Austenitic Stainless Steel – Density – Strength – Hardness – Melting Point

About Austenitic Stainless Steel – Type 304

Type 304 stainless steel (containing 18%-20% chromium and 8%-10.5% nickel) is the most common stainless steel. It is also known as “18/8” stainless steel because of its composition, which includes 18% chromium and 8% nickel. This alloy resists most types of corrosion. It is an austenitic stainless steel and it has also excellent cryogenic properties, and good high-temperature strength as well as good forming and welding properties. It is less electrically and thermally conductive than carbon steel and is essentially non-magnetic.

Type 304L stainless steel, which is widely used in nuclear industry, is an extra-low carbon version of the 304 steel alloy. This grade has slightly lower mechanical properties than the standard 304 grade, but is still widely used thanks to its versatility. The lower carbon content in 304L minimizes deleterious or harmful carbide precipitation as a result of welding. 304L can, therefore, be used “as welded” in severe corrosion environments, and it eliminates the need for annealing. Grade 304 has also good oxidation resistance in intermittent service to 870 °C, and in continuous service to 925 °C.

The body of the reactor vessel is constructed of a high-quality low-alloy carbon steel, and all surfaces that come into contact with reactor coolant are clad with a minimum of about 3 to 10 mm of austenitic stainless steel in order to minimize corrosion. Since grade 304L does not require post-weld annealing, it is extensively used in heavy gauge components.

austenitic steel properties density strength price

Summary

Name Austenitic Stainless Steel
Phase at STP solid
Density 7850 kg/m3
Ultimate Tensile Strength 515 MPa
Yield Strength 205 MPa
Young’s Modulus of Elasticity 193 GPa
Brinell Hardness 201 BHN
Melting Point 1450 °C
Thermal Conductivity 20 W/mK
Heat Capacity 500 J/g K
Price 2 $/kg

Density of Austenitic Stainless 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 Austenitic Stainless Steel is 7850 kg/m3.

 

Example: Density

Calculate the height of a cube made of Austenitic Stainless 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:

density of material - equation

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

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Austenitic Stainless Steel

Strength of Austenitic Stainless 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. 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 Austenitic Stainless Steel

Ultimate tensile strength of Austenitic Stainless Steel is 280 MPa.

Yield Strength of Austenitic Stainless Steel

Yield strength of Austenitic Stainless Steel is 145 MPa.

Modulus of Elasticity of Austenitic Stainless Steel

The Young’s modulus of elasticity of Austenitic Stainless Steel is 45 GPa.

Hardness of Austenitic Stainless Steel

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Austenitic Stainless Steel is approximately 70 BHN(converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Austenitic Stainless 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 = 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:

strength of material - equation

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

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Austenitic Stainless Steel

Austenitic Stainless Steel – Melting Point

Melting point of Austenitic Stainless Steel 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.

Austenitic Stainless Steel – Thermal Conductivity

Thermal conductivity of Austenitic Stainless Steel 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:

thermal conductivity - definition

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.

Austenitic Stainless Steel – Specific Heat

Specific heat of Austenitic Stainless Steel 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 cv and cp 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 cv and cp 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.

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

 

Aluminium Bronze – Density – Strength – Hardness – Melting Point

About Aluminium Bronze

The aluminum bronzes are a family of copper-based alloys offering a combination of mechanical and chemical properties unmatched by any other alloy series. They contain about 5 to 12% of aluminium. In addition, aluminium bronzes also contain nickel, silicon, manganese, and iron. They have excellent strength, similar to that of low alloy steels, and excellent corrosion resistance especially in seawater and similar environments, where the alloys often outperform many stainless steels. Their excellent resistance to corrosion results from the aluminium in the alloys, which reacts with atmospheric oxygen to form a thin, tough surface layer of alumina (aluminium oxide) which acts as a barrier to corrosion of the copper-rich alloy. They are found in wrought and cast form. Aluminium bronzes are usually golden in color.

aluminium bronze properties density strength price

Summary

Name Aluminium Bronze
Phase at STP solid
Density 7640 kg/m3
Ultimate Tensile Strength 550 MPa
Yield Strength 250 MPa
Young’s Modulus of Elasticity 110 GPa
Brinell Hardness 170 BHN
Melting Point 1030 °C
Thermal Conductivity 59 W/mK
Heat Capacity 380 J/g K
Price 9 $/kg

Density of Aluminium Bronze

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 Aluminium Bronze is 7640 kg/m3.

