3.1 – Density and specific gravity (S.G.)

3.1.0 – Learning objectives

By the end of this notebook you should be able to:

Differentiate density and specific gravity.
Utilize density and specific gravity to interchangeably find the mass and/or volumetric flow rates.

3.1.1 – Introduction

Density is the amount of mass per unit volume. Specific gravity (S.G.) is the ratio of the density of the object to the density of a standard, usually water for a liquid or solid, and air for a gas. Both density and S.G. are common units in the determination of how much mass is in a chemical process based on the volumetric flow rate of the substance in the process.

$D e n s i t y = ρ$

$S p e c i f i c g r a v i t y = \frac{ρ_{s a m p l e}}{ρ_{H_{2} O}}$

3.1.2 – Example 1

Natural gas is volumetrically made up of 94.44% Methane ( $C H_{4}$ ), 4.24% Ethane ( $C_{2} H_{6}$ ), 0.22% Propane ( $C_{3} H_{8}$ ), 0.78% Nitrogen ( $N_{2}$ ), and 0.32% Carbon Dioxide ( $C O_{2}$ ). What is the density of this natural gas mixture?

Let's assume a total volume of 1 litre. This means there will be:

$1 L \cdot 0.9444 = 0.9444 L [C H_{4}]$

$1 L \cdot 0.0424 = 0.0424 L [C_{2} H_{6}]$

$1 L \cdot 0.0022 = 0.0022 L [C_{3} H_{8}]$

$1 L \cdot 0.0078 = 0.0078 L [N_{2}]$

$1 L \cdot 0.0032 = 0.0032 L [C O_{2}]$

Note we cannot use volumetric fractions directly to calculate the mixture density. Looking up the density of each component, the total mass and the mass fractions would be:

$944.4 L [C H_{4}] \cdot \frac{0.72 g}{L} = 679.97 g$

$42.4 L [C_{2} H_{6}] \cdot \frac{1.34 g}{L} = 0.05682 g$

$2.2 L [C_{3} H_{8}] \cdot \frac{1.97 g}{L} = 0.004334 g$

$7.8 L [N_{2}] \cdot \frac{1.251 g}{L} = 0.009758 g$

$3.2 L [C O_{2}] \cdot \frac{1.977 g}{L} = 0.006326 g$

The total mass of the 1 litre mixture would be the sum of these masses, which comes to $680.05 g$ . The mass fractions of the components then are:

$\frac{679.97}{680.05} [C H_{4}] = 0.99988$

$\frac{0.05682}{680.05} [C_{2} H_{6}] = 0.00008355$

$\frac{0.004334}{680.05} [C_{3} H_{8}] = 0.00000637$

$\frac{0.009758}{680.05} [N_{2}] = 0.00001435$

$\frac{0.006326}{680.05} [C O_{2}] = 0.0000093$

The density of the mixture can be approximated by just the methane alone, the mass fraction is substantially more than the others.

${\overline{ρ}}_{n a t u r a l g a s} = \sum_{i = 1}^{n} x_{i} ρ_{i} = 0.99988 \cdot \frac{0.72 g}{L} = 0.72 \frac{g}{L}$

A common usage of densities and S.G. is the calculation of mass or volumetric flowrates, given one of the two factors, since:

$ρ = \frac{\dot{m}}{\dot{V}} = \frac{m}{V}$

Note: The dot above the variable means that the unit is the variable per unit time. (e.g. $\dot{m}$ = $m a s s / t i m e$ )

3.1.2 – Example 2

The volumetric flow rate of $C C l_{4}$ ( $ρ = 1.595 g / c m^{3}$ ) in a pipe is 100.0 cm $3$ /min. What is the mass flow rate of the $C C l_{4}$ ?

${\dot{m}}_{C C l_{4}} = 100.0 \frac{c m^{3}}{m i n} \times 1.595 \frac{g}{c m^{3}} = 159.5 \frac{g}{m i n}$