## About Sand

Sand is a granular material composed of finely divided rock and mineral particles. The composition of sand varies, depending on the local rock sources and conditions, but the most common constituent of sand in inland continental settings and non-tropical coastal settings is silica (silicon dioxide, or SiO2), usually in the form of quartz. Silica is one of the most complex and most abundant families of materials, existing as a compound of several minerals and as synthetic product.

### Summary

Name | Sand |

Phase at STP | solid |

Density | 1500 kg/m3 |

Ultimate Tensile Strength | N/A |

Yield Strength | N/A |

Young’s Modulus of Elasticity | N/A |

Brinell Hardness | N/A |

Melting Point | 1577 °C |

Thermal Conductivity | 0.25 W/mK |

Heat Capacity | 830 J/g K |

Price | 0.03 $/kg |

## Density of Sand

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/m ^{3}**). The Standard English unit is

**pounds mass per cubic foot**(

**lbm/ft**).

^{3}Density of Sand is **1500 kg/m ^{3}.**

### Example: Density

Calculate the height of a cube made of Sand, 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 = a^{3}), the height of this cube can be calculated:

The height of this cube is then **a = 0.874 m**.

### Density of Materials

## Thermal Properties of Sand

### Sand – Melting Point

**Melting point of Sand is 1577 ****°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.

### Sand – Thermal Conductivity

Thermal conductivity of Sand is **0.25** **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.

### Sand – Specific Heat

**Specific heat of Sand is ****830 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

**are defined for pure, simple compressible substances as partial derivatives of the**

*c*_{p}**internal energy**and

*u(T, v)***enthalpy**, respectively:

*h(T, p)*where the subscripts **v** and **p** denote the variables held fixed during differentiation. The properties **c _{v} **and

**c**are referred to as

_{p}**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 m^{2}). The wall is 15 cm thick (L_{1}) and it is made of Sand with the __thermal conductivity__ of k_{1} = 0.25 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 h_{1} = 10 W/m^{2}K and h_{2} = 30 W/m^{2}K, 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 + 0.15/0.25 + 1/30) = 1.36 W/m^{2}K

The heat flux can be then calculated simply as: q = 1.36 [W/m^{2}K] x 30 [K] = 40.91 W/m^{2}

The total heat loss through this wall will be: **q _{loss} **= q . A = 40.91 [W/m

^{2}] x 30 [m

^{2}] =

**1227.27 W**