Nick Connor

About Inconel

In general, Inconel is a registered trademark of Special Metals for a family of austenitic nickel-chromium-based superalloys. Inconel 718 is a nickel-based superalloy that possesses high strength properties and resistance to elevated temperatures. It also demonstrates remarkable protection against corrosion and oxidation. Inconel’s high temperature strength is developed by solid solution strengthening or precipitation hardening, depending on the alloy. Inconel 718 is composed of 55% nickel, 21% chromium, 6% iron, and small amounts of manganese, carbon, and copper.

Common uses of superalloys are in the aerospace and some other high-technology industries. With the combination of corrosion resistance and material strength in the face of extreme heat, this kind of superalloy works well in the nuclear industry. Some nuclear plants use nickel-based superalloys for the reactor core, control rod, and similar parts.​ In nuclear industry, especially low-cobalt superalloys (due to possible activation of cobalt-59) are used. Some of structural parts of nuclear fuel assemblies, such as top and bottom nozzle, may be produced from superalloys such as Inconel. Spacing grids are usually made of a corrosion-resistant material with low absorption cross section for thermal neutrons, usually zirconium alloy (~ 0.18 × 10^–24 cm²). First and last spacing grid may be also made of low-cobalt Inconel, which is a superalloy well suited for service in extreme environments subjected to pressure and heat.

Summary

Name	Inconel
Phase at STP	solid
Density	8200 kg/m3
Ultimate Tensile Strength	1200 MPa
Yield Strength	1030 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	330 BHN
Melting Point	1400 °C
Thermal Conductivity	6.5 W/mK
Heat Capacity	460 J/g K
Price	40 $/kg

Density of Inconel

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Inconel is 8200 kg/m³.

 

Example: Density

Calculate the height of a cube made of Inconel, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.496 m.

Mechanical Properties of Inconel

Materials are frequently chosen for various applications because they have desirable combinations of mechanical characteristics. For structural applications, material properties are crucial and engineers must take them into account.

Strength of Inconel 718

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of Inconel 718 depends on heat treatment process, but it is about 1200 MPa.

The ultimate tensile strength is the maximum on the engineering stress-strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Ultimate tensile strength is often shortened to “tensile strength” or even to “the ultimate.”  If this stress is applied and maintained, fracture will result. Often, this value is significantly more than the yield stress (as much as 50 to 60 percent more than the yield for some types of metals). When a ductile material reaches its ultimate strength, it experiences necking where the cross-sectional area reduces locally. The stress-strain curve contains no higher stress than the ultimate strength. Even though deformations can continue to increase, the stress usually decreases after the ultimate strength has been achieved. It is an intensive property; therefore its value does not depend on the size of the test specimen. However, it is dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material. Ultimate tensile strengths vary from 50 MPa for an aluminum to as high as 3000 MPa for very high-strength steels.

Yield Strength

Yield strength of Inconel 718 depends on heat treatment process, but it is about 1030 MPa.

The yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning plastic behavior. Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. Prior to the yield point, the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible. Some steels and other materials exhibit a behaviour termed a yield point phenomenon. Yield strengths vary from 35 MPa for a low-strength aluminum to greater than 1400 MPa for very high-strength steels.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of Inconel 718 is 200 GPa.

The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests. Up to a limiting stress, a body will be able to recover its dimensions on removal of the load. The applied stresses cause the atoms in a crystal to move from their equilibrium position. All the atoms are displaced the same amount and still maintain their relative geometry. When the stresses are removed, all the atoms return to their original positions and no permanent deformation occurs. According to the Hooke’s law, the stress is proportional to the strain (in the elastic region), and the slope is Young’s modulus. Young’s modulus is equal to the longitudinal stress divided by the strain.

Hardness of Inconel 718

Brinell hardness of Inconel 718 depends on heat treatment process, but it is approximately 330 MPa.

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Hardness is probably the most poorly defined material property because it may indicate resistance to scratching, resistance to abrasion, resistance to indentation or even resistance to shaping or localized plastic deformation. Hardness is important from an engineering standpoint because resistance to wear by either friction or erosion by steam, oil, and water generally increases with hardness.

Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested. The typical test uses a 10 mm (0.39 in) diameter  hardened steel ball as an indenter with a 3,000 kgf (29.42 kN; 6,614 lbf) force. The load is maintained constant for a specified time (between 10 and 30 s). For softer materials, a smaller force is used; for harder materials, a tungsten carbide ball is substituted for the steel ball.

The test provides numerical results to quantify the hardness of a material, which is expressed by the Brinell hardness number – HB. The Brinell hardness number is designated by the most commonly used test standards (ASTM E10-14[2] and ISO 6506–1:2005) as HBW (H from hardness, B from brinell and W from the material of the indenter, tungsten (wolfram) carbide). In former standards HB or HBS were used to refer to measurements made with steel indenters.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

There are a variety of  test methods in common use (e.g. Brinell, Knoop, Vickers and Rockwell). There are tables that are available correlating the hardness numbers from the different test methods where correlation is applicable. In all scales, a high hardness number represents a hard metal.

