Nick Connor

About Hastelloy

Hastelloy is a nickel-molybdenum-chromium wrought superalloy that is generally considered a versatile corrosion-resistant alloy. This alloy resists the formation of grain-boundary precipitates in the weld heat affected zone, thus making it suitable for most chemical process applications in the as-welded condition. Super alloys have good oxidation and creep resistance and can be strengthened by precipitation hardening, solid-solution hardening and work hardening methods. They can also function under high mechanical stress and high temperatures and also in places that require high surface stability.

Summary

Name	Hastelloy
Phase at STP	solid
Density	8890 kg/m3
Ultimate Tensile Strength	600 MPa
Yield Strength	300 MPa
Young’s Modulus of Elasticity	205 GPa
Brinell Hardness	180 BHN
Melting Point	1327 °C
Thermal Conductivity	10.2 W/mK
Heat Capacity	420 J/g K
Price	50 $/kg

Density of Hastelloy

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 Hastelloy is 8890 kg/m³.

 

Example: Density

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

Solution:

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

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

The height of this cube is then a = 0.483 m.

Mechanical Properties of Hastelloy

Strength of Hastelloy

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 Hastelloy

Ultimate tensile strength of Hastelloy is 600 MPa.

Yield Strength of Hastelloy

Yield strength of Hastelloy is 300 MPa.

Modulus of Elasticity of Hastelloy

The Young’s modulus of elasticity of Hastelloy is 205 GPa.

Hardness of Hastelloy

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

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

Brinell hardness of Hastelloy is approximately 180 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Hastelloy. 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 = 600 MPa.

Solution:

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

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

F = UTS x A = 600 x 10⁶ x 0.0001 = 60 000 N

Thermal Properties of Hastelloy

Hastelloy – Melting Point

Melting point of Hastelloy is 1327 °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.

Hastelloy – Thermal Conductivity

Thermal conductivity of Hastelloy is 10.2 W/(m·K).

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

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

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

Hastelloy – Specific Heat

Specific heat of Hastelloy is 420 J/g K.

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Hastelloy with the thermal conductivity of k₁ = 10.2 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/10.2 + 1/30) = 6.75 W/m²K

The heat flux can be then calculated simply as: q = 6.75 [W/m²K] x 30 [K] = 202.65 W/m²

The total heat loss through this wall will be: q_loss = q . A = 202.65 [W/m²] x 30 [m²] = 6079.47 W

Properties and prices of other materials

material-table-in-8k-resolution

About Mo-25 Re Alloy

Molybdenum 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. The molybdenum-rhenium alloys exhibiting good low-temperature ductility. The Mo-35Re alloy also has a low ductile-to-brittle transition temperature in the as-cast condition. The alloy is extremely difficult to fracture by hammering as-cast material at temperatures down to -196°C.

Summary

Name	Mo-25 Re Alloy
Phase at STP	solid
Density	11200 kg/m3
Ultimate Tensile Strength	1100 MPa
Yield Strength	N/A
Young’s Modulus of Elasticity	360 GPa
Brinell Hardness	350 BHN
Melting Point	2527 °C
Thermal Conductivity	70 W/mK
Heat Capacity	220 J/g K
Price	3000 $/kg

Density of Mo-25 Re 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 Mo-25 Re Alloy is 11200 kg/m³.

 

Example: Density

Calculate the height of a cube made of Mo-25 Re Alloy, which weighs one metric ton.

Solution:

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

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

The height of this cube is then a = 0.447 m.

Mechanical Properties of Mo-25 Re Alloy

Strength of Mo-25 Re 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 Mo-25 Re Alloy

Ultimate tensile strength of Mo-25 Re Alloy is 1100 MPa.

Yield Strength of Mo-25 Re Alloy

Yield strength of Mo-25 Re Alloy is N/A.

Modulus of Elasticity of Mo-25 Re Alloy

The Young’s modulus of elasticity of Mo-25 Re Alloy is 360 GPa.

Hardness of Mo-25 Re Alloy

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

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

Brinell hardness of Mo-25 Re Alloy is approximately 350 BHN (converted).

