Air Standard Cycle
TITLE: Air standard cycle
AIM: To write a matlab program for solving otto cycle and create a plot for otto cycle.
OBJECTIVE:
(i) To write a matlab program to create a PV diagram for otto cycle.
(ii) To calculate the thermal efficiency of the engine.
THEORY:
- The air-standard-Otto cycle is an idealized cycle for the spark-ignition internal combustion engines.
- Otto cycle describes how an internal combustion engine which firstly converts the fuel’s chemical energy into thermal energy and then thermal energy into mechanical power, which is further used to move a vehicle.
- The air standard otto cycle has four processes to form a complete cycle, and the following assumptions are considered:-
(1). Air which is a working fluid throughout the cycle acts as an ideal gas.
(2). Combustion processes in otto cycle are replaced by heat addition processes at constant volume.
(3). Exhaust process is replaced by a heat rejection process at constant volume.
(4). As air is assumed to be an working fluid in the cycle ,constant values of specific heat at constant volume and pressure are assumed.
Figure.(1)- P-V Diagram of otto cycle
Otto cycle is made up of the following four internally reversible processes:
- Process 0-1 : Intake stroke
- Process 1-2 : Isentropic compression
- Process 2-3 : Constant-volume heat addition
- Process 3-4 : Isentropic expansion
- Process 4-1 : Constant-volume heat rejection
- The above PV diagram represents that the ideal Otto cycle represents how the volume and pressure of the air-fuel mixture changes due to the compression and expansion process.
- It clearly represents that when the piston performs work on the air-fuel mixture, the volume of the gas is compressed and substantially the pressure increases. Similarly, after the combustion process the heat released by the fuel/air mixture performs work on the piston due to which it expands.
- The above animation shows the working of otto cycle.
- Process 0-1 is an intake stroke in which the mass of working fluid is drawn into the cylinder at constant atmospheric pressure through the intake valve and the exhaust valve is kept closed during this process.
- Process 1-2 is an adiabatic compression stroke in which the piston moves from bottom dead centre to top dead centre.This is an isentropic compression process which states that there is no loss of mechanical energy due to friction.
- Process 2-3 is an ignition phase in which the piston is momentarily at rest position at top dead centre and this process represents the ignition of the fuel-air mixture and the subsequently burning rapidly.
- Process 3–4 is an adiabatic expansion (power stroke) in which the increased high pressure exerts a force on the piston and pushes it towards the bottom dead centre.
- Process 4–1 is an heat rejection phase in which the cycle is completed at constant volume and the heat is rejected while the piston is at bottom dead center.
- Process 1–0 is an exhaust stroke in which the gaseous mixture is emitted out to the atmosphere and the process starts again.
How to increase the efficiency of the Otto cycle?
- In order to increase the efficiency of otto cycle it has to be operated on lean fuel which is a fuel/air mixture containing relatively low proportion of fuel.
- It can also be improved by increasing the compression ratio which allows an engine to extract more mechanical energy from a given mass of air/fuel mixture.
PROCEDURE FOR MATLAB PROGRAM:
- Initially to start with the matlab program we need to assume temperature(t1) and pressure(p1) which is 500k and 101325 Pascal respectively.Gamma is a constant assumed as 1.4.
- Now engine geometric parameters which are bore,stroke,length of connecting rod and compression ratio are to be assumed.
- Next, we need to calculate the swept volume which is the volume swept by the engine piston during one stroke and the clearance volume which is the volume between the cylinder head and the piston when the piston is at top dead centre by using a formula. (i). Swept Volume=π/4*bore^2*stroke (ii). Clearance volume=swept volume/(compression ratio -1)
- We have v1 which is sum of swept volume and clearance volume and v2 which represents volume at top dead centre.
- To calculate the pressure at point 2 we have the equation for isentropic compression (i.e) p1v1^gamma=p2v2^gamma through which we can calculate the p2.
- After knowing the value for pressure at point 2 we can calculate the temperature at point 2 by using ideal gas law (P1V1/T1=P2V2/T2).
- Next the variables at state point 3 are calculated which represents the end of heat addition process at constant volume.As this process is at constant volume ,hence (V3=V2).Temperature at point 3 is assumed as 2300k.
- Now we can calculate pressure at point 3 by using ideal gas equation.
- Finally we need to find out the values for variables at state point 4, which represents the constant volume heat rejection process.Therefore (V4=V1).
- Now we can find pressure at state point 4 by using the formula (P4V4^Gamma=P3V3^Gamma) and similarly temperature at point 4 can be found by using (P3V3/T3=P4V4/T4).
- To calculate thermal efficiency of engine we can find it by using the formula Thermal efficiency=(1-1/(cr)^(gamma-1))*100 where ,cr represents the compression ratio.
