Wide Open Throttle

WOT

Wide open throttle (WOT) can be explained as maximum intake of air and fuel that occurs when the throttle plates inside the carburetor or throttle body are "wide open", providing the least resistance to the incoming air. In the case of an automobile, WOT is when the accelerator is depressed fully, sometimes referred to as "flooring it."

In the case of a diesel engine, which does not have a throttle valve, WOT is the point at which the maximum amount of fuel is being injected relative to the amount of air pumped by the engine, generally in order to bring the fuel-air mixture up to the stoichiometric point.

At wide open throttle, manifold vacuum decreases. The higher manifold pressure in turn allows more air to enter the combustion cylinders, and thus additional fuel is required to balance the combustion reaction. The additional air and fuel reacting together produce more power.

PROGRAM FOR CALCULATION OF WOT:

close all
clear all
clc

%Wide Open Throttle 

% REFERENCES
  %[WEBSITE:  https://avt.inl.gov/sites/default/files/pdf/fsev/genmot.pdf]
  %[BOOK:ELECTRIC VEHICLE TECHNOLOGY EXPLAINED, IInd EDITION 2012]

% Inputs
Mass=1546.75 %kg
Air_Density=1.25 %kg/m^3
% Linear_Velocity=19.8 %m/s
Tyre_Width=175 %mm 
Aspect_Ratio=65 %
Height_Tyre=113.75 %mm
Rim_Size=355.6 %mm
r=((2*Height_Tyre+Rim_Size)/2)/1000 %mm Radius Of Tyre
% Angular_Velocity=(Linear_Velocity)/(2*3.14*r) %rpm
Drag_Coeff=0.19
Length=4.310 %m
Width=1.765 %m
Frontal_Area=1.8 %m^2
Rolling_Resistance_Coeff=0.0048
G=11 %11:1, Gear Ratio
T_max = 140 %N-m
Wc = 733 %rad/s
Gear_Eff=0.95
P=102000 %W

% Formula,
% I*(G^2/(Gear_Eff*r^2))= 5% of Mass since we are not involving rotating part inetia forces in the calculation
% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+77.3375)*(dv/dt)

% d=(G/r)*T_max*Gear_Eff
% a=(Rolling_Resistance_Coeff*Mass*9.81)=72.83
% b=(0.625*Frontal_Area*Drag_Coeff*1)=0.213V^2
% c=(Mass+77.3375)

% (4971.99=0.213V^2+1624*(dv/dt))
% (4971.99/1624)=0.00013V^2+(dv/dt))
% (dv/dt)=3.06-0.00013V^2

% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
%  T=P/v
%  T=102000/(37.93*v)= 2756.75/V
%  T=2756.75/V

% Gear_Eff*(G/r)*(T)=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
% 0.95*37.93*/(2756.75/v)=72.83+0.213V^2+1624.08*(dv/dt)
% (1624.08*(dv/dt))*v=((36.03*2756.75/V)-(72.83-0.213V^2))
% 1624.08*(dv/dt)=((96899.76/v)-(72.83-0.213V^2))
% (dv/dt)=((58.08/v)-0.04-0.00013V^2)

% Matlab conversion of formula,
% (dv/dt)=3.06-0.0.00013V^2
% (V(n+1)-Vn)/dt)=3.06-0.00013*(Vn)^2
% V(n+1)=V(n)+dt*(3.06-(0.00013*(V(n))^2))


% (dv/dt)=((58.08/v)-0.04-0.00013V^2)
% (V(n+1)-Vn)/dt)=((58.08/v)-0.04-0.00013*(Vn)^2)
% V(n+1)=V(n)+dt*((58.08/V(n))-0.04-(0.00013*(V(n)^2)))


% To Find

Time=linspace(0,25,501)
V(1)=0
dt=0.05;
for n=1:500
if V(n)<19.8
V(n+1)=V(n)+dt*(3.06-(0.00013*(V(n))^2))
elseif V(n)<35.8
V(n+1)=V(n)+dt*((58.08/V(n))-0.04-(0.00013*(V(n)^2)))
else
V(n+1)=V(n)
end
end
V=(V*3.6);
figure(1)
plot(Time,V)
hold on
% REDUCING MASS BY 5% TO ACHIEVE FASTER WIDE OPEN THROTTLE

Mass=1469.4125 %kg

% Formula,
% I*(G^2/(Gear_Eff*r^2))= 5% of Mass since we are not involving rotating part inetia forces in the calculation
% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+77.3375)*(dv/dt)

% d=(G/r)*T_max*Gear_Eff=5016.76
% a=(Rolling_Resistance_Coeff*Mass*9.81)=69.1916
% b=(0.625*Frontal_Area*Drag_Coeff*1)=0.213V^2
% c=(Mass+77.3375)=1546.75

% (5016.76-69.1916)=0.213V^2+1546.75*(dv/dt))
% 4947.5684=0.213V^2+1546.75*(dv/dt))
% (4947.5684/1546.75)=0.00013V^2+(dv/dt))
% (dv/dt)=3.19-0.00013V^2

% (G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
%  T=P/v
%  T=102000/(37.72*v)= 2704.1357/V
%  T=2704.1357/V

% Gear_Eff*(G/r)*(T)=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)
% 0.95*37.72*/(2704.1357/v)=69.1916+0.0.213V^2+1546.75*(dv/dt)
% (1546.75*(dv/dt))*v=((35.83*2704.1357/V)-(69.1916-0.213V^2))
% 1546.75*(dv/dt)=((96889.1821/v)-(69.1916-0.213V^2))
% (dv/dt)=((62.64/v)-0.04-0.00013V^2)

% Matlab conversion of formula,
% (dv/dt)=3.19-0.0.00013V^2
% (V(n+1)-Vn)/dt)=3.19-0.00013*(Vn)^2
% V(n+1)=V(n)+dt*(3.19-(0.00013*(V(n))^2))


% (dv/dt)=((62.64/v)-0.04-0.00013V^2)
% (V(n+1)-Vn)/dt)=((62.64/v)-0.04-0.00013*(Vn)^2)
% V(n+1)=V(n)+dt*((62.64/V(n))-0.04-(0.00013*(V(n)^2)))
    
Time=linspace(0,25,501)
V(1)=0
dt=0.05;
for n=1:500
if V(n)<19.8
V(n+1)=V(n)+dt*(3.19-(0.00013*(V(n))^2))
elseif V(n)<35.8
V(n+1)=V(n)+dt*((62.64/V(n))-0.04-(0.00013*(V(n)^2)))
else
V(n+1)=V(n)
end
end
V=(V*3.6);
plot(Time,V)
legend({'Original Data','5% Mass Reduced Data'},'Location','southeast')


   

PROCEDURE:

• Basic Procedure to clear exisiting data in the workspace, command and temp datas cleared.
• Input data required are given
• Reference Taken from are given
• The governing Equation can be explained as,

(G/r)*T_max=((Rolling_Resistance_Coeff*Mass*9.81))+(0.625*Frontal_Area*Drag_Coeff*v^2)+((Mass+I*(G^2/(Gear_Eff*r^2))*(dv/dt)

Gear_Eff*(G/r)*T_max, In simpler terms is newtons law

F=M.a

which

T(Torque)=F(Force).r(Distance, in here radius)

Gear is used and so its efficiency and Ratio is included. Total tractive force,(i.e)

• Frr(Rolling Resistance),
• Fad(Aero Dynamic Drag),
• Fi(Inertia),
• Facc(Accerational Force)=Dv/Dt

In First Case of the program the original data's are used to measure WOT Value of

* General Motor's EV-1

And in second case, the weight reduction is being considered as the mass is reduced 5% from the data which is to be plotted in relationship of Velocity Vs Time, that explains the WOT, Maximum speed achieved in certain period of time.

RESULT:

The graph clearly depicts that reduction of mass helps in achieving maximum speed faster than the original data.