Power difference of TATA ultra truck, Modification of Driver glider model, WOT simulation of GM EV1 using MATLAB
Objective
> Ratio of hill-climbing power required by fully loaded tata ultra truck to the half-loaded one.
> Prepare a new drive cycle excel file with hill climbing angle and include them to modify the Driver_glider.slx file to calculate gradient force according.
> To make a WOT simulation of GM EV1 car and compare it with a 5% change in vehicle parameters
Ratio of hill climbing power required for a fully loaded truck anf a half loaded truck
Hill climbing force
Hill climbing force is one of the forces acting on a vehicle, when the vehicle is moving upwards on a slope. This force is dependent on the maas of the vehicle and the slope angle of the hill or flyover.
To aquire the ratio of power required for fully loaded vehicle and a half loaded vehicle the specifications of TATA ULTRA T.7 is considered with the following specification.
Kerb weight (Kg) - 2990
Payload (Kg) - 4310
Width (m) - 2.05
Height(m) - 2.68
Gradeability (rad) - 0.2617
The reference of the truck specifications is given below
An excel calculator is used to calculate the ratio as show below.
The excel sheet link is attached below
Therefore the ratio of hill climbing power of a fully loaded Tata ultra truck to the half loaded one is = 1.41
Modification in Driver Glider model
To modify the Driver_glider.slx file to calculate gradient force according to the user defined drive cycle which is in the form of excel file. The drive cycle as excel file with time and speed is given in the below link.
The drive cycle can be imported from the workspace and also directly by using from spreadsheet option, then the US06 block is replaced with the drive cycle block as shown in the image below
Simulation of driver glider model with zero inclination angle
The simulation is made to run without giving the input of gradient values. Since, the gradient angle is zero the gradient force is also zero. The resultant drive cycle of the vehicle with zero inclination is shown below
Simulation of driver glider model with constant inclination angle
The simulation can be made to run with constant gradient angle and also with variying angles. To get the results of a constant gradient angle, the value of inclination angle is given as 15 degrees in the model workspace window.
A scope is attached to the grade force block which is in the subsystem of the glider block, to get the value of the gradient force. The resultant value will be a constant value as the inclination angle is also constant.
The results of the constant inclination angle with a user defined driver cycle from spreadsheet of the driver glider model is as follows.
The gradient force for constant inclination = 14000 N
By comparing the speed vs time plot of both the condtions (zero inclination and constant inclination) it is clearly seen that the vheicle with gradient force cannot reach 60mph at t=60.
Simulation of driver glider model with variying inclination angle
A variying slope angle data is created in the excel sheet and is imported using the from spreadsheet block into the subsystem of the glider model. The slope angle data is attached with the link below
(https://docs.google.com/spreadsheets/d/1ZODqP6eVqyQF58qoNRTJ2TNfLxGCqTAY6yZ1Dg51mLk/edit?usp=sharing)
The grade force calculation block is modified in the glider model as show below. A scope is attached to get the value of the grade force applied to the vehicle. The resulting grade force value will be variying since the slope angle is also variying.
The results of the simulation is as follows.
The speed vs time plot shows the change in variation of speed due to the variying inclination angle induced to the vehicle.
The gradient force varies from = 0 - 25,000 N
The WOT (Wide Open Throttle) response of GM EV1 car
The parameters of the GM EV1 vehicle is refered, which are then changed with a 5% increase in value. The comparison between the original value to the modified value is tabulated. The comparison table is given below.
The original parameters are refered from the below link.
(https://avt.inl.gov/sites/default/files/pdf/fsev/genmot.pdf)
The objective is to find the accleration time of the vehicle to reach the top speed of 35.8m/s with, original parameters and the modified parameters.
Since, the GM EV1 is a pure electric vehicle the motor charecteristics are taken into consideration. In an electric motor the torque is maximum till the vehicle reaches the critical speed, after the critical speed the torque starts reducing. Therefore, in a wide open throttle condition both states of the motor gets involved.
The formula to calculate the two stages of accleration is
(G/r)*T = Frr + Fad + [m+[I*[G^2/(ng*r)]]. (dv/dt)
where,
G = Gear ratio
r = Radius of tyre (m)
T = Torque (Nm)
Frr = Rolling resistance force (N)
Fad = Air drag force (N)
m = Mass (kg)
I = moment of inertia
ng = Gear efficiency
dv/dt = Rate of change of velocity with respective to time
Since we do not know the moment of inertia(I) we assume that [I*[G^2/(ng*r)] is 5% of mass of the vehicle with payload. In other word the angular accleration effect is considered to be 5% of mass of the vehicle.
For original parameters of the vehicle the equations are as follows
The equation below the ctitical speed is
dv = dt*[3.51-(0.000169*V^2)]
dv = V(n+1) - V(n)
Therefore,
V(n+1) - V(n) = dt*[3.51-(0.000169*V^2)]
V(n+1) = V(n) + dt*[3.51-(0.000169*V^2)]
The equation above the critical speed is
dv = dt*[(65.7/V)-0.044-(0.000169*V^2)]
dv = V(n+1) - V(n)
Therefore,
V(n+1) - V(n) = dt*[(65.7/V(n))-0.044-(0.000169*V(n)^2)]
V(n+1) = V(n) + dt*[(65.7/V(n))-0.044-(0.000169*V(n)^2)]
For the modified parameters of the vehicle the equations are as follows
The equation below the ctitical speed is
dv = dt*[3.109-(0.000187*V^2)]
dv = V(n+1) - V(n)
Therefore,
V(n+1) - V(n) = dt*[3.109-(0.000187*V(n)^2)]
V(n+1) = V(n) + dt*[3.109-(0.000187*V(n)^2)]
The equation above the critical speed is
dv = dt*[(62.67/V)-0.0467-(0.000187*V^2)]
dv = V(n+1) - V(n)
Therefore,
V(n+1) - V(n) = dt*[(62.67/V(n))-0.0467-(0.000187*V(n)^2)]
V(n+1) = V(n) + dt*[(62.67/V(n))-0.0467-(0.000187*V(n)^2)]
Matlab code
With all the four equations a matlab program is created as show below, to get the accleration curve plots and to get the accleration timings of both the vehicles
The following program can simulate the vehicles for 50 seconds with a wide open throttle condition using the calculated equations.
% To_plot_WOT_of_GM_EV1_with_orignal_and_changed_parameters
clear all
close all
clc
% input_and_storage_of time_and_velocities
t = linspace(0,50,501);
v = zeros(1,501);
v2= zeros(1,501);
dT= 0.1;
% Wide_open_throttle_response_of GM_EV1_with_original_parameters
for n = 1:500
if v(n)<19.8
v(n+1) = v(n)+dT*(3.51-(0.000169*(v(n))^2));
elseif v(n)>= 35.8
v(n+1)= v(n);
else
v(n+1) = v(n)+dT*((-0.044)+(65.7/v(n))-(0.000169*(v(n))^2));
end
end
% To_diplay_the_time_taken_for_accleration
for n=1:500
if v(n)>= 35.8
v(n+1) = v(n);
disp('Time required to acclerate with original parameters is')
disp(t(n))
break
end
end
% Wide_open_throttle_response_of GM_EV1_with_modified_parameters
for n = 1:500
if v2(n)<19.8
v2(n+1)= v2(n)+dT*((3.109)-(0.000187*(v2(n))^2));
elseif v2(n)>=35.8
v2(n+1) = v2(n);
else
v2(n+1) = v2(n)+dT*((62.67/v2(n)-0.0467-(0.000187*(v2(n))^2)));
end
end
% To_diplay_the_time_taken_for_accleration
for n=1:500
if v2(n)>= 35.8
v2(n+1) = v2(n);
disp('Time required to acclerate with modified parameters is')
disp(t(n))
break
end
end
% To_convert_m/s_to_kmph_3.6_is_multiplied
v = v.*3.6;
v2 = v2.*3.6;
% To_plot_time_vs_velocities
plot (t,v);
hold on
plot (t,v2)
grid on
xlabel('Time in (sec)')
ylabel('speed of the vehicle in (kmph)')
legend('Default parameters','Modified parameters')
Results of the code
When the code is made to run a plot with accleration curves of both the vehicles with original parameters and the modified parameters is generated as shown below.
From the plot it is clearly seen that, the vehicle with the default parameters can acclerate to its top speed quicker than the vehicle with change in plus 5% parameters.
The accleration timings are shown in the command window
Time taken to reach the cut off speed with a wide open throttle condition of both the vehicles are shown in which, the vehicle with original parameters is 1.2 s faster than the vehicle with modified parameters.
Therefore, when the parameters of the vehicle such as mass, frontal area, rolling resistance, tyre radius and air drag coefficient insreases the accleration time increases. By sacrificing the accleration time, a vehicle can give advantages like more comfort due to the increase in area and increase in payload capacity.