Chapter 2.8, Problem 116RP

To determine

The mass flow rate of sulphur in the exhaust.

The mass flow rate of sulphurous acid added to the environment.

Answer to Problem 116RP

The mass flow rate of sulphur in the exhaust is $0.014\text{kg}/\text{h}$.

The mass flow rate of sulphurous acid added to the environment is $0.036\text{kg}/\text{h}$.

Explanation of Solution

Write the formula to calculate mass flow rate of fuel in the exhaust $\left({\dot{m}}_{\text{fuel}}\right)$.

${\dot{m}}_{\text{fuel}}=\frac{{\dot{m}}_{\text{air}}}{\text{AF}}$ (I)

Here, air-fuel ratio is $\text{AF}$ and mass flow rate of air is ${\dot{m}}_{\text{air}}$.

Write the formula to calculate mass flow rates of sulfur $\left({\dot{m}}_{\text{sulfur}}\right)$.

${\dot{m}}_{\text{sulfur}}=\left(\text{ppm}\text{of}\text{sulfur}\text{by}\text{mass}\right){\dot{m}}_{\text{fuel}}$ (II)

Write the formula to calculate rate of sulphurous acid added to the environment $\left({\dot{m}}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}\right)$.

${\dot{m}}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}=\frac{M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}}{M_{\text{S}}}{\dot{m}}_{\text{sulfur}}$ (III)

Here, molar mass of sulphurous acid is $M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}$ and molar mass of sulphur is $M_{\text{S}}$.

Conclusion:

Substitute $336\text{kg air}/\text{h}$ for ${\dot{m}}_{\text{air}}$ and $18\text{kg air}/\text{kg fuel}$ for $\text{AF}$ in Equation (I).

$\begin{array}{c}{\dot{m}}_{\text{fuel}}=\frac{336\text{kg air}/\text{h}}{18\text{kg air}/\text{kg fuel}}\\ =18.67\text{kg fuel}/\text{h}\end{array}$

Substitute $18.67\text{kg fuel}/\text{h}$ for ${\dot{m}}_{\text{fuel}}$ and $750\text{ppm}$ for $\text{ppm}\text{of}\text{sulfur}\text{by}\text{mass}$ in Equation (II).

$\begin{array}{c}{\dot{m}}_{\text{sulfur}}=\left(750\text{ppm}\right)\left(18.67\text{kg fuel}/\text{h}\right)\\ =\left(750\text{ppm}\times \frac{\text{1}\times {\text{10}}^{-6}}{\text{1}\text{ppm}}\right)\left(18.67\text{kg fuel}/\text{h}\right)\\ =0.014\text{kg}/\text{h}\end{array}$

Thus, the mass flow rate of sulphur in the exhaust is $0.014\text{kg}/\text{h}$.

The molar masses of hydrogen, sulphur and oxygen are as $1\text{kg}/\text{mol}$, $32\text{kg}/\text{mol}$ and $16\text{kg}/\text{mol}$.

$\begin{array}{l}{M}_{\text{H}}=1\text{kg}/\text{mol}\\ M_{\text{S}}=32\text{kg}/\text{mol}\\ M_{\text{O}}=16\text{kg}/\text{mol}\end{array}$

Calculate the molar mass of sulphurous acid $\left(M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}\right)$.

$\begin{array}{c}{M}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}=2M_{\text{H}}+M_{\text{S}}+3M_{\text{O}}\\ =2\left(1\text{kg}/\text{mol}\right)+32\text{kg}/\text{mol}+3\left(16\text{kg}/\text{mol}\right)\\ =82\text{kg}/\text{mol}\end{array}$

Substitute $32\text{kg}/\text{mol}$ for $M_{\text{S}}$, $82\text{kg}/\text{mol}$ for $M_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}$ and $0.014\text{kg}/\text{h}$ for ${\dot{m}}_{\text{sulfur}}$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{{\text{H}}_{\text{2}}{\text{SO}}_{\text{3}}}=\frac{82\text{kg}/\text{mol}}{32\text{kg}/\text{mol}}\left(0.014\text{kg}/\text{h}\right)\\ =0.036\text{kg}/\text{h}\end{array}$

Thus, the mass flow rate of sulphurous acid added to the environment is $0.036\text{kg}/\text{h}$.