AIR STANDARD CYCLE

AIR STANDARD CYCLE

Introduction:

In this project we here going to draw the ottocycle using the matlab program.\

Ottocycle: Otto cycle is an idealized thermodynamic cycle describes the functioning of typical spark iginition piston engine(petrol engine),this thermodynamic cycle mostly found in all automobiles.

The otto cycle is the description of changes takes in the gas when it is subjected to pressure ,volume ,temperature and heat addition and heat rejection.The mass that changes respect to the parameters is called the system.In otto cycle we assume the working fluid is to be air(gas).In this cycle describing the changes takes place in system it will also describes the inverse and system effects on the environment. in case of otto cycle the system effect will create the enough power to move the automobile.

Assumptions:

• working fluid is air and behaves as idea gas
• it was a closed cycle
• all the cycles are internally reversible
• it was quasi-static process
• isentropic process(friction less,adiabatic reversible)
• isochoric process(constant volume)

In addition, if specific heats are assumed constants at their ambient temperature, this assumption is called a cold air-standard assumption.

Ideal Otto Cycle - Ideal Cycle for Spark-ignition Engines:

The Otto cycle is the ideal cycle for spark-ignition engines, in honor of Nikolaus Otto, who invented it in 1867. In ideal Otto cycles, air-standard assumption is used. The ideal Otto cycle consists of four internal reversible processes:

• 1-2 Isentropic compression

• 2-3 Constant volume heat addition

• 3-4 Isentropic expansion

• 4-1 Constant volume heat rejection

The Otto cycle is executed in a closed system and the working fluid is air according to the air-standard assumption. Also, changes in kinetic and potential energies are negligible. No heat transfer is involved in the two isentropic processes. The energy balances for these two processes are:

-w12 = u2 - u1

-w34 = u4 - u3

w12 is negative since work is needed to compress the air in the cylinder and w34 is positive since air does work to the surroundings during its expansion.

In the constant volume heat addition and heat rejection process, no work interaction is involved since no volume change occurs. The energy balances for these two processes are:

q23 = u3 - u2

q41 = u1 - u4

q23 is positive since heat is added to the air and q41 is negative since heat is rejected to the surroundings.

The thermal efficiency for an ideal Otto cycle is

ηth, Otto = wnet/qin

According to the analysis above, the net work output is

wnet = w34 + w12 = q23 + q41

qin = q23

ηth, Otto = 1+ q41/q23

Under the cold air-standard assumption, the thermal efficiency of the ideal Otto cycle is

Process 1-2 and process 3-4 are isentropic. Thus,

Since v2 = v3 and v4 = v1,

Considering all the relations above, the thermal efficiency becomes

`eta th=1-(T1)/(T2)=1-((V2)/(V1))^(K-1)`

`=1-1/R^(K-1)`

Where 'r' is the compression ratio and k is the specific heat ratio

program for matlab:

clear all
close all
clc

%inputs
gamma=1.4;

%engine parameters
bore=0.1;
stroke=0.1;
con_rod=0.15;
cr=12;
%state variables
p1=10325;
t1=500;
t3=2300;
%calculating the clearance volume and the swept volume
v_s=(pi/4)*bore^2*stroke;
v_c=v_s/(cr-1);

v1=v_s+v_c
v2=v_c

%state variables at state point 2
%p2v2^gamma=p1v1^gamma
p2=p1*cr^gamma;
t2=p2*v2*t1/(p1*v1);

constant_c=p1*v1^gamma;
V_compression=piston_kinematics2(bore,stroke,con_rod,cr,180,0);
P_compression=constant_c./V_compression.^gamma;

%state variables at point 3
v3=v2

p3=p2*t3/t2
constant_c=p3*v3^gamma;
V_expansion=piston_kinematics2(bore,stroke,con_rod,cr,0,180);
P_expansion=constant_c./V_expansion.^gamma;

%state variables at 4
v4=v1

p4=p3*(v3/v4)^gamma

figure(1)

hold on
plot(v1,p1,'*','color','r')
plot(V_compression,P_compression)
plot(v2,p2,'*','color','r')
plot(v3,p3,'*','color','r')
plot(V_expansion,P_expansion)
plot(v4,p4,'*','color','r')
plot(v2,p2)
plot([v2,v3],[p2,p3],'color','r')
plot([v4,v1],[p4,p1],'color','r')
xlabel('volume')
ylabel('pressure')

program for piston kinematics function:

function [v]=piston_kinematics2(bore,stroke,con_rod,cr,start_cranck,end_cranck)

%input parameters

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

theta=linspace(start_cranck,end_cranck)
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

here the output for the code can obtained by studying the piston kinematics ,so to get the output of the program i have written a piston kinematics function so that ,by using this function we can calculate the volume and pressure for the program.