See also: Hardness of Materials

 

Example: Strength

Assume a plastic rod, which is made of Mo-25 Re 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 = 1100 MPa.

Solution:

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

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

F = UTS x A = 1100 x 10⁶ x 0.0001 = 110 000 N

Thermal Properties of Mo-25 Re Alloy

Mo-25 Re Alloy – Melting Point

Melting point of Mo-25 Re Alloy is 2527 °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.

Mo-25 Re Alloy – Thermal Conductivity

Thermal conductivity of Mo-25 Re 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:

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.

Mo-25 Re Alloy – Specific Heat

Specific heat of Mo-25 Re Alloy is 220 J/g K.

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Mo-25 Re 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 coefficient, known as a U-factor. The U-factor is defined by an expression analogous to Newton’s law of cooling:

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/70 + 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: q_loss = q . A = 221.44 [W/m²] x 30 [m²] = 6643.23 W

Properties and prices of other materials

material-table-in-8k-resolution

About White Iron – White Cast Iron

As was written, cast irons are one of the most complex alloys used in industry. Because of the higher carbon content, the structure of cast iron, as opposed to that of steel, exhibits a carbon-rich phase. Depending primarily on composition, cooling rate, and melt treatment, the carbon-rich phase can solidify with formation of either a stable (austenite-graphite) or a metastable (austenite-Fe₃C) eutectic.

With a lower silicon content (containing less than 1.0 wt% Si – graphitizing agent) and faster cooling rate, the carbon in cast iron precipitates out of the melt as the metastable phase cementite, Fe₃C, rather than graphite. The product of this solidification is known as white cast iron (also known as chilled irons). White cast irons are hard, brittle, and unmachinable, while gray irons with softer graphite are reasonably strong and machinable. A fracture surface of this alloy has a white appearance, and thus it is termed white cast iron. It is difficult to cool thick castings fast enough to solidify the melt as white cast iron all the way through. However, rapid cooling can be used to solidify a shell of white cast iron, after which the remainder cools more slowly to form a core of gray cast iron. This type of casting, sometimes referred to as a “chilled casting” has a harder outer surface and a tougher inner core.

Summary

Name	White Iron
Phase at STP	solid
Density	7770 kg/m3
Ultimate Tensile Strength	350 MPa
Yield Strength	N/A MPa
Young’s Modulus of Elasticity	175 GPa
Brinell Hardness	470 BHN
Melting Point	1260 °C
Thermal Conductivity	15-30 W/mK
Heat Capacity	540 J/g K
Price	1.5 $/kg

Density of White Iron

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

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

Density of White Iron is 7770 kg/m³.

 

Example: Density

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

Solution:

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

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

The height of this cube is then a = 0.505 m.

Mechanical Properties of White Cast Iron

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

Strength of White Cast Iron

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

Ultimate Tensile Strength of White Cast Iron

Ultimate tensile strength of martensitic white cast iron (ASTM A532 Class 1 Type A) is 350 MPa.

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

Young’s Modulus of Elasticity

Young’s modulus of elasticity of martensitic white cast iron (ASTM A532 Class 1 Type A) is 175 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 White Cast Iron

Brinell hardness of white cast iron (ASTM A532 Class 1 Type A) is approximately 470 MPa.

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

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

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

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

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

 

Example: Strength

Assume a plastic rod, which is made of Boron Carbide. This plastic rod has a cross-sectional area of 1 cm². Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 350 MPa.

Solution:

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

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

F = UTS x A = 350 x 10⁶ x 0.0001 = 35 000 N

Thermal Properties of White Cast Iron

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

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

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

Melting Point of White Cast Iron

Melting point of martensitic white cast iron (ASTM A532 Class 1 Type A) steel is around 1260°C.

