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Air standard Cycle

Aim: To demonstrate the working of an otto cycle and write a MATLAB code to plot the graphs and calculate the thermal efficiency. Theory: The Air-Standard cycles work on the principle that the working medium is air. These cycles gained became more and more popular in the theoretical analysis of an Internal Combustion Engine.…

MATLAB

ARTH SOJITRA
updated on 12 Jun 2020

Aim:

To demonstrate the working of an otto cycle and write a MATLAB code to plot the graphs and calculate the thermal efficiency.

Theory:

The Air-Standard cycles work on the principle that the working medium is air. These cycles gained became more and more popular in the theoretical analysis of an Internal Combustion Engine. The Otto cycle is the one in which the working fluid inside the engine is Petrol but is assumed as air for theoretical aspects i.e. to calculate the thermal efficiency and engine parameters.

Before diving into the detail of the analysis of the Otto cycle we need to be familiar with some of the terms related to the cylinder of an I.C. Engine. Consider the basic cylinder of a petrol engine.

The diameter of the cylinder is referred to as bore
The Volume of the cylinder when the piston is at the topmost position is called the clearance volume.
The ratio of the total volume of the cylinder to the Clearance volume is called the Compression Ratio ( C.R.).
The length which the piston travels in moving from the bottom-most position to the topmost position is called the stroke.

The basic Otto cycle comprises of four main strokes.

1.) The intake Stroke: Which comprises of the piston moving from the topmost position to the bottommost position which sucking in the fresh charge.

2.) The Compression Stroke: It is the stroke in which the cylinder moves from the Bottom Dead Center (BDC) to the Top Dead Center (TDC) such that it compresses the fresh fuel charge mixture so that it's temperature rises sufficiently to ignite the fuel via a spark plug. Also in the compression stroke, the head of the piston plays a crucial role in creating turbulence for the fresh charge so as to ensure uniform mixing.

3.) The power Stoke: With the compressed fuel, the spark is initiated causing a huge release of energy which forces the piston down delivering the power to the engine.

4.) The Exhaust Stroke: After the transfer of power to the crankshaft the residual gases remain which needs to be excreted out so as to allow a new charge to come inside and the combustion to take place which is done in the expansion stroke.

Consider the following basic P-V Diagram representing the Ideal Otto cycle of an IC Engine.

Process 1-2 is the compression stroke which takes the state of the fuel from point 1 to 2 with an isentropic compression. Process 2-3 is the combustion process, in which the spark is assumed to be initiated and completed in an infinitesimal amount of time such that the combustion of the fuel is instantaneous without the loss of time. Process 3-4 is the Expansion process which delivers the energy obtained via the fuel burnt to the engine crankshaft. This process is also assumed to be isentropic. Process 4-1 is the exhaust stroke with throws away the residual gases out of the engine so as to allow the fresh intake of the new charge.

IDEA

The main idea behind coding this challenge is to understand how to find the state points 1,2,3 and 4 and how to join them with appropriate plots.

State point 1 is fairly simple as it is the initial condition of the engine at the time of intake and is known. So we know P1, T1 from the initial conditions. To calculate the state point 2 we use the principle of the isentropic process i.e.

$P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ $P_1V_1^gamma = P_2V_2^gamma$ where $γ = \frac{C_{p}}{C_{v}}$ $gamma = frac{C_p}{C_v}$ .

