Assignment - Week 2
Question - 1
Let us take TATA Ultra 1518 under consideration, as per the name and the variant details it depicts that the truck has the carriage capacity of 15 tonnes and has the power of 180 ps.
The Truck Specifications are mentioned in below,
We need to calculate the Hill Climibing Power for 2 different weight loaded conditions.
Gross Vehicle Weight - 15710 kg
Kerb Vehicle Weight - 5860 kg
Loading Capacity - 9850 kg
Width - 2330 mm
Height - 575 mm
Frontal Area - 1.33 m2
Coefficient of Rolling Resistance - It may range between (0.0044 to 0.0072)
Considering 0.0058 for this example.
Air Density (kg/m3) - 1.25
Drag Coefficient - 0.8
Speed - 40 kmph
Angles (Degree) - 10
Speed During Hill Climbing - 25 kmph
Situation 1 -
Kerb Weight - 5860 kg
Pay Load (Full) - 9850 kg
Situation 2 -
Kerb Weight - 5860 kg
Pay Load (Half) - 4925 kg
Power Required for Full Loaded Truck = 252.37 ps
Power Required for Half Loaded Truck = 173.25 ps
Ratio = 3:2
Question - 2
Considering the Heavy Capacity Carriage Carriers in Truck the minimum and the best graded angle should be 10 Degrees. As per the overall Degree Rating that all Fly Over are built are between 10 - 20 Degrees depending upon the area of usage.
Minimum angle/gradeability of the bridge - 10°
Minimum Length of the Plain Road - 16.5 m
Minimum Length of the Fly Over - 20.9 m
Carrying the load for constant duration longer than usual where the Average Required Power would be higher. If the slope is steeper then the vehicle has to be configured with higher engine capacity or either there are chances that the vehicle is unable to reach the top and back decrest in ease. Suggested Inclined Angle should not be more than that of 10 - 15 Degrees.
Question - 3
The Drive Cycle Formed is mentioned as below,
Where the Blue and the Green Curves depicts the Driving Cycle at the Flat Surface and the Inclined Angle.
The step wise procedure is attached as a File.
Question - 4.1
Actual Specifications of the General Motors EV1
Power - 102 kW
Torque - 149 Nm
Cd - 0.19
μrr - 0.0048
Mass - 1470 kg (Inclusive of a Driver 70kg)
G - 11
r - 0.30 m
Area - 1.8 m2
ηg - 0.95 (Torque would be reduced to the factor of 0.95)
For the motor Tmax = 140 Nm and wc = 733 rad sec-1
This means T = Tmax till the v = 19.8 m sec-1
Above 19.8 m sec-1 the motor operates at a constant 102 kW so T - 102000/37*v = 2756/v
Substituting the values in the formula below,
Equation 1
G/r*T = μrr*m*g+0.625*A*Cd+(m+I G2/ngr2) dv/dt
(As we do not have the value of motor rotor inertia we would consider the value to be 5% more as per the angular velocity)
0.95*11/0.30*140 = 0.0048*1400*11+(0.625*1.8*0.19)v2+1470 dv/dt
dv/dt = 3.266 - 0.000145 v2 Equation (1)
Equation 2
Once the speed has reached 19.8ms−1 the velocity would be as given under
G/r*T = μrr*m*g+0.625*A*Cd+(m+I G2/ngr2) dv/dt
(As we do not have the value of motor rotor inertia we would consider the value to be 5% more as per the angular velocity)
0.95*11/0.30*2756/v = 0.0048*1400*11+(0.625*1.8*0.19)v2+1470 dv/dt
dv/dt = 65.30 / v - 0.0502 - 0.00015 v2 Equation (2)
Finite Difference Form
v n+1 = vn + dt (3.266 - 0.000145 v2n)
v n+1 = vn + dt (65.30/vn - 0.0502 - 0.000145 v2n)
Question 4.2
Changing the vehicle parameters by adding 5%
Power - 107 kW
Torque - 156 Nm
Cd - 0.20
μrr - 0.0050
Mass - 1545 kg (Inclusive of a Driver 73.5 kg)
G - 12
r - 0.31 m
Area - 1.9 m2
ηg - 0.99 (Torque would be reduced to the factor of 0.99)
For the motor Tmax = 147 Nm and wc = 769 rad sec-1
This means T = Tmax till the v = 20.79 m sec-1
Above 20.79 m sec-1 the motor operates at a constant 107 kW so T - 107000/39*v = 2744/v
Substituting the values in the formula below,
Equation 1
G/r*T = μrr*m*g+0.625*A*Cd+(m+I G2/ngr2) dv/dt
(As we do not have the value of motor rotor inertia we would consider the value to be 5% more as per the angular velocity)
0.99*12/0.31*147 = 0.0050*1545*12+(0.625*1.9*0.20)v2+1545 dv/dt
dv/dt = 3.585 - 0.00015 v2 Equation (1)
Equation 2
Once the speed has reached 19.8ms−1 the velocity would be as given under
G/r*T = μrr*m*g+0.625*A*Cd+(m+I G2/ngr2) dv/dt
(As we do not have the value of motor rotor inertia we would consider the value to be 5% more as per the angular velocity)
0.99*12/0.31*2744/v = 0.0050*1545*12+(0.625*1.9*0.20)v2+1545 dv/dt
dv/dt = 68 / v - 0.060 - 0.00015 v2 Equation (2)
Finite Difference Form
v n+1 = vn + dt (3.585 + 0.00015 v2n)
v n+1 = vn + dt (68/vn - 0.060 - 0.00015 v2n)
Plotting the Graph
The Excel File for the same is attached.