 

Example: Density

Calculate the height of a cube made of Aluminium Bronze, 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:

density of material - equation

The height of this cube is then a = 0.508 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Aluminium Bronze

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 Aluminium Bronze

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 aluminium bronze – UNS C95400 is about 550 MPa.

Yield Strength - Ultimate Tensile Strength - Table of MaterialsThe 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 aluminium bronze – UNS C95400 is about 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 aluminium bronze – UNS C95400 is about 110 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 Bronze

Brinell hardness of aluminium bronze – UNS C95400 is approximately 170 MPa. The hardness of aluminum bronzes increases with aluminum (and other alloy) content as well as with stresses caused through cold working.

Brinell hardness number

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.

 

Example: Strength

Assume a plastic rod, which is made of Aluminium Bronze. 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 = 550 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:

strength of material - equation

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

F = UTS x A = 550 x 106 x 0.0001 = 55 000 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Aluminium Bronze

Thermal properties of materials refer to the response of materials to changes in their temperature and to the application of heat. As a solid absorbs 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 Bronze

Melting point of aluminium bronze – UNS C95400 is around 1030°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 Bronze

The thermal conductivity of aluminium bronze – UNS C95400 is 59 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:

thermal conductivity - definition

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

Aluminium Bronze - Thermal ConductivityThermal 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 Aluminium Bronze with the thermal conductivity of k1 = 59 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/59 + 1/30) = 7.36 W/m2K

The heat flux can be then calculated simply as: q = 7.36 [W/m2K] x 30 [K] = 220.79 W/m2

The total heat loss through this wall will be: qloss = q . A = 220.79 [W/m2] x 30 [m2] = 6623.701 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Tungsten – Density – Strength – Hardness – Melting Point

About Pure Tungsten

Tungsten is a rare metal found naturally on Earth almost exclusively in chemical compounds. Tungsten is an intrinsically brittle and hard material, making it difficult to work.

pure tungsten properties density strength price

Summary

Name Pure Tungsten
Phase at STP solid
Density 19250 kg/m3
Ultimate Tensile Strength 980 MPa
Yield Strength 750 MPa
Young’s Modulus of Elasticity 750 GPa
Brinell Hardness 3695 BHN
Melting Point 1687 °C
Thermal Conductivity 170 W/mK
Heat Capacity 130 J/g K
Price 110 $/kg

Density of Pure Tungsten

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 Pure Tungsten is 19250 kg/m3.

 

Example: Density

Calculate the height of a cube made of Pure Tungsten, 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:

density of material - equation

The height of this cube is then a = 0.373 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Pure Tungsten

Strength of Pure Tungsten

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 Pure Tungsten

Ultimate tensile strength of Pure Tungsten is 980 MPa.

Yield Strength of Pure Tungsten

Yield strength of Pure Tungsten is 750 MPa.

Modulus of Elasticity of Pure Tungsten

The Young’s modulus of elasticity of Pure Tungsten is 750 GPa.

Hardness of Pure Tungsten

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Pure Tungsten is approximately 2570 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Pure Tungsten. 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 = 980 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:

strength of material - equation

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

F = UTS x A = 980 x 106 x 0.0001 = 98 000 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Pure Tungsten

Pure Tungsten – Melting Point

Melting point of Pure Tungsten is 3695 °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.

Pure Tungsten – Thermal Conductivity

Thermal conductivity of Pure Tungsten is 170 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:

thermal conductivity - definition

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.