We know $V_{2}$ $V_2$ as the clearance volume of the engine will be provided and hence $P_{2}$ $P_2$ can be calculated from the equation above. Hence knowing $P_{2} & V_{2}$ $P_2 & V_2$ we can easily calculate the Temperature at state point 2 i.e. $T_{2}$ $T_2$

Process 2-3 which is assumed as a constant volume heat addition and knowing the Volume at state point 2 as $V_{2}$ $V_2$ we can assert that $V_{3} = V_{2}$ $V_3 = V_2$ . Now generally in the engine thermodynamics, the temperature of the combusted products is provided or is set as an independent variable to facilitate the calculation of other variables. Hence assuming that $T_{3}$ $T_3$ or $P_{3}$ $P_3$ is known we can calculate the other by :

$\frac{P_{3}}{T_{3}} = \frac{P_{2}}{T_{2}}$ $P_3/T_3 = P_2/T_2$

Now that we know the calculation of $P_{3} and T_{3}$ $P_3 and T_3$ we can apply the isentropic governing equation between state points 3 and 4, and calculate the state variables at point 4.

Process 4-1 is a simple isochoric process hence joining points 4 and 1 with a straight line would do the job.

Now enough with the theoretical calculation of the state variables, we must also understand how to calculate the thermal efficiency of the engine as it plays a crucial role in the engine analysis and development.

Again consider the graph below:

Process 1-2:

Work done = $\frac{R}{γ - 1} (P_{1} V_{1} - P_{2} V_{2})$ $R/(gamma-1)(P_1V_1 - P_2V_2)$

Heat transfer = 0. ( Isentropic process )

Process 2-3:

Work done = 0 ( $Δ V = 0$ $DeltaV = 0$ )

Heat transfer = $C_{v} (T_{3} - T_{2})$ $C_v(T_3-T_2)$ . [ This is the energy that the fuel releases when it is burnt ]

Process 3-4:

Work done = $\frac{R}{γ - 1} (P_{3} V_{3} - P_{4} V_{4})$ $R/(gamma-1)(P_3V_3 - P_4V_4)$

Heat transfer = 0. ( Isentropic process )

Process 2-3:

Work done = 0 ( $Δ V = 0$ $DeltaV = 0$ )

Heat transfer = $C_{v} (T_{1} - T_{4})$ $C_v(T_1-T_4)$ . [ This is the energy that the residual gases reject ]

One important thing to note in our observation is that we assume constant volume heat addition and rejection. However, the reality is far beyond that. There are so many factors to be taken into consideration like the ignition delay, variable specific heat, etc. However, the scope of all this is beyond the scope of this project and we shall limit our observation and calculations to only the super ideal cases.

So in the overall process work output = $\frac{R}{γ - 1} (P_{3} V_{3} - P_{4} V_{4})$ $R/(gamma-1)(P_3V_3 - P_4V_4)$

And the overall heat addition = $- C_{v} (T_{4} - T_{1}) + C_{v} (T_{3} - T_{2})$ $-C_v(T_4-T_1) + C_v(T_3-T_2)$

So the overall thermal efficiency can be written as :

$η = \frac{Work done}{Energy input}$ $eta = \text{Work done}/\text{Energy input}$

Substituting into the formula we get:

$η = \frac{Q_{2} - Q_{1}}{Q_{2}}$ $eta = (Q_2-Q_1)/Q_2$

where $Q_{2}$ $Q_2$ is the energy input and $Q_{1}$ $Q_1$ is the energy output

$η = 1 - \frac{Q_{1}}{Q_{2}}$ $eta = 1-Q_1/Q_2$

Substituing the values we get:

$η = 1 - \frac{C_{v} (T_{4} - T_{1})}{C_{v} (T_{3} - T_{2})}$ $eta = 1 - (C_v(T_4-T_1))/(C_v(T_3-T_2))$

Simplifying we get:

$η = 1 - \frac{T_{4}}{T_{3}} [\frac{1 - \frac{T_{1}}{T_{4}}}{1 - \frac{T_{2}}{T_{3}}}]$ $eta =1 -T_4/T_3[(1 - T_1/T_4)/(1-T_2/T_3)]$

Now using the isentropic process state equation we get :

