# Pure Tungsten | Density, Strength, Hardness, Melting Point (2023)

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

### Summary

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

## Density of Pure Tungsten

Typical densities of various substances are at atmospheric pressure.Densityis defined as themass per unit volume. It is anintensive 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 iskilograms per cubic meter(kg/m3). The Standard English unit ispounds mass per cubic foot(lbm/ft3).

Density of Pure Tungsten is 19250 kg/m3.

### Example: Density

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

Solution:

Densityis defined as themass 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 = 0.373 m.

### Density of Materials (Video) HIDRAULIC PRESS VS THE HARDEST STONES VS TUNGSTEN

## Mechanical Properties of Pure Tungsten

### Strength of Pure Tungsten

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

Strength of a material is its ability to withstand this applied load without failure or plastic deformation. For tensile stress, the capacity of a material or structure to withstand loads tending to elongate is known as ultimate tensile strength (UTS). Yield strength or yield stress is the material property defined as the stress at which a material begins to deform plastically whereas yield point is the point where nonlinear (elastic + plastic) deformation begins. In case of tensional stress of a uniform bar (stress-strain curve), theHooke’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.

### Ultimate Tensile Strength of Pure Tungsten

Ultimate tensile strength of Pure Tungsten is 980 MPa.

### Yield Strength of Pure Tungsten

Yield strength of Pure Tungstenis 750 MPa.

### Modulus of Elasticity of Pure Tungsten

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

### Hardness of Pure Tungsten

In materials science,hardnessis the ability to withstandsurface indentation(localized plastic deformation) andscratching.Brinell hardness testis 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.

TheBrinell 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 Pure Tungsten is approximately 2570 BHN (converted).

### Example: Strength

Assume a plastic rod, which is made of Pure Tungsten. This plastic rod has a cross-sectional area of 1 cm2. Calculate the tensile force needed to achieve the ultimate tensile strength for this material, which is: UTS = 980 MPa.

Solution:

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

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

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

## Thermal Properties of Pure Tungsten

### Pure Tungsten – Melting Point

Melting point of Pure Tungsten is 3695 °C.

Note that, these points are associated with the standard atmospheric pressure. In general,meltingis aphase changeof a substance from the solid to the liquid phase. Themelting pointof a substance is the temperature at which this phase change occurs. Themelting pointalso 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.

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### Pure Tungsten – Thermal Conductivity

Thermal conductivity of Pure Tungsten is 170 W/(m·K).

The heat transfer characteristics of a solid material are measured by a property called thethermal conductivity, k (or λ), measured inW/m.K. It is a measure of a substance’s ability to transfer heat through a material byconduction. Note thatFourier’s lawapplies for all matter, regardless of its state (solid, liquid, or gas), therefore, it is also defined for liquids and gases.

Thethermal conductivityof 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 writek = k (T). Similar definitions are associated with thermal conductivities in the y- and z-directions (ky, kz), but for an isotropic material the thermal conductivity is independent of the direction of transfer, kx = ky = kz = k.

### Pure Tungsten – Specific Heat

Specific heat of Pure Tungsten is 130J/g K.

Specific heat, or specific heat capacity,is a property related tointernal energythat is very important in thermodynamics. Theintensive propertiescvandcpare defined for pure, simple compressible substances as partial derivatives of theinternal energyu(T, v)andenthalpyh(T, p), respectively:

where the subscriptsvandpdenote the variables held fixed during differentiation. The propertiescvandcpare referred to asspecific heats(orheat 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 areJ/kg KorJ/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 ofheat fluxthrough a wall 3 m x 10 m in area (A = 30 m2). The wall is 15 cm thick (L1) and it is made of Pure Tungsten with the thermal conductivityof k1 = 170 W/m.K (poor thermal insulator). Assume that, the indoor and the outdoor temperaturesare 22°C and -8°C, and theconvection heat transfer coefficientson 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 bothconductionandconvection. With these composite systems, it is often convenient to work with anoverall heat transfer coefficient,known as aU-factor. The U-factor is defined by an expression analogous toNewton’s law of cooling:

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Theoverall heat transfer coefficientis related to thetotal thermal resistanceand depends on the geometry of the problem.

Assuming one-dimensional heat transfer through the plane wall and disregarding radiation, theoverall heat transfer coefficientcan be calculated as:

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

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

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

### Heat Capacity of Materials

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