Question

Flue gas from a furnace is 30 mol% SO₂, 25 mol% O₂, and remainder CO₂. To clean 1155 mol/hr of flue gas and remove SO₂, it is to be reacted with pure methane (natural gas) such that for every 65 moles of SO₂ in the flue gas, 67 moles of methane is used (supplied).

The primary reaction between methane and SO₂ is as follows

      CH₄    +     SO₂ →     H₂S   +    CO₂ +    H₂O

A side reaction also occurs as follows

      H₂S +    SO₂ →    S    + H₂O

Due to the risk of ignition of methane, nitrogen is to be added to the flue gas and methane mixture. It is known that the lower flammability limit (LFL) for methane in a mixture is 5 mol% and upper flammability limit (UFL) is 15 mol%. Only between the LFL and UFL, the gas mixture is flammable.

In a steady state process, the flue gas and methane is mixed with X moles/hr of nitrogen and then sent to a reactor where the two reactions shown above occur. It is found that 67% conversion is achieved for methane and no SO₂ exits the reactor. The elemental sulfur (S) formed exits as a solid product and the remainder species exit in a gas mixture.

Determine the following quantities using a material balance.

(a) % excess of methane in the input to the reactor based on the primary reaction:

(b) mass flow rate (in kg/hr) of solid sulfur in the reactor output

(c) The value of X (mol/hr) has to range between a minimum value X_min and a maximum value X_max. The minimum value corresponds to a nitrogen addition that leads to methane concentration of 5 mol% (LFL) in the exit gas mixture from the reactor. The maximum value corresponds to a nitrogen addition that leads to methane concentration of 15% (UFL) in the input gas mixture to the reactor. Find these two values and complete the following sentence.

The amount of nitrogen to be added should be either kept below moles/hr or above moles/hr.

d)

The presence of water vapor in the gas mixture exiting from the reactor affects the UFL value for methane. A scientific study has determined that the modified UFL can be found using mol% of water vapor, Y, in a mixture with the correlation:

UFL_new = 15 - 830830 *Y

Determine the new UFL value when X=X_min is used for the nitrogen addition:

(I have an idea of how to do parts A and B and are mostly concerned with parts C and D. For A and B, I calculated 37.4 and 16.6).

Answer 1

c) Flowrate of Inlet flue gas = 1155 mol/hr

In Inlet flue gas, SO2 = 30%

Therefore, In Inlet, SO2 = .3*1155 = 346.5 mol/hr

Therefore, In Inlet, O2 = .25*1155 = 288.75 mol/hr

Therefore, In Inlet, CO2 = (1-.25-.3)*1155=529.75 mol/hr

For every 65 moles, 67 moles Methane are supplied

Therefore, In Inlet, moles of Methane = 67/65*346.5 = 357.16 moles/hr

In Let, Assume Nitrogen flowrate = n moles/hr

In Outlet, Vapor phase consists of Methane, CO2, O2, Nitrogen, H2O

Since, 67% Methane consumed, Therefore, moles of Methane consumed = (.67)*357.16 = 203.58 moles/hr

In main reaction, one mole SO2 reacts with one mole Methane

Therefore, SO2 consumes in main reaction = 203.58 moles/hr

In main reaction, one mole SO2 gives 1 mole each of CO2, H2S and H2O

Therefore, in main reaction, CO2 produce=H2Sproduce=H2O produce = 203.58 moles/hr

Since, whole SO2 will be consumed, Therefore, SO2 reacts in Side reaction = SO2 in Inlet - SO2 reacts in main reaction

Therefore, SO2 reacts in Side reaction = 346.5-203.58 = 142.92 moles/hr

One mole SO2 consume in side reaction = one mole H2S consume in Side reaction = one mole S produce in Side reaction =one mole H2O produce in Side reaction

Therefore, H2S consumes in side reaction = 142.92 moles/hr

Therefore, H2O produce in side reaction = 142.92 moles/hr

Moles of Methane in Outlet = (1-.67)*357.16 = 117.86 moles/hr

In the reaction, one mole SO2 consume produce 1 mole CO2

CO2 moles in outlet = CO2 moles in Inet + CO2 moles produce in main reaction+CO2 moles produce in side reaction= 529.75+0+203.58=733.33 moles/hr

H2S moles in outlet = H2S moles in Inet + H2S moles produce in main reaction+H2S moles produce in side reaction= 0+203.58-142.92=60.66 moles/hr

H2O moles in outlet = H2O moles in Inet + H2O moles produce in main reaction+H2O moles produce in side reaction= 0+203.58+142.92=346.5 moles/hr

Since, O2 is not reacting, Therefore, Outlet O2 moles = Inlet O2 moles =288.75 moles/hr

Since, N2 is not reacting, Therefore, Outlet N2 moles = Inlet N2 moles =n moles/hr

Total Number of moles in Intlet = n+1155+357.16 = 1512.16 + n moles/hr

Concentration of Methane in Inlet = 357.16/(1512.16+n)

Given, For Xmax, Concentration of Methane in Inlet = 15%

Therefore, 0.15=357.16/(1512.16+n)

herefore, Xmax = n = 868 moles/hr

Total Number of moles in Outlet = n+288.75+346.5+60.66+733.33+117.86 = 1547.1 + n moles/hr

Concentration of Methane in Outlet = 117.86/(n+288.75+346.5+60.66+733.33+117.86)

Given, For Xmin, Concentration of Methane in Outlet = 5%

Therefore, .05=117.86/(n+288.75+346.5+60.66+733.33+117.86)

Therefore, Xmin = n = 810.1 moles/hr

d) Given X=Xmin

Therefore, n = 810.1 moles/hrb

Y = 346.5/(n+288.75+346.5+60.66+733.33+117.86) = 0.14 = 14%

Therefore, UFL new = 15-0.830830*14 = 3.36%