$T V^{γ - 1} = c$ $TV^(gamma-1) = c$

$T_{3} V_{3}^{γ - 1} = T_{4} V_{4}^{γ - 1} \to \frac{T_{4}}{T_{3}} = {(\frac{V_{3}}{V_{4}})}^{γ - 1} = r_{c}^{γ - 1}$ $T_3V_3^(gamma-1) = T_4V_4^(gamma-1) rightarrow T_4/T_3 = (V_3/V_4)^(gamma-1) = r_c^(gamma-1)$

so then we have ,

$\frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{γ - 1}$ $T_2/T_1 = (V_2/V_1)^(gamma-1)$ and

$\frac{T_{3}}{T_{4}} = {(\frac{V_{4}}{V_{3}})}^{γ - 1}$ $T_3/T_4 = (V_4/V_3)^(gamma-1)$

Now as :

$\frac{V_{2}}{V_{1}} = \frac{V_{3}}{V_{4}} = r_{c} (Compression ratio)$ $V_2/V_1 = V_3/V_4 = r_c (\text{Compression ratio})$ we get

$\frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{4}}$ $T_2/T_1 = T_3/T_4$ and hence

$\frac{T_{1}}{T_{4}} = \frac{T_{2}}{T_{3}}$ $T_1/T_4 = T_2/T_3$

Substituting into the formula we get:

$η = 1 - \frac{1}{{(r_{c})}^{γ - 1}} [\frac{1 - \frac{T_{1}}{T_{4}}}{1 - \frac{T_{1}}{T_{4}}}] = 1 - \frac{1}{{(r_{c})}^{γ - 1}}$ $eta = 1 - 1/(r_c)^(gamma-1)[ (1 - T_1/T_4)/(1 - T_1/T_4)] = 1 - 1/(r_c)^(gamma-1)$

Which is the thermal efficiency of the Engine:

Now we shall see how to implement the concepts discussed above in MATLAB Program:

First of all we create a basic script file defining all the initial parameters:

clear all;
close all;
clc;

% 
% In this program I shall be demonstrating the Methodological Plotting of
% an Otto cycle
%

% Define the initial variable at point 1
P1 = 101325;
T1 = 500;

% Defining the engine geometric parameters
bore = 0.1;                 % bore
stroke = 0.1;               % Stroke
con_rod_length = 0.15;      % Connecting rod length
r_c = 12;                   % Compression ratio

v_swept = pi/4*bore^2;
v_clearance = v_swept/(r_c-1);

% Volume at point 1 is nothing but the total volume
V1 = v_clearance + v_swept ;

% Volume at state point 2 is nothing but the clearance volume
V2 = v_clearance;

Explanation: This MATLAB Code basically creates the initial variables of the engine parameters and the state variables

Now to facilitate the calculation and the plotting of curves between points 1 and 2 we must create a function that accurately describes the variation of the volume with respect to the crank angle. The following MATLAB function does this job:

function V_variation = V_with_theta(start_angle,end_angle,...
    stroke,con_rod_length,r_c,v_clearance)

%
% THis function basically finds the variation of the Volume with respect to
% theta 
%

% Creating a linearly spaced vector for the theta values
theta = linspace(start_angle,end_angle,100);

% Defining the variables of the equation
a = stroke/2;
R = con_rod_length/a;

% Defining the individual terms of the equation
term1 = 0.5*(r_c-1);
term2 = R + 1 - cosd(theta);
term3 = R^2 - (sind(theta)).^2;
term3 = term3.^(0.5);

% Calculating a vector of the volume elements
V_variation = ( 1 + term1 *(term2 - term3) ) * v_clearance;

 

end

Explanation: This code takes in input the engine parameters and the start of the crank angle and the end of the crank angle and creates a vector distribution of the volume variables as per the linear variation of theta which is a reasonable assumption to make. One crucial thing to note here is that the variation of V itself is not linear but is dependent upon the variation of the crank angle.

Now after the creation of the artilleries needed to attack this problem we are now ready to compute the state variables and the distributions. The following MATLAB Code does this job:

%
% Creating the variation of theta from the BDC to TDC for the compression
% process
%
%              BDC - Bottom Dead Center
%              TDC - Top Dead Center
%
V_variation_1_to_2 = V_with_theta(180,0,...
    stroke,con_rod_length,r_c,v_clearance);

%
% Now using this variation I will calculate the values of the state
% variables at each individual points\
%
% using the isentropic governing equation
%                 PV^(gamma) = constant
%
P_variation_1_to_2 = (P1 * V1.^gamma ./ V_variation_1_to_2.^(gamma));

% Now that we knwo the state variables at point 2
P2 = P_variation_1_to_2(end);   % last element
V2 = V_variation_1_to_2(end);   % last element
T2 = T1*P2*V2/P1/V1;            % using the ideal gas law

% To calculate the state variables at point 3 
T3 = 2000;                % Assume 
V3 = V2;                  % Constant volume process
P3 = P2*T3/T2;            % Assuming the ideal gas law with V = constant

%
% Creating the variation of theta from the TDC to BDC for the Expansion
% Process
%
%              BDC - Bottom Dead Center
%              TDC - Top Dead Center
%
V_variation_3_to_4 = V_with_theta(0,180,...
    stroke,con_rod_length,r_c,v_clearance);


%
% Now using this variation I will calculate the values of the state
% variables at each individual points\
%
% using the isentropic governing equation
%                 PV^(gamma) = constant
%
P_variation_3_to_4 = (P3 * V3.^gamma ./ V_variation_3_to_4.^(gamma));

% state variables at point 4
P4 = P_variation_3_to_4(end);           % last element
V4 = V_variation_3_to_4(end);           % last element
T4 = P4*V4*T3/P3/V3;                    % Using the Ideal gas law

Explanation: The above MATLAB Code basically computes the state variables at all the 4 points and computes a vector of state variables between the points 1-2 and 3-4 as the variation between them is not linear and the variation between the points 2-3 and 4-1 is linear and hence doesn't require any extra computation of the points between them.

Now after the computation of the points we are now ready to plot them in a graph. This is achieved by the following MATLAB Code.

%
% Now all of out state variables have been calculated
%
% It is time to join them using appropriate curves
figure(1);clf;

% isentropic variation between 1-2
plot(V_variation_1_to_2 , P_variation_1_to_2 , 'linew' , 3);
hold on;

% Straight line with constant volume between 2-3
plot([V2 V3],[P2 P3], 'linew' , 3);

% isentropic variation between 3-4
plot(V_variation_3_to_4 , P_variation_3_to_4 , 'linew' , 3);
hold on;

% Straight line with constant volume between 4-1
plot([V4 V1],[P4 P1], 'linew' , 3);

hold off;

% Making the plot look a bit nicer
grid on;
grid minor;

% adding the labels
xlabel('Volume in m^3');
ylabel('Pressure in Pa');

% adding the title
title({'PV diagram of an IC engine';...
    ['Using the Compression ratio : ' num2str(r_c)]});
legend('Compression','Combustion','Expansion','Exhaust');
set(gca,'Fontsize',20);

Explanation: As the state variables are discreet functions and points we must use a clever technique of plotting them using the MATLAB hold on command which plots all the curves in a single graph with smooth variations locally. Legends are added to better visualize the processes occurring in the Engine.

The following plot was obtained:

To output the thermal efficiency of the engine the following code is implemented which is direct usage of the formula as derived above in the theory section:

% calculating the thermal Efficiency
thermal_efficiency = 1 - 1/r_c^(gamma-1);

% printing the thermal efficiency:
fprintf(['The thermal efficiency of the cycle is : ' ...
    num2str(thermal_efficiency)]);
fprintf('\n');

The following output was obtained in the command window:

The above were the outputs considering the Compression ratio as 12. Let us look at a few different cases by changing the CR.

1.) Cr = 6

The following outputs were obtained: