Otto cycle

Aim: To solve an Otto Cycle and make plots

Objectives:

1) Piston Kinematics for volume function of the isentropic process

2) P V Diagram

3) Thermal Efficiency

Procedure:

The Otto cycle was given by Dr. Nikolaus August Otto. It is a gas power cycle that is used in spark ignition engine(i.e. petrol engine) for its working. The entire modern petrol engine works on Otto cycle. It consist of four processes, Two isentropic (reversible adiabatic) processes and two isochoric (constant volume) processes. It has low compression ratio ranges from 7:1 to 10:1.

The four processes of this cycle is as follows:

Process 1-2:Isentropic ( reversible adiabatic) compression
Process 2-3:Constant volume (Isochoric) heat addition
Process 3-4:Isentropic (reversible adiabatic) Expansion
Process 4-1:Constant volume heat rejection.

Engine Properties:

Bore=0.1m
stroke=0.1m
Compression ratio(cr)=12
Connencting rod length(cond_rod)=0.15

Initial Conditions::

Pressure(p1)=103125 pa
temperature(t1)=500 k
temperature(t3)=2300 k
gamma=1.4

Formulas Used:

Swept volume(v_swept)=(pi/4)*bore^2*stroke
clearance volume(v_clearance)=v_swept/(cr-1)
v1=v_swept+v_clearance
v2=v_clearance

To find p2 and t2:
p2= p1*(cr)^gamma
t2=p2*v2*t1/(p1*v1)

v2=v3(constant volume heat addition)

To find p3:
p3=(p2*t3)/t2

v4=v1(constant volume heat rejection)

To find p4:
p4=p3*(v3/v4)^gamma

Engine Efficiency:

Thermodynamic Efficiency of an engine is the ration between net work done by the engine and the net heat supplied to the engine.

Efficiency = Work done by the process/heat added to the system

work done = heat added - heat rejected
efficiency=1-(heat rejected/heat added)
heat added= m*cv*(t3-t2)
heat rejected= m*cv*(t4-t1)

E=1-(t4-t1)/(t3-t2)

E=1-t1((t4/t1)-1)/t2((t3/t2)-1)

E=1-(t1/t2)*((t4/t1)-1)/((t3/t2)-1)

But from cycle properties we know

t4/t1=t3/t2

Hence,

E=1-(t1/t2)

clear all
close all
clc

%defining engine parametere
bore=0.1
stroke=0.1
cr=12
cond_rod=0.15

%defining state variables
p1=103125
t1=500
gamma=1.4
t3=2300

%1-2 isentopic compression
v_swept=(pi/4)*bore^2*stroke;
v_clearance= v_swept/(cr-1);

v1=v_swept+v_clearance
v2=v_clearance

% to find p2 and t2

p2= p1*(cr)^gamma
t2=p2*v2*t1/(p1*v1)

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


% 2-3 constant volume heat addition
v3=v2

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


%3-4 isentropic expansion

v4=v1
p4=p3*(v3/v4)^gamma

% To find efficieny
E=1-(t1/t2)

% marking the points
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 v3],[p2 p3])
plot([v4 v1],[p4,p1])
xlabel('Volume')
ylabel('Pressure')
title('PV-Diagram Otto Cycle')

Volume of the chamber during the compression and expansion strokes is governed by the crank rotation and is related to the crank angle and crank ratio(R).

V=Vc*(1+0.5(cr-1)[R+1-cosθ-(R^2-sin)^0.5])

Where R=cond_rod/Crank pin radius
Crank pin radius(a)=stroke/2

function [V]=engine_kinematics(bore, stroke, cond_rod, cr, start_crank, end_crank)

a=stroke/2;
R=cond_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

Results:

V_compression Plot:

PV Graph:

Volume in combustion chamber during two cycle:

Efficiency: