Introduction to Hybrid Electric Vehicle using MATLAB and Simulink : Forces Acting on a Vehicle

Problem Statement 1:

1. What should be the maximum speed of the motor used in an electric scooter capable to run at 90 kmph, if the fixed gear ratio is 7 and tire size is 90/100 R10 53J? Assume the following arrangement from the motor to the wheel.

Solution:

Given,

Max. Speed: 90 km/h

1st convert it into m/s

1. Max. Speed (in m/s): 90⋅518ms90⋅518ms = 25 m/s

Gear ratio= 7

Tire size = 90/100 R10 53J

The tire nomenclature image is shown below:

So as per the given data:

90/100 R10 53J

90 - width of the tire in mm

100- % of the tire width which gives the tire height

to find tire height= 185100⋅100185100⋅100 = 185mm (height of the tire)

R- Radial Construction (R means radial tier)

10- rim diameter in inches = 10*25.4mm = 254mm

53- Load Index of the tire = 206 kg load capacity

J- Speed rating = 100kmph

the maximum speed of the motor used in an electric scooter:

Maximum Speed = Angular Speed generated by the Motor

Angular speed (w) = G⋅VrG⋅Vr

Where,

G = Gear Ratio

V= Linear Velocity (m/s)

r= radius of the tire (m)

G, V is given so we have to find r.

Diameter = Tire height + Rim Diameter

= 185mm + 254mm

= 439mm

radius = 439/2 mm

= 219.5mm

= 0.2195m

Angular speed (w) = G⋅VrG⋅Vr

= 7⋅250.21957⋅250.2195

= 797.26 rad/sec

= 797.26 * 9.5493 RPM

= 7613 RPM

b. Recalculate the same value if the tire code is 90/90 R18 51P.

So as per the given data:

90/90 R18 51P

90 - width of the tire in mm

90 - % of the tire width which gives the tire height

to find tire height= 185100⋅90185100⋅90 = 166.5mm (height of the tire)

R- Radial Construction (R means radial tier)

18- rim diameter in inches = 18*25.4mm = 457.2mm

51- Load Index of the tire = 195 kg load capacity

P- Speed rating = 160kmph

the maximum speed of the motor used in an electric scooter:

Maximum Speed = Angular Speed generated by the Motor

Angular speed (w) = G⋅VrG⋅Vr

Where,

G = Gear Ratio

V= Linear Velocity (m/s)

r= radius of the tire (m)

G, V is given so we have to find r.

Diameter = Tire height + Rim Diameter

= 166.5mm + 457.2mm

= 623.7 mm

radius = 623.7/2 mm

= 311.85mm

= 0.31185m

Angular speed (w) = G⋅VrG⋅Vr

= 7⋅250.311857⋅250.31185

= 561.16 rad/sec

= 561.16 * 9.5493 RPM

= 5358 RPM

PS2. Prepare a simple excel calculator to identify vehicle propulsion power based on given inputs and outputs. Implement formulas in cells.

Inputs:

1. Kerb Weight (kg),
2. Payload (kg),
3. Coefficient of rolling resistance,
4. Air density (kg/m3),
5. Width (m),
6. Height (m),
7. Drag coefficient,
8. Acceleration (0 to top speed in specified seconds),
9. Hill climbing angle,
10. Speed

Output: Total power in kW

What are the limitations of this calculation?

Solution:

The formula used for calculations:

1) Rolling Resistance Force (FrrFrr):

FrrFrr = UrrUrr * m * g

Where,

UrrUrr = coefficient of rolling resistance

UrrUrr = 0.01( 1 + (0.001v))

m = mass

g = gravitational force

2) Air drag force (Fad):

Fad=12⋅p⋅a⋅Cd⋅v2Fad=12⋅p⋅a⋅Cd⋅v2

Where,

p = air density

a = frontal area

CdCd=drag coefficient

v = velocity

3) Hill Climbing Force (FhcFhc):

Fhc=m⋅g⋅sinΘFhc=m⋅g⋅sinΘ

where,

m= total mass of the vehicle

g= gravitational force

Θ= slop angle (in radian)

4) Total Power(P):

P=(Frr+Fad).vP=(Frr+Fad).v

Assumption:

Acceleration =0 (Moving with the constant velocity)

so,

Fac= 0.

a simple excel calculator to identify vehicle propulsion power based on given inputs and outputs <-- Click here

What are the limitations of this calculation?

1. The limitation was the decimal point and approximation.
2. This calculation does not consider all forces acting on the vehicle
3. decimal errors are there
4. SIN(Numbers) is taking value only in radian
5. we are taking acceleration is 0, so no acceleration forces are acting on the vehicle.

PS3. Assuming that tire pressure reduction by 15% results in an increase of rolling resistance effect by two times, how much will be the difference in total traction power keeping all other parameters the same?

Solution:

as given Urr is increased two times by a decrease in 15% of tire pressure.

tire pressure reduction by 15% sheet <-- Click here

the difference in total traction power keeping all other parameters the same

Vehicle 1 = 46.7824 - 44.5441

= 2.23kW

Vehicle 2 = 74.6038 - 70.837

= 3.7668kW

PS4.See the video of the Range rover sport dragon challenge.

1. How much is the gradeability?
2. What was the average speed?
3. Which are the forces acting on the car?

Link to the video: https://www.youtube.com/watch?v=EUwzWHD3Htg

Solution:

1. How much is the gradeability?

Below is the scale to measure the gradeablility.

as mentioned in the video angle is 45 degrees.

so gradeablility=100%

1. What was the average speed?

https://www.landroverusa.com/experiences/stories/dragon-challenge.html

total time taken is 22min 41sec

total distance: 11.3km

avg. speed = Total Distance / Total time taken

= 11.3/0.3735

= 30.254 km/h

3. Which are the forces acting on the car?

1. Rolling Resistance Force
2. Aerodynamic Drag Force
3. Hill Climbing Force
4. Acceleration Force

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