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

Thermal Conductivity of White Cast Iron

The thermal conductivity of martensitic white cast iron (ASTM A532 Class 1 Type A) is 15 – 30 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

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

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/30 + 1/30) = 7.23 W/m²K

The heat flux can be then calculated simply as: q = 7.23 [W/m²K] x 30 [K] = 216.87 W/m²

The total heat loss through this wall will be: q_loss = q . A = 216.87 [W/m²] x 30 [m²] = 6506.02 W

About High-Carbon Steel

High-carbon steel has approximately 0.60 to 1.00% carbon content. Hardness is higher than the other grades but ductility decreases. They are almost always used in a hardened and tempered condition and, as such, are especially wear resistant and capable of holding a sharp cutting edge. Therefore, high carbon steels could be used for springs, rope wires, hammers, screwdrivers, wrenches and knives. The 10xx series (e.g. 1095 steel) is the most popular choice for carbon steel used in knives as well as katanas. AISI 1095 carbon steel is brittle, and has high hardness and strength. 1095 steel, when used in knives, holds a great edge and is very easy to sharpen. However, the properties of this type of steel give it a tendency to easily rust if not oiled and deliberately cared for.

Summary

Name	High-carbon Steel
Phase at STP	solid
Density	7850 kg/m3
Ultimate Tensile Strength	685 MPa
Yield Strength	525 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	200 BHN
Melting Point	1515 °C
Thermal Conductivity	50 W/mK
Heat Capacity	490 J/g K
Price	1 $/kg

 

Density of High-Carbon 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 High-Carbon Steel is 7850 kg/m³.

 

Example: Density

Calculate the height of a cube made of High-Carbon Steel, which weighs one metric ton.

Solution:

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

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

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

Mechanical Properties of High-Carbon Steel

Strength of High-Carbon Steel

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

Ultimate Tensile Strength

Ultimate tensile strength of High-Carbon Steel is 685 MPa.

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

Yield Strength

Yield strength of High-Carbon Steel is 525 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 High-Carbon Steel is 200 GPa.

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

Hardness of High-carbon Steel

Brinell hardness of High-Carbon Steel is approximately 200 MPa.

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

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

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

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

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

 

Example: Strength

Assume a plastic rod, which is made of High-Carbon 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 = 685 MPa.

Solution:

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

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

F = UTS x A = 685 x 10⁶ x 0.0001 = 68 500 N

Thermal Properties of High-Carbon Steel

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

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

Melting Point of High-Carbon Steel

Melting point of High-Carbon Steel is around 1515°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 High-Carbon Steel

The thermal conductivity of High-Carbon Steel is 50 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of High-Carbon Steel with the thermal conductivity of k₁ = 50 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/50 + 1/30) = 7.33 W/m²K

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

The total heat loss through this wall will be: q_loss = q . A = 220.05 [W/m²] x 30 [m²] = 6601.47 W

About Martensitic Stainless Steel

Martensitic stainless steels are similar to ferritic steels in being based on chromium but have higher carbon levels up as high as 1%. They are sometimes classified as low-carbon and high-carbon martensitic stainless steels. They contain 12 to 14% chromium, 0.2 to 1% molybdenum, and no significant amount of nickel. Higher amounts of carbon allows them to be hardened and tempered much like carbon and low-alloy steels. They have moderate corrosion resistance, but are considered hard, strong, slightly brittle. They are magnetic and they can be nondestructively tested using the magnetic particle inspection method, unlike austenitic stainless steel.  A common martensitic stainless is AISI 440C, which contains 16 to 18% chromium and 0.95 to 1.2% carbon. Grade 440C stainless steel is used in the following applications: gage blocks, cutlery, ball bearings and races, molds and dies, knives.

Summary

Name	Martensitic Stainless Steel
Phase at STP	solid
Density	7650 kg/m3
Ultimate Tensile Strength	760 MPa
Yield Strength	450 MPa
Young’s Modulus of Elasticity	200 GPa
Brinell Hardness	270 BHN
Melting Point	1450 °C
Thermal Conductivity	24 W/mK
Heat Capacity	460 J/g K
Price	5 $/kg

Density of Martensitic 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 Martensitic Stainless Steel is 7650 kg/m³.

