Air Standard Cycle in MATLAB

OBJECTIVES: To write a code in Matlab for :

1. The PV diagram of an Air Standard Otto Cycle
2. To calculate the Thermal Efficiency of the engine.

INTRODUCTION:

Otto cycle is a gas power cycle that is used in the spark-ignition internal combustion engine. An Otto cycle consists of two isentropic (reversible adiabatic) processes and two isochoric (constant volume) processes.

p-V Diagram

T-s Diagram

Here,

p - Pressure ; V - Volume ; T - Temperature ; s - Entropy ;

V_c - Clearance Volume ; V_s - Stroke Volume.

Process 1-2: Isentropic compression

In this process, the piston moves from the bottom dead centre (BDC) to top dead centre (TDC) position. Air undergoes reversible adiabatic (isentropic) compression. We know that compression is a process in which volume decreases and pressure increases. Hence, in this process, the volume of air decreases from V₁ to V₂ and pressure increases from p₁ to p₂. Temperature increases from T₁ to T₂. As this an isentropic process, entropy remains constant (i.e., s₁=s₂).

Process 2-3: Constant Volume Heat Addition:

Process 2-3 is isochoric (constant volume) heat addition process. Here, piston remains at the top dead centre for a moment. Heat is added at constant volume (V₂ = V₃) from an external heat source. Temperature increases from T₂ to T₃, pressure increases from p₂ to p₃ and entropy increases from s₂ to s₃.

In this process,

Heat Supplied = mC_v(T₃ – T₂)

where,

m - Mass

C_v - Specific heat at constant volume

Process 3-4: Isentropic expansion

In this process, air undergoes isentropic (reversible adiabatic) expansion. The piston is pushed from the top dead centre (TDC) to bottom dead centre (BDC) position. Here, pressure decreases from p₃ to p₄, volume rises from v₃ to v₄, the temperature falls from T₃ to T₄ and entropy remains constant (s₃=s₄).

Process 4-1: Constant Volume Heat Rejection

The piston rests at BDC for a moment and heat is rejected at constant volume (V₄=V₁). In this process, the pressure falls from p₄ to p₁, temperature decreases from T₄ to T₁ and entropy falls from s₄ to s₁.

In process 4-1,

Heat Rejected = mC_v(T₄ – T₁)

Thermal efficiency (air-standard efficiency) of the Otto Cycle is given by,

PROCEDURE:

1. At first, we input the engine parameters and assign and derive the state variables.
2. We cannot connect the points linearly and thus create another function [V] to derive the Compression and the Expansion curves.
3. We calculate the Volume Trace inside the function by splitting the formula into smaller terms.
4. We name the output function of the Volume Trace as V_compression and V_expansion and with the help of this and constant 'c', we calculate the P_compression and P_expansion.
5. We connect the points 1-2 and 3-4 using the values from above.
6. We connect the Isochoric (Constant Volume) lines with straight lines using the plot command.
7. We calculate the Thermal Efficiency of the engine by using the code "E = (1-(1/cr)^(gamma-1))*100"
8. We use the legend command to introduce multiple legends into the figure to name each of the graphs (processes).

CODE:

1. Main code

clear all
close all
clc

% Inputs
gamma = 1.4;
t3 = 2300;

% Engine geometric parameters
bore = 0.1;
stroke = 0.1;
con_rod = 0.15;
cr = 12;

% Calculating the swept volume and the clearance volume
v_swept = (pi/4)*bore^2*stroke;
v_clearance = v_swept/(cr-1);

% State variables at point (1)
p1 = 101325;
t1 = 500;

v1 = v_swept + v_clearance
v2 = v_clearance;

% State variables at point 2
% p2v2^gamma = p1v1*gamma
p2 = p1*cr^gamma;

%p1v1/T1 = p2v2/t2
t2 = p2*v2*t1/(p1*v1);

% constant
c = p1*v1^gamma;
V_compression = engine_kinematics_otto_cycle1(bore, stroke, con_rod, cr,180,0);
P_compression = c./(V_compression.^gamma);

% State variables at point 3
v3 = v2
%p3v3/t3 = p2v2/t2
p3 = p2*t3/t2
% constant
c = p3*v3^gamma;
V_expansion = engine_kinematics_otto_cycle1(bore, stroke, con_rod, cr,0,180);
P_expansion = c./(V_expansion.^gamma);



%State variables at state point 4
v4 = v1
%p3v3^gamma = p4v4^gamma
p4 = p3*(v3/v4)^gamma;

%Efficiency
disp('Efficiency of OTTO CYCLE') 
E = (1-(1/cr)^(gamma-1))*100

%plotting
figure(1)
hold on
plot(v1,p1,'*','color','r')
aa = plot(V_compression,P_compression,'linewidth',2,'color','b');

plot(v2,p2,'*','color','r');
plot(v3,p3,'*','color','r');

cc = plot(V_expansion, P_expansion,'linewidth',2,'color','black');
plot(v4,p4,'*','color','r');

bb = plot([v2 v3],[p2 p3],'linewidth',2,'color','g');
dd = plot([v4 v1],[p4 p1],'linewidth',2,'color','yellow');
title('OTTO CYCLE')
xlabel('VOLUME')
ylabel('PRESSURE')
legend([aa,bb,cc,dd],'Isentropic Compression','Isochoric Heat Addition','Isentropic Expansion','Isochoric Heat Rejection')

2. Function

function [V] = engine_kinematics_otto_cycle1(bore, stroke, con_rod, cr, start_crank, end_crank)


a = stroke/2;
R =con_rod/a;

v_s = (pi/4)*bore^2*stroke;
v_c = v_s/(cr-1);

theta = linspace(start_crank,end_crank,180);

term1 = 0.5*(cr-1);
term2 = R + 1 - cosd(theta);
term3 = (R^2 - sind(theta).^2).^0.5;

V = (1 + term1*(term2 - term3))*v_c;

end

OUTPUT:

1.

2.

The thermal efficiency of the engine was found to be '62.98%'.

ERRORS:

1.

Case Sensitive Error: yLabel('PRESSURE')

Correction: ylabel('PRESSURE')

2.

Error using plot: plot(V_compression,P_compression,'inewidth',3)

Correction: plot(V_compression,P_compression,'linewidth',2,'color','b')

3.

Error in plotting color: cc = plot(V_expansion, P_expansion,'linewidth',2,'color','p')

The specified character vector 'p' is an invalid color value.

Correction: cc = plot(V_expansion, P_expansion,'linewidth',2,'color','black')