Problems
1. The average home in India uses about 11,000 kWh/yr. What is the
electricity-based carbon footprint of the average home, in tonnes?
Assume 1.5 lbs CO2/kWh.
Ans: 11,000×1.5 = (16500×0.45)/1000 = 7.425 tonnes
There are about 115,000,000 households in the Delhi. What is the total
electricity-based carbon footprint of Delhi (in tonnes)?
Ans: 115,000,000×7.425= 8.538e8 tonnes
2. A corporate vehicle fleet consumes 90,000 gallons of LNG each year. How
many tonnes of greenhouse gases is the fleet responsible for each year?
One thousand cubic feet (MCF) = 12.1 gallons of LNG
Emission Factor for LNG = 0.12 lbs/cubic feet
Ans: (90000/12.1)*1000 = 7438017 cu. Ft LNG
Now, 7438017*0.12 = 892,562 lbs CO2
Then, (892,562*0.45)/1000 = 401.65 Tonnes
3. A company uses fuel purchase records to determine its scope I
emissions. Last Year, the company purchased 714 gallons of gasoline, 212
gallons of diesel fuel, and 3 million cubic feet of natural gas. What were
the company’s scope 1 emissions this year?
Emission factor for Gasoline = 19.37 lbs CO2/gallon
Emission factor for Diesel = 22.23 lbs CO2/gallon
mission factor for Natural Gas = 0.12 lbs CO2/Cubic Foot
Ans:
Gasolin: 714*19.37 = 13830 lbs CO2
Diesel: 212*22.23 = 4713 lbs CO2
NG: 3000000*0.12 = 360000 lbs CO2
Total = ((13830+4713+360000)*0.45)/1000 = 170.34 tonne
4. A water treatment plant consumes one-kilowatt hour of electricity for
every 900 gallons of water it treats. The plant gets its electricity from a
coal-fired power plant with an emission factor of 1500 lbs CO2/MWh. If
the facility treats 3.4 million gallons of water each month, how many
tonnes of CO2 does it emit each year?
Ans:
First, evaluate how many gallons of water it treats per year i.e.
3.4*1000000*12 = 40,800,000 gallons per year
Each 900 gallons of water it consumes 1 KWh of electricity
For 40,800,000 it will consume, 40,800,000/900 = 45,333 kWh/Year
i.e. 45.33 mWh
Emission factor is given as: 1500 lbs CO2/mWh
In total we get 1500*45.33 = 67995 lbs CO2/year
In Tonne = (67995*0.45)/1000 = 30.60 Tonne.
5. A power plant’s generation mix consists of coal (electric utility coal) and
natural gas. Every day the plant combusts 1600 tonne of coal and 27,500
Mcf (thousand cubic feet) of natural gas. How many tonnes of carbon
dioxide does the plant emit in one year?
Emission factor for coal = 4728.82 lbs CO2/tonne
Emission factor for natural gas = 0.12 lbs CO2/cubic feet
Ans:
Coal:
1600 tonne consumption/day
i.e. per year consumption is 1600*365 = 5,84,000 tonnes/year
So by consuming 5,84,000 tonnes/year, it emits 5,84,000*4728.82 =
2761630880 lbs/year
So in tonne = (2761630880*0.45)/1000 = 1242734 tonne
Natural Gas:
Every day it consumes 27500 Mcf i.e. 27500 thousand cubic foot
i.e. 27500*1000 cubic foot 27500000
so in a year, it will consume = 27500000*365 = 10037500000 cubic foot/year
so it will emit 10037500000*0.12 = 1204500000 lbs
so in tonne (1204500000*0.45)/1000 = 542025 tonne
Total
1242734+542025 = 1784759 tonne.
6. The distance between Atlanta airport and Narita airport in Tokyo, Japan
is 6876 miles. A Boeing 747 is estimated to consume around 5 gallons
per mile on a long-haul flight. The emission coefficient for jet fuel is 9644
gCO2/gal. what is the carbon footprint for this trip?
Ans:
6876 miles per trip * 5 gallons per mile = 34380 gallons/trip
So 34380 gallons/trip * 9644 = 331560720 g of CO2/trip i.e. 331.56
Tonne of CO2/trip
7. A 66-acre lined cell landfill contains 2.4 million tons of waste. At its
current level, the landfill emits 2000 metric tons of CO2 and 2000 metric
tons of methane per year, with traces of other compounds. Looking only
at the methane and CO2 emissions, what are the total emissions per
year in terms of CO2e? in this example, assume methane has a GWP of
21.
Ans:
2000*1 (GWP) = 2000 mtCO2e/year
2000*21 = 42000 mtCO2e/year
So, in total 44000 mtCO2e/year
8.
Ans:
21.18 Tonnes per Year.
Ans:
Process Emission:
180000*925 = 166500000 lbs CO2 = 74925 Tonnes of CO2
Combustion Emission:
Carbon Dioxide = 205 lbs CO2/MMBtu * 1 (GWP =1) = 205 lbs CO2e/MMBtu
Methane = 5 * 25 (GWP) = 125 lbs CO2e/MMBtu
N2O = 0.36*298 (GWP) = 107.28 lbs CO2e/MMBtu
Total = 205+125+107.28 = 437.28 lbs Co2e/MMBtu
Now, 1.7 MMBtu is used per metric ton of cement
Therefore, 180000*1.7*437.28 = 133807680 lbs of CO2e/year
So, 60213 tonne of CO2e/year
So, grand total will be
74925+60213 = 136219 tonnes of CO2e/year.