 

Example: Density

Calculate the height of a cube made of Martensitic 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:

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

Mechanical Properties of Martensitic Stainless Steel

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

Ultimate Tensile Strength

Ultimate tensile strength of Martensitic Stainless Steel – Grade 440C is 760 MPa.

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

Yield Strength

Yield strength of Martensitic Stainless Steel – Grade 440C is 450 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 Martensitic Stainless Steel – Grade 440C is 200 GPa.

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

Hardness of Martensitic Stainless Steel

Brinell hardness of Martensitic Stainless Steel – Grade 440C is approximately 270 MPa.

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

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

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

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

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

 

Example: Strength

Assume a plastic rod, which is made of Martensitic 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 = 760 MPa.

Solution:

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

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

F = UTS x A = 760 x 10⁶ x 0.0001 = 76 000 N

Thermal Properties of Martensitic Stainless Steel

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

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

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

Melting Point of Martensitic Stainless Steel

Melting point of Martensitic Stainless Steel – Grade 440C steel is around 1450°C.

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

Thermal Conductivity of Martensitic Stainless Steel

The thermal conductivity of Martensitic Stainless Steel – Grade 440C is 24 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Martensitic Stainless Steel with the thermal conductivity of k₁ = 24 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/24 + 1/30) = 7.16 W/m²K

The heat flux can be then calculated simply as: q = 7.16 [W/m²K] x 30 [K] = 214.93 W/m²

The total heat loss through this wall will be: q_loss = q . A = 214.93 [W/m²] x 30 [m²] = 6447.76 W

About 6061 Alloy

In general, 6000 series aluminium alloys are alloyed with magnesium and silicon. Alloy 6061 is one of the most widely used alloys in the 6000 Series. It has good mechanical properties, it is easy to machine, it is weldable, and can be precipitation hardened, but not to the high strengths that 2000 and 7000 can reach. It has very good corrosion resistance and very good weldability although reduced strength in the weld zone. The mechanical properties of 6061 depend greatly on the temper, or heat treatment, of the material. In comparison to 2024 alloy, 6061 is more easily worked and remains resistant to corrosion even when the surface is abraded.

This standard structural alloy, one of the most versatile of the heat-treatable alloys, is popular for medium to high strength requirements and has good toughness characteristics. Applications range from aircraft components (aircraft structures, such as wings and fuselages) to automotive parts such such as the chassis of the Audi A8. 6061-T6 is widely used for bicycle frames and components.

Summary

Name	6061 Alloy
Phase at STP	solid
Density	2700 kg/m3
Ultimate Tensile Strength	290 MPa
Yield Strength	240 MPa
Young’s Modulus of Elasticity	69 GPa
Brinell Hardness	105 BHN
Melting Point	600 °C
Thermal Conductivity	150 W/mK
Heat Capacity	896 J/g K
Price	6 $/kg

Density of 6061 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 6061 Alloy is 2700 kg/m³.

 

Example: Density

Calculate the height of a cube made of 6061 Alloy, which weighs one metric ton.

Solution:

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

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

The height of this cube is then a = 0.718 m.

Mechanical Properties of Boron Carbide

Strength of 6061 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 6061 aluminium alloy depends greatly on the temper of the material, but for T6 temper it is about 290 MPa.

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

Yield Strength

Yield strength of 6061 aluminium alloy depends greatly on the temper of the material, but for T6 temper it is about 240 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 6061 aluminium alloy is about 69 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 6061 Alloy

Brinell hardness of 6061 aluminium alloy depends greatly on the temper of the material, but for T6 temper it is approximately 105 MPa.

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

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

 

Example: Strength

Assume a plastic rod, which is made of 6061 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 = 290 MPa.

