Part b determine the thermal efficiency Part c determine the

e previous | 4 of 4 return to assig An air standard Otto cycle is has a compression ratio of 9. At the beginning of the compression, pressure is 95 kPa and t , and the maxihum temperature in the cycle is 750 C. Use the IG model for air. Assume mass of air as 0 005 kg emperature is 30 °C. Heat addition to the air is 1kJ/kg Part A Determine the net work Express your answer to three significant figures and include the appropriate units. 146 /kJ Submit My Answers Give Up Incorrect: One attempt remaining; Try Again Part B Determine thermal efficiency Express your answer to the nearest tenth.

Solution

An air standard otto cycle is considered that means the air is considered as ideal or perfect gas which are both same.

Compression ratio is given i.e r_k=9

As we know compression ratio is the ratio of intial volume (before compression )and final volume (end of compression .

As we know the efficiency in air standard cycle is given by:   1-(1r_k)^1.4-1

Thus efficiency of otto cycle becomes      =1-(1/9)^0.4=0.584

And efficiency=work/heat supplied

Thus, work done =.584*1kj/kg*.005kg=2.924*10^-3 kJ

Mean effective pressure =work done/swept volume

Where swept volume is the difference between volumes at the starting of compression and at the end of compression in the internal combustion engine.i.e. V₁-V₂

To calculate V₁ ,we use ideal gas equation PV=mRT

Thus, V₁=mRT₁/P₁

M=mass=.005kg

R=characterstic gas constant=0.287kJ/kgK

T₁=temperature at the starting of compression which is given in question=30 celcius=(30+273)K=303 K

Here degree Celsius is converted into Kelvin as all units are in Kelvin

Therefore, V₁=0.005*0.287*303/95=0.00457 m³

And V₁/V₂ is given as the compression ratio=9

Therefore,V₂=V₁/9=0.0005085 m³

Hence swept volume=V₁-V₂=0.00457-0.0005085=0.0040615 m³

Now we can easily calculate mean effective pressure=work done/swept volume=

2.924*10^-3 kJ/0.0040615 m³   =0.7199 kPa