Question

At the beginning of the compression process of an airstandard Otto cycle, \(p_{1}=1\) bar, \(T_{1}=290 \mathrm{~K}, V_{1}=400 \mathrm{~cm}^{3}\). The maximum temperature in the cycle is \(2200 \mathrm{~K}\) and the compression ratio is 8. Determine (a) the heat addition, in \(\mathrm{kJ}\). (b) the net work, in \(\mathrm{kJ}\). (c) the thermal efficiency. (d) the mean effective pressure, in bar. (e) Develop a full accounting of the exergy transferred to the air during the heat addition, in \(\mathrm{kJ}\). (f) Devise and evaluate an exergetic efficiency for the cycle. Let \(T_{0}=290 \mathrm{~K}, p_{0}=1\) bar.

Short answer

q_in = 1120.25 kJ, W_net = 618.10 kJ, η_th = 55.2%, MEP ≈ 17.72 bar, e_q = 973.13 kJ, η_exergetic ≈ 63.54%

Step-by-step solution

Define given parameters

Given values are: \[ p_{1}=1 \text{ bar}, T_{1}=290 \text{ K}, V_{1}=400 \text{ cm}^3 \text{ or } 0.0004 \text{ m}^3 \] \[ T_{max}=2200 \text{ K}, r=8 \] The environmental conditions are \[ T_{0} = 290 \text{ K} \text{ and } p_{0} = 1 \text{ bar} \text{ or } 100 \text{ kPa} \]

Calculate intermediate state parameters

First, determine state parameters at the end of compression (state 2): Use the relation \[ V_{2} = \frac{V_{1}}{r} = \frac{0.0004}{8} = 0.00005 \text{ m}^3 \] For an air-standard Otto cycle, we also use the isentropic relations for temperature and pressure during the compression process from state 1 to state 2. For an ideal gas with specific heat ratio \( \gamma = 1.4 \): \[ T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=290\left(\frac{0.0004}{0.00005}\right)^{0.4}=645.84 \text{ K} \] \[ \frac{p_{2}}{p_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{\frac{\gamma}{\gamma-1}}=\left(\frac{645.84}{290}\right)^{\frac{1.4}{0.4}}=\frac{p_{2}}{1}\] Therefore, \[ p_{2}\approx 15.69 \text{ bar} \]

Calculate parameters at maximum temperature

Using the relation for the heat addition process (isochoric process from state 2 to 3): \[ V_{3} = V_{2}=V_{1}/r \] \[ T_{3}=2200 \text{ K} \text{ (given)} \text{ and thus from the ideal gas law: } \frac{p_{3}}{p_{2}}= \frac{T_{3}}{T_{2}} \] \[ p_{3}=p_{2}\left(\frac{T_{3}}{T_{2}}\right)=15.69\left( \frac{2200}{645.84} \right) \text { bar } \approx 53.42 \text{ bar} \]

Calculate parameters at state 4

At state 4, we assume the isentropic expansion process: \( T_{4} \) and \( p_{4} \) can be found using similar relations as state 2: \[ T_{4}=T_{3}\left(\frac{V_{3}}{V_{4}}\right)^{\gamma-1} = T_{3}\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}\] Substituting values: \[ T_{4}=2200 \left( \frac{1}{8} \right)^{0.4} \approx 989.37 \text{K} \] \[ \frac{p_{4}}{p_{3}}=\left(\frac{T_{4}}{T_{3}}\right)^{\frac{\gamma}{\gamma-1}} \] \[ p_{4} = p_{3}\left(\frac{T_{4}}{T_{3}}\right)^{\frac{\gamma}{\gamma-1}} = 53.42\left( \frac{989.37}{2200} \right)^{3.5} \approx 3.41 \text{ bar} \]

Compute the heat addition

Using specific heat values for air (constant values): \( c_{v} = 0.718 \text{ kJ/kg.K} \) and noting that \[ q_{in} = c_{v}(T_{3}-T_{2}) \] \[ q_{in}=(0.718)(2200-645.84) \text{ kJ/kg} \text{ } = 1120.25 \text{ kJ} \]