 

Example: Strength

Assume a plastic rod, which is made of Inconel. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 1200 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 1200 x 10⁶ x 0.0001 = 120 000 N

Thermal Properties of Inconel

Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.

Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.

Melting Point of Inconel 718

Melting point of Inconel 718 steel is around 1400°C.

In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.

Thermal Conductivity of Inconel 718

The thermal conductivity of Inconel 718 is 6.5 W/(m.K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

 

Example: Heat transfer calculation

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.

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 Inconel with the thermal conductivity of k₁ = 6.5 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/6.5 + 1/30) = 6.39 W/m²K

The heat flux can be then calculated simply as: q = 6.39 [W/m²K] x 30 [K] = 191.8 W/m²

The total heat loss through this wall will be: q_loss = q . A =191.8 [W/m²] x 30 [m²] = 5754.1 W

About TZM Alloy

Molybdenum-titanium-zirconium (TZM) alloys contain small amounts of titanium and zirconium doped with small amounts of very fine carbides. This alloy belongs to refractory alloys, which are well known for their extraordinary resistance to heat and wear. Key requirement to withstand high temperatures is a high melting point and stable mechanical properties (e.g. high hardness) even at high temperatures. This alloy functions most efficiently in temperature ranges of 700-1400°C. The alloy exhibits a higher creep resistance and strength at high temperatures, making service temperatures of above 1060 °C possible for the material. It is typically manufactured by powder metallurgy or arc-casting processes.

Summary

Name	TZM Alloy
Phase at STP	solid
Density	10220 kg/m3
Ultimate Tensile Strength	800 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	320 GPa
Brinell Hardness	220 BHN
Melting Point	2597 °C
Thermal Conductivity	126 W/mK
Heat Capacity	305 J/g K
Price	100 $/kg

Density of TZM Alloy

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of TZM Alloy is 10220 kg/m³.

 

Example: Density

Calculate the height of a cube made of TZM Alloy, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.461 m.

Mechanical Properties of TZM Alloy

Strength of TZM Alloy

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests.

See also: Strength of Materials

Ultimate Tensile Strength of TZM Alloy

Ultimate tensile strength of TZM Alloy is 800 MPa.

Yield Strength of TZM Alloy

Yield strength of TZM Alloy is N/A.

Modulus of Elasticity of TZM Alloy

The Young’s modulus of elasticity of TZM Alloy is 320 GPa.

Hardness of TZM Alloy

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

Brinell hardness of TZM Alloy is approximately 220 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of TZM Alloy. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 800 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 800 x 10⁶ x 0.0001 = 80 000 N

Thermal Properties of TZM Alloy

TZM Alloy – Melting Point

Melting point of TZM Alloy is 2597 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

TZM Alloy – Thermal Conductivity

Thermal conductivity of TZM Alloy is 126 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

TZM Alloy – Specific Heat

Specific heat of TZM Alloy is 305 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 TZM Alloy with the thermal conductivity of k₁ = 126 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/126 + 1/30) = 7.43 W/m²K

The heat flux can be then calculated simply as: q = 7.43 [W/m²K] x 30 [K] = 223.01 W/m²

The total heat loss through this wall will be: q_loss = q . A = 223.01 [W/m²] x 30 [m²] = 6690.27 W

About Gray Iron

Gray cast iron is the oldest and most common type of iron in existence and probably what most people think of when they hear the term “cast iron”. The carbon and silicon contents of gray cast irons vary between 2.5 and 4.0 wt% and 1.0 and 3.0 wt%, respectively.

Gray cast iron is characterised by its graphitic microstructure, which causes fractures of the material to have a gray appearance. This is due to the presence of graphite in its composition. In gray cast iron the graphite forms as flakes, taking on a three dimensional geometry.

Gray cast iron has less tensile strength and shock resistance than steel, but its compressive strength is comparable to low- and medium-carbon steel. Gray cast iron has good thermal conductivity and specific heat capacity,  therefore it is often used in cookware and brake rotors.

Summary

Name	Gray Iron
Phase at STP	solid
Density	7150 kg/m3
Ultimate Tensile Strength	395 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	124 GPa
Brinell Hardness	235 BHN
Melting Point	1260 °C
Thermal Conductivity	53 W/mK
Heat Capacity	460 J/g K
Price	1.2 $/kg

Density of Gray Iron

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Gray Iron is 7150 kg/m³.

 

Example: Density

Calculate the height of a cube made of Gray Iron, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.519 m.

Mechanical Properties of Gray Iron

Materials are frequently chosen for various applications because they have desirable combinations of mechanical characteristics. For structural applications, material properties are crucial and engineers must take them into account.

Strength of Gray Iron

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of Gray Iron (ASTM A48 Class 40) is 295 MPa.

