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PhD Thesis - Energy Systems Research Unit - University of Strathclyde

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The percentage load is then calculated using Equation 5.23,<br />

Percentage Load = 100 x Required Engine Power (kW) (5.23)<br />

Derated Maximum Power<br />

The specific fuel consumption or efficiency is calculated by interpolating<br />

between the known values for 100%, 75%, 50%, and 25%, as shown in Figure<br />

5.9. If values are known for other percentages, these may also be input and used<br />

for interpolation, however, the value for 100% load must be given.<br />

5.2.5 Multiple Engine Generator Sets<br />

In order to work more efficiently, provide a better range <strong>of</strong> loads that may be<br />

supplied, and reduce engine damage, it is common practice to use a number <strong>of</strong><br />

lower rated engine generating sets rather than one higher rated one. This also<br />

means that, if one fails, essential power may still be provided. This possibility<br />

has been incorporated into this model, by allowing a maximum <strong>of</strong> five<br />

generating sets, <strong>of</strong> equal rated power, in the one engine definition. These engine<br />

sets share the load between them in a manner that ensures they are working as<br />

efficiently as possible. As the engine efficiency has a low rate <strong>of</strong> decrease at<br />

higher loadings and a greater rate <strong>of</strong> decrease at much lower loadings (see<br />

Figure 5.7), the best performance is obtained from the plant by running the<br />

lowest possible number <strong>of</strong> individual engines at equal percentage loadings.<br />

The number <strong>of</strong> available engine sets is defined, and if the required engine power<br />

is less than the maximum derated power, one engine only is required, and its<br />

operating parameters are calculated as described previously. All other engines<br />

are not used. If the required engine power is between one and two times the<br />

maximum derated power, the load is divided equally between the two engines,<br />

to allow them to work as efficiently as possible. If this calculated load is below<br />

the minimum load, one engine will work at full load, while the other is not used,<br />

and there will be some unmet demand. This will only happen, however, if the<br />

minimum load is above 50%. This is repeated for between two and three times<br />

the maximum derated power, between three and four times, and between four<br />

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