Question

Natural gas at \(23^{\circ} \mathrm{C}, 1\) bar enters a furnace with the following molar analysis: \(40 \%\) propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right), 40 \%\) ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right), 20 \%\) methane \(\left(\mathrm{CH}_{4}\right)\). Determine (a) the analysis in terms of mass fractions. (b) the partial pressure of each component, in bar. (c) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), for a volumetric flow rate of \(20 \mathrm{~m}^{3} / \mathrm{s}\).

Short answer

Mass fractions are 0.5366 for propane, 0.3659 for ethane, and 0.0976 for methane. Partial pressures are 0.4 bar for propane, 0.4 bar for ethane, and 0.2 bar for methane. Mass flow rate is 29.87 kg/s.

Step-by-step solution

- Calculate Molar Masses

First, find the molar masses of each gas. For propane (\text{C}_3\text{H}_8): \[ \text{C}_3\text{H}_8 = 3 \times 12 + 8 \times 1 = 44 \text{ g/mol} \] For ethane (\text{C}_2\text{H}_6): \[ \text{C}_2\text{H}_6 = 2 \times 12 + 6 \times 1 = 30 \text{ g/mol} \] For methane (\text{CH}_4): \[ \text{CH}_4 = 12 + 4 \times 1 = 16 \text{ g/mol} \]

- Determine Mass Fractions

Use the molar fractions to find the mass fraction of each component. For propane: \[ \text{Mass fraction of propane} = \frac{40 \times 44}{40 \times 44 + 40 \times 30 + 20 \times 16} \] For ethane: \[ \text{Mass fraction of ethane} = \frac{40 \times 30}{40 \times 44 + 40 \times 30 + 20 \times 16} \] For methane: \[ \text{Mass fraction of methane} = \frac{20 \times 16}{40 \times 44 + 40 \times 30 + 20 \times 16} \] Simplifying the denominators: \[ (\text{denominator}) = 40 \times 44 + 40 \times 30 + 20 \times 16 = 1760 + 1200 + 320 = 3280 \text{ g} \]

- Calculate Mass Fractions

Now, calculate the mass fraction of each component. For propane: \[ \text{Mass fraction of propane} = \frac{40 \times 44}{3280} = \frac{1760}{3280} = 0.5366 \] For ethane: \[ \text{Mass fraction of ethane} = \frac{40 \times 30}{3280} = \frac{1200}{3280} = 0.3659 \] For methane: \[ \text{Mass fraction of methane} = \frac{20 \times 16}{3280} = \frac{320}{3280} = 0.0976 \]

- Determine Partial Pressures

Given the total pressure is 1 bar, the partial pressure of each gas is calculated using its mole fraction. For propane: \[ \text{Partial pressure of propane} = 0.4 \times 1 \text{ bar} = 0.4 \text{ bar} \] For ethane: \[ \text{Partial pressure of ethane} = 0.4 \times 1 \text{ bar} = 0.4 \text{ bar} \] For methane: \[ \text{Partial pressure of methane} = 0.2 \times 1 \text{ bar} = 0.2 \text{ bar} \]

- Determine Mass Flow Rate

Use the ideal gas law to determine the total mass flow rate. The volume flow rate is given as \(20 \text{ m}^3/\text{s}\), and the molar mass of the mixture is the weighted average: \[ \text{Molar mass of mixture} = 0.4 \times 44 + 0.4 \times 30 + 0.2 \times 16 = 36.8 \text{ g/mol}\] Convert the flow rate to moles per second: \[ \text{n} = \frac{PV}{RT} \] Given: \(P = 1 \text{ bar} = 10^5 \text{ Pa}\), \(V = 20 \text{ m}^3/\text{s}\), \(R = 8.314 \text{ J/mol K}\), \(T = (23+273)K = 296K\), \[ \text{n} = \frac{10^5 \times 20}{8.314 \times 296} = 811.8 \text{ moles/s} \] Finally, calculate the mass flow rate: \[ \text{Mass flow rate} = 811.8 \times 36.8 \text{ g/s} = 2.987 \times 10^4 \text{ g/s} = 29.87 \text{ kg/s} \]

Key concepts

Molar Mass Calculation

Knowing the molar mass of each gas component in a mixture is crucial for further calculations. The molar mass (M) is the mass of one mole of a substance. It can be derived from the periodic table, where the atomic masses of the elements are listed. For example:
- For propane (C₃H₈):
\[ M_{C₃H₈} = (3 \times 12 \text{ g/mol}) + (8 \times 1 \text{ g/mol}) = 44 \text{ g/mol} \]
- For ethane (C₂H₆):
\[ M_{C₂H₆} = (2 \times 12 \text{ g/mol}) + (6 \times 1 \text{ g/mol}) = 30 \text{ g/mol} \]
- For methane (CH₄):
\[ M_{CH₄} = 12 \text{ g/mol} + (4 \times 1 \text{ g/mol}) = 16 \text{ g/mol} \]
Understanding these values allows us to convert mole fractions into mass fractions and proceed with other thermodynamic calculations.

Mass Fraction Determination

The mass fraction represents the ratio of the mass of a specific component to the total mass of the mixture. First, convert molar fractions to mass fractions using the molar masses calculated. This is done by multiplying each component's molar fraction by its molar mass and then dividing by the total mass of the mixture.

For propane:
\[ \text{Mass fraction of propane} = \frac{40 \times 44}{40 \times 44 + 40 \times 30 + 20 \times 16} = \frac{40 \times 44}{3280} = 0.5366 \]
For ethane:
\[ \text{Mass fraction of ethane} = \frac{40 \times 30}{3280} = 0.3659 \]
For methane:
\[ \text{Mass fraction of methane} = \frac{20 \times 16}{3280} = 0.0976 \]
These fractions reflect the proportion of each gas by weight in the mixture.

Partial Pressure

Partial pressure is the pressure a gas in a mixture would exert if it alone occupied the entire volume. Given the total pressure of the gas mixture, each component's partial pressure can be found by multiplying its mole fraction by the total pressure. With a total pressure of 1 bar, the partial pressures are:
- For propane:
\[ \text{Partial pressure of propane} = 0.4 \times 1 \text{ bar} = 0.4 \text{ bar} \]
- For ethane:
\[ \text{Partial pressure of ethane} = 0.4 \times 1 \text{ bar} = 0.4 \text{ bar} \]
- For methane:
\[ \text{Partial pressure of methane} = 0.2 \times 1 \text{ bar} = 0.2 \text{ bar} \]
These values are essential for various applications, including gas law equations and chemical reactions involving gases.

Ideal Gas Law

The ideal gas law relates the pressure (P), volume (V), and temperature (T) of an ideal gas with its amount in moles (n) through the equation:
\[ PV = nRT \]
where R is the universal gas constant (8.314 \text{ J/(mol K)} ). To find the number of moles in a gas flow, use:
\[ n = \frac{PV}{RT} \]
Given: \[ P = 1 \text{ bar} = 10^5 \text{ Pa}, \ V = 20 \text{ m}^3/\text{s}, \ T = 296 \text{ K} \]
The calculation becomes:
\[ n = \frac{10^5 \times 20}{8.314 \times 296} = 811.8 \text{ moles/s} \]
This formula is fundamental for converting between different gas properties and conditions.

Mass Flow Rate Calculation

The mass flow rate is critical in determining how much mass of a gas passes through a point per unit time. To calculate it from a known volumetric flow rate, use the formula:
\[ \text{Mass flow rate} = n \times \text{Molar mass} \]
With the molar mass of the gas mixture calculated as:
\[ \text{Molar mass of mixture} = 0.4 \times 44 + 0.4 \times 30 + 0.2 \times 16 = 36.8 \text{ g/mol} \]
And the number of moles per second (n) using the ideal gas law:
\[ n = 811.8 \text{ moles/s} \]
Finally, the mass flow rate is:
\[ \text{Mass flow rate} = 811.8 \times 36.8 = 2.987 \times 10^4 \text{ g/s} = 29.87 \text{ kg/s} \]
Understanding this process allows for accurate measurement of gas movement in industrial and laboratory settings.