Pure Tungsten – Specific Heat

Specific heat of Pure Tungsten is 130 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 cv and cp 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 cv and cp 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

Pure Tungsten - Thermal ConductivityThermal 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 Pure Tungsten with the thermal conductivity of k1 = 170 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/170 + 1/30) = 7.45 W/m2K

The heat flux can be then calculated simply as: q = 7.45 [W/m2K] x 30 [K] = 223.52 W/m2

The total heat loss through this wall will be: qloss = q . A = 223.52 [W/m2] x 30 [m2] = 6705.63 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Tungsten-Rhenium Alloy – Density – Strength – Hardness – Melting Point

About Tungsten-Rhenium Alloy

Tungsten and rhenium are both refractory metals. These metals 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. These metals are usually combined together to obtain desired fabricability, thermal and mechanical properties. Powder-metallurgy methods can be used in consolidating tungsten-rhenium alloys. Up to 22% rhenium is alloyed with tungsten to improve its high temperature strength and corrosion resistance. The hardness of as-cast W – 30Re alloy is around 500 BHN. This hardness strongly depends on rhenium content.

tungsten rhenium alloy properties density strength price

Summary

Name Tungsten-Rhenium Alloy
Phase at STP solid
Density 19700 kg/m3
Ultimate Tensile Strength 2100 MPa
Yield Strength N/A
Young’s Modulus of Elasticity 400 GPa
Brinell Hardness 500 BHN
Melting Point 3027 °C
Thermal Conductivity 70 W/mK
Heat Capacity 140 J/g K
Price 3000 $/kg

Density of Tungsten-Rhenium 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 Tungsten-Rhenium Alloy is 19700 kg/m3.

 

Example: Density

Calculate the height of a cube made of Tungsten-Rhenium 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:

density of material - equation

The height of this cube is then a = 0.37 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Tungsten-Rhenium Alloy

Strength of Tungsten-Rhenium 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 Tungsten-Rhenium Alloy

Ultimate tensile strength of Tungsten-Rhenium Alloy is 2100 MPa.

Yield Strength of Tungsten-Rhenium Alloy

Yield strength of Tungsten-Rhenium Alloy is N/A.

Modulus of Elasticity of Tungsten-Rhenium Alloy

The Young’s modulus of elasticity of Tungsten-Rhenium Alloy is 400 GPa.

Hardness of Tungsten-Rhenium Alloy

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Tungsten-Rhenium Alloy is approximately 500 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Tungsten-Rhenium 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 = 2100 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:

strength of material - equation

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

F = UTS x A = 2100 x 106 x 0.0001 = 210 000 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Tungsten-Rhenium Alloy

Tungsten-Rhenium Alloy – Melting Point

Melting point of Tungsten-Rhenium Alloy is 3027 °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-Rhenium Alloy – Thermal Conductivity

Thermal conductivity of Tungsten-Rhenium Alloy is 70 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:

thermal conductivity - definition

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-Rhenium Alloy – Specific Heat

Specific heat of Tungsten-Rhenium Alloy is 140 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 cv and cp 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 cv and cp 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

Tungsten-Rhenium Alloy - Thermal ConductivityThermal 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-Rhenium Alloy with the thermal conductivity of k1 = 70 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/70 + 1/30) = 7.38 W/m2K

The heat flux can be then calculated simply as: q = 7.38 [W/m2K] x 30 [K] = 221.44 W/m2

The total heat loss through this wall will be: qloss = q . A = 221.44 [W/m2] x 30 [m2] = 6643.23 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Zirconium-Niobium Alloy – Density – Strength – Hardness – Melting Point

About Zirconium-Niobium Alloy

Zirconium alloys with niobium are used as claddings of fuel elements of VVER and RBMK reactors. These alloys are the basis material of assembly channel of RBMK reactor. The Zr + 1% Nb alloy of type N-1 E-110 is used for fuel element claddings, the Zr + 2.5% Nb alloy of type E-125 is applied for tubes of assembly channels.

The cladding material for the new 17×17 fuel designs is based also on the zirconium-niobium alloys (e.g. Optimized ZIRLO material), which has been demonstrated to have improved corrosion resistance compared with prior fuel cladding materials. The optimized tin level provides a reduced corrosion rate while maintaining the benefits of mechanical strength and resistance to accelerated corrosion from abnormal chemistry conditions.

zirconium niobium alloy properties density strength price

Summary

Name Zirconium-Niobium Alloy
Phase at STP solid
Density 6560 kg/m3
Ultimate Tensile Strength 514 MPa
Yield Strength 381 MPa
Young’s Modulus of Elasticity 99 GPa
Brinell Hardness 89 BHN
Melting Point 1850 °C
Thermal Conductivity 18 W/mK
Heat Capacity 285 J/g K
Price 25 $/kg

Density of Zirconium-Niobium 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 Zirconium-Niobium Alloy is 6560 kg/m3.

 

Example: Density

Calculate the height of a cube made of Zirconium-Niobium 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:

density of material - equation

The height of this cube is then a = 0.534 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Zirconium-Niobium Alloy

Strength of Zirconium-Niobium 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. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of Zirconium-Niobium Alloy is about 514 MPa.

Yield Strength - Ultimate Tensile Strength - Table of MaterialsThe 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 Zirconium-Niobium Alloy is about 381 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 Zirconium-Niobium Alloy is about 99 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 Zirconium-Niobium Alloy

Rockwell hardness of Zirconium-Niobium Alloy is approximately 89 HRB.

Brinell hardness number

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.

 

Example: Strength

Assume a plastic rod, which is made of Zirconium-Niobium 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 = 514 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:

strength of material - equation

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

F = UTS x A = 514 x 106 x 0.0001 = 51 400 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Zirconium-Niobium Alloy

Thermal properties of materials refer to the response of materials to changes in their temperature and to the application of heat. As a solid absorbs 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 Zirconium-Niobium Alloy

Melting point of Zirconium-Niobium Alloy is around 1850°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 Zirconium-Niobium Alloy

Zirconium alloys have lower thermal conductivity (about 18 W/m.K) than pure zirconium metal (about 22 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:

thermal conductivity - definition

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

Zirconium-Niobium Alloy - Thermal ConductivityThermal 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 Zirconium-Niobium Alloy with the thermal conductivity of k1 = 18 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/18 + 1/30) = 7.06 W/m2K

The heat flux can be then calculated simply as: q = 7.06 [W/m2K] x 30 [K] = 211.77 W/m2

The total heat loss through this wall will be: qloss = q . A = 211.77 [W/m2] x 30 [m2] = 6352.94 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Zirconium-Tin Alloy – Density – Strength – Hardness – Melting Point

About Zirconium-Tin Alloy

Zirconium alloys, in which tin is the basic alloying element, provides improvement of their mechanical properties, have a wide distribution in the USA. A common subgroup has the trade mark Zircaloy. In case of zirconium-tin alloys, the decrease of corrosion resistance in water and steam is taken place that resulted in the need for additional alloying.

Zirconium alloys with niobium are used as claddings of fuel elements of VVER and RBMK reactors. These alloys are the basis material of assembly channel of RBMK reactor. The Zr + 1% Nb alloy of type N-1 E-110 is used for fuel element claddings, the Zr + 2.5% Nb alloy of type E-125 is applied for tubes of assembly channels.

zirconium tin alloy properties density strength price

Summary

Name Zirconium-Tin Alloy
Phase at STP solid
Density 6560 kg/m3
Ultimate Tensile Strength 514 MPa
Yield Strength 381 MPa
Young’s Modulus of Elasticity 99 GPa
Brinell Hardness 89 BHN
Melting Point 1850 °C
Thermal Conductivity 18 W/mK
Heat Capacity 285 J/g K
Price 25 $/kg

Density of Zirconium-Tin 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 Zirconium-Tin Alloy is 6560 kg/m3.