The ultimate tensile strength is the maximum on the engineering stress-strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Ultimate tensile strength is often shortened to “tensile strength” or even to “the ultimate.”  If this stress is applied and maintained, fracture will result. Often, this value is significantly more than the yield stress (as much as 50 to 60 percent more than the yield for some types of metals). When a ductile material reaches its ultimate strength, it experiences necking where the cross-sectional area reduces locally. The stress-strain curve contains no higher stress than the ultimate strength. Even though deformations can continue to increase, the stress usually decreases after the ultimate strength has been achieved. It is an intensive property; therefore its value does not depend on the size of the test specimen. However, it is dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material. Ultimate tensile strengths vary from 50 MPa for an aluminum to as high as 3000 MPa for very high-strength steels.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of Gray Iron (ASTM A48 Class 40) is 124 GPa.

The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests. Up to a limiting stress, a body will be able to recover its dimensions on removal of the load. The applied stresses cause the atoms in a crystal to move from their equilibrium position. All the atoms are displaced the same amount and still maintain their relative geometry. When the stresses are removed, all the atoms return to their original positions and no permanent deformation occurs. According to the Hooke’s law, the stress is proportional to the strain (in the elastic region), and the slope is Young’s modulus. Young’s modulus is equal to the longitudinal stress divided by the strain.

Hardness of Gray Cast Iron – ASTM A48 Class 40

Brinell hardness of Gray Iron (ASTM A48 Class 40) is approximately 235 MPa.

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Hardness is probably the most poorly defined material property because it may indicate resistance to scratching, resistance to abrasion, resistance to indentation or even resistance to shaping or localized plastic deformation. Hardness is important from an engineering standpoint because resistance to wear by either friction or erosion by steam, oil, and water generally increases with hardness.

Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested. The typical test uses a 10 mm (0.39 in) diameter  hardened steel ball as an indenter with a 3,000 kgf (29.42 kN; 6,614 lbf) force. The load is maintained constant for a specified time (between 10 and 30 s). For softer materials, a smaller force is used; for harder materials, a tungsten carbide ball is substituted for the steel ball.

The test provides numerical results to quantify the hardness of a material, which is expressed by the Brinell hardness number – HB. The Brinell hardness number is designated by the most commonly used test standards (ASTM E10-14[2] and ISO 6506–1:2005) as HBW (H from hardness, B from brinell and W from the material of the indenter, tungsten (wolfram) carbide). In former standards HB or HBS were used to refer to measurements made with steel indenters.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

There are a variety of  test methods in common use (e.g. Brinell, Knoop, Vickers and Rockwell). There are tables that are available correlating the hardness numbers from the different test methods where correlation is applicable. In all scales, a high hardness number represents a hard metal.

 

Example: Strength

Assume a plastic rod, which is made of Gray Iron. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 295 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 295 x 10⁶ x 0.0001 = 29 500 N

Thermal Properties of Gray Iron

Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.

Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.

Melting Point of Gray Gray Iron

Melting point of Gray Iron is around 1260°C.

In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.

Thermal Conductivity of Gray Iron

The thermal conductivity of Gray Iron is 53 W/(m.K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

 

Example: Heat transfer calculation

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.

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 Gray Iron with the thermal conductivity of k₁ = 53 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/53 + 1/30) = 7.34 W/m²K

The heat flux can be then calculated simply as: q = 7.34 [W/m²K] x 30 [K] = 220.32 W/m²

The total heat loss through this wall will be: q_loss = q . A = 220.32 [W/m²] x 30 [m²] = 6609.7 W

See above:
Cast Iron

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About Mild Steel

Low-carbon steel, also known as mild steel is now the most common form of steel because its price is relatively low while it provides material properties that are acceptable for many applications. Low-carbon steel contains approximately 0.05–0.25% carbon making it malleable and ductile. Mild steel has a relatively low tensile strength, but it is cheap and easy to form; surface hardness can be increased through carburizing.

Summary

Name	Mild Steel
Phase at STP	solid
Density	7850 kg/m3
Ultimate Tensile Strength	400-550 MPa
Yield Strength	250 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	120 BHN
Melting Point	1450 °C
Thermal Conductivity	50 W/mK
Heat Capacity	510 J/g K
Price	0.5 $/kg

Density of Mild Steel

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Mild Steel is 7850 kg/m³.

 

Example: Density

Calculate the height of a cube made of Mild Steel, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.503 m.

Mechanical Properties of Mild Steel

Materials are frequently chosen for various applications because they have desirable combinations of mechanical characteristics. For structural applications, material properties are crucial and engineers must take them into account.

Strength of Mild Steel

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of Mild Steel is between 400 – 550 MPa.