Solution:

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

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

F = UTS x A = 290 x 10⁶ x 0.0001 = 29 000 N

Thermal Properties of 6061 Alloy

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

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

Melting Point of 6061 Alloy

Melting point of 6061 aluminium alloy is around 600°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 6061 Alloy

The thermal conductivity of 6061 aluminium alloy is 150 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of 6061 Alloy with the thermal conductivity of k₁ = 150 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/150 + 1/30) = 7.44 W/m²K

The heat flux can be then calculated simply as: q = 7.44 [W/m²K] x 30 [K] = 223.33 W/m²

The total heat loss through this wall will be: q_loss = q . A = 223.33 [W/m²] x 30 [m²] = 6699.75 W

About Pure Titanium

Titanium is a lustrous transition metal with a silver color, low density, and high strength. Titanium is resistant to corrosion in sea water, aqua regia, and chlorine. In power plants, titanium can be used in surface condensers. The metal is extracted from its principal mineral ores by the Kroll and Hunter processes. Kroll’s process involved reduction of titanium tetrachloride (TiCl4), first with sodium and calcium, and later with magnesium, under an inert gas atmosphere. Pure titanium is stronger than common, low-carbon steels, but 45% lighter. It is also twice as strong as weak aluminium alloys but only 60% heavier. The two most useful properties of the metal are corrosion resistance and strength-to-density ratio, the highest of any metallic element. The corrosion resistance of titanium alloys at normal temperatures is unusually high. Titanium’s corrosion resistance is based on the formation of a stable, protective oxide layer. Although “commercially pure” titanium has acceptable mechanical properties and has been used for orthopedic and dental implants, for most applications titanium is alloyed with small amounts of aluminium and vanadium, typically 6% and 4% respectively, by weight. This mixture has a solid solubility which varies dramatically with temperature, allowing it to undergo precipitation strengthening.

Titanium alloys are metals that contain a mixture of titanium and other chemical elements. Such alloys have very high tensile strength and toughness (even at extreme temperatures). They are light in weight, have extraordinary corrosion resistance and the ability to withstand extreme temperatures.

Commercially pure titanium grade 2 is very similar to grade 1, but it has higher strength than grade 1 and excellent cold forming properties. It provides excellent welding properties and has excellent resistance to oxidation and corrosion. This grade of titanium is the most common grade of the commercially pure titanium industry.

Summary

Name	Pure Titanium
Phase at STP	solid
Density	4510 kg/m3
Ultimate Tensile Strength	293 MPa
Yield Strength	380 MPa
Young’s Modulus of Elasticity	116 GPa
Brinell Hardness	150 BHN
Melting Point	1941 °C
Thermal Conductivity	22 W/mK
Heat Capacity	520 J/g K
Price	61 $/kg

Density of Pure Titanium

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 Titanium is 4510 kg/m³.

 

Example: Density

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

Solution:

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

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

The height of this cube is then a = 0.605 m.

Mechanical Properties of Pure Titanium

Strength of Pure Titanium

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 commercially Pure Titanium is about 293 MPa.

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

Yield Strength

Yield strength of commercially Pure Titanium is about 380 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 commercially Pure Titanium is about 116 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 Titanium Grade 2

Rockwell hardness of commercially Pure Titanium – Grade 2 is approximately 150 HRB.

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 Pure Titanium. 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 = 293 MPa.

Solution:

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

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

F = UTS x A = 293 x 10⁶ x 0.0001 = 29 300 N

Thermal Properties of Pure Titanium

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

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

Melting Point of Pure Titanium

Melting point of commercially Pure Titanium  is around 1941°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 Pure Titanium

The thermal conductivity of commercially Pure Titanium is 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:

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

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

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/22 + 1/30) = 7.14 W/m²K

The heat flux can be then calculated simply as: q = 7.14 [W/m²K] x 30 [K] = 214.05 W/m²

The total heat loss through this wall will be: q_loss = q . A = 214.05 [W/m²] x 30 [m²] = 6421.62 W

About Zamak

Zamak is a family of alloys with a base metal of zinc and alloying elements of aluminium, magnesium, and copper. Alloys of zinc with small amounts of copper, aluminium, and magnesium are useful in die casting as well as spin casting, especially in the automotive, electrical, and hardware industries. Zinc alloys have low melting points, require relatively low heat input, do not require fluxing or protective atmospheres. Because of their high fluidity, zinc alloys can be cast in much thinner walls than other die castings alloys, and they can be die cast to tighter dimensional tolerances. These zinc alloys are marketed under the name Zamak. The name zamak is an acronym of the German names for the metals of which the alloys are composed: Zink (zinc), Aluminium, Magnesium and Kupfer (copper). The low melting point together with the low viscosity of the alloy makes possible the production of small and intricate shapes.

