Question

Methane enters the furnace at a rate of 100 mol/min and is burned with air in a furnace. Dry excess air enters the furnace at 25°C and 1 atm. The stack gas leaves the furnace at 473 K and 1 atm and contains CO2 at a partial pressure of 61.56 mmHg, CO at partial pressure of 6.84 mmHg, water, oxygen, and nitrogen.

Calculate

a. the percentage of excess air.

b. the composition of stack gas in dry basis.

c. the dew point of the stack gas.

d. the volumetric rates of air and stack gas

P* [mmHg], T[°C]

logP ∗ = A − (B/T+C)

A = 8.10765

B = 1750.286

C = 235

​

Answer 1

Answer:

The percentage of excess air is 0.35%

The composition of stack gas in dry basis is 0.7154

The dew point of the stack gas is 50.10 degree C

The volumetric rates of air and stack gas is 43.13

Explanation:

Step 1: Pencentage of excess air, (x)

$\begin{aligned}& N=10.524 \times 90+0.643 \times 10+\frac{4.762}{100} \times\left(3 \varepsilon_1+\frac{3}{2} \varepsilon_2\right) \\& \text { butting } \varepsilon_\alpha=10 \text {, \&E90 } \\& N=947.16+06.43+9.2859 x \\& \text { uring equation (13), } \frac{90}{947.16+06.43+9.2859 x}=0.001 \\& \frac{1000}{9}=1033.59+9.2859 x \\& x=0.35 \% \text { \& } N=1111.11 \text { m.les } \\&\end{aligned}$

Step 2: composition-

$y_{\text {Co2 }_2}=\frac{\varepsilon_2}{N}=\frac{90}{111.11} \Rightarrow\left(y_{1_2}=0.001\right)$

$y_{c_0}=\frac{\varepsilon_1}{N}=\frac{1 .}{111.11} \Rightarrow\left(y_{c_0}=0.009\right)$

$y_{\ell 22}=\frac{\left(2 \varepsilon_1+1.5 \varepsilon_2\right) \frac{x}{100}}{N}=\frac{(2 \times 90+1.5 \times 10) \times \frac{0.35}{100}}{1111.11}$

$\left(y_{t_2}=0.01465\right)$

$y_{H_2 0}=\frac{2\left(\varepsilon_1+\varepsilon_2\right)}{N}=\frac{2 \times 100}{1111.11} \Rightarrow\left(y_{H_2 \mathrm{C}}=0.18\right)$

$y_{N_2}=\frac{3.762\left(2 \varepsilon_1+1.5 \varepsilon_2\right)\left(1+\frac{x}{1 \omega 0}\right)}{1111.11} \Rightarrow\left(y_{N_2}=0.7154\right)$

Step 3: Dew point -

Dew point, partial pressured of water vapor is the mixture equal to vapaur pressure of water at dewpoint, so,

$\begin{aligned}& P_{H_1 \mathrm{O}}=P_{\mathrm{H}_1 \mathrm{O}} \\& P_{H_1 0}=y_{H_2} 0 \times 760 \mathrm{mmHg}=0.18 \times 760 \\& P_{H_2 0}=136.8 \mathrm{mmHg} \\& \text { So, } \log (136.8)=0.10765-\frac{1750.286}{T+235} \\& \frac{1750.286}{T+235}=8.10765-2.13600 \\& T+235=293.1034 \\& T=50.10^{\circ} \mathrm{C} \text { (dewpeint) } \\&\end{aligned}$

Step 4: volumetric rates of air \& stack gas

$\text { nolan flat rate of ais }=\left(2 \varepsilon_1+7.5 \varepsilon_2\right)\left(\frac{x}{10}+1\right) 4.762 \frac{\text { mol }}{\text { min }}$

$\text { fats }=1006.13 \frac{\mathrm{m} / \mathrm{m}}{\mathrm{min}}$

$\text { volumetric flow rat of air }=\frac{\text { fair } \times R T}{P}$

$=\frac{1006.13 \times 92.06 \times 26 \Omega}{1}$

$\text { Vail }=24.60 \times 10^6 \mathrm{~cm}^3 / \mathrm{min}$

$=24.6 \mathrm{~m}^3 / \mathrm{min}$

$\text { The molar flow rate of stack gas }=1111.11 \frac{\mathrm{molar}}{\mathrm{men}}$

$\text { volumetric flow rall }=\frac{1111.11 \times 82.06 \times 473}{1}$

$=43.13 \times 10^6 \mathrm{~cm}^3 / \mathrm{men}$

$=43.13 \mathrm{~m}^3 / \mathrm{min}$

Learn more about similar questions visit:

https://brainly.in/question/10002322?referrer=searchResults

https://brainly.in/question/54099605?referrer=searchResults

#SPJ1