The ultimate tensile strength is the maximum on the engineering stress-strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Ultimate tensile strength is often shortened to “tensile strength” or even to “the ultimate.”  If this stress is applied and maintained, fracture will result. Often, this value is significantly more than the yield stress (as much as 50 to 60 percent more than the yield for some types of metals). When a ductile material reaches its ultimate strength, it experiences necking where the cross-sectional area reduces locally. The stress-strain curve contains no higher stress than the ultimate strength. Even though deformations can continue to increase, the stress usually decreases after the ultimate strength has been achieved. It is an intensive property; therefore its value does not depend on the size of the test specimen. However, it is dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material. Ultimate tensile strengths vary from 50 MPa for an aluminum to as high as 3000 MPa for very high-strength steels.

Yield Strength

Yield strength of Mild Steel is 250 MPa.

The yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning plastic behavior. Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. Prior to the yield point, the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible. Some steels and other materials exhibit a behaviour termed a yield point phenomenon. Yield strengths vary from 35 MPa for a low-strength aluminum to greater than 1400 MPa for very high-strength steels.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of Mild Steel is 200 GPa.

The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests. Up to a limiting stress, a body will be able to recover its dimensions on removal of the load. The applied stresses cause the atoms in a crystal to move from their equilibrium position. All the atoms are displaced the same amount and still maintain their relative geometry. When the stresses are removed, all the atoms return to their original positions and no permanent deformation occurs. According to the Hooke’s law, the stress is proportional to the strain (in the elastic region), and the slope is Young’s modulus. Young’s modulus is equal to the longitudinal stress divided by the strain.

Hardness of Mild Steel

Brinell hardness of Mild Steel is approximately 120 MPa.

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Hardness is probably the most poorly defined material property because it may indicate resistance to scratching, resistance to abrasion, resistance to indentation or even resistance to shaping or localized plastic deformation. Hardness is important from an engineering standpoint because resistance to wear by either friction or erosion by steam, oil, and water generally increases with hardness.

Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested. The typical test uses a 10 mm (0.39 in) diameter  hardened steel ball as an indenter with a 3,000 kgf (29.42 kN; 6,614 lbf) force. The load is maintained constant for a specified time (between 10 and 30 s). For softer materials, a smaller force is used; for harder materials, a tungsten carbide ball is substituted for the steel ball.

The test provides numerical results to quantify the hardness of a material, which is expressed by the Brinell hardness number – HB. The Brinell hardness number is designated by the most commonly used test standards (ASTM E10-14[2] and ISO 6506–1:2005) as HBW (H from hardness, B from brinell and W from the material of the indenter, tungsten (wolfram) carbide). In former standards HB or HBS were used to refer to measurements made with steel indenters.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

There are a variety of  test methods in common use (e.g. Brinell, Knoop, Vickers and Rockwell). There are tables that are available correlating the hardness numbers from the different test methods where correlation is applicable. In all scales, a high hardness number represents a hard metal.

 

Example: Strength

Assume a plastic rod, which is made of Mild Steel. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 500 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 500 x 10⁶ x 0.0001 = 50 000 N

Thermal Properties of Mild Steel

Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.

Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.

Melting Point of Mild Steel

Melting point of Mild Steel is around 1450°C.

In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.

Thermal Conductivity of Mild Steel

Mild Steel is a multi-element substance, principally of iron, with additions of carbon and impurities. The thermal conductivity of wrought iron is around 50 W/(m.K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

 

Example: Heat transfer calculation

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.

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 Mild Steel with the thermal conductivity of k₁ = 50 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/50 + 1/30) = 7.33 W/m²K

The heat flux can be then calculated simply as: q = 7.33 [W/m²K] x 30 [K] = 220.05 W/m²

The total heat loss through this wall will be: q_loss = q . A = 220.05 [W/m²] x 30 [m²] = 6601.48 W

See above:
Steels

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About Duralumin

Aluminium alloys of 2000 series are alloyed with copper, they can be precipitation hardened to strengths comparable to steel. Formerly referred to as duralumin, they were once the most common aerospace alloys, but were susceptible to stress corrosion cracking and are increasingly replaced by 7000 series in new designs. In addition to aluminium, the main materials in duralumin are copper, manganese and magnesium.

Duralumin (also called duraluminum, duraluminium, duralum, dural(l)ium, or dural) is a strong, lightweight alloy of aluminium discovered in 1910 by Alfred Wilm, a German metallurgist. He discovered that after quenching, an aluminium alloy containing 4% copper would slowly harden when left at room temperature for several days. This process is now known as natural aging. He also designed an alloy (Duralumin) suitable for strengthening by this process in what is now known as preciptitation hardening.  Although an explanation for the phenomenon was not provided until 1919, duralumin was one of the first “age hardening” alloys used.

