Week 2 Air standard Cycle

OBJECTIVE:

To solve an Otto cycle and plot the PV diagram, we'll first define the necessary thermodynamic processes. The Otto cycle consists of four key processes:

1. Isentropic Compression (Process 1-2): Compression of the air-fuel mixture.
2. Constant Volume Heat Addition (Process 2-3): The fuel ignites, raising the pressure at constant volume.
3. Isentropic Expansion (Process 3-4): Expansion of gases.
4. Constant Volume Heat Rejection (Process 4-1): Heat is rejected, lowering the pressure at constant volume.

Requirements:

1. PV Diagram: A plot of pressure vs. volume for the entire cycle.
2. Thermal Efficiency: Calculate the thermal efficiency of the Otto cycle.

Key Equations:

1. Thermal Efficiency:

   $$\eta = 1 - \frac{1}{r^{\gamma - 1}}$$​

   where $r$ is the compression ratio and $\gamma$ is the specific heat ratio (usually 1.4 for air).

2. Isentropic Relation: For an isentropic process:

   $$P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma}$$
   

Assumptions:

• Air behaves as an ideal gas.
• The specific heat ratio $\gamma =\mathrm{1.4\; for\; air.}$
• The compression ratio $r = \frac{V_1}{V_2}$.

MATLAB PROGRAM:

% Constants

gamma = 1.4; % Specific heat ratio (Cp/Cv for air)

r = 8; % Compression ratio

T1 = 300; % Initial temperature in Kelvin

P1 = 101325; % Initial pressure in Pa

V1 = 1; % Initial volume in m^3

Qc_in = 500; % Heat added during combustion (J)

% State 1

V2 = V1 / r; % Compression volume

% Process 1 -> 2 (Isentropic Compression)

T2 = T1 * (V1 / V2)^(gamma - 1);

P2 = P1 * (V1 / V2)^gamma;

% Process 2 -> 3 (Constant volume heat addition)

% Applying energy balance: Q_in = Cv*(T3 - T2)

Cv = Qc_in / (T2 - T1); % Find Cv based on known Q_in

T3 = T2 + Qc_in / Cv; % T3 after heat addition

P3 = P2 * (T3 / T2); % Constant volume, P3 = P2 * T3 / T2

% Process 3 -> 4 (Isentropic Expansion)

V3 = V2; % Constant volume

V4 = V1; % Expansion back to original volume

T4 = T3 * (V3 / V4)^(gamma - 1);

P4 = P3 * (V3 / V4)^gamma;

% Process 4 -> 1 (Constant volume heat rejection)

P1 = P4 * (T1 / T4); % P1 returns to initial state (ideal)

% Thermal efficiency of Otto cycle

efficiency = 1 - 1 / r^(gamma - 1);

% Plotting PV Diagram

V = [V1, V2, V2, V1];

P = [P1, P2, P3, P4];

figure;

plot(V, P, '-o', 'LineWidth', 2);

title('Otto Cycle PV Diagram');

xlabel('Volume (m^3)');

ylabel('Pressure (Pa)');

grid on;

% Display thermal efficiency

fprintf('Thermal Efficiency of the Otto Cycle: %.2f%%\n', efficiency * 100);

Explanation of the Code:

1. Isentropic Compression (Process 1-2):

   • We calculate the pressure $p_2$ using the isentropic relation: $P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma}$.

2. Constant Volume Heat Addition (Process 2-3):

   • At this point, the volume stays constant, and we assume a temperature ratio $T_3/T_2$​. The new pressure $p_3$​ is calculated using the ideal gas law.

3. Isentropic Expansion (Process 3-4):

   • The gas expands isentropically, and we calculate the pressure $p_4$ using the same isentropic relation for expansion.

4. Constant Volume Heat Rejection (Process 4-1):

   • The heat is rejected, and we return to the initial state.

5. Thermal Efficiency:

   • The efficiency of the Otto cycle is calculated using the formula: $$\eta = 1 - \frac{1}{r^{\gamma - 1}}$$​

PV Diagram:

• We plot the pressure vs. volume curve, which shows the four processes of the Otto cycle.

Results:

1. The thermal efficiency of the Otto cycle is printed based on the compression ratio and specific heat ratio.
2. The PV diagram displays the four stages of the cycle: isentropic compression, constant volume heat addition, isentropic expansion, and constant volume heat rejection.

Therefore, Thermal efficiency of the Otto cycle: 56.47%

Enhancements:

• You can change the compression ratio, initial pressure, or temperature ratio to simulate different Otto cycles.
• More advanced models could include detailed heat addition and rejection models instead of fixed temperature ratios.