Fuel oil calculation Basics-SFOC-Speed - consumption

Energy efficiency training 2012

How much is one gram?

If SFOC of a single MPC vessel is increased by 1g/kWh, how much does the annual fuel consumption increase?

1. A) The effect is negligible

2. B) 10t

3. C) 25t

4. D) 45t

How much is one gram?

How much is one gram / kWh?

> Annual energy production MPC Class:

> Main engines ~ 40,000 MWh

> Auxiliary engines ~ 5,000 MWh

Total ~ 45,000 MWh

> 45,000,000 kWh/a * 1 g/kWh = 45,000 kg/a

> 45 t/a * 750$/t = 33,750 $/a

Value of 1 g/kWh is about 35,000 $/a

(per each MCC7S60 ME)

Worst case scenario

The diesels could be +25..30 g/kWh from optimum

25 g/kWh * 45 000 000 kWh/a = 1125 t/a

1125 t/a * 750 $/t = 843 750 $/a

Calculation Example

Consider a Superflex Heavy MPC vessel is to travel a distance of2160nm on design load on normal sea conditions.

Which option is more fuel efficient?

A) 18kn for 5 days and then wait 2 days.

B) Constant speed for 7 days.

Calculation Exercise, source data

SFOC ME 175g/kWh

15% sea margin

AE fuel consumption 3t/d

Speed-Power curve ->

Calculation Exercise, solution – Step 1

Required ME power in case A (including sea-margin):

6750 * 1,15 = 7762,5kW

Required kWh in case A:

7762,5kW * 24h * 5 = 931 500 kWh

Calculation Exercise, solution – Step 2

In case all 7 days would be used for sailing, the required speed would be:

2160nm / (24h * 7) = 12,9 kn

Required ME power in case B (including sea-margin):

2400 * 1,15 = 2760kW

Required kWh in case B:

2760kW * 24h * 7 = 463 680 kWh

Calculation Exercise, solution – Step 3

Consumed fuel in case A:931 500 kWh * 190g/kWh / 1 000 000 = 177 t

Consumed fuel in case B:463 680 kWh * 200g/kWh / 1 000 000 = 93 t

Difference: 84t (or 47%, or $ 59 000)

Auxiliary Engine consumption similar in both cases

The end