Summary

Name	Duralumin
Phase at STP	solid
Density	2780 kg/m3
Ultimate Tensile Strength	450 MPa
Yield Strength	300 MPa
Young’s Modulus of Elasticity	76 GPa
Brinell Hardness	120 BHN
Melting Point	570 °C
Thermal Conductivity	140 W/mK
Heat Capacity	900 J/g K
Price	6 $/kg

In terms of age hardening, solution annealed aluminium-copper alloys can be aged naturally at room temperature for four days or more to obtain maximum properties such as hardness and strength. This process is known as natural aging. At room temperature, the solubility of copper in aluminium drops to a small fraction of 1%. At this point, the copper solute is locked inside the aluminium lattice (matrix), but must “precipitate” out of the supersaturated aluminium lattice. The aging process also can be accelerated to a matter of hours after solution treatment and quenching by heating the supersaturated alloy to a specific temperature and holding at that temperature for a specified time. This process is called artificial aging.

Duralumin is relatively soft, ductile and easily workable under normal temperature. The alloy can be rolled, forged and extruded into various forms and products. The light weight and high strength of duralumin when compared to steel enabled its application in aircraft construction. Although the addition of copper improves strength, it also makes these alloys susceptible to corrosion. The electrical and heat conductivity of duralumin is less than that of pure aluminium and more than that of steel.

Density of Duralumin

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Duralumin is 2780 kg/m³.

Example: Density

Calculate the height of a cube made of Duralumin, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.711 m.

Mechanical Properties of Duralumin

Strength of Duralumin

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength – Duralumin

Ultimate tensile strength of 2024 aluminium alloy depends greatly on the temper of the material, but it is about 450 MPa.

The ultimate tensile strength is the maximum on the engineering stress-strain curve. This corresponds to the maximum stress that can be sustained by a structure in tension. Ultimate tensile strength is often shortened to “tensile strength” or even to “the ultimate.”  If this stress is applied and maintained, fracture will result. Often, this value is significantly more than the yield stress (as much as 50 to 60 percent more than the yield for some types of metals). When a ductile material reaches its ultimate strength, it experiences necking where the cross-sectional area reduces locally. The stress-strain curve contains no higher stress than the ultimate strength. Even though deformations can continue to increase, the stress usually decreases after the ultimate strength has been achieved. It is an intensive property; therefore its value does not depend on the size of the test specimen. However, it is dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material. Ultimate tensile strengths vary from 50 MPa for an aluminum to as high as 3000 MPa for very high-strength steels.

Example: Strength

Assume a plastic rod, which is made of Duralumin. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 450 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 450 x 10⁶ x 0.0001 = 45 000 N

Yield Strength

Yield strength of 2024 aluminium alloy depends greatly on the temper of the material, but it is about 300 MPa.

The yield point is the point on a stress-strain curve that indicates the limit of elastic behavior and the beginning plastic behavior. Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. Prior to the yield point, the material will deform elastically and will return to its original shape when the applied stress is removed. Once the yield point is passed, some fraction of the deformation will be permanent and non-reversible. Some steels and other materials exhibit a behaviour termed a yield point phenomenon. Yield strengths vary from 35 MPa for a low-strength aluminum to greater than 1400 MPa for very high-strength steels.

Young’s Modulus of Elasticity

Young’s modulus of elasticity of 2024 aluminium alloy is about 76 GPa.

The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests. Up to a limiting stress, a body will be able to recover its dimensions on removal of the load. The applied stresses cause the atoms in a crystal to move from their equilibrium position. All the atoms are displaced the same amount and still maintain their relative geometry. When the stresses are removed, all the atoms return to their original positions and no permanent deformation occurs. According to the Hooke’s law, the stress is proportional to the strain (in the elastic region), and the slope is Young’s modulus. Young’s modulus is equal to the longitudinal stress divided by the strain.

Hardness of Aluminium Alloys – Duralumin

Brinell hardness of 2024 aluminium alloy depends greatly on the temper of the material, but it is approximately 110 MPa.

Rockwell hardness test is one of the most common indentation hardness tests, that has been developed for hardness testing. In contrast to Brinell test, the Rockwell tester measures the depth of penetration of an indenter under a large load (major load) compared to the penetration made by a preload (minor load). The minor load establishes the zero position. The major load is applied, then removed while still maintaining the minor load. The difference between depth of penetration before and after application of the major load is used to calculate the Rockwell hardness number. That is, the penetration depth and hardness are inversely proportional. The chief advantage of Rockwell hardness is its ability to display hardness values directly. The result is a dimensionless number noted as HRA, HRB, HRC, etc., where the last letter is the respective Rockwell scale.

The Rockwell C test is performed with a Brale penetrator (120°diamond cone) and a major load of 150kg.

Thermal Properties of Aluminium Alloys – Duralumin

Thermal properties of materials refer to the response of materials to changes in their thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature

and to the application of heat. As a solid absorbs thermodynamics/what-is-energy-physics/”>energy in the form of heat, its temperature rises and its dimensions increase. But different materials react to the application of heat differently.

Heat capacity, thermal expansion, and thermal conductivity are properties that are often critical in the practical use of solids.

Melting Point of Aluminium Alloys

Melting point of 2024 aluminium alloy is around 570°C.

