality between the neutron flux and the reactor thermal power. This proporcionality is determined by the fission reaction rate per unit volume (RR = Ф . Σ). The fission reaction rate within a nuclear reactor is controlled by several factors. For simplicity let assume that the fissionable material is uniformly distributed in the reactor. In this case, the macroscopic cross-sections are independent of position. Multiplying the fission reaction rate per unit volume (RR = Ф . Σ) by the total volume of the core (V) gives us the total number of reactions occurring in the reactor core per unit time. But we also know the amount of energy released per one fission reaction to be about 200 MeV/fission. Now, it is possible to determine the rate of energy release (power) due to the fission reaction. It is given by following equation:

P = RR . E_r . V = Ф . Σ_f . E_r . V = Ф . N_U235 . σ_f²³⁵ . E_r . V

where:

P – reactor power (MeV.s^-1)

Ф – neutron flux (neutrons.cm^-2.s^-1)

σ – microscopic cross section (cm²)

N – atomic number density (atoms.cm^-3)

Er – the average recoverable energy per fission (MeV / fission)

V – total volume of the core (m³)

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.

As can be seen from the table, 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 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.

As can be seen from the description of the individual components of the total energy energy released during the fission reaction, there is significant amount of energy generated outside the nuclear fuel (outside fuel rods). Especially the kinetic energy of prompt neutrons is largely generated in the coolant (moderator). This phenomena needs to be included in the nuclear calculations.

For LWR, it is generally accepted that about 2.5% of total energy is recovered in the moderator. This fraction of energy depends on the materials, their arrangement within the reactor, and thus on the reactor type.

It must be also added, also the other reactor internals must be cooled sufficiently in order to prevent overheating of thir construction materials. One of most exposed components is the neutron reflector, especially the heavy reflector. While acting as a neutron shield, the heavy reflector is heated due to absorption of the gamma radiation. In order to avoid overheating, the heat in the reflector is removed by water flowing through cooling channels drilled through the reflector.

See also: Energy release per fission

See also: Residual Heat

Temperature Profile – Nuclear Fuel

Most of PWRs use the uranium fuel, which is in the form of uranium dioxide. Uranium dioxide is a black semiconducting solid with very low thermal conductivity. On the other hand the uranium dioxide has very high melting point and has well known behavior. The UO₂ is pressed into cylindrical pellets, these pellets are then sintered into the solid.

These cylindrical pellets are then loaded and encapsulated within a fuel rod (or fuel pin), which is made of zirconium alloys due to its very low absorption cross-section (unlike the stainless steel). The surface of the tube, which covers the pellets, is called fuel cladding.

See also: Thermal Conduction of Uranium Dioxide

Thermal and mechanical behavior of fuel pellets and fuel rods constitute one of three key core design disciplines. Nuclear fuel is operated under very inhospitable conditions (thermal, radiation, mechanical) and must withstand more than normal conditions operation. For example temperatures in the centre of fuel pellets reach more than 1000°C (1832°F) accompanied by fission-gas releases. Therefore detailed knowledge of temperature distribution within a single fuel rod is essential for safe operation of nuclear fuel. In this section we will study heat conduction equation in cylindrical coordinates using Dirichlet boundary condition with given surface temperature (i.e. using Dirichlet boundary condition). Comprehensive analysis of fuel rod temperature profile will be studied in separate section.

Temperature in the centerline of a fuel pellet

Consider the fuel pellet of radius r_U = 0.40 cm, in which there is uniform and constant heat generation per unit volume, q_V [W/m³]. Instead of volumetric heat rate q_V [W/m³], engineers often use the linear heat rate, q_L [W/m], which represents the heat rate of one meter of fuel rod. The linear heat rate can be calculated from the volumetric heat rate by:

The centreline is taken as the origin for r-coordinate. Due to symmetry in z-direction and in azimuthal direction, we can separate of variables and simplify this problem to one-dimensional problem. Thus, we will solve for the temperature as function of radius, T(r), only. For constant thermal conductivity, k, the appropriate form of the cylindrical heat equation, is:

The general solution of this equation is:

where C₁ and C₂ are the constants of integration.

Calculate the temperature distribution, T(r), in this fuel pellet, if:

• the temperatures at the surface of the fuel pellet is T_U = 420°C
• the fuel pellet radius r_U = 4 mm.
• the averaged material’s conductivity is k = 2.8 W/m.K (corresponds to uranium dioxide at 1000°C)
• the linear heat rate is q_L = 300 W/cm and thus the volumetric heat rate is q_V = 597 x 10⁶ W/m³

In this case, the surface is maintained at given temperatures T_U. This corresponds to the Dirichlet boundary condition. Moreover, this problem is thermally symmetric and therefore we may use also thermal symmetry boundary condition. The constants may be evaluated using substitution into the general solution and are of the form:

The resulting temperature distribution and the centerline (r = 0) temperature (maximum) in this cylindrical fuel pellet at these specific boundary conditions will be:

The radial heat flux at any radius, q_r [W.m^-1], in the cylinder may, of course, be determined by using the temperature distribution and with the Fourier’s law. Note that, with heat generation the heat flux is no longer independent of r.

The following figure shows the temperature distribution in the fuel pellet at various power levels.

______

The temperature in an operating reactor varies from point to point within the system. As a consequence, there is always one fuel rod and one local volume, that are hotter than all the rest. In order to limit these hot places the peak power limits must be introduced. The peak power limits are associated with a boiling crisis and with the conditions which could cause fuel pellet melt. However, metallurgical considerations place an upper limits on the temperature of the fuel cladding and the fuel pellet. Above these temperatures there is a danger that the fuel may be damaged. One of the major objectives in the design of a nuclear reactors is to provide for the removal of the heat produced at the desired power level, while assuring that the maximum fuel temperature and the maximum cladding temperature are always below these predetermined values.

See also:

Heat Generation