Project 1
Conventional vehicles use internal combustion engine and in this the heat engine uses the principle of heat release through combustion of a fuel a combustion chamber to create expanding gases which pushes against a piston creating force and torque output. Vehicle engine converts chemical energy stored in the fuel into heat which is used to produce useful work.
performance of vehicle
power demand on the wheel need to instant supplied by the powertrain. i.e. word the power coming from the powertrain decides the power demand on the wheels(transmission). This is done by the transmission and gear sets to be operated at the set operating point.
the powertrain is consisting of an engine connected to the wheels through a transmission therefore the power comes from the engine
powertrain efficiency: -
we get power by converting chemical energy into the mechanical energy by the process of combustion. This process of energy conversion also has an efficiency and the energy input is in the form of fuel consumption.
____Calculation___
nt = efficiency of transmission(eta t)
ne = efficiency of engine (eta engine)
QLHV = it is the calorific value/ lower heating value of fuel.
mf = mass flow rate in [kg]
pe = Power of engine
pt = power of transmission
System Dynamics: -
Each system has are two generalised variables-
1] Efforts: - It is generalised force variable. ex- torque, force, voltage, heating value.
2] Flow: - It is the generalised flow/velocity variable. ex:- velocity, current, mass flow rate.
3] Power:- It is the product of the effort and flow variable. It is interchangeable between energy domains.
Energy based mode:-
1] Map based models: - Elements are modeled using maps which characterized the relationship between the input and output , output and losses etc.
2] Analytical models: - Elements are modeled using physical equation that represent the internal dynamics. These are the two extremes most models are a combination of the two with some dynamics models analytically and for some using maps.
in this model We use Forward Model:-
////////program script structure///////
%% Vehicle Parameters
M = 1500; % Vehicle weight [kg]
Cr = 0.015; % Rolling resistance coefficient
Rw = 0.35; % Rolling radius [m]
Cd = 0.33; % Drag coefficient
Af = 2; % Frontal area [m2]
rhoa = 1.225; % Air density [kg/m3]
Vwind = 0; % Wind velocity [kmph]
g = 9.81; % Gravitational acceleration [m/s2]
theta = 0; % Slope [deg]
myu_road_tire = 1; % Coefficient of friction between tire - road interface
%% Powertrain Parameters
FDR = 3.73; % Final Drive Ratio / Diff Ratio
DiffEff = 0.99; % Differential efficiency (assumed independent of operating conditions)
NumGear = 4; % Number of gears in transmission
TransGear = [1:NumGear]; % Transmission Gears
TransRatio = [3.06, 1.62, 1, 0.7]; % Corresponding Transmission ratios
TransEff = 0.90; % Transmission efficiency (assumed independent of operating conditions)
TC_SR = [0, .5, .6, .7, .8, .87, .92, .94, .96, .97, 1]; % Torque Converter Speed Ratio (Matlab default) -> TurbSpd/PumpSpd
TC_TR = [2.232, 1.5462, 1.4058, 1.2746, 1.1528, 1.0732, 1.0192, 1, 1, 1, 1]; % Torque Converter Torque Ratio (Matlab default) -> TurbTrq/PumpTrq
TC_Cf = 1e-4 * [.72558, .66322, .63463, .59042, .51331, .4144, .29287, .22444, .12186, .04386, 0]; % Torque Converter Capacity Factor (Matlab default) -> PumpTrq/PumpSpd^2 [Nm/rpm^2]
load Vehicle_Data/Engine_data_1_9L_Saturn_95kW
load Vehicle_Data/Trans_Shift_map
%% Driver Parameters
Kp = 10; % P gain for driver PI model
Ki = 1; % I gain for driver PI model
Drive_Cycle_Select = 1; % Drive cycle selection
% Drive cycle values in m/s
if Drive_Cycle_Select == 1
load DriveCycles/UDDS;
elseif Drive_Cycle_Select == 2
load DriveCycles/HWFET;
elseif Drive_Cycle_Select == 3
load DriveCycles/FTP;
elseif Drive_Cycle_Select == 4
load DriveCycles/US06;
elseif Drive_Cycle_Select == 5
load DriveCycles/NEDC;
elseif Drive_Cycle_Select == 6
load DriveCycles/WLTP;
end
time_end = cyc_time(end);
vel_end = cyc_vel(end);
-:Simulink model of the drivetrain:-