In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium.

Thermal Conductivity of Aluminium Alloys

The thermal conductivity of 2024 aluminium alloy is 140 W/(m.K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Example: Heat transfer calculation

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.

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 Duralumin with the thermal conductivity of k₁ = 140 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/140 + 1/30) = 7.44 W/m²K

The heat flux can be then calculated simply as: q = 7.44 [W/m²K] x 30 [K] = 223.21 W/m²

The total heat loss through this wall will be: q_loss = q . A = 223.21 [W/m²] x 30 [m²] = 6696.19 W

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Elektron 21 – UNS M12310

In general, Elektron is the registered trademark of a wide range of magnesium alloys manufactured by a British company Magnesium Elektron Limited. Elektron 21, designated by UNS M12310, is one of alloys with excellent corrosion resistance and castability. Cast products possess a fine-grained microstructure and pressure tightness. This alloy can be easily machined. Application include motorsports and aerospace, since it possess high strength, light weight and it has excellent vibration damping characteristics.

Summary

Name	Elektron 21
Phase at STP	solid
Density	1800 kg/m3
Ultimate Tensile Strength	280 MPa
Yield Strength	145 MPa
Young’s Modulus of Elasticity	45 GPa
Brinell Hardness	70 BHN
Melting Point	550-640 °C
Thermal Conductivity	116 W/mK
Heat Capacity	900 J/g K
Price	40 $/kg

Density of Elektron 21

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Elektron 21 is 1800 kg/m³.

 

Example: Density

Calculate the height of a cube made of Elektron 21, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.822 m.

Mechanical Properties of Elektron 21 – UNS M12310

Strength of Elektron 21

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests.

See also: Strength of Materials

Ultimate Tensile Strength of Elektron 21

Ultimate tensile strength of Elektron 21 is 280 MPa.

Yield Strength of Elektron 21

Yield strength of Elektron 21 is 145 MPa.

Modulus of Elasticity of Elektron 21

The Young’s modulus of elasticity of Elektron 21 is 45 GPa.

Hardness of Elektron 21

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

Brinell hardness of Elektron 21 is approximately 70 BHN(converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Elektron 21. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 280 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A =280 x 10⁶ x 0.0001 = 28 000 N

Thermal Properties of Elektron 21 – UNS M12310

Elektron 21 – Melting Point

Melting point of Elektron 21 is 550-640 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Elektron 21 – Thermal Conductivity

Thermal conductivity of Elektron 21 is 116 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Elektron 21 – Specific Heat

Specific heat of Elektron 21 is 900 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 Elektron 21 with the thermal conductivity of k₁ = 116 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/116 + 1/30) = 7.43 W/m²K

The heat flux can be then calculated simply as: q = 7.43 [W/m²K] x 30 [K] = 222.84 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.84 [W/m²] x 30 [m²] = 6685.17 W

About Uranium Dioxide

Uranium Dioxide is a ceramic refractory uranium compound, in many cases used as a nuclear fuel. Most of LWRs use the uranium fuel, which is in the form of uranium dioxide (chemically UO2). 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. Uranium dioxide has significantly lower density than uranium in the metal form. Uranium dioxide has a density of 10.97 g/cm3, but this value may vary with fuel burnup, because at low burnup densification of pellets can occurs and at higher burnup swelling occurs.

Summary

Name	Uranium Dioxide
Phase at STP	solid
Density	10970 kg/m3
Ultimate Tensile Strength	N/A
Yield Strength	N/A
Young’s Modulus of Elasticity	N/A
Brinell Hardness	N/A
Melting Point	2847 °C
Thermal Conductivity	8.68 W/mK
Heat Capacity	235 J/g K
Price	60 $/kg

Density of Uranium Dioxide

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Uranium Dioxide is 10970 kg/m³.

 

Example: Density

Calculate the height of a cube made of Uranium Dioxide, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.45 m.

Thermal Properties of Uranium Dioxide

Uranium Dioxide – Melting Point

Melting point of Uranium Dioxide is 2847 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Uranium Dioxide – Thermal Conductivity

Thermal conductivity of Uranium Dioxide is 8.68 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Uranium Dioxide – Specific Heat

Specific heat of Uranium Dioxide is 235 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 Uranium Dioxide with the thermal conductivity of k₁ = 8.68 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/8.68 + 1/30) = 6.64 W/m²K

The heat flux can be then calculated simply as: q = 6.64 [W/m²K] x 30 [K] = 199.18 W/m²

The total heat loss through this wall will be: q_loss = q . A = 199.18 [W/m²] x 30 [m²] = 5975.52 W

About Graphene

Graphene is an allotrope of carbon consisting of a single layer of atoms hexagonally arranged in a two-dimensional lattice. Graphene is a substance with very interesting properties. Graphene has high thermal conductivity, high electrical conductivity, high elasticity and flexibility, high hardness and strength. These properties, together with the abundance of carbon in nature, have made graphene a very studied material with great possibilities.