- In order to plot the P-V diagram we have to use plot command and input the variables as required for plotting.
- For providing the name for x-axis and y-axis we use the command xlabel and ylabel respectively.
- As we cannot join the curve of compression and expansion stage in the P-V diagram.we have to understand the concept of piston kinematics.
- Now we have to write a new matlab program to determine the volume information during the compression and expansion stroke.
- To plot the compression and expansion stroke correctly we should know the thermodynamic relationship during compression and expansion stroke and the piston kinematics to find the volume change from tdc to bdc.
- Now we will use the formula to know the change in volume inside the combustion chamber. `V/V_c=1+1/2*(cr-1)[R+1-cos(theta)-(R^2-sin^2(theta))^0.5]`
- Now we will use function command which can be used whenever required and will save the script by the name engine kinematics(bore,stroke,con_rod,cr,start_crank,end_crank).
- Next we will call the function in command window and input the value of variables.
- We can input the value of v_compression - p_compression and and v_expansion -p_expansion.
- Finally we can plot the exact PV diagram for otto cycle.
MATLAB PROGRAM:
% writing a matlab program for calculating the thermal efficiency and plotting pv diagram for otto cycle
clear all
close all
clc
% state variables at state point 1
p1=101325
t1=500
gamma=1.4;
% Engine geometric parameteres
bore=0.1;
stroke=0.1;
con_rod=0.15;
cr=12;
% Formula for 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
% p1v1^gamma=p2v2^gamma which is isentropic compression process
p2=p1*(v1/v2)^gamma
% Temperature at point 2 can be calculated by using ideal gas law
t2=p2*v2*t1/(p1*v1)
c_constant=p1*v1^gamma;
v_compression=engine_kinematics(bore,stroke,con_rod,cr,180,0);
p_compression=c_constant./v_compression.^gamma;
% state variables at state point 3
% at state point 3 it is constant volume combustion process .hence (v3=v2)
v3=v2;
t3=2300;
% pressure at state point 3 can be calculated using ideal gas law
p3=p2*(t3/t2);
c_constant=p3*v3^gamma
v_expansion=engine_kinematics(bore,stroke,con_rod,cr,0,180)
p_expansion=c_constant./v_expansion.^gamma
% state variable at state point 4
% constant volume heat rejection.hence(v4=v1)
v4=v1;
%p3v3^gamma=p4v4^gamma
p4=p3*(v3/v4)^gamma
t4=p4*v4*t3/(p3*v3)
% efficiency for otto cycle
thermal_efficiency=(1-1/(cr)^(gamma-1))*100
figure(1)
hold on
plot(v_compression,p_compression,'LineWidth',1,'Color','r')
plot(v_expansion,p_expansion,'LineWidth',1,'Color','r')
plot(v1,p1,'linewidth',1,'Color','r')
plot(v2,p2,'linewidth',1,'Color','r')
plot([v2 v3],[p2 p3],'linewidth',1)
plot([v4 v1],[p4 p1],'linewidth',1)
plot(v3,p3,'linewidth',1,'Color','r')
plot(v4,p4,'linewidth',1,'Color','r')
title('P-V Diagram for otto cycle')
xlabel('Volume')
ylabel('Pressure')
grid onOUTPUT:
p1 =
101325
t1 =
500
v_swept =
7.8540e-04
v_clearance =
7.1400e-05
v1 =
8.5680e-04
v2 =
7.1400e-05
p2 =
3.2853e+06
t2 =
1.3510e+03
R =
3.0000
theta =
Columns 1 through 15
180.0000 178.4874 176.9748 175.4622 173.9496 172.4370 170.9244 169.4118 167.8992 166.3866 164.8739 163.3613 161.8487 160.3361 158.8235
Columns 16 through 30
157.3109 155.7983 154.2857 152.7731 151.2605 149.7479 148.2353 146.7227 145.2101 143.6975 142.1849 140.6723 139.1597 137.6471 136.1345
Columns 31 through 45
134.6218 133.1092 131.5966 130.0840 128.5714 127.0588 125.5462 124.0336 122.5210 121.0084 119.4958 117.9832 116.4706 114.9580 113.4454
Columns 46 through 60
111.9328 110.4202 108.9076 107.3950 105.8824 104.3697 102.8571 101.3445 99.8319 98.3193 96.8067 95.2941 93.7815 92.2689 90.7563
Columns 61 through 75
89.2437 87.7311 86.2185 84.7059 83.1933 81.6807 80.1681 78.6555 77.1429 75.6303 74.1176 72.6050 71.0924 69.5798 68.0672
Columns 76 through 90
66.5546 65.0420 63.5294 62.0168 60.5042 58.9916 57.4790 55.9664 54.4538 52.9412 51.4286 49.9160 48.4034 46.8908 45.3782
Columns 91 through 105
43.8655 42.3529 40.8403 39.3277 37.8151 36.3025 34.7899 33.2773 31.7647 30.2521 28.7395 27.2269 25.7143 24.2017 22.6891
Columns 106 through 120
21.1765 19.6639 18.1513 16.6387 15.1261 13.6134 12.1008 10.5882 9.0756 7.5630 6.0504 4.5378 3.0252 1.5126 0
c_constant =
8.7666
R =
3.0000
theta =
Columns 1 through 15
0 1.5126 3.0252 4.5378 6.0504 7.5630 9.0756 10.5882 12.1008 13.6134 15.1261 16.6387 18.1513 19.6639 21.1765
Columns 16 through 30
22.6891 24.2017 25.7143 27.2269 28.7395 30.2521 31.7647 33.2773 34.7899 36.3025 37.8151 39.3277 40.8403 42.3529 43.8655
Columns 31 through 45
45.3782 46.8908 48.4034 49.9160 51.4286 52.9412 54.4538 55.9664 57.4790 58.9916 60.5042 62.0168 63.5294 65.0420 66.5546
Columns 46 through 60
68.0672 69.5798 71.0924 72.6050 74.1176 75.6303 77.1429 78.6555 80.1681 81.6807 83.1933 84.7059 86.2185 87.7311 89.2437
Columns 61 through 75
90.7563 92.2689 93.7815 95.2941 96.8067 98.3193 99.8319 101.3445 102.8571 104.3697 105.8824 107.3950 108.9076 110.4202 111.9328
Columns 76 through 90
113.4454 114.9580 116.4706 117.9832 119.4958 121.0084 122.5210 124.0336 125.5462 127.0588 128.5714 130.0840 131.5966 133.1092 134.6218
Columns 91 through 105
136.1345 137.6471 139.1597 140.6723 142.1849 143.6975 145.2101 146.7227 148.2353 149.7479 151.2605 152.7731 154.2857 155.7983 157.3109
Columns 106 through 120
158.8235 160.3361 161.8487 163.3613 164.8739 166.3866 167.8992 169.4118 170.9244 172.4370 173.9496 175.4622 176.9748 178.4874 180.0000
v_expansion =
1.0e-03 *
Columns 1 through 15
0.0714 0.0716 0.0721 0.0730 0.0743 0.0760 0.0779 0.0803 0.0830 0.0861 0.0895 0.0932 0.0973 0.1017 0.1065
Columns 16 through 30
0.1116 0.1170 0.1227 0.1287 0.1350 0.1416 0.1485 0.1557 0.1631 0.1708 0.1787 0.1869 0.1953 0.2040 0.2128
Columns 31 through 45
0.2219 0.2311 0.2406 0.2502 0.2600 0.2699 0.2800 0.2902 0.3005 0.3109 0.3214 0.3320 0.3427 0.3535 0.3643
Columns 46 through 60
0.3752 0.3860 0.3970 0.4079 0.4188 0.4297 0.4407 0.4515 0.4624 0.4732 0.4840 0.4947 0.5053 0.5158 0.5263
Columns 61 through 75
0.5366 0.5469 0.5571 0.5671 0.5770 0.5868 0.5965 0.6060 0.6154 0.6247 0.6338 0.6427 0.6515 0.6601 0.6685
Columns 76 through 90
0.6768 0.6849 0.6928 0.7006 0.7081 0.7155 0.7227 0.7297 0.7366 0.7432 0.7497 0.7559 0.7620 0.7679 0.7736
Columns 91 through 105
0.7791 0.7844 0.7895 0.7945 0.7992 0.8037 0.8081 0.8123 0.8163 0.8201 0.8237 0.8271 0.8303 0.8333 0.8362
Columns 106 through 120
0.8389 0.8413 0.8436 0.8457 0.8477 0.8494 0.8510 0.8523 0.8535 0.8545 0.8553 0.8560 0.8564 0.8567 0.8568
p_expansion =
1.0e+06 *
Columns 1 through 15
5.5931 5.5732 5.5141 5.4180 5.2884 5.1297 4.9468 4.7451 4.5298 4.3061 4.0783 3.8505 3.6258 3.4070 3.1959
Columns 16 through 30
2.9941 2.8025 2.6216 2.4517 2.2927 2.1444 2.0064 1.8783 1.7596 1.6496 1.5479 1.4538 1.3669 1.2865 1.2122
Columns 31 through 45
1.1435 1.0799 1.0211 0.9666 0.9161 0.8693 0.8259 0.7855 0.7480 0.7131 0.6806 0.6504 0.6222 0.5958 0.5712
Columns 46 through 60
0.5482 0.5267 0.5065 0.4876 0.4699 0.4532 0.4376 0.4229 0.4091 0.3961 0.3838 0.3722 0.3613 0.3510 0.3413
Columns 61 through 75
0.3321 0.3234 0.3152 0.3074 0.3000 0.2930 0.2864 0.2801 0.2741 0.2685 0.2631 0.2580 0.2532 0.2485 0.2442
Columns 76 through 90
0.2400 0.2360 0.2323 0.2287 0.2253 0.2220 0.2189 0.2160 0.2132 0.2105 0.2080 0.2056 0.2033 0.2011 0.1990
Columns 91 through 105
0.1971 0.1952 0.1934 0.1918 0.1902 0.1887 0.1872 0.1859 0.1846 0.1834 0.1823 0.1812 0.1803 0.1793 0.1785
Columns 106 through 120
0.1777 0.1770 0.1763 0.1757 0.1751 0.1746 0.1742 0.1738 0.1734 0.1732 0.1729 0.1727 0.1726 0.1725 0.1725
p4 =
1.7251e+05
t4 =
851.2465
thermal_efficiency =
62.9893
PROGRAM FOR PISTON KINEMATICS:
function [v]=engine_kinematics(bore,stroke,con_rod,cr,start_crank,end_crank)
% a is crank pin radius
a=stroke/2;
% R is the ratio of connecting rod to crank pin radius
R=con_rod/a;
cr=12;
% Formula for swept volume and clearance volume
v_swept=(pi/4)*bore^2*stroke;
v_clearance=v_swept/(cr-1);
theta= linspace(start_crank,end_crank,120);
% formula for change in volume inside the engine combustion chamber
% V/Vc=1+(1/2)*(cr-1)[R+1-cos(theta)-(R^2-sin^2(theta)^0.5]
term1=0.5*(cr-1);
term2=R+1-cosd(theta);
term3=(R^2-sind(theta).^2).^0.5;
v=(1+term1*(term2-term3)).*v_clearance;
end
OUTPUT:
v_compression=engine_kinematics(0.1,0.1,0.15,12,0,180)
v_compression =
1.0e-03 *
Columns 1 through 15
0.0714 0.0716 0.0721 0.0730 0.0743 0.0760 0.0779 0.0803 0.0830 0.0861 0.0895 0.0932 0.0973 0.1017 0.1065
Columns 16 through 30
0.1116 0.1170 0.1227 0.1287 0.1350 0.1416 0.1485 0.1557 0.1631 0.1708 0.1787 0.1869 0.1953 0.2040 0.2128
Columns 31 through 45
0.2219 0.2311 0.2406 0.2502 0.2600 0.2699 0.2800 0.2902 0.3005 0.3109 0.3214 0.3320 0.3427 0.3535 0.3643
Columns 46 through 60
0.3752 0.3860 0.3970 0.4079 0.4188 0.4297 0.4407 0.4515 0.4624 0.4732 0.4840 0.4947 0.5053 0.5158 0.5263
Columns 61 through 75
0.5366 0.5469 0.5571 0.5671 0.5770 0.5868 0.5965 0.6060 0.6154 0.6247 0.6338 0.6427 0.6515 0.6601 0.6685
Columns 76 through 90
0.6768 0.6849 0.6928 0.7006 0.7081 0.7155 0.7227 0.7297 0.7366 0.7432 0.7497 0.7559 0.7620 0.7679 0.7736
Columns 91 through 105
0.7791 0.7844 0.7895 0.7945 0.7992 0.8037 0.8081 0.8123 0.8163 0.8201 0.8237 0.8271 0.8303 0.8333 0.8362
Columns 106 through 120
0.8389 0.8413 0.8436 0.8457 0.8477 0.8494 0.8510 0.8523 0.8535 0.8545 0.8553 0.8560 0.8564 0.8567 0.8568OUTPUT PLOTS:
PLOT(1)
Crank angle vs volume Crank angle vs volume
PLOT(2) PLOT(3)
Workspace Output
Errors during programming:
Error occurred because of not using element-wise operator.It was rectified by using element-wise operator.
CONCLUSION/RESULTS:
- Using the Matlab program,PV diagram for otto cycle has been plotted for the input variable.
- The thermal efficiency of otto cycle is 62.98%.
- The thermal efficiency can be improved by operating the engine on lean fuel and by increasing the compression ratio.
REFERENCES:
- Figure(1) is obtained from https://www.nuclear-power.com/nuclear-engineering/thermodynamics/thermodynamic-cycles/otto-cycle-otto-engine/otto-cycle-pv-ts- diagram/
- Animation of otto cycle is obtained from https://www.youtube.com/watch?v=CaVXe3Gxomo