CALCULATION OF THERMAL EFFICIENCY AND PLOTTING PRESSURE V/S VOLUME OF OTTO CYCLE USING MATLAB
AIM: To determine the thermal efficiency and to plot pressure v/s volume of an Otto cycle.
OBJECTIVE: To execute program for simulating an air standard cycle or Otto cycle using MATLAB.
THEORY:
The Otto cycle is a cycle of engine operation which requires four strokes of the piston for induction, compression, ignition, and exhaust. The fuel and air mixture is compressed before combustion is started by an electrical spark or other means. Named after German engineer Nikolaus Otto.
The Otto Cycle, describes how heat engines turn gasoline into motion. Like thermodynamic cycles, this cycle turns chemical energy into thermal energy and then into motion. The Otto cycle describes how internal combustion engines (that use gasoline) work, like automobiles etc....
As in an ideal Diesel cycle four distinct processes are also followed in the Otto cycle: reversible adiabatic compression, addition of heat at constant volume, reversible adiabatic expansion, and rejection of heat at constant volume as follows:
i. ISENTROPIC COMPRESSION:
Air enters the cycle at point 1 when compression starts and it ends at point 2. The air is compressed in an isentropic process through a volume ratio v1/v2, known as the compression ratio Cr (the ratio of the volume of the working fluid before the compression process to its volume after).
ii. CONSTANT VOLUME HEAT ADDITION:
Heat addition starts at point 2 and ends at point 3. A quantity of heat Qin is added at constant volume, and the working fluid temperature rises.
iii. ISENTROPIC EXPANSION:
The isentropic expansion process starts at point 3 and it ends at point 4 to the original volume.
iv. CONSTANT VOLUME HEAT REJECTION:
The heat Qout is rejected at constant volume from point 4 to point 1. Air gets cooled and the working fluid temperature goes down at a constant volume until the cycle is completed.
FIG:1: Shows an P-V Diagram of an ideal otto cycle.
INPUTS WHICH ARE INITIALLY ASSUMED:
1. Pressure at state-1 = 1atmp
2. Temperature at state-1 = 500K
3. Temperature at state-3 = 2300k
4.`gamma`= 1.4
5. Compression ratio `(C_r) = 12`
6. Bore (d)= 0.1m
7. Stroke = 0.1m
8. Length of connecting rod (l)= 0.15m
FORMULAE:
1. Swept volume `(V_S)` = `(pi/4)*d^2*l`
2. Clearance volume `(V_c)`= `V_s/(C_r-1)`
3. Volume at state-1 (v1) = `V_s+V_c`
4. Volume at state-2 (v2) = `V_c`
5. Isentropic equation = `p1v1^gamma=p2v2^gamma`
4. Compression ratio = v1/v2
6. Ideal gas law = (p1v1)/t1 = (p2v2)/t2 = Constant
PISTON KINEMATIC EQUATION:
`V/V_c=1+1/2(C_r-1)[R+1-cos(theta)-(R^2-sin^2(theta))^(1/2)]`
R = Is the ratio of length of connecting rod and Crank pin radius = l/a
a = stroke/2
PROGRAM-1 :FOR SIMULATING OTTO CYCLE WITHOUT USING PRE-DETERMINED FUNCTION
clear all
close all
clc
% input
gamma= 1.4
t3= 2300
% state variables
p1= 101325
t1= 500
% engine geometric parameters
bore= 0.1;
stroke= 0.1;
con_rod = 0.15;
cr = 12;
% calculating the swept volume and clearance volume
v_swept = (pi/4)*bore^2*stroke;
v_clearance = v_swept/(cr-1);
v1 = v_swept + v_clearance
v2 = v_clearance
% state variables at state point 2
% p2v2^gamma = p1v1^gamma
% p2 = p1*(v1/v2)^gamma = p1*(cr)^gamma
p2 = p1*(cr)^gamma;
% p1v1/t1 = p2v2/t2| t2 = p2*v2*t1/(p1*v1)
t2 = p2*v2*t1/(p1*v1);
% state variables at state point 3
v3 = v2
% p3v3/t3=p2v2/t2| p3=p2*t3/t2
p3 = p2*(t3/t2)
% state variables at state point 4
v4 = v1
% p3v3^gamma=p4v4^gamma| p4=p3v3/v4
p4 = p3*(v3/v4)^gamma
figure(1)
hold on
plot(v1,p1,'*','color','r')
plot(v2,p2,'*','color','r')
plot(v3,p3,'*','color','r')
plot(v4,p4,'*','color','r')
figure(2)
plot([v1 v2 v3 v4 v1],[p1 p2 p3 p4 p1],'color','b')RESULT OF PROGRAM-1
This above plot result is wrong, because there should be an exponential curve for adiabatic expansion and adiabatic compression
PROGRAM-2: FUNCTION TO DETERMINE CYLINDER VOLUME AT ANY CRANK ANGLE
function [V]= engine_kinematics(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,100);
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"function [V]= engine_kinematics(bore, stroke, con_rod, cr, start_crank, end_crank)" This is used to call the program using the name of the function engine_kinematics.
FINAL PROGRAM BY CALLING FUNCTION:
clear all
close all
clc
% input
gamma= 1.4
t3= 2300
% state variables
p1= 101325
t1= 500
% engine geometric parameters
bore= 0.1;
stroke= 0.1;
con_rod = 0.15;
cr = 12;
% calculating the swept volume and clearance volume
v_swept = (pi/4)*bore^2*stroke;
v_clearance = v_swept/(cr-1);
v1 = v_swept + v_clearance
v2 = v_clearance
% state variables at state point 2
% p2v2^gamma = p1v1^gamma
% p2 = p1*(v1/v2)^gamma = p1*(cr)^gamma
p2 = p1*(cr)^gamma;
% p1v1/t1 = p2v2/t2| t2 = p2*v2*t1/(p1*v1)
t2 = p2*v2*t1/(p1*v1);
constant_c = p1*v1^gamma;
V_compression = engine_kinematics(bore, stroke, con_rod, cr,180,0)
P_compression = constant_c./V_compression.^gamma;
% state variables at state point 3
v3 = v2
% p3v3/t3=p2v2/t2| p3=p2*t3/t2
p3 = p2*(t3/t2)
constant_c = p3*v3^gamma;
V_expansion = engine_kinematics(bore, stroke, con_rod, cr,0,180)
P_expansion = constant_c./V_expansion.^gamma;
% state variables at state point 4
v4 = v1
% p3v3^gamma=p4v4^gamma| p4=p3v3/v4
p4 = p3*(v3/v4)^gamma
% Thermal efficiency
thermal_efficiency = 1-(1/(cr^(gamma-1)))
figure(1)
hold on
xlabel('volume')
ylabel('pressure')
plot(v1,p1,'*','color','r')
plot(V_compression,P_compression)
plot(v2,p2,'*','color','r')
line([v2 v3],[p2 p3])
plot(v3,p3,'*','color','r')
plot(V_expansion,P_expansion)
plot(v4,p4,'*','color','r')
line([v4 v1],[p4 p1])
RESULT OF FINAL PROGRAM:
VALUES OF EACH VARIABLES:
Thus thermal efficiency is = 0.6299 OR 62.99%