Summary

Name	Graphene
Phase at STP	solid
Density	2270 kg/m3
Ultimate Tensile Strength	130000 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	1000 GPa
Brinell Hardness	N/A
Melting Point	3697 °C
Thermal Conductivity	4000 W/mK
Heat Capacity	N/A
Price	100 $/kg

Density of Graphene

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Graphene is 2270 kg/m³.

 

Example: Density

Calculate the height of a cube made of Graphene, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.761 m.

Mechanical Properties of Graphene

Strength of Graphene

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests.

See also: Strength of Materials

Ultimate Tensile Strength of Graphene

Ultimate tensile strength of Graphene is 130000 MPa.

Yield Strength of Graphene

Yield strength of Graphene is N/A.

Modulus of Elasticity of Graphene

The Young’s modulus of elasticity of Graphene is 1000 GPa.

 

Example: Strength

Assume a plastic rod, which is made of Graphene. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 130000 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 130000 x 10⁶ x 0.0001 = 13 000 000 N

Thermal Properties of Graphene

Graphene – Melting Point

Melting point of Graphene is 3697 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Graphene – Thermal Conductivity

Thermal conductivity of Graphene is 4000 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Graphene – Specific Heat

Specific heat of Graphene is 509 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 Graphene with the thermal conductivity of k₁ = 4000 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/4000 + 1/30) = 7.5 W/m²K

The heat flux can be then calculated simply as: q = 7.5 [W/m²K] x 30 [K] = 224.94 W/m²

The total heat loss through this wall will be: q_loss = q . A = 224.94 [W/m²] x 30 [m²] = 6748.1 W

About Tungsten Carbide

Tungsten carbide is a very dense carbide containing equal parts of tungsten and carbon atoms. In its most basic form, tungsten carbide is a fine gray powder, but it can be pressed and formed into shapes through a process called sintering for use in industrial machinery, cutting tools, abrasives, armor-piercing shells and jewellery. Tungsten carbide is approximately twice as stiff (high modulus of elasticity) as steel. Tungsten carbide has very high impact resistance and it has very high strength for a material so hard and rigid. Compressive strength is higher than virtually all melted and cast or forged metals and alloys. With temperature increase to 1400°F, tungsten carbide retains much of its room temperature hardness.

Summary

Name	Tungsten Carbide
Phase at STP	solid
Density	14500 kg/m3
Ultimate Tensile Strength	370 MPa
Yield Strength	330 MPa
Young’s Modulus of Elasticity	600 GPa
Brinell Hardness	25000 BHN
Melting Point	2867 °C
Thermal Conductivity	110 W/mK
Heat Capacity	292 J/g K
Price	300 $/kg

Density of Tungsten Carbide

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Tungsten Carbide is 14 500 kg/m³.

 

Example: Density

Calculate the height of a cube made of Tungsten Carbide, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.41 m.

Mechanical Properties of Tungsten Carbide

Strength of Tungsten Carbide

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests.

See also: Strength of Materials

Ultimate Tensile Strength of Tungsten Carbide

Ultimate tensile strength of Tungsten Carbide is 370 MPa.

Yield Strength of Tungsten Carbide

Yield strength of Tungsten Carbide is 330 MPa.

Modulus of Elasticity of Tungsten carbide

The Young’s modulus of elasticity of Tungsten Carbide is 600 MPa.

Hardness of Tungsten Carbide

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

Brinell hardness of tungsten carbide is approximately 25000 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Tungsten Carbide. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 370 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 370 x 10⁶ x 0.0001 = 37 000 N

Thermal Properties of Tungsten Carbide

Tungsten Carbide – Melting Point

Melting point of Tungsten Carbide is 2867 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Tungsten Carbide – Thermal Conductivity

Thermal conductivity of Tungsten Carbide is 110 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Tungsten Carbide – Specific Heat

Specific heat of Tungsten Carbide is 292 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 Tungsten Carbide with the thermal conductivity of k₁ =110 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/110 + 1/30) = 7.42 W/m²K

The heat flux can be then calculated simply as: q = 7.42 [W/m²K] x 30 [K] = 222.72 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.72 [W/m²] x 30 [m²] = 6681.67 W

About Carbon Fiber

Carbon fiber is a polymer that is is a very strong material that is also very lightweight. Carbon fibers have several advantages including high stiffness, high tensile strength, low weight, high chemical resistance, high temperature tolerance and low thermal expansion. Carbon fiber is five-times stronger than steel and twice as stiff. Though carbon fiber is stronger and stiffer than steel, it is lighter than steel; making it the ideal manufacturing material for many parts. Carbon fibers are usually combined with other materials to form a composite.

