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    A simplified representation of the diesel cycle, with just air as the working substance. Show that the efficiency of the engine is:

    n = 1 - 1/y((1/rye-1/ryc)/(1/re-1/rc))

    where re = V3/V2, the expansion ratio, and rc = V3/V1, the compression ratio.
    If re = 5, rc = 15 and y = 1.4, evaluate n.
    Notice that the compression ratio can be very much higher in a diesel engine than in a petrol engine because diesels do not suffer from pre-ignition, or pinking, as the fuel is sprayed in at the end of the compression stroke; this allows an increase in re. This is one reason why diesels are more efficient than petrol engines.

    Please see attachment for further clarification, if the equation above is not already clear.

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    Solution Preview

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    We start by drawing the diesel engine cycle on a pressure/volume digram and labelling the key features. This is shown as below

    The cycle consists of the following:

    1. Injection and release of the air stroke a-e-a as shown
    2. An adiabatic compression (a zero heat exchange process) of the piston from a volume to (where )
    3. A isobaric (constant pressure) change between b and c where the diesel fuel is taken into the chamber and combusted. This process takes in an amount of heat as shown between b and c results in an increase in the charge volume, increasing from to a volume as shown. Note that this also results in an increase in temperature from to

    The amount of specific heat taken in can be determined from the thermodynamic relation shown by {1} as the product of the isobaric specific heat capacity and the temperature change for the process b to c,


    As the temperature change we can write {1} more formally as {2}


    4. An adiabatic expansion (a zero heat exchange process) of the piston from a volume to (where )
    5. An iso-volumetric process ...

    Solution Summary

    The expert simplifies representation of the diesel cycle. Expansion and compression ratios are provided.