Summary

Name	Zamak 3
Phase at STP	solid
Density	6600 kg/m3
Ultimate Tensile Strength	268 MPa
Yield Strength	208 MPa
Young’s Modulus of Elasticity	96 GPa
Brinell Hardness	82 BHN
Melting Point	385 °C
Thermal Conductivity	113 W/mK
Heat Capacity	420 J/g K
Price	3 $/kg

Density of Zamak 3

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 Zamak 3 is 6600 kg/m³.

 

Example: Density

Calculate the height of a cube made of Zamak 3, which weighs one metric ton.

Solution:

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

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

The height of this cube is then a = 0.533 m.

Mechanical Properties of Zamak 3

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 Zinc Alloys – Zamak 3

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 Zinc Alloy – Zamak 3 is about 268 MPa.

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

Yield Strength

Yield strength of Zinc Alloy – Zamak 3 is about 208 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 Zinc Alloy – Zamak 3 is about 96 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 Zinc Alloys – Zamak 3

Brinell hardness of Zinc Alloy – Zamak 3 is approximately 82 HB.

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 Zamak 3. 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 = 268 MPa.

Solution:

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

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

F = UTS x A = 268 x 10⁶ x 0.0001 = 26 800 N

Thermal Properties of Zinc Alloys – Zamak 3

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

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

Melting Point of Soft Zinc Alloys – Zamak 3

Melting point of Zinc Alloy – Zamak 3 is around 385°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 Zinc Alloys – Zamak 3

The thermal conductivity of Zinc Alloy – Zamak 3 is 113 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Zamak 3 with the thermal conductivity of k₁ = 113 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/113 + 1/30) = 7.43 W/m²K

The heat flux can be then calculated simply as: q = 7.43 [W/m²K] x 30 [K] = 222.78 W/m²

The total heat loss through this wall will be: q_loss = q . A = 222.78 [W/m²] x 30 [m²] = 6683.46 W

About Cupronickel

Cupronickels are copper-nickel alloys that contain typically from 60 to 90 percent of copper and nickel as the main alloying element. The two main alloys are 90/10 and 70/30. Other strengthening elements, such as manganese and iron, may be also contained. Cupronickels have excellent resistance to corrosion caused by sea water. Despite its high copper content, cupronickel is silver in colour. The addition of nickel to copper also improves strength and corrosion resistance, but good ductility is retained.

Summary

Name	Cupronickel
Phase at STP	solid
Density	8940 kg/m3
Ultimate Tensile Strength	275 MPa
Yield Strength	105 MPa
Young’s Modulus of Elasticity	135 GPa
Brinell Hardness	100 BHN
Melting Point	1100 °C
Thermal Conductivity	40 W/mK
Heat Capacity	400 J/g K
Price	8 $/kg

Density of Cupronickel

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 Cupronickel is 8940 kg/m³.

 

Example: Density

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

Solution:

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

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

The height of this cube is then a = 0.482 m.

Mechanical Properties of Cupronickel

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 Cupronickel

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 Cupronickel – UNS C70600 is about 275 MPa.

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

Yield Strength

Yield strength of Cupronickel – UNS C70600 is about 105 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 Cupronickel – UNS C70600 is about 135 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 Cupronickel

Brinell hardness of Cupronickel – UNS C70600 is approximately HB 100.

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 Cupronickel. 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 = 275 MPa.