 

Example: Density

Calculate the height of a cube made of Zirconium-Tin 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:

density of material - equation

The height of this cube is then a = 0.534 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Zirconium-Tin Alloy

Strength of Zirconium-Tin 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. Strength of a material is its ability to withstand this applied load without failure or plastic deformation.

Ultimate Tensile Strength

Ultimate tensile strength of Zirconium-Tin Alloy is about 514 MPa.

Yield Strength - Ultimate Tensile Strength - Table of MaterialsThe 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 Zirconium-Tin Alloy is about 381 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 Zirconium-Tin Alloy is about 99 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 Zirconium-Tin Alloy

Rockwell hardness of Zirconium-Tin Alloy is approximately 89 HRB.

Brinell hardness number

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.

 

Example: Strength

Assume a plastic rod, which is made of Zirconium-Tin 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 = 514 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:

strength of material - equation

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

F = UTS x A = 514 x 106 x 0.0001 = 51 400 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Zirconium-Tin Alloy

Thermal properties of materials refer to the response of materials to changes in their temperature and to the application of heat. As a solid absorbs 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 Zirconium-Tin Alloy

Melting point of Zirconium-Tin Alloy is around 1850°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 Zirconium-Tin Alloy

Zirconium alloys have lower thermal conductivity (about 18 W/m.K) than pure zirconium metal (about 22 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:

thermal conductivity - definition

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

Zirconium-Tin Alloy - Thermal ConductivityThermal 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 Zirconium-Tin Alloy with the thermal conductivity of k1 = 18 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/18 + 1/30) = 7.06 W/m2K

The heat flux can be then calculated simply as: q = 7.06 [W/m2K] x 30 [K] = 211.77 W/m2

The total heat loss through this wall will be: qloss = q . A = 211.77 [W/m2] x 30 [m2] = 6352.94 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Aerogel – Density – Heat Capacity – Thermal Conductivity

About Aerogel

Aerogel is a synthetic porous ultralight solid material derived from a gel, in which the liquid component of the gel has been replaced with a gas (during a supercritical drying process). Aerogels can be made from a variety of chemical compounds, but the base material for aerogel is usually silicon.

aerogel properties density strength price

Summary

Name Aerogel
Phase at STP solid
Density 10 kg/m3
Ultimate Tensile Strength 0.08 MPa
Yield Strength N/A
Young’s Modulus of Elasticity 0.005 GPa
Brinell Hardness N/A
Melting Point 1197 °C
Thermal Conductivity 0.01 W/mK
Heat Capacity 1900 J/g K
Price 12 $/kg

Density of Aerogel

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 Aerogel is 10 kg/m3.

 

Example: Density

Calculate the height of a cube made of Aerogel, 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:

density of material - equation

The height of this cube is then a = 4.642 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Aerogel

Strength of Aerogel

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 Aerogel

Ultimate tensile strength of Aerogel is 0.08 MPa.

Yield Strength of Aerogel

Yield strength of Aerogel is N/A.

Modulus of Elasticity of Aerogel

The Young’s modulus of elasticity of Aerogel is 0.005 GPa.

Hardness of Aerogel

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Aerogel is approximately N/A.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Aerogel. 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 = 0.08 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:

strength of material - equation

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

F = UTS x A = 0.08 x 106 x 0.0001 = 0.08 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Aerogel

Aerogel – Melting Point

Melting point of Aerogel is 1197 °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.

Aerogel – Thermal Conductivity

Thermal conductivity of Aerogel is 0.01 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:

thermal conductivity - definition

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.

Aerogel – Specific Heat

Specific heat of Aerogel is 1900 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 cv and cp 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 cv and cp 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

Aerogel - Thermal ConductivityThermal 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 Aerogel with the thermal conductivity of k1 = 0.01 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/0.01 + 1/30) = 0.066 W/m2K

The heat flux can be then calculated simply as: q = 0.066 [W/m2K] x 30 [K] = 1.98 W/m2

The total heat loss through this wall will be: qloss = q . A = 1.98 [W/m2] x 30 [m2] = 59.47 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Glass Wool – Density – Heat Capacity – Thermal Conductivity

About Glass Wool

Glass wool (originally known also as fiberglass) is an insulating material made from fibres of glass arranged using a binder into a texture similar to wool. Glass wool and stone wool are produced from mineral fibres and are therefore often referred to as ‘mineral wools’. Mineral wool is a general name for fiber materials that are formed by spinning or drawing molten minerals. Glass wool is a furnace product of molten glass at a temperature of about 1450 °C.

glass wool properties density strength price

Summary

Name Glass Wool
Phase at STP solid
Density 20 kg/m3
Ultimate Tensile Strength 0.02 MPa
Yield Strength N/A
Young’s Modulus of Elasticity N/A
Brinell Hardness N/A
Melting Point 1227 °C
Thermal Conductivity 0.03 W/mK
Heat Capacity 840 J/g K
Price 3 $/kg

Density of Glass Wool

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 Glass Wool is 20 kg/m3.

 

Example: Density

Calculate the height of a cube made of Glass Wool, 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:

density of material - equation

The height of this cube is then a = 3.684 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Glass Wool

Strength of Glass Wool

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 Glass Wool

Ultimate tensile strength of Glass Wool is 0.02 MPa.

Yield Strength of Glass Wool

Yield strength of Glass Wool is N/A.

Modulus of Elasticity of Glass Wool

The Young’s modulus of elasticity of Glass Wool is N/A.

Hardness of Glass Wool

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Glass Wool is approximately N/A.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Glass Wool. 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 = 0.02 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:

strength of material - equation

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

F = UTS x A = 0.02 x 106 x 0.0001 = 2 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Glass Wool

Glass Wool – Melting Point

Melting point of Glass Wool is 1227 °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.

Glass Wool – Thermal Conductivity

Thermal conductivity of Glass Wool is 0.03 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:

thermal conductivity - definition

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.

Glass Wool – Specific Heat

Specific heat of Glass Wool is 840 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 cv and cp 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 cv and cp 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

Glass Wool - Thermal ConductivityThermal 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 Glass Wool with the thermal conductivity of k1 = 0.03 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/0.03 + 1/30) = 0.19 W/m2K

The heat flux can be then calculated simply as: q = 0.19 [W/m2K] x 30 [K] = 5.84 W/m2

The total heat loss through this wall will be: qloss = q . A = 5.84 [W/m2] x 30 [m2] = 175.32 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Quartz – Density – Heat Capacity – Thermal Conductivity

About Quartz

Quartz is very abundant mineral of many varieties that consists primarily of silica, or silicon dioxide (SiO2). There are many different varieties of quartz, several of which are semi-precious gemstones. Amethyst is a form of quartz that ranges from a bright vivid violet to dark or dull lavender shade. Citrine is a variety of quartz whose color ranges from a pale yellow to brown due to ferric impurities. Quartz is the mineral defining the value of 7 on the Mohs scale of hardness, a qualitative scratch method for determining the hardness of a material to abrasion.

quartz properties density strength price

Summary

Name Quartz
Phase at STP solid
Density 2650 kg/m3
Ultimate Tensile Strength 48 MPa
Yield Strength N/A
Young’s Modulus of Elasticity 37 GPa
Brinell Hardness 7 Mohs
Melting Point 1667 °C
Thermal Conductivity 3 W/mK
Heat Capacity 741 J/g K
Price 20 $/kg

Density of Quartz

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 Quartz is 2650 kg/m3.

 

Example: Density

Calculate the height of a cube made of Quartz, 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:

density of material - equation

The height of this cube is then a = 0.723 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Quartz

Strength of Quartz

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 Quartz

Ultimate tensile strength of Quartz is 48 MPa.

Yield Strength of Quartz

Yield strength of Quartz is N/A.

Modulus of Elasticity of Quartz

The Young’s modulus of elasticity of Quartz is 37 GPa.

Hardness of Quartz

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Hardness of Quartz is approximately 7 Mohs.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Quartz. 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 = 48 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:

strength of material - equation

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

F = UTS x A = 48 x 106 x 0.0001 = 4 800 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Quartz

Quartz – Melting Point

Melting point of Quartz is 1667 °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.

Quartz – Thermal Conductivity

Thermal conductivity of Quartz is 3 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:

thermal conductivity - definition

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.

Quartz – Specific Heat

Specific heat of Quartz is 741 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 cv and cp 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 cv and cp 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

Quartz - Thermal ConductivityThermal 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 Quartz with the thermal conductivity of k1 = 3 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/3 + 1/30) = 5.45 W/m2K

The heat flux can be then calculated simply as: q = 5.45 [W/m2K] x 30 [K] = 163.63 W/m2

The total heat loss through this wall will be: qloss = q . A = 163.63 [W/m2] x 30 [m2] = 4909.09 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity

Stone Wool – Density – Heat Capacity – Thermal Conductivity

About Stone Wool

Stone Wool, also known as rock wool, is based on natural minerals present in large quantities throughout the earth, e.g. volcanic rock, typically basalt or dolomite. Next to raw materials, also recycled rock wool can be added to the process as well as slag residues from the metal industry. It combines mechanical resistance with good thermal performance, fire safety and high temperature suitability.

stone wool properties density strength price

Summary

Name Stone Wool
Phase at STP solid
Density 20 kg/m3
Ultimate Tensile Strength 0.02 MPa
Yield Strength N/A
Young’s Modulus of Elasticity N/A
Brinell Hardness N/A
Melting Point 997 °C
Thermal Conductivity 0.03 W/mK
Heat Capacity 700 J/g K
Price 3 $/kg

Density of Stone Wool

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 Stone Wool is 20 kg/m3.

 

Example: Density

Calculate the height of a cube made of Stone Wool, 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:

density of material - equation

The height of this cube is then a = 3.684 m.

Density of Materials

Material Table - Density of Materials

 

Mechanical Properties of Stone Wool

Strength of Stone Wool

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 Stone Wool

Ultimate tensile strength of Stone Wool is 0.02 MPa.

Yield Strength of Stone Wool

Yield strength of Stone Wool is N/A.

Modulus of Elasticity of Stone Wool

The Young’s modulus of elasticity of Stone Wool is N/A.

Hardness of Stone Wool

In materials science, hardness is the ability to withstand surface indentation (localized plastic deformation) and scratchingBrinell 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 number - definition

Brinell hardness of Stone Wool is approximately N/A.

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Stone Wool. 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 = 0.02 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:

strength of material - equation

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

F = UTS x A = 0.02 x 106 x 0.0001 = 2 N

Strength of Materials

Material Table - Strength of Materials

Elasticity of Materials

Material Table - Elasticity of Materials

Hardness of Materials

Material Table - Hardness of Materials  

Thermal Properties of Stone Wool

Stone Wool – Melting Point

Melting point of Stone Wool is 997 °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.

Stone Wool – Thermal Conductivity

Thermal conductivity of Stone Wool is 0.03 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:

thermal conductivity - definition

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.

Stone Wool – Specific Heat

Specific heat of Stone Wool is 700 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 cv and cp 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 cv and cp 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

Stone Wool - Thermal ConductivityThermal 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 Stone Wool with the thermal conductivity of k1 = 0.03 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 coefficientknown as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

Heat transfer calculation - 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:

Heat transfer calculation - U-factor

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/0.03 + 1/30) = 0.19 W/m2K

The heat flux can be then calculated simply as: q = 0.19 [W/m2K] x 30 [K] = 5.84 W/m2

The total heat loss through this wall will be: qloss = q . A = 5.84 [W/m2] x 30 [m2] = 175.33 W

Melting Point of Materials

Material Table - Melting Point

Thermal Conductivity of Materials

Material Table - Thermal Conductivity

Heat Capacity of Materials

Material Table - Heat Capacity