In an air standard Otto cycle, the compression ratio is 7, and the compression begins at 1 bar and 40°C. The heat added is 2500 kJ/kg. Find: (a) work done per kg of air; (b) cycle efficiency; (c) maximum pressure and temperature of the cycle; and (d) mean effective pressure.
For air, cV = 0.712 kJ/kgK ; and R = 0.288 kJ/kgK.
The Correct Answer and Explanation is :
To solve the given problem, we need to understand the key parameters and equations involved in the Otto cycle, which operates as an idealized internal combustion engine cycle. The Otto cycle consists of two adiabatic processes (compression and expansion) and two constant-volume processes (heat addition and rejection).
Given Data:
- Compression ratio (( r )) = 7
- Initial pressure (( P_1 )) = 1 bar = 100 kPa
- Initial temperature (( T_1 )) = 40°C = 313.15 K
- Heat added (( Q_{in} )) = 2500 kJ/kg
- Specific heat at constant volume (( c_V )) = 0.712 kJ/kg·K
- Gas constant (( R )) = 0.288 kJ/kg·K
Now, let’s calculate the requested parameters one by one.
(a) Work Done per kg of Air
The work done in an Otto cycle is equal to the net work, which is the heat added (( Q_{in} )) minus the heat rejected (( Q_{out} )).
For an ideal Otto cycle:
[
\text{Net work} = Q_{in} – Q_{out}
]
Using the relation for heat rejected in the Otto cycle:
[
Q_{out} = c_V (T_4 – T_3)
]
Where ( T_4 ) is the temperature at the end of the expansion stroke, and ( T_3 ) is the temperature after heat addition.
Since heat added equals the increase in internal energy, we can assume the work done is the amount of energy transferred during the compression and expansion phases.
(b) Cycle Efficiency
The efficiency of an Otto cycle is given by the formula:
[
\eta = 1 – \frac{1}{r^{\gamma – 1}}
]
Where ( \gamma ) is the ratio of specific heats (( \gamma = \frac{c_P}{c_V} )). For air, ( \gamma ) is approximately 1.4.
(c) Maximum Pressure and Temperature
The maximum pressure and temperature in an Otto cycle occur at the end of the compression stroke. Using the adiabatic relation for the temperature and pressure:
[
T_2 = T_1 \cdot r^{\gamma – 1}
]
[
P_2 = P_1 \cdot r^{\gamma}
]
Where ( T_2 ) and ( P_2 ) are the temperature and pressure at the end of the compression stroke.
(d) Mean Effective Pressure (MEP)
The mean effective pressure is defined as the average pressure in the cylinder during the power stroke. It is calculated as:
[
MEP = \frac{W_{\text{net}}}{V_{\text{displaced}}}
]
Where ( W_{\text{net}} ) is the net work done per unit mass of air, and ( V_{\text{displaced}} ) is the displaced volume of the cylinder.
Let me now calculate these parameters and generate a relevant image to represent the Otto cycle.
Here are the results of the calculations:
(a) Work Done per kg of Air:
- The work done per kg of air is approximately 2500 kJ/kg, which is equal to the heat added in the ideal Otto cycle.
(b) Cycle Efficiency:
- The efficiency of the Otto cycle is approximately 54.08%.
(c) Maximum Pressure and Temperature:
- The maximum temperature at the end of the compression stroke is 682.01 K.
- The maximum pressure at the end of the compression stroke is 1524.53 kPa.
(d) Mean Effective Pressure (MEP):
- The mean effective pressure (MEP) is approximately 2916.67 kPa.
This is a simplified analysis of the Otto cycle, based on the assumptions of ideal gas behavior and perfect processes. The key takeaway is the direct relationship between the heat added and the work done in an ideal cycle, and the effectiveness of the compression ratio in increasing both temperature and pressure to improve efficiency.
I will now generate an image to visually represent the Otto cycle.
Here is the diagram of the Otto cycle, showing the four main processes: isentropic compression, isochoric heat addition, isentropic expansion, and isochoric heat rejection. The diagram also includes pressure vs volume, with the labeled states corresponding to the start and end of each phase of the cycle.