Solution:

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

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

F = UTS x A = 275 x 10⁶ x 0.0001 = 27 500 N

Thermal Properties of Cupronickel

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

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

Melting Point of Cupronickel

Melting point of Cupronickel – UNS C70600 is around 1100°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 Cupronickel

The thermal conductivity of Cupronickel – UNS C70600 is 40 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of Cupronickel with the thermal conductivity of k₁ = 40 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/40 + 1/30) = 7.29 W/m²K

The heat flux can be then calculated simply as: q = 7.29 [W/m²K] x 30 [K] = 218.85 W/m²

The total heat loss through this wall will be: q_loss = q . A = 218.85 [W/m²] x 30 [m²] = 6565.35 W

About Electrolytic-tough pitch (ETP) copper

Electrolytic tough pitch copper, UNS C11000, is pure copper (with a maximum of 0.0355% of impurities) refined by electrolytic refining process and it is the most widely used grade of copper all over the world. ETP has a minimum conductivity rating of 100% IACS and is required to be 99.9% pure. It has 0.02% to 0.04% oxygen content (typical). Electrical wiring is the most important market for the copper industry. This includes structural power wiring, power distribution cable, appliance wire, communications cable, automotive wire and cable, and magnet wire. Roughly half of all copper mined is used for electrical wire and cable conductors. Pure copper has the best electrical and thermal conductivity of any commercial metal. The conductivity of copper is 97% that of silver. Due to its much lower cost and greater abundance, copper has traditionally been the standard material used for electricity transmission applications.

Summary

Name	ETP
Phase at STP	solid
Density	8890 kg/m3
Ultimate Tensile Strength	250 MPa
Yield Strength	60-300 MPa
Young’s Modulus of Elasticity	120 GPa
Brinell Hardness	70 BHN
Melting Point	1085 °C
Thermal Conductivity	394 W/mK
Heat Capacity	380 J/g K
Price	10 $/kg

Density of ETP

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 ETP is 8890 kg/m³.

 

Example: Density

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

Solution:

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

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

The height of this cube is then a = 0.483 m.

Mechanical Properties of ETP

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 Copper Alloys

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 electrolytic-tough pitch (ETP) copper it is about 250 MPa.

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

Yield Strength

Proof strength of electrolytic-tough pitch (ETP) copper is between 60-300 MPa.

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

Young’s Modulus of Elasticity

Young’s modulus of elasticity of electrolytic-tough pitch (ETP) copper is about 120 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 Electrolytic-tough pitch (ETP) copper

Vickers hardness of electrolytic-tough pitch (ETP) copper depends greatly on the temper of the material, but it is between 50 – 150 HV.

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 EPT. 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 = 250 MPa.

Solution:

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

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

F = UTS x A = 250 x 10⁶ x 0.0001 = 25 000 N

Thermal Properties of ETP

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

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

Melting Point of ETP

Melting point of electrolytic-tough pitch (ETP) copper is around 1085°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 EPT

The thermal conductivity of electrolytic-tough pitch (ETP) copper is 394 W/(m.K).

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

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

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

 

Example: Heat transfer calculation

Thermal conductivity is defined as the amount of heat (in watts) transferred through a square area of material of given thickness (in metres) due to a difference in temperature. The lower the thermal conductivity of the material the greater the material’s ability to resist heat transfer.

Calculate the rate of heat flux through a wall 3 m x 10 m in area (A = 30 m²). The wall is 15 cm thick (L₁) and it is made of EPT with the thermal conductivity of k₁ = 394 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperatures are 22°C and -8°C, and the convection heat transfer coefficients on the inner and the outer sides are h₁ = 10 W/m²K and h₂ = 30 W/m²K, respectively. Note that, these convection coefficients strongly depend especially on ambient and interior conditions (wind, humidity, etc.).

Calculate the heat flux (heat loss) through this wall.

Solution:

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

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

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

 

The overall heat transfer coefficient is then: U = 1 / (1/10 + 0.15/394 + 1/30) = 7.48 W/m²K

The heat flux can be then calculated simply as: q = 7.48 [W/m²K] x 30 [K] = 224.36 W/m²

The total heat loss through this wall will be: q_loss = q . A = 224.36 [W/m²] x 30 [m²] = 6730.78 W