Question

Belchatow in Poland is the largest coal-burning plant in the world. It produces 27.5 terawatt-hours of electricity each year (1 terrawatt = 1 billion kilowatts). Show all of your calculation as you answer the following questions.

1. Determine the number of BTUs of heat requird to generate the electricity produced by Belchatow every day. Assume that 10,000 BTUs produce 1 kW-h-of electricity

2. How many pounds of coal are consumed by the power plant every year? Assume that 1 lb of coal produces 10,000 BTus of heat.

3. How many pounds of sulfur are released by the power plant every day, assuming the plant has nothing that reduces pollution output and that coal used by the plant contains 1.5% sulfur, by weight?

4. The United Nations signs a new sulfur protocol that requires coal-fired power plants to emit no more thatn 1.2 lb of sulfur per million BTUs of heat generated. Does the Belachatow plant meet this new standard? If not, describe one way that the plant could reduce sulfur emissions.

5. Identify a technology that the Belachatow plant can use to reduce the amount of sulfur released. Explain how this technology functions to reduce air pollution.

Answer 1

Electricity produced by Belchatow each year = 27.5 TW-h = 2.75 X 10¹⁰ kW-h

1. Electricity produced by Belchatow every day = 2.75 X 10¹⁰ kW-h / 365 = 7.53 X 10⁷  kW-h

1 kW-h of electricity is produced by 10,000 BTUs.

7.53 X 10⁷  kW-h of electricity is produced by = 7.53 X 10⁷ X 10,000 = 7.53 X 10¹¹ BTUs

The number of BTUs of heat required to generate the electricity produced by Belchatow

every day = 7.53 X 10¹¹ BTUs

2. The number of BTUs of heat required every year = 2.75 X 10¹⁰ kW-h X 10,000 BTUs = 2.75 X 10¹⁴ BTUs

10,000 BTUs of heat is produced by 1 lb or 1 pound of coal.

No. of pounds of coal consumed by the power plant every year = 2.75 X 10¹⁴ BTUs / 10,000

= 2.75 X 10¹⁰ pounds

3. No. of pounds of coal consumed by the power plant every year = 2.75 X 10¹⁰ pounds

No. of pounds of coal consumed by the power plant every day = 2.75 X 10¹⁰ pounds / 365 = 7.53 X 10⁷ pounds

Coal used by the plant contains 1.5% sulfur by weight.

No. of pounds of sulfur released by the power plant every day = (7.53 X 10⁷ X1.5 ) /100 = 1.13 X 10⁶ pounds

4. No. of pounds of sulfur released every day i.e., for 7.53 X 10¹¹ BTUs = 1.13 X 10⁶ pounds of sulfur

No. of pounds of sulfur released for 1 million BTUs = (1.13 X 10⁶ X 10⁶) /(7.53 X 10¹¹) = 1.5 pounds or 1.5 lb

The Belachatow plant does not meet this new standard as 1.5 lb / million BTUs is more than

1.2 lb / million BTUs

5. Reducing the amount of sulfur from coal before combustion by switching to a fuel that has a lower sulphur content

Removal of Sulphur during Combustion by

1. Fluidised Bed Combustion (FBC) process

2. Integrated Gasification Combined Cycle (IGCC) system.

Removal of Sulphur after Combustion by

1. Limestone/Gypsum System

2. Spray Dry System

3. Seawater Scrubbing Process

4. Wellman-Lord Process

This technology funtions to reduce air pollution by reducing the amount of SO₂ emitted into the atmosphere.