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.
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% sulphur
SFOC = 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/bhph
3 x 8,130
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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| 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% sulphur
SFOC = 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/bhph
3 x 8,130
Correction of LCV of Fuel Oil
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 bedetermined 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
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 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.
ReplyDeleteWhen crude oil is heated, then which of the following hydrocarbon is given off first oil funding
ReplyDeleteSFOC is a measure of the amount of fuel consumed by a marine engine to produce a unit of power output, usually expressed in grams per kilowatt-hour (g/kWh). It is a function of the engine design, operating conditions, fuel quality, and maintenance practices. by DG Marine
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