Week 2 - Assignment Solutions

Q 2.1 Compute Forces due to drag, rolling resistance and gradient for the following
vehicles assuming ρ = 1.2 (kg/m3) and θ = 8°. For the three vehicles given in the table
1, find Aerodynamic drag at velocity V1 and V2; also find rolling resistance at two
velocities.
Table 1

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

2-wheeler 200 0.9 0.6 0.015 30 80 0.28

3-wheeler 600 0.45 1.6 0.015 30 80 0.2

4-wheeler 1500 0.3 2.5 0.015 30 80 0.3

If,
AVi -> Aerodynamic drag force at velocity Vi
RVi -> Rolling resistance force at velocity Vi
GF -> Gradient force
For the three vehicles given in Table 1, find the missing values in Table 2 (Correct up
to 2 decimal places):
Table 2

Vehicle AV1 AV2 RV1 RV2 GF

(N) (N) (N) (N) (N)

2-wheeler 22.5 160 29.4 29.4 273

3-wheeler [AV13w] [AV23w] [RV13w] [RV23w] [GF3w]

4-wheeler [AV14w] [AV24w] [RV14w] [RV24w] [GF4w]

 Solution:

1
Aerodynamic drag = ∗ ρ * 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2
2
For a 3-Wheeler:

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

3-wheeler 600 0.45 1.6 0.015 30 80 0.2

At 30kmph = 8.33 mps; Aerodynamic drag = 30 N

At 80kmph = 22.22 mps; Aerodynamic drag = 213.33 N
Gradient force = 𝐹𝑔 = mg Sinɵ = 819.17 N

Rolling resistance = µ *m*g* Cosɵ = 87.43 N (independent of velocity)

For a 4-Wheeler:

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

4-wheeler 1500 0.3 2.5 0.015 30 80 0.3

At 30kmph = 8.33 mps; Aerodynamic drag = 31.25 N

At 80kmph = 22.22 mps; Aerodynamic drag = 222.22 N
Gradient force = 𝐹𝑔 = mg Sinɵ = 2047.93 N

Rolling resistance = µ *m*g* Cosɵ = 218.57 N (independent of velocity)

 Q 2.2 a) For the three vehicles in Assignment 2.1, compute total traction force
assuming pick-up from 0 to 50 kmph in 20 sec, with linear acceleration and zero
slope.

The total traction force for the 2-wheeler is given as A*vx + B

The total traction force for the 3-wheeler is C*vx + D

The total traction force for the 4-wheeler is E*vx + F

Compute the values of x, A, B, C, D, E, F

For the three vehicles also compute the power and torque at v = 30 km/h, 50 km/h and 80 km/h

Solution:
For a 2W

Acceleration a = dv/dt = 0.694 𝑚𝑠 −2

V = at = 0.694t

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

2-wheeler 200 0.9 0.6 0.015 30 80 0.28

1
Aerodynamic drag = 𝐹𝑑 = 2 ∗ 𝜌 ∗ 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.324𝑣 2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 29.4 N

Acceleration force = 𝐹𝑎 = m*a = 138.8 N

𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 0.324𝒗𝟐 + 168.2

At v = 30kmph = 8.33mps; t=v/a = 12 sec; 𝐹𝑡𝑟𝑎𝑐 = 190.66N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 = 𝟏. 𝟓𝟖𝟕𝒌𝑾 ,
Torque = r* 𝐹𝑡𝑟𝑎𝑐 =53.38Nm
At v = 50kmph = 13.89mps; t=v/a = 20 sec; 𝐹𝑡𝑟𝑎𝑐 = 230.6N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 = 𝟑. 𝟐𝟎𝟐𝒌𝑾 ,
Torque = r* 𝐹𝑡𝑟𝑎𝑐 = 𝟔𝟒. 𝟓𝟔Nm
At v = 80kmph = 22.22mps; t=v/a = 32 sec; 𝐹𝑡𝑟𝑎𝑐 = 328.17N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 =
𝟕. 𝟐𝟗𝟏 𝒌𝑾 , Torque = r* 𝐹𝑡𝑟𝑎𝑐 = 𝟗𝟏. 𝟖8Nm

For a 3W

Acceleration a = dv/dt = 0.694 𝑚𝑠 −2

V = at = 0.694t
 Vehicle GVW CD Area μ V1 V2 Tyre Radius
(kg) (m2) (km/h) (km/h) (m)

3-wheeler 600 0.45 1.6 0.015 30 80 0.2

1
Aerodynamic drag = 𝐹𝑑 = 2 ∗ 𝜌 ∗ 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.432𝑣 2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 87.318 N