Compute the net work

The net work output \( W_{net} \) is the difference between heat added and rejected. Heat rejected \( q_{out} = c_{v}(T_{4}-T_{1}) \): \[ q_{out} = c_{v}(989.37-290) = 0.718 \times 699.37 = 502.15 \text{ kJ} \] Therefore, \[ W_{net}= q_{in} - q_{out} = 1120.25 - 502.15 = 618.10 \text{ kJ} \]

Calculate thermal efficiency

The thermal efficiency \( \eta_{th} \) for the Otto cycle is given by: \[ \eta_{th} = 1- \frac{q_{out}}{q_{in}} = 1 - \frac{502.15}{1120.25} = 0.552 \text{ or } 55.2% \]

Mean effective pressure

Mean effective pressure (MEP) is given by: \[ MEP = \frac{W_{net}}{V_{1}-V_{2}} = \frac{618.10}{0.0004-0.00005}\, \text{ Pa}\approx 1,771,714 \text{ Pa} \text{ or } 17.72 \text{ bar} \]

Exergy transferred during heat addition

Exergy transferred: \[ e_{q} = q_{in} \left(1 - \frac{T_{0}}{T_{3}} \right) = 1120.25 \left(1-\frac{290}{2200} \right) = 973.13 \text{ kJ} \]

Exergetic efficiency

Exergy transferred to work output: \[ \eta_{exergetic} = \frac{W_{net}}{e_{q}} = \frac{618.10}{973.13} \approx 63.54% \]

Key concepts

Heat Addition in Otto Cycle

In the Otto cycle, heat addition occurs during the combustion process. This takes place at constant volume. Here, the air-fuel mixture is ignited, raising the temperature within the cylinder. The amount of heat added, denoted as \( q_{in} \), can be calculated using the specific heat capacity at constant volume \( c_v \), and the temperature change during the process. In our example: \[ q_{in} = c_v (T_3 - T_2) \]. This ensures we have all required parameters to determine the precise heat addition in kilojoules (kJ). \[ q_{in} = 1120.25 \text{ kJ} \].

Net Work in Thermodynamic Cycles

Net work in thermodynamic cycles is crucial for understanding the efficiency of an engine. The net work output \( W_{net} \) is the difference between the work done during heat addition and the work lost during heat rejection. This is represented formally as: \[ W_{net} = q_{in} - q_{out} \]. In our example, the heat rejected \( q_{out} \) is calculated using the specific heat capacity and temperature difference during the heat rejection phase. Subsequently, the net work can be computed: \[ W_{net} = 618.10 \text{kJ} \].

Thermal Efficiency Calculation

Thermal efficiency measures how effectively an engine converts heat into work. For an Otto cycle, the thermal efficiency \( \eta_{th} \) can be determined by the formula: \[ \eta_{th} = 1 - \frac{q_{out}}{q_{in}} \]. The higher the thermal efficiency, the more efficient the engine is. In our case, substituting the values of heat added and rejected yields: \[ \eta_{th} = 55.2% \].

Mean Effective Pressure

Mean effective pressure (MEP) represents the average pressure in the cylinder over one cycle, and it's an important indicator of an engine's performance. Calculated as: \[ \text{MEP} = \frac{W_{net}}{V_{1} - V_{2}} \], where \( W_{net} \) is the net work, and \( V_{1} - V_{2} \) is the change in volume. Here, the MEP is found to be: \[ 17.72 \text{ bar} \]. This helps engineer gauge how effectively the engine can perform useful work.

Exergy in Thermodynamic Cycles

Exergy quantifies the maximum useful work possible during a process that brings the system into equilibrium with its environment. During the heat addition process in the Otto cycle, exergy transferred, denoted as \( e_q \) is calculated by: \[ e_q = q_{in} \(1 - \frac{T_0}{T_3} \) \]. Using given and determined values, the exergy transferred during heat addition is: \[ 973.13 \text{ kJ} \]. Understanding exergy helps optimize the thermodynamic process by evaluating losses and inefficiencies.

Exergetic Efficiency

Exergetic efficiency assesses how efficiently a system converts available energy (exergy) into useful work. This efficiency is crucial for definitive performance improvement and environmental considerations. It is determined by: \[ \eta_{exergetic} = \frac{W_{net}}{e_q} \]. For our specific Otto cycle, the exergetic efficiency is: \[ \eta_{exergetic} = 63.54% \]. This indicates how efficiently the system transforms exergy into work. Enhancements to this metric can lead to significant improvements in overall engine performance and sustainability.