A 500 MW coal based power station is operating at an efficiency of 30%. If the coal has 1% of sulphur content and 1 tonne of coal produces 8000 kWh energy, how much SO2 will be emitted daily by the plant?
100 tonne
This problem requires us to calculate the daily emission of sulfur dioxide (SO2) from a coal-based power station given its capacity, efficiency, the sulfur content in the coal, and the energy density of the coal.
We are provided with the following information about the coal power station:
We need to find the total amount of SO2 emitted daily.
The plant operates at a capacity of 500 MW. We need to find the energy produced in 24 hours.
Capacity = 500 MW
Daily operating time = 24 hours
Total daily energy output = Capacity \(\times\) Time
First, convert MW to kW (since energy density is in kWh):
\(500 \text{ MW} = 500 \times 10^3 \text{ kW}\)
Daily Energy Output (kWh) = \(500 \times 10^3 \text{ kW} \times 24 \text{ hours}\)
Daily Energy Output = \(12,000,000 \text{ kWh}\)
The plant has an efficiency of 30%. This means that for every unit of energy input from coal, only 30% is converted into electrical energy output.
Efficiency \(= \frac{\text{Energy Output}}{\text{Energy Input}}\)
Energy Input \(= \frac{\text{Energy Output}}{\text{Efficiency}}\)
Daily Energy Input from coal = \(\frac{12,000,000 \text{ kWh}}{0.30}\)
Daily Energy Input from coal = \(40,000,000 \text{ kWh}\)
We know that 1 tonne of coal produces 8000 kWh of energy.
Daily Coal Consumption = \(\frac{\text{Daily Energy Input}}{\text{Energy per tonne of coal}}\)
Daily Coal Consumption = \(\frac{40,000,000 \text{ kWh}}{8000 \text{ kWh/tonne}}\)
Daily Coal Consumption = \(5000 \text{ tonnes}\)
The coal has a sulfur content of 1% by weight.
Daily Sulfur Consumption = Daily Coal Consumption \(\times\) Sulfur Content
Daily Sulfur Consumption = \(5000 \text{ tonnes} \times 0.01\)
Daily Sulfur Consumption = \(50 \text{ tonnes}\)
When sulfur (S) burns, it reacts with oxygen (O2) to form sulfur dioxide (SO2).
The chemical reaction is: \(S + O_2 \rightarrow SO_2\)
The molecular weight of Sulfur (S) is approximately 32 g/mol.
The molecular weight of Sulfur Dioxide (SO2) is approximately 32 (for S) + 2 \(\times\) 16 (for O) = 64 g/mol.
From the reaction, 1 mole of S produces 1 mole of SO2. By mass, 32 units of mass of S produce 64 units of mass of SO2. This means the mass of SO2 produced is double the mass of sulfur burned.
Mass of SO2 emitted = Mass of Sulfur consumed \(\times\) 2
Daily SO2 Emission = \(50 \text{ tonnes} \times 2\)
Daily SO2 Emission = \(100 \text{ tonnes}\)
Thus, the power plant will emit 100 tonnes of SO2 daily.
| Parameter | Value |
|---|---|
| Power Plant Capacity | 500 MW |
| Efficiency | 30% |
| Coal Sulfur Content | 1% |
| Coal Energy Content | 8000 kWh/tonne |
| Daily Energy Output | 12,000,000 kWh |
| Daily Energy Input | 40,000,000 kWh |
| Daily Coal Consumption | 5000 tonnes |
| Daily Sulfur Consumption | 50 tonnes |
| Daily SO2 Emission | 100 tonnes |
Reviewing the steps involved in calculating SO2 emissions from a coal power plant:
Sulfur dioxide (SO2) is a major air pollutant produced primarily from the burning of fossil fuels, especially coal, that contain sulfur. It has significant environmental and health impacts.
Power plants often use technologies like Flue Gas Desulfurization (FGD), also known as "scrubbers," to remove SO2 from the exhaust gases before they are released into the atmosphere, reducing their environmental impact.
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