Acceleration force = 𝐹𝑎 = m*a = 416.4 N

𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 0.𝟒𝟑𝟐𝒗𝟐 + 503.718

At v = 30kmph = 8.33mps; t=v/a = 12 sec; 𝐹𝑡𝑟𝑎𝑐 = 533.69N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 = 𝟒. 𝟒𝟒𝟓𝒌𝑾 ,
Torque = r* 𝐹𝑡𝑟𝑎𝑐 = 𝟏𝟎𝟔. 𝟕𝟒Nm
At v = 50kmph = 13.89mps; t=v/a = 20 sec; 𝐹𝑡𝑟𝑎𝑐 = 587.06N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 = 𝟖. 𝟏𝟓𝟒𝒌𝑾 ,
Torque = r* 𝐹𝑡𝑟𝑎𝑐 =117.41Nm
At v = 80kmph = 22.22mps; t=v/a = 32 sec; 𝐹𝑡𝑟𝑎𝑐 = 717.01N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 =
𝟏𝟓. 𝟗𝟑𝟏𝒌𝑾 , Torque = r* 𝐹𝑡𝑟𝑎𝑐 =143.40Nm

For a 4W

Acceleration a = dv/dt = 0.694 𝑚𝑠 −2

V = at = 0.694t

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

4-wheeler 1500 0.3 2.5 0.015 30 80 0.3

1
Aerodynamic drag = 𝐹𝑑 = ∗ 𝜌 ∗ 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.45𝑣 2
2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 218.29 N

Acceleration force = 𝐹𝑎 = m*a = 1041 N

𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 𝟎. 𝟒𝟓𝒗𝟐 +1259.29

At v = 30kmph = 8.33mps; t=v/a = 12 sec; 𝐹𝑡𝑟𝑎𝑐 = 1290.51N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 = 𝟏𝟎. 𝟕𝟓𝒌𝑾 ,
Torque = r* 𝐹𝑡𝑟𝑎𝑐 =387.15Nm
At v = 50kmph = 13.89mps; t=v/a = 20 sec; 𝐹𝑡𝑟𝑎𝑐 = 1346.11N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 =
𝟏𝟖. 𝟔𝟗𝟕𝒌𝑾 , Torque = r* 𝐹𝑡𝑟𝑎𝑐 = 𝟒𝟎𝟑. 𝟖𝟑Nm
 At v = 80kmph = 22.22mps; t=v/a = 32 sec; 𝐹𝑡𝑟𝑎𝑐 = 1481.47N, Power = 𝐹𝑡𝑟𝑎𝑐 ∗ 𝑣 =
𝟑𝟐. 𝟗𝟏𝟖𝒌𝑾 , Torque = r* 𝐹𝑡𝑟𝑎𝑐 =444.44Nm

2.2 b) Assume vehicle acceleration is some value for first ten seconds, and half as much
for next ten seconds to still reach 50 kmph in 20 seconds. Now again compute traction
force, Power and Torque at 50 kmph.

Vehicle Traction Force Power Toque

(N) (kW) (Nm)

2-wheeler A D G

3-wheeler B E H

4-wheeler C F I

Solution.
Let a be the acceleration for first 10 sec, v1 the velocity at end of 10sec and a/2 the acceleration
for next 10 sec, v2 the velocity at end of 20sec.
v1 = 10a and v2 = v1 + (a/2)*10 = v1 + 5a = 15a ; a=v2/15

Solving we get a=0.926 𝑚𝑠 −2

1
𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = ∗ ρ * 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 + µ *m*g + m*a
2
For a 2W:

Acceleration = a/2 = 0.463 𝑚𝑠 −2

V = 13.89mps

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

2-wheeler 200 0.9 0.6 0.015 30 80 0.28

1
Aerodynamic drag = 𝐹𝑑 = ∗ 𝜌 ∗ 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.324𝑣 2 = 62.51 N
2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 29.4 N

Acceleration force = 𝐹𝑎 = m*a = 92.6 N
𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 184.51 N
Power = 2.562 kW
Torque = 51.66 Nm
For a 3W:

Acceleration = a/2 = 0.463 𝑚𝑠 −2

V = 13.89mps

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

3-wheeler 600 0.45 1.6 0.015 30 80 0.2

1
Aerodynamic drag = 𝐹𝑑 = ∗ ρ * 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.432𝑣 2 = 83.34 N
2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 88.29 N

Acceleration force = 𝐹𝑎 = m*a = 277.8 N
𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 450.06 N
Power = 6.251 kW
Torque = 90.012 Nm

For a 4W:

Acceleration = a/2 = 0.463 𝑚𝑠 −2

V = 13.89mps

Vehicle GVW CD Area μ V1 V2 Tyre Radius

(kg) (m2) (km/h) (km/h) (m)

4-wheeler 1500 0.3 2.5 0.015 30 80 0.3

1
Aerodynamic drag = 𝐹𝑑 = ∗ ρ * 𝐶𝐷 ∗ 𝐴 ∗ 𝑣 2 = 0.45𝑣 2 = 86.82 N
2

Rolling resistance = 𝐹𝑟𝑟 =µ *m*g = 220.725 N

Acceleration force = 𝐹𝑎 = m*a = 694.5 N
𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑑 + 𝐹𝑟𝑟 + 𝐹𝑎 = 1002.045 N
Power = 13.918 kW
Torque = 300.6 Nm
2.2 c) To reach maximum speed vf in T seconds, if a vehicle accelerates at a rate “a” for
first T/2 time and at a rate “a/2” from T/2 to T. It therefore gives (vf-v0) = a*T/2 +
(a/2)*T/2 = (3/4)*aT and a = (vf-v0)* (4/3T).
(Consider the vehicle starts from rest while answering below questions)

If the average power required

During acceleration is expressed as: K1* mx1 * vfy1 * Tz1
At the end of time T is expressed as: K2* mx2 * vfy2 * Tz2
At the end of time T, assuming linear acceleration, is expressed as: K3* mx3 * vfy3 *
Tz3
Compute the values of K1, K2, K3, x1, x2, x3, y1, y2, y3, z1, z2 and z3
Ans.
Average Power required during the acceleration is (m * vf2 )/2T = 0.5 * m * vf2 * T-1
(i.e K1= 0.5, x1 = 1, y1= 2, z1 = -1)
Average power required at end time T is 2*(m * vf2 )/3T = 0.67 * m * vf2 * T-1
(i.e K2= 0.6 to 0.7, x2 = 1, y2= 2, z2 = -1)
Average power required at end time T, assuming linear acceleration, is: m * vf2/T = 1 * m * vf2 *
T-1 (i.e K3= 1, x3 = 1, y3= 2, z3 = -1)
Solution:
To reach maximum speed vf in T seconds, if a vehicle accelerates at a rate “a” for first T/2 time
and at a rate “a/2” from T/2 to T. It therefore gives (vf-v0) = a*T/2 + (a/2)*T/2 = (3/4)*aT and
a = (vf-v0)* (4/3T). Find the average power during acceleration and the power required at the
end of time T. Compare it if there was linear acceleration.
o s1 = 0.5*a*(T/2)^2 = aT^2/8
o v1= v0+ a*T/2
o w1= m*a*s1 = m*a*a*T^2/8
o s2 = v1*T/2+ 0.5*(a/2)*(T/2)^2 = (v0+aT/2)*T/2+ (a/16)*T^2 =
(T/2)*(v0+5aT/8]
o w2 = m*a/2*s2 = m*aT/4*[v0+5aT/8]
o Total work w=w1+w2
= maT/4*[aT/2+v0+5aT/8] = m*a*T/4*[v0+a*T*9/8]
o Average Power during acceleration= w/T
o If v0 is zero, average power
= ma/4*[v0+9aT/8]
o =(m*vf/3T)*9vf/6 = m*vf2 /2*T
▪ Same as that for linear velocity
For linear acceleration, Power required at end is m*vf^2/T
◦ In case of this acceleration, Power required at end is m*(2/3)*vf^2/T
Q 2.3. a. For a 2-wheeler, e-rickshaw and sedan with specifications as given below,
compute total traction force, Power and Torque required at 30 kmph and 80 kmph.
Consider the pickup time to attain 30kmph and 80kmph to be 20 seconds. What
would be the power and torque required if the 4-wheeler sedan goes to 160 kmph.
Assume slope to be zero.
Compute the average acceleration power Pa as (Acceleration force * v)/2 and for
computation of climbing power use Pg = Climbing Force*(v/3)

Vehicle  (kg/m3) CD A  weight Tyre

(m2) (kg) radius
(m)
2-wheeler 1.2 0.9 0.5 0.013 180 0.28
e-rickshaw 1.2 0.44 1.6 0.013 680 0.2
Sedan 1.2 0.35 2.5 0.013 1200 0.31

