Calculation of the specific fuel oil consumption (g/kWh, g/bhph) requires that engine power, and the consumed fuel oil amount, are known for a certain period of time. The engine power (in bhp or KW) can be calculated from the Indicator diagram or from Fuel Pump Index method or from Turbocharger speed. The oil amount is measured for a few hours to avoid calculation mistakes. The engine parameters shoudnot be changed during this period. Since quantity measurements will be in volume units (m3), it will be necessary to know the oil density, in order to convert to weight units (gram). The density is to correspond to the temperature at the measuring point. Density can be calculated on the basis of bunker specifications. Density at 15 deg is available in the BDN (Bunker Delivery Note). The density at required temperature can be calculated with the density correction factor equation. But normally graphs are provided in the manual to find the density at required temperature, where the change in density is shown as a function of temperature.

SFOC =

h x Pe

where:

Co = Fuel oil consumption over the period (m^3)

D 119 = Corrected gravity (t/m3)

h = Measuring period, hours

Pe =Brake horse power, bhp

10^6 is multiplied to convert the fuel oil unit in tonnes to gram

3 x 8,130

In order to be able to compare consumption measurements carried out for various types of fuel oil, allowance must be made for the differences in the lower calorific value (LCV) of the fuel concerned. Normally, on the testbed, gas oil will have been used, having a lower calorific value of approx. 42,707 kJ/kg (corresponding to 10,200 kcal/kg). If no other instructions have been given by the shipowner, it is recommended to convert to this value. Usually, the lower calorific value of a bunker oil is not specified by the oil companies. However, by means of the graph given in the manual as a function of sulphur content and density at 15 deg C, the LCV can be determined . The corrected consumption can then be

determined by multiplying the "measured consumption

OR

LCV2 = the specific lower calorific value, in kcal/kg, of the bunker oil concerned

Calculating LCV from the graph

LCV1, = 40,700 kJ/kg, derived from the graph given in manual

Consumption corrected for calorific value:

Note - The ambient conditions (blower inlet temperature and pressure and scavenge air coolant temperature) will also influence the fuel consumption. Correction for ambient conditions is not considered important when comparing service measurements.

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Density correction graph |

__Calculation with an example__**Effective Engine Power, Po - Say 8,130 bhp**(Engine power is calculated from the performance sheet. It can be calculated from the Indicator digram or with fuel pump index or turbocharger method. All methods are explained below)**Consumption, Co**- 3.83 m3 over 3 hours**Measuring point temperature**- 119°C**Density at 119°C**- Specific gravity 0.9364 t/m^3, 3% sulphurSFOC =

__Co x D119 x 10^6__h x Pe

where:

Co = Fuel oil consumption over the period (m^3)

D 119 = Corrected gravity (t/m3)

h = Measuring period, hours

Pe =Brake horse power, bhp

10^6 is multiplied to convert the fuel oil unit in tonnes to gram

__3.83 x 0.8684 x 10^6__= 136.4 g/bhph3 x 8,130

__Correction of LCV of Fuel Oil__

determined by multiplying the "measured consumption

**LCV 1****42,707**__LCV1 = the specific lower calorific value, in kJ/kg, of the bunker oil concerned__

OR

**LCV2****10,200**LCV2 = the specific lower calorific value, in kcal/kg, of the bunker oil concerned

Calculating LCV from the graph

LCV1, = 40,700 kJ/kg, derived from the graph given in manual

Consumption corrected for calorific value:

__136.4 x 40,700__= 130.0 g/bhph**42,707**Note - The ambient conditions (blower inlet temperature and pressure and scavenge air coolant temperature) will also influence the fuel consumption. Correction for ambient conditions is not considered important when comparing service measurements.

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When someone writes an article he/she keeps the thought of a user in his/her brain that how a user can be aware of it. Therefore that’s why this post is outstanding.Thanks!

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When the LCV of fuel reduces, the SFOC should increase. The last formulae should be rearranged to suit this !!!1 I suppose. Do correct me if i am wrong.

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