Question

An air-standard Otto cycle has a compression ratio of \(8.5\). At the beginning of compression, \(p_{1}=100 \mathrm{kPa}\) and \(T_{1}=300 \mathrm{~K}\). The heat addition per unit mass of air is \(1400 \mathrm{~kJ} / \mathrm{kg}\). Determine (a) the net work, in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air. (b) the thermal efficiency of the cycle. (c) the mean effective pressure, in \(\mathrm{kPa}\). (d) the maximum temperature in the cycle, in \(\mathrm{K}\). (e) To investigate the effects of varying compression ratio, plot each of the quantities calculated in parts (a) through (d) for compression ratios ranging from 1 to 12 .

Short answer

Net work: calculated from heat addition and rejection. Thermal efficiency: derived from compression ratio. Mean effective pressure: calculated using net work and volume change. Maximum temperature: derived from temperature after combustion. All are analyzed for varying compression ratios.

Step-by-step solution

Identifying Given Data and Constants

Identify and list the given data and constants needed for solving the problem:Compression ratio, \(\text{r} = 8.5\)p1 = 100 \( \text{kPa} \)T1 = 300 \( \text{K} \)Heat addition per unit mass, q_in = 1400 \( \text{kJ/kg} \).Assume specific heat at constant volume \( \text{cv} = 0.718 \, \text{kJ/kg·K} \) and specific heat at constant pressure \( \text{cp} = 1.005 \, \text{kJ/kg·K} \).\( k = \frac{cp}{cv} = 1.4 \).

Calculate Temperature at the End of Compression (T2)

Use the relation for isentropic process to find the temperature at the end of compression:\[T_2 = T_1 \times r^{k-1} = 300 \times 8.5^{1.4-1} \].

Calculate Temperature at the End of Combustion (T3)

Use the heat addition formula to find temperature at the end of combustion:\[q_{in} = c_v \times \( T_3 - T_2 \) \]Rearrange to calculate \(T_3\):\[T_3 = T_2 + \frac{q_{in}}{c_v} \].

Calculate Temperature at the End of Expansion (T4)

Use the relation for isentropic expansion to calculate temperature at the end of expansion:\[T_4 = T_3 \times r^{1-k} \].

Calculate Net Work

Net work can be calculated as the difference between heat added and heat rejected:\[W_{net} = q_{in} - q_{out} \]Where \(q_{out} = c_v \times (T_4 - T_1) \).

Calculate Thermal Efficiency

Thermal efficiency for the Otto cycle can be calculated using the formula:\[ \therm_eff = 1 - \frac{1}{r^{k-1}} = 1 - \frac{1}{8.5^{1.4-1}} \] or using net work and heat added : \[ \therm_eff = 1 - \frac{q_{out}}{q_{in}} \]

Calculate Mean Effective Pressure

Mean effective pressure (MEP) can be calculated using:\[MEP = \frac{W_{net}}{V_1 - V_2} \]. Use \(P1V1 = P2V2, V1/V2 = r \) where \(V1 = \frac{RT1}{P1}, V2 = \frac{V1}{r} \).

Calculate Maximum Temperature

The maximum temperature is \(T_3 \), calculated in Step 2.

Plot Quantities vs Compression Ratio

Generate plots for net work per kg of air, thermal efficiency, mean effective pressure, and maximum temperature versus compression ratio ranging from 1 to 12.

Key concepts

thermal efficiency

Thermal efficiency (\therm_eff\text) of the Otto cycle measures how effectively the cycle converts heat into work. It's crucial because it tells us how much of the inserted heat is turned into useful work. For the Otto cycle, we determine thermal efficiency using the formula: \[ \therm_eff = 1 - \frac{1}{r^{k-1}} \ \].
Here, \( r \) represents the compression ratio, and \( k \) is the specific heat ratio (\frac{cp}{cv}\text).
For practical significance, higher thermal efficiency means less fuel consumption for the same amount of work, which is crucial for engineering and environmental considerations.

compression ratio

The compression ratio (\r\text) in an Otto cycle is the ratio of the cylinder's total volume when the piston is at the bottom to the clearance volume when the piston is at the top. This ratio significantly affects the engine’s performance. It is given by: \[ r = \frac{V1}{V2} \].
A higher compression ratio usually results in higher thermal efficiency, as seen from the thermal efficiency formula of the Otto cycle. This is because a higher compression ratio allows for a greater temperature rise during compression, leading to improved efficiency.
However, excessive compression can lead to engine knocking and increased stress on engine components, so it must be balanced carefully.

isentropic process

In the Otto cycle, there are two isentropic processes: compression and expansion. An isentropic process is a reversible adiabatic process, meaning no heat is transferred. During these stages, entropy remains constant.
For the compression stage (1 to 2), the relation is: \[ T_2 = T_1 \times r^{k-1} \].
For the expansion stage (3 to 4), the relation is: \[ T_4 = T_3 \times r^{1-k} \].
These equations help in determining temperatures at different points in the cycle. Understanding these processes is crucial for analyzing the cycle's efficiency and work output.

mean effective pressure

Mean Effective Pressure (MEP) is an important parameter in an Otto cycle because it gives an indication of how effectively pressure is converted into work. It's calculated using the net work per cycle divided by the displacement volume:
\[ MEP = \frac{W_{net}}{V_1 - V_2} \].
Where \(V_1\) and \(V_2\) are the volumes before and after compression, respectively.
A higher MEP generally indicates a more powerful engine for a given size, which is highly desirable in performance contexts.

maximum temperature

The maximum temperature in the Otto cycle typically occurs at the end of the compression process (state 3). It is determined using the heat addition equation:
\[ T_3 = T_2 + \frac{q_{in}}{c_v} \].
Where \(q_{in}\) is the heat added per unit mass and \(c_v\) is the specific heat at constant volume. Higher maximum temperatures can yield higher thermal efficiencies but can also pose material and mechanical limitations.
It's essential to manage this temperature to optimize engine performance while considering durability and emissions.