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PhD Thesis - Energy Systems Research Unit - University of Strathclyde

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where Max = maximum energy available from the fuel (kWh/kg)<br />

CV = calorific value <strong>of</strong> the fuel (kJ/kg)<br />

1 kWh = 3610.3 kJ.<br />

The actual fuel consumption at 100% load is given by<br />

Actual FC = SFC x Power (5.25)<br />

1000<br />

where Actual FC = actual fuel consumption at 100% load (kg/h)<br />

SFC = specific fuel consumption at 100% load (g/kWh)<br />

Power = derated maximum power (kW).<br />

If the maximum energy available from the fuel and the actual fuel consumption<br />

at 100% load are multiplied, the result is the total energy potential <strong>of</strong> the fuel<br />

used in kW. The percentage <strong>of</strong> this energy that is available for use at 100% load<br />

is given by<br />

Percentage = (Power + (Power x Ratio)) x 100 (5.26)<br />

Potential<br />

where Percentage = percentage <strong>of</strong> energy available for use at 100% loading<br />

Power = derated maximum power (kW)<br />

Ratio = heat to electricity ratio at 100% loading<br />

Potential = total energy potential <strong>of</strong> the fuel (kW).<br />

The actual fuel used at the partial load is calculated using Equation 5.25,<br />

however, in this case, the specific fuel consumption must be that for the<br />

percentage load being considered, and the power is the required power at which<br />

the engine is operating, in kW. Assuming that the same percentage <strong>of</strong> the<br />

energy is available for use as at 100% loading (i.e. that the overall efficiency<br />

(heat and electricity) is constant), the total useful energy that this amount <strong>of</strong> fuel<br />

can produce is given by<br />

135

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