Solution is Excel Sheet

b. Assume the sedan is stuck on a climb 12° slope. It needs to start and have a
acceleration of 0.5 m/sec2. What is the starting Torque required?
Solution:
Sedan

 = 1.2 kg/m3, CD = 0.35, A= 2.5 sqm,  = 0.013, weight = 1200 Kg, r=0.31m, Gradient of 12°,
Acceleration = 0.5 𝑚𝑠 −2
Rolling resistance = 𝐹𝑟𝑟 =µ *m*g* cos ɵ = 149.69 N
Acceleration force = 𝐹𝑎 = ma = 600 N
Gradient force = 𝐹𝑔 = mg Sinɵ = 2447.5 N

𝐹𝑡𝑟𝑎𝑐 = 𝐹𝑟𝑟 + 𝐹𝑎 + 𝐹𝑔 = 3194.19 𝑁

Starting Torque required = r* 𝐹𝑡𝑟𝑎𝑐 = 990.198 Nm

 Q 2.4. a. A sedan, with specs as given below, accelerates from 0 to 50 kmph in 20
seconds. It then travels at constant speed of 50 kmph for five minutes. It then
decelerates to 0 kmph in 20 seconds. Compute the energy used, assuming R=1 and R
=0.3 and the distance travelled. What is Wh/km? Consider the gradient as 0°.

Vehicle  (kg/m3) CD A  weight Tyre

(m2) (kg) radius
(m)
Sedan 1.2 0.35 2.5 0.013 1200 0.31

Solution:
Method 1: Using excel Sheets (Attached)

Method 2:

 = 1.2 kg/m3, CD = 0.35, A= 2.5 sqm,  = 0.013, weight = 1200 Kg, Gradient of 0°

For first 20 sec

V = (13.89/20)t = 0.694 t

Fd = 0.525*𝑣 2 = 0.253 𝑡 2 N
Frr = 153.036 N
Fa = ma = 832.8 N

Ftrac = Fd + Frr + Fa = 0.253 𝑡 2 + 984.84

Ptrac = Ftrac*v = 0.1756 𝑡 3 + 683.47 t

Energy consumed(Wh) = Ptrac*(dt/3600) = (0.1756 𝑡 3 + 683.47 t)/3600 --
20 0.1756 𝑡 3 + 683.47 t
Total energy consumed in 20 sec = ∫0 3600
dt = 39.96 Wh

Distance s =0+0.5*a* 𝑡 2 = 0.5*0.694*20*20 = 138.8 m

For next 5 min

v = 13.89 𝑚𝑠 −1

Fd = 0.525*𝑣 2 = 101.29 N
Frr = 153.036 N
Fa = ma = 0 N
Ftrac = Fd + Frr + Fa = 254.32 N
Ptrac = Ftrac*v = 3532.58 N
Energy consumed(Wh) = Ptrac*(300/3600) = 294.38 Wh
Distance s =13.89*300+0 = 4167 m

For last 20 sec

The aero drag and rolling resistance remain same as first 20 sec

Fd = 0.525*𝑣 2 = 0.525 (13.89 − 0.694𝑡)2 N

Frr = 153.036 N
Fa = ma = - 832.8 N
Ftrac = Fd + Frr + Fa = 0.252 𝑡 2 -10.11 t +674.6

Ptrac = R*Ftrac*v = −0.175 𝑡 3 + 10.516 𝑡 2 + 261.18 𝑡1 − 8038

Regen Energy (Wh) = Ptrac*(dt/3600)
20 (−0.175 𝑡 3 + 10.516 𝑡 2 + 261.18 𝑡 1 −8038)
Regen energy in 20 sec = R* ∫0 3600
dt = R * 24.3 Wh

At R = 1, Regenerative Energy = 24.3 Wh ; At R = 0.3, Regenerative Energy = 7.29 Wh;

Distance s =0+0.5*a* 𝑡 2 = 0.5*0.694*20*20 = 138.8 m

For entire trip

Total distance travelled = 4.444 km
At R = 1 :
Energy Consumed = 39.96+294.38-24.3 = 310.04 Wh
Wh/km = 69.76 Wh/km
At R = 0.3 :
Energy Consumed = 39.96+294.38-7.29 = 327.05 Wh
Wh/km = 73.59 Wh/km
b. The sedan now goes from 0 to 25 kmph in 15 seconds, travel at 25 kmph for 2 minutes,
speeds up to 50 kmph in another 15 seconds, travel for 4 minutes at 50 kmph and then
decelerates to 0 kmph in 20 seconds. C Compute the energy used (Wh), assuming R=1 and
R =0.3 and the distance travelled (km). What is Wh/km?
Solution:
Method 1: Using excel Sheets (Attached)

