For a heat input of 13,000 KJ/KWhr to a coal-fired electric power plant: Assume

Question

For a heat input of 13,000 KJ/KWhr to a coal-fired electric power plant: Assume a coal sulfur content between 1.5 and 4 percent. Choose a coal energy content from these choices: 20 KJ/g; 28 kJ/g. or 32 kJ/g. Make one assumption for each of these variables and work the problem. Use only one set of as - sumptions to produce one answer. Calculate the uncontrolled sulfur dioxide emission rate from the plant. Calculate the sulfur dioxide control equipment efficiency required to limit the plant's sulfur di - oxide emissions to 260 g sulfur dioxide per million kJ heat input.

Explanation / Answer

Heat Input of 13000kJ/KWhr

coal sulfur content between 1.5 and 4 percent

Coal energy content 20 kJ/g, 28 kJ/g or 32 kJ/g

Let us assume 20 kJ/g energy content and 1.5 percent sulfur content

for heat input of 13000kJ/KWhr

amount of coal required 13000 kJ/20 kJ/g = 650 g of coal will produce 13000 kJ heat

1.5 percent sulfur content will mean 1.5/100 x 650 = 9.75 g of sulfur in 650 g of coal

MW of Sulfur dioxide is 64.066 g/mol

we get  64.066 g of Sulfur dioxide from 32.06 g of sulfur

so from 9.75 g of sulfur we will get 19.48 g of Sulfur dioxide

so rate of sulfur dioxide emission will be 19.48 g of Sulfur dioxide/KWhr

For 1 million kJ heat input Coal required will be 50000g

Sulfur will be 1.5/100 x 50000 = 750 g of sulfur

Sulfur dioxide will be (750 x 64.06)/32.06 = 1498 g

%Rg = SO2 removal efficiency of the control device, percent.

where %Rg = 100 x (1-Eo/Ei)

Eo and Ei are the  output and input to the control device.

To limit sulfur dioxide to 260 g efficient of sulfur dioxide control equipment should be

%Rg = 100 x (1-Eo/Ei)

%Rg = 100 x (1- 260/1498)

%Rg = 82.64%

Now Let us assume 32 kJ/g energy content and 4.0 percent sulfur content

amount of coal required 13000 kJ/32 kJ/g = 406.25 g of coal will produce 13000 kJ heat

4.0 percent sulfur content will mean 4.0/100 x 406.25 = 16.25 g of sulfur in 406.25 g of coal

MW of Sulfur dioxide is 64.066 g/mol

we get  64.066 g of Sulfur dioxide from 32.06 g of sulfur

so from 16.25 g of sulfur we will get 32.47 g of Sulfur dioxide

so rate of sulfur dioxide emission will be 32.47 g of Sulfur dioxide/KWhr

For 1 million kJ heat input Coal required will be 31250g

Sulfur will be 4.0/100 x 31250 = 1250 g of sulfur

Sulfur dioxide will be (1250 x 64.06)/32.06 = 2497 g

To limit sulfur dioxide to 260 g efficient of sulfur dioxide control equipment should be

%Rg = 100 x (1-Eo/Ei)

%Rg = 100 x (1- 260/2497)

%Rg = 89.59%

Now Let us assume 20 kJ/g energy content and 4.0 percent sulfur content

for heat input of 13000kJ/KWhr

amount of coal required 13000 kJ/20 kJ/g = 650 g of coal will produce 13000 kJ heat

4.0 percent sulfur content will mean 4.0/100 x 650 = 26 g of sulfur in 650 g of coal

MW of Sulfur dioxide is 64.066 g/mol

we get  64.066 g of Sulfur dioxide from 32.06 g of sulfur

so from 26 g of sulfur we will get 51.95 g of Sulfur dioxide

so rate of sulfur dioxide emission will be 51.95 g of Sulfur dioxide/KWhr

For 1 million kJ heat input Coal required will be 50000g

Sulfur will be 4.0/100 x 50000 = 2000 g of sulfur

Sulfur dioxide will be (2000 x 64.06)/32.06 = 3996 g

To limit sulfur dioxide to 260 g efficient of sulfur dioxide control equipment should be

%Rg = 100 x (1-Eo/Ei)

%Rg = 100 x (1- 260/3996)

%Rg = 93.49%

Now Let us assume 32 kJ/g energy content and 1.5 percent sulfur content

amount of coal required 13000 kJ/32 kJ/g = 406.25 g of coal will produce 13000 kJ heat

1.5 percent sulfur content will mean 1.5/100 x 406.25 = 6.09 g of sulfur in 406.25 g of coal

MW of Sulfur dioxide is 64.066 g/mol

we get  64.066 g of Sulfur dioxide from 32.06 g of sulfur

so from 6.09 g of sulfur we will get 12.17 g of Sulfur dioxide

so rate of sulfur dioxide emission will be 12.17 g of Sulfur dioxide/KWhr

For 1 million kJ heat input Coal required will be 31250g

Sulfur will be 1.5/100 x 31250 = 468.7g of sulfur

Sulfur dioxide will be (468.7 x 64.06)/32.06 = 936 g

To limit sulfur dioxide to 260 g efficient of sulfur dioxide control equipment should be

%Rg = 100 x (1-Eo/Ei)

%Rg = 100 x (1- 260/936)

%Rg = 72.24%

Now Let us assume 28 kJ/g energy content and 1.5 percent sulfur content

amount of coal required 13000 kJ/28 kJ/g = 464.28 g of coal will produce 13000 kJ heat

1.5 percent sulfur content will mean 1.5/100 x 464.28 = 6.96 g of sulfur in 464.28 g of coal

MW of Sulfur dioxide is 64.066 g/mol

we get  64.066 g of Sulfur dioxide from 32.06 g of sulfur

so from 6.96 g of sulfur we will get 13.91 g of Sulfur dioxide

so rate of sulfur dioxide emission will be 13.91 g of Sulfur dioxide/KWhr

For 1 million kJ heat input Coal required will be 35714g

Sulfur will be 1.5/100 x 35714 = 535.7g of sulfur

Sulfur dioxide will be (535.7 x 64.06)/32.06 = 1070 g

To limit sulfur dioxide to 260 g efficient of sulfur dioxide control equipment should be

%Rg = 100 x (1-Eo/Ei)

%Rg = 100 x (1- 260/1070)

%Rg = 75.71%

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