The displacement volume of an internal combustion engine is 2.2 liters. The processes within each cylinder of the engine are mod
eled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by p1 = 1.6 bar, T1 = 325 K and V1 = 2.36 liters.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.
1) If the cycle is executed 2100 times per min, determine:
a) the compression ratio.
b) the net work per cycle, in kJ.
c) the maximum temperature, in K.
d) the power developed by the engine, in kW.
e) and the thermal efficiency.
1 answer:
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Answer:
See explaination
Explanation:
Lets first consider the term Isentropic efficiency. The isentropic efficiency of a compressor or pump is defined as the ratio of the work input to an isentropic process, to the work input to the actual process between the same inlet and exit pressures. IN practice, compressors are intentionally cooled to minimize the work input.
Please kindly check attachment for the step by step solution of the given problem.
Answer:
18 kJ
Explanation:
Given:
Initial volume of air = 0.05 m³
Initial pressure = 60 kPa
Final volume = 0.2 m³
Final pressure = 180 kPa
Now,
the Work done by air will be calculated as:
Work Done = Average pressure × Change in volume
thus,
Average pressure = = 120 kPa
and,
Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³
Therefore,
the work done = 120 × 0.15 = 18 kJ