Transport maths

I have just spent around 4 hours calculating the exact emissions of the cargo ship I took to South America last summer, and how much more efficient it is than flying. It actually came pretty close to my original estimate, so I’m fairly chuffed..

If you don’t like maths, look away now, but this is pretty rigorous:

All figures are based on the MSC Monterey, a ship with:

  • 59300 DWT deadweight
  • 4860 TEU capacity
  • operating 4 B & W diesel engines, with combined output of 39,352 kW
  • 54,549t tonnage

Appendix 1

 

CO2 Emissions Calculations for MSC Monterey on voyage Antwerpen-Callao

150.6 t d-1 daily fuel consumption[i]

496.1˙ nm d -1 rate of travel (based on 8930 nautical miles at average speed of 14 knots)

150.6 t d-1 / 496.1˙ nm d -1 = 0.30356 t fuel/nm

= 0.3 t/nm

0.3 t/nm * 8930 nm = 2710.86071 t

=2710.9 t total fuel for trip

0.3 t/nm / 4151 t (based on 70% capacity of 59300 DWT) = 7.33079 x 10-5

= 7.3 x 10-5 nm fuel/ t cargo

7.3 x 10-5 nm/t * 8930 =0.65464 t fuel/t cargo

= 0.7 tf /tc

 

0.7 tf /tc * 3.1 (CO2 conversion factor[ii]) = 2.11318 t/tc

= 2.1 t/tc

2.1 t/tc * 8930nm = 18870.67157 t/tc

18870.7 t/tc

18870.7 t/tc * 0.000058t (my weight) = 1094.49895 kg CO2 equivalent

= 1094.5 kg CO2e

= 1.1 t CO2e


[i] DEFRA 2011 Guidelines to Defra / DECC’s GHG Conversion Factors for Company Reporting (July 2011) Produced by AEA for the Department of Energy and Climate Change (DECC) and the Department for Environment, Food and Rural Affairs (Defra)

[ii] CE Delft, Germanischer Lloyd, MARINTEK & Det Norske Veritas Greenhouse Gas Emissions for Shipping and Implementation of the Marine Sulphur Directive (2006) Delft

Appendix II

 

C13H28 + O2 => CO2 + H2O

Cruising distance – distance travelled

Cruising fuel = cruising distance x 10.1

Landing and take-off fuel = 7840kg

Occupancy (~80%)

Therefore number passengers = % occupancy * maximum number passengers

Fuel pp = (cruising fuel + take-off/landing fuel) / number passengers

CO2 emissions = fuel pp * 3 * (44/12) * (156/184)[i]

Lima – Madrid

LAN airlines 19:45 flight daily using Airbus A340-600

Capacity 419

At 90% occupancy 419/100 * 90 = 377.1

Distance 5905 mil

Cruising distance = 5905 – 250 = 5655 mil

5655 mil * 10.1 = 57115.5 kg

57115.5 kg + 7840 kg = 64955.5 kg

64955.5 kg / 377.1 = 172.3 kg fuel

172.3 * 3 * (44/12) * (156/184) = 1606.41910

= 1606.4 kg CO2e

Madrid – London

Iberia airlines flight using Airbus A320

Capacity 150

At 80% occupancy 150/100 * 80 = 120

Distance 785 mil

Cruising distance 785 – 250 = 535 mil

535 mil *10.1 = 5403.5 kg

5403.5 + 7840 = 13243.5 kg

13243.5 / 120 = 110.3625

= 110.4 kg

110.4 * 3 * (44/12) * (156/184) = 1029.250272

= 1029.3 kg CO2e

1029.3 + 1606.4 = 2635.7 kg CO2e


[i] CO2 conversion factor obtained from World Resources Institute; Safe Climate project

2635.7 / 1094.5 = 2.40813

Therefore taking the cargo ship was 2.4 times more efficient than flying.

And that’s a roundabout way, taking detours through the Caribbean and stuff.

Oh, and did I forget the SHEER INCREDIBLENESS OF TAKING THAT SHIP??

Added on top of this will be the 3.5kg for the Eurostar, and 6.16g for the train from Brussels to Antwerp, and the 21.7kg emitted by buses in South America, taking the entire contribution of transport for my trip to

3.76 t CO2e

My head hurts from all the numbers now so I’m gonna take a break before I try and figure out my carbon footprint for when I live in Ingerland.

ship

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About shakkka

Londoner, climate scientist, extremist. All views are my own.

2 responses to “Transport maths”

  1. Michael says :

    there’s a study going on installing sails on this large modern cargo ships taking the design from the old wooden ships which enable to move with the help of wind force so if this come through then i can safely say that sailing is the best way to go. i have seen the concept design from a magazine try to see for yourself self if this is possible

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