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(Solved): The compression ratio of an Otto cycle as shown in Figure 22.13 is VA/VB=8.00. At the beginn ...


The compression ratio of an Otto cycle as shown in Figure 22.13 is \( V_{A} / V_{B}=8.00 \). At the beginning \( A \) of the(b) Fill in this table to follow the processes: (c) Identify the energy input \( \left|Q_{h}\right| \), (d) the energy exhaus

The compression ratio of an Otto cycle as shown in Figure 22.13 is . At the beginning of the compression process, of gas is at and . At the beginning of the adiabatic expansion, the temperature is . Model the working fluid as an ideal gas with . (a) Fill in this table to follow the states of the gas: (b) Fill in this table to follow the processes: (c) Identify the energy input , (d) the energy exhaust , and (e) the net output work (f) Calculate the thermal efficiency. (g) Find the number of crankshaft revolutions per minute required for a onecylinder engine to have an output power of . Note: The thermodynamic cycle involves four piston strokes.


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(a) The complete table is shown below

Given Info: The compression ratio of an Otto cycle is VA/VB = 8, at the beginning of A, the initial volume of the gas is 500 cm3 and the initial temperature is 20 degree C and 100kPa. At the beginning of the adiabatic expansion, the temperature is TC = 750 degree C. The adiabatic index is 1.4.

The expression to calculate the quantity of the gas is

Here,
PA is the pressure of gas at point A
VA is the volume of gas at point A
TA is the temperature of gas at point A
R is the universal gas constant.

Substitute 100kPa for PA , 500cm3 for VA , 20 deg C for TA and 8.314 J/mol . K for R in above equation to find n.


= 0.0205 mol

In process A   B ,
Pressure at point B
PB = PA   
Substitute 8 for   , 1.40 for    and 100 KPa for   in the above equation


The compression ratio is,   = 8
Substitute 500 cm3 for VA in the above equation.
  
The expression to calculate the temperature at point B is

  

Substitute 1.84 x 106 Pa for PB , 62.5cm3 for VB , 0.0205 for n and 8.314 J/mol . K for R in above equation.



= 673 K

At state C:
VC = VB
Pressure at point C.


Substitute 62.5cm3 for VC , 0.0205 for n and 8.314 J/mol . K for R and 750 dec C fot TC in above equation.



= 2 .79 x 106 Pa

= 2 .79 x 103 kPa

State D:

VD = VA and VC = VB

  
=   

Pressure at point D

PD = PC   
Substitute 2 .79 x 106 Pa for PC and 1/8 for   in the above equation



Temperature at D
  

Substitute 1.52 x 105 Pa for PD , 500cm3 for VD , 0.0205 for n and 8.314 J/mol K for R in the above equation



From the above explanation, the complete table is given below.











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