Method 2:

 = 1.2 kg/m3, CD = 0.35, A= 2.5 sqm,  = 0.013, weight = 1200 Kg, Gradient of 0°

0 to 25 kmph in 15 seconds
V = (6.94/15)t = 0.463 t

Fd = 0.525*𝑣 2 = 0.112 𝑡 2 N
Frr = 153.036 N
Fa = ma = 555.56 N

Ftrac = Fd + Frr + Fa = 0.112𝑡 2 + 708.59

Ptrac = Ftrac*v = 0.052𝑡 3 + 328.08 t

Energy consumed (Wh) = Ptrac*(dt/3600) = (0.052𝑡 3 + 328.08 t)dt/3600 --

15 0.052 𝑡 3 + 328.08 t
Total energy consumed in 15 sec = E1 = ∫0 3600
dt = 37566/3600 =10.43 Wh

Distance s1 =0+0.5*a* 𝑡 2 = 0.5*0.463*15*15 = 52.08 m

at 25 kmph for 2 minutes

v = 6.94 𝑚𝑠 −1

Fd = 0.525*𝑣 2 = 25.32 N
Frr = 153.036 N
Fa = ma = 0 N
Ftrac = Fd + Frr + Fa = 178.35 N
Ptrac = Ftrac*v = 1237.8 W
Energy consumed(Wh) = E2 = Ptrac*(120/3600) = 41.26 Wh
Distance = s2 =6.94*120+0 = 833 m

25 to 50 kmph in 15 seconds
a=((25/3.6)/15) = 0.463; v = 6.94 + 0.463 t

Fd = 0.525*𝑣 2 = 0.525 * (48.16 + 0.214 𝑡 2 + 6.426t) N = 25.28 + 0.112 𝑡 2 + 3.37t

Frr = 153.036 N
Fa = ma = 555.56 N

Ftrac = Fd + Frr + Fa = 0.112 𝑡 2 + 3.37t + 733.9

Ptrac = Ftrac*v = 0.052 𝑡 3 + 2.337 𝑡 2 + 363.18 t + 5093.2

Energy consumed(Wh) = E3 = Ptrac*(dt/3600)

= (0.052 𝑡 3 + 2.337 𝑡 2 + 363.18 t + 5093.2)dt/3600

15 0.052 𝑡 3 + 2.337 𝑡 2 + 363.18 t + 5093.2
Total energy consumed in 15 sec = ∫0 dt
3600

= 120543/3600 = 33.48 Wh

Distance s3 = (𝑣 2 - 𝑢2 ) / (2*a)) = 156 m

at 50 kmph for 4 minutes

v = 13.89 𝑚𝑠 −1

Fd = 0.525*𝑣 2 = 101.27 N
Frr = 153.036 N
Fa = ma = 0 N
Ftrac = Fd + Frr + Fa = 254.31 N
Ptrac = Ftrac*v = 3532.4 W
Energy consumed(Wh) = E4 = Ptrac*(240/3600) = 235.49 Wh
Distance s4 =13.89*240+0 = 3333 m

Decelerates 50 to 0 kmph in 20 seconds

Fd = 0.525*𝑣 2 = 0.525 (13.89 − 0.694𝑡)2 N

Frr = 153.036 N
Fa = ma = - 832.8 N

Ftrac = Fd + Frr + Fa = 0.252 𝑡 2 -10.11 t +674.6

Ptrac = R*Ftrac*v = −0.175 𝑡 3 + 10.516 𝑡 2 + 261.18 𝑡1 − 8038

Regen Energy (Wh) = Ptrac*(dt/3600)
20 (−0.175 𝑡 3 + 10.516 𝑡 2 + 261.18 𝑡 1 −8038)
Regen energy in 20 sec = R* ∫0 3600
dt = R * 24.3 Wh

At R = 1, Regenerative Energy =E5= 24.3 Wh

At R = 0.3, Regenerative Energy = E5 = 7.29 Wh;
Distance s5 = 139 m

Entire trip
Total distance travelled = s1+s2+s3+s4+s5 = 4.513 km
At R = 1:
Total Energy consumed= E = E1+E2+E3+E4-E5 = 296.36 Wh
Wh/km = 65.66
At R = 0.3:
Total Energy consumed = E = E1+E2+E3+E4-E5 = 313.37 Wh
Wh/km = 69.43