Summary

Name	Carbon Fiber
Phase at STP	solid
Density	2000 kg/m3
Ultimate Tensile Strength	4000 MPa
Yield Strength	2500 MPa
Young’s Modulus of Elasticity	500 GPa
Brinell Hardness	N/A
Melting Point	3657 °C
Thermal Conductivity	100 W/mK
Heat Capacity	800 J/g K
Price	22 $/kg

Density of Carbon Fiber

Typical densities of various substances are at atmospheric pressure. Density is defined as the mass per unit volume. It is an intensive property, which is mathematically defined as mass divided by volume:  ρ = m/V

In words, the density (ρ) of a substance is the total mass (m) of that substance divided by the total volume (V) occupied by that substance. The standard SI unit is kilograms per cubic meter (kg/m³). The Standard English unit is pounds mass per cubic foot (lbm/ft³).

Density of Carbon Fiber is 2000 kg/m³.

 

Example: Density

Calculate the height of a cube made of Carbon Fiber, which weighs one metric ton.

Solution:

Density is defined as the mass per unit volume. It is mathematically defined as mass divided by volume: ρ = m/V

As the volume of a cube is the third power of its sides (V = a³), the height of this cube can be calculated:

The height of this cube is then a = 0.793 m.

Mechanical Properties of Carbon Fiber

Strength of Carbon Fiber

In mechanics of materials, the strength of a material is its ability to withstand an applied load without failure or plastic deformation. Strength of materials basically considers the relationship between the external loads applied to a material and the resulting deformation or change in material dimensions. In designing structures and machines, it is important to consider these factors, in order that the material selected will have adequate strength to resist applied loads or forces and retain its original shape.

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), the Hooke’s law describes behaviour of a bar in the elastic region. The Young’s modulus of elasticity is the elastic modulus for tensile and compressive stress in the linear elasticity regime of a uniaxial deformation and is usually assessed by tensile tests.

See also: Strength of Materials

Ultimate Tensile Strength of Carbon Fiber

Ultimate tensile strength of Carbon Fibre is 4000 MPa.

Yield Strength of Carbon Fiber

Yield strength of Carbon Fibre is 2500 MPa.

Modulus of Elasticity of Carbon Fiber

The Young’s modulus of elasticity of Carbon Fibre is 500 MPa.

Hardness of Carbon Fiber

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratching. Brinell hardness test is one of indentation hardness tests, that has been developed for hardness testing. In Brinell tests, a hard, spherical indenter is forced under a specific load into the surface of the metal to be tested.

The Brinell hardness number (HB) is the load divided by the surface area of the indentation. The diameter of the impression is measured with a microscope with a superimposed scale. The Brinell hardness number is computed from the equation:

Brinell hardness of Carbon Fiber is approximately N/A.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Carbon Fiber. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 4000 MPa.

Solution:

Stress (σ) can be equated to the load per unit area or the force (F) applied per cross-sectional area (A) perpendicular to the force as:

therefore, the tensile force needed to achieve the ultimate tensile strength is:

F = UTS x A = 4000 x 10⁶ x 0.0001 = 400 000 N

Thermal Properties of Carbon Fiber

Carbon Fiber – Melting Point

Melting point of Carbon Fiber is 3657 °C.

Note that, these points are associated with the standard atmospheric pressure. In general, melting is a phase change of a substance from the solid to the liquid phase. The melting point of a substance is the temperature at which this phase change occurs. The melting point also defines a condition in which the solid and liquid can exist in equilibrium. For various chemical compounds and alloys, it is difficult to define the melting point, since they are usually a mixture of various chemical elements.

Carbon Fiber – Thermal Conductivity

Thermal conductivity of Carbon Fiber is 100 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called the thermal conductivity, k (or λ), measured in W/m.K. It is a measure of a substance’s ability to transfer heat through a material by conduction. Note that Fourier’s law applies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

The thermal conductivity of most liquids and solids varies with temperature. For vapors, it also depends upon pressure. In general:

Most materials are very nearly homogeneous, therefore we can usually write k = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

Carbon Fiber – Specific Heat

Specific heat of Carbon Fiber is 800 J/g K.

Specific heat, or specific heat capacity, is a property related to internal energy that is very important in thermodynamics. The intensive properties c_v and c_p are defined for pure, simple compressible substances as partial derivatives of the internal energy u(T, v) and enthalpy h(T, p), respectively:

where the subscripts v and p denote the variables held fixed during differentiation. The properties c_v and c_p are referred to as specific heats (or heat capacities) because under certain special conditions they relate the temperature change of a system to the amount of energy added by heat transfer. Their SI units are J/kg K or J/mol K.

 

Example: Heat transfer calculation

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.

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 Carbon Fiber with the thermal conductivity of k₁ = 100 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.).

Calculate the heat flux (heat loss) through this 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.

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 + 0.15/100 + 1/30) = 7.42 W/m²K

The heat flux can be then calculated simply as: q = 7.42 [W/m²K] x 30 [K] = 222.5 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.5 [W/m²] x 30 [m²] = 6674.91 W