In an ideal Otto cycle, air is compressed from 1.15 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 620 kJ/kg. The mean effective pressure (MEP) for this cycle isQuestion 8Select one:a.713 kPab.807 kPac.638 kPad.96 kPae.367 kPaClear my choiceCheckQuestion 8
Question
In an ideal Otto cycle, air is compressed from 1.15 kg/m3 and 2.2 L to 0.26 L, and the net work output of the cycle is 620 kJ/kg. The mean effective pressure (MEP) for this cycle isQuestion 8Select one:a.713 kPab.807 kPac.638 kPad.96 kPae.367 kPaClear my choiceCheckQuestion 8
Solution
The mean effective pressure (MEP) in an Otto cycle can be calculated using the formula:
MEP = (net work output) / (displacement volume)
First, we need to calculate the displacement volume, which is the difference between the initial and final volumes. Given that the initial volume is 2.2 L and the final volume is 0.26 L, the displacement volume is:
Displacement volume = 2.2 L - 0.26 L = 1.94 L
We need to convert this volume from liters to cubic meters, since the net work output is given in kJ/kg and we want the MEP in kPa. There are 1000 liters in a cubic meter, so:
Displacement volume = 1.94 L * (1 m^3 / 1000 L) = 0.00194 m^3
Now we can calculate the MEP:
MEP = (620 kJ/kg) / (0.00194 m^3) = 319587.63 kPa
However, this value is too high. The error comes from the fact that the net work output is given per unit mass, and we haven't taken into account the mass of the air. The density of the air at the beginning of the compression is given as 1.15 kg/m^3, so the mass of the air is:
Mass = density * volume = 1.15 kg/m^3 * 0.00194 m^3 = 0.002231 kg
Now we can calculate the MEP taking into account the mass of the air:
MEP = (620 kJ/kg * 0.002231 kg) / (0.00194 m^3) = 709.5 kPa
So, the closest answer is a. 713 kPa.
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