Question

Coal with 40% C, 30% H and 30% ash by weight at 1000 kg / h flow (school
the last digit of your number) is burned with dry air fed by kmol / second. Chimney from the oven
In addition to its gas, there is a solid waste, 95% of this waste is all of the ash in coal, back
the rest is unburned carbon. While all the hydrogen in the fuel is oxidized to water;
carbon burns so that the selectivity of carbon dioxide in carbon monoxide is 10.
a. calculate Percentage of excess air used,
b. Calculate what percentage of oxygen increases

if you can help me quickly that would be great thanks

.

Answer 1

Given: The coal contains 40% Carbon, 30% Hydrogen and 30% Ash.

This type of coal undergoes combustion at a mass flow rate of 1000 kg/hr

Therefore in a feed of stream of 1000 kg / hr,

Mass flow rate of Carbon = 40% of 1000 kg / hr = 400 kg / hr

Mass flow rate of Hydrogen = 30% of 1000 kg / hr = 300 kg / hr

Mass flow rate of Ash = 30% of 1000 kg / hr = 300 kg / hr

The coal on combustion produces Gases and also leaves a solid product that cannot be burnt anymore.

According the question, The gases produced are CO₂ and CO with the selectivity or the molar ratio being 10 :1

Ash remains as Ash does not change during the combustion reaction unless told otherwise in the problem.

Therefore if you, for example, have 10 kg of ash in the coal, the solid residue will leave behind all the 10 kg of ash as it is after combustion.

Therefore, the solid residue left behind will contain 300 kg of ashe and also contains some carbon as per the question.

The coal combusts in dry air to give Solid residue and gases.

Let us denote coal flow rate as F, Solid residue as R and Gases as G

The problem states that the ash produces accounts for 95% of the solid waste produced.

The mass of ash is 300 kg / hr as we already know.

Therefore we can say: 95% of R = 300 kg / hr

0.95 x R = 300

R = 315.789 kg

Therefore, mass of carbon unburnt in the ash = 315.789 - 300 = 15.789 kg /hr

Therefore total mass of carbon burnt = Total carbon content - unburnt carbon

= 400 - 15.789 = 384.211 kg /hr

When the selectivity of a reaction is given, it is always with respect to the number of moles formed and NOT the weights formed as the reaction occurs based on molar concentrations and NOT weights.

Molar mass of Carbon is 12 kg / kmol , Therefore molar rate of carbon being burnt = 384.211 / 12 = 32. 07 kmol / hr

The combustion reactions:

We see that 1 mole of carbon required 1 mole of O₂ to produce 1 mole of CO₂ and 0.5 moles of O₂ to produce 1 mole of CO.

In both cases we see that 1 mole of carbon produces 1 mole of the respective gas. Therefore we can say that the total number of moles of gas formed as per the question would be 32.07 kmol / hr

Let X be the number of moles of Carbon Monoxide formed.

Then the question says that the number of moles of Carbon dioxide formed is 10X

We know that the number of moles carbon being burnt is 32.07 kmol / hr .

Therefore: X + 10X = 32.07

11X = 32.07 kmol / hr

X = 2.915 kmol / hr; 2.915 kmol/hr of Carbon monoxide is being produced

Moles of Carbon dioxide present = 10X = 10 x 2.915 = 29.15 kmol / hr

As 1 mole of C needs 0.5 mole of O₂ to form 1 mole of CO, 2.915 kmol / hr of Carbon will need 2.915/2 = 1.457 kmol/hr .

Similarly 1 mole of C needs 1 mole of O₂ to produce 1 mole of CO₂, therefore 29.15 kmol / hr of CO₂

Hydrogen Balance:

All the hydrogen is said to convert to water:

The reaction:

We see that 1 mole of H₂ produces 1 mole of H₂O on reacting with 0.5 moles of O₂

Mass flow rate of Hydrogen = 300 kg / hr

Molar mass of H₂ = 2 kg / kmol

Molar flow rate of Hydrogen = 300 / 2 = 150 kmol / hr

therefore, 150 kmol / hr of Water is being produced.

As 1 mole of H₂ needs 0.5 mole of O₂, 150 kmol / hr will need (150/2)= 75 kmol / hr

Therefore total moles of O₂ needed = 29.15 + 1.457 + 75 = 105.6 kmol / hr

1 mole of dry air contains 0.79 moles of N₂ and 0.21 moles of O₂.

Let X kmol of air contain 105.6 kmol / hr

Therefore we get:

1 / 0.21 = X / 105.6

X = 502.89 kmol / hr