Question

An air-standard Otto cycle has a compression ratio of $7.5$. At the beginning of compression, $p_1=85\mathrm{kPa}$ and $T_1={32}^{\circ }\mathrm{C}$. The mass of air is $2\mathrm{g}$, and the maximum temperature in the cycle is $960\mathrm{K}$. Determine (a) the heat rejection, in $\mathrm{kJ}$. (b) the net work, in $\mathrm{kJ}$. (c) the thermal efficiency. (d) the mean effective pressure, in $\mathrm{kPa}$.

Short answer

Heat rejection: 0.3026 kJ, Net work: 0.626 kJ, Thermal efficiency: 52.6%, Mean effective pressure: 618.27 kPa.

Step-by-step solution

- Convert Initial Temperature to Kelvin

Convert the initial temperature from Celsius to Kelvin: equation: T_1 = 32 + 273.15 = 305.15 K

- Determine Specific Heats

For air, specific heats at constant pressure and volume can approximate as: c_v = 0.718 kJ/kg·K c_p = 1.005 kJ/kg·K Also, c_p - c_v = R R = 0.287 kJ/kg·K

- Calculate State Variables at Points 1 and 2

Using the given compression ratio r = 7.5: PV = nRT T_2 = T_1 * (r)^{(k-1)}= 305.15 * (7.5)^{0.2857} = 523.7137 K P_2 = P_1 * (V_1 / V_2)^k = 85 * (7.5)^{1.4} = 1636.109 kPa

- Maximum Temperature and Pressure

Given maximum temperature T_3 = 960 K: We use isochoric process, relation: T_3/T_2 = P_3 / P_2 P_3 = P_2 * (T_3/T_2) = 1636.109 kPa * (960 /523.7137) = 3000 kPa

- Calculate P4, T4 Using Adiabatic Relations

Adiabatic process relationship between T3, T4 and compression ratio: T_4 = T_3 / (r^{k-1}) = 515.92 K Using the relation of pressure and volume: P_4 = P_3 / (r^k) = 155.36 kPa

- Heat Rejection Calculation -as per Isochoric processes

Calculate heat rejection between T4 and T1 using the specific heat, c_v: Q_out = m*c_V*(T_4 -T_1) =0.002 * 0.718 * (515.92-305.15) = 0.3026 kJ

- Work Calculation

The net work done by the cycle can be calculated using the heat absorbed and rejected: Net Work = (m * c_V * (T_3 - T_2 )) - (Q_out) = [0.002*0.718*(960-523.7137)]- 0.3026= 0.626 kJ

- Thermal Efficiency

Thermal Efficiency can be calculated using the formula: Efficiency = 1 - ( 1 / r ^ (k - 1)) =1- 1/(7.5^0.4) = 0.5260 or 52.6%

- Mean Effective Pressure

The Mean effective pressure is calculated using Total Work done and volumes: MEP= W /V_max -V_min= W *2 / ( mass *R*T1)*(1- 1/ compression ratio) = 0.627 kJ. Converting unit 618.27 kPa

Key concepts

Compression Ratio

The compression ratio is a crucial aspect of the Otto cycle. It is defined as the ratio of the volume of the cylinder when the piston is at the bottom (largest volume, known as V₁) to the volume when the piston is at the top (smallest volume, known as V₂). In formulas, it is denoted by **r** and calculated as:
$$r=\frac{V_1}{V_2}$$
In our example, the compression ratio is 7.5. This number tells us how much the air-fuel mixture is compressed before ignition.

• Higher compression ratios typically result in better thermal efficiency.
• Higher ratios can also cause engine knocking if not managed correctly.

Understanding the compression ratio helps in determining other state variables using the adiabatic process equations, which we'll discuss next.

Specific Heats

Specific heats at constant pressure (c_p) and constant volume (c_v) are vital in thermodynamics. They represent the amount of heat per unit mass required to raise the temperature by one degree under constant pressure or volume, respectively. For air, these values are:
$$c_v=0.718\text{kJ/kg·K}$$
$$c_p=1.005\text{kJ/kg·K}$$
The difference between these two values gives the specific gas constant **R** for air:
$$R=c_p-c_v=0.287\text{kJ/kg·K}$$
Having these values allows us to calculate the energy changes in each part of the cycle. For instance, during the constant volume processes, heat addition and rejection are determined using c_v.

Adiabatic Process

An adiabatic process is one during which no heat is transferred to or from the system. This means all the work done changes the internal energy of the system. In the Otto cycle, two such processes occur:

• Compression (from state 1 to 2)
• Expansion (from state 3 to 4)

For an ideal gas undergoing an adiabatic process, the relationship between temperature and volume (or pressure) is given by:
$$T{V}^{k-1}=\text{constant}$$
$$P{V}^k=\text{constant}$$
where **k** is the adiabatic index or specific heat ratio (**k = c_p / c_v**). Using these relationships, we can find temperatures T₂ and T₄, and pressures P₂ and P₄ during the compression and expansion phases, respectively.

Thermal Efficiency

The thermal efficiency of the Otto cycle measures the effectiveness of the cycle in converting heat into work. It is calculated using the formula:
$$\text{Efficiency}=1-\frac{1}{r^{k-1}}$$
In this context, **r** is the compression ratio, and **k** is the specific heat ratio. For our cycle with a compression ratio of 7.5, the thermal efficiency is:
$$1-\frac{1}{{7.5}^{0.4}}=52.6\mathrm{\%}$$
This indicates that 52.6% of the heat energy is converted to work, with the remainder being lost as waste heat. Generally, higher compression ratios lead to higher efficiencies, but practical limits exist due to factors like engine knocking.