Week 2 : Calculation of vehicle and trailer Value in Matlab (DHEV)
Q1) Consider a vehicle and trailer having the following parameters:
Using the above values, calculate, the force in [N], torque in [Nm] and power in [kW] requirement at the wheel as a function of vehicle speed in [kmph] for the following scenarios:
a. No acceleration, flat ground, no headwind, no trailer
b. No acceleration, 7% grade, no headwind, with trailer
c. Acceleration of 0.2g, no grade, headwind of 5 kmph in the vehicle direction, no trailer
This will give an idea of power and torque requirements for a typical vehicle for different driving conditions.
Show sample calculations for 50 kmph, tabulate the force components, total force, along with total torque and power for 25, 50, 75, 100, 125 kmph, and plot the total power and torque requirements for all 3 scenarios as a function of speed (0 to 150 kmph). Use Matlab for calculations and plotting.
Comment on the results, comparing the differences and outlining certain observations.
Note: For calculating rolling radius, assume rolling radius to be 97% of the tire calculated radius, due to tire compression. For case with trailer, only consider the aerodynamic drag of the trailer, as it has much
larger projected area and higher drag coefficient.
Optional: Plot the components too, this will help in observing the trends
Your Answers
Given :
M = 2400 kg
C_d = 0.35
A_f = 2.4 m^2
rho_a = 1.225 kg/m^3
C_r = 0.013
Tire Size = 245/60R18
A_ft = 5 m^2
C_dt = 0.8
M_t = 1250 kg
V_veh = 50 kmph
To Find :
F = ? [N]
T = ? [Nm]
P = ? [kW ]
Condition :
a. No acceleration, flat Ground, no headwind, no trailer.
b. No acceleration, 7% grade, no headwind, with trailer
c. Acceleration og 0.2g, no grade, headwind of 5 kmph in vehicle direction, no trailer
Formulas :
Fr=Cr⋅M⋅g⋅cosθ">Fr=Cr⋅M⋅g⋅cosθFr=Cr⋅M⋅g⋅cosθ
Fw=12⋅ρa⋅Af⋅Cd⋅(Vveh+Vwind)2">Fw=12⋅ρa⋅Af⋅Cd⋅(Vveh+Vwind)2Fw=12⋅ρa⋅Af⋅Cd⋅(Vveh+Vwind)2
Fg=M⋅g⋅sinθ">Fg=M⋅g⋅sinθFg=M⋅g⋅sinθ
Ftmax=M⋅dVvehdt">Ftmax=M⋅dVvehdtFtmax=M⋅dVvehdt
Ptrac=Ftrac⋅Vveh">Ptrac=Ftrac⋅VvehPtrac=Ftrac⋅Vveh
Pgr=Cr⋅M⋅g⋅Vveh">Pgr=Cr⋅M⋅g⋅VvehPgr=Cr⋅M⋅g⋅Vveh
Pw=12⋅Pa⋅Af⋅Cd⋅(Vveh)3">Pw=12⋅Pa⋅Af⋅Cd⋅(Vveh)3Pw=12⋅Pa⋅Af⋅Cd⋅(Vveh)3
Ptmax=M⋅Vveh⋅(d⋅Vvehdt)">Ptmax=M⋅Vveh⋅(d⋅Vvehdt)Ptmax=M⋅Vveh⋅(d⋅Vvehdt)
Ptrac=Pgr+Pw+Ptmax">Ptrac=Pgr+Pw+PtmaxPtrac=Pgr+Pw+Ptmax
T=r⋅Ftrac⋅sinθ">T=r⋅Ftrac⋅sinθT=r⋅Ftrac⋅sinθ
Calculation :
a). no acceleration, flat ground, no headwind, no trailer
F_r = 0.013*2400*9.81*cos 0
= 306.1 N
F_w = 0.5*1.225*2.4*0.35*(13.89+0)^2
= 99.25 N
F_g = 2400 * 9.81* sin 0
= 0
Ft_max = 2400*(d(13.89))/dt
= 2400*0
= 0 N
F_trac = F_r + F_w + F_g + Ft_max
= 306.1 + 99.25 + 0 + 0
= 405.4 N
P_trac = F_trac * V_veh
= 405.4 * 13.89
= 5631 kW
T = 405.4 * 0.388 *sin 0
= 0 Nm
b). no acceleration, 7% grade , no heatwind, with trailer
F_r = 0.013*2400*9.81*cos 7
= 230.7 N
F_rt = 0.013*1250*9.81*cos7
= 120.2 N
F_w = 0.5*1.225*2.4*0.35*13.89^2
= 183.6 N
F_wt = 0.5*1.225*5*0.8*13.89^2
= 472.7 N
F_g = 2400*9.81*sin 7
= 1.547e+04 N
F_gt = 1250*9.81*sin 7
= 8056 N
Ft_max = 2400*0
= 0 N
Ft_max_t = 1250*0
= 0 N
F_trac_total = ((F_r + F_rt) + (F_w + F_wt) + (F_g + F_gt) + (Ft_max + Ft_max_t))
= ((230.7 + 120.2) + (183.6 + 472.7) + (1.547e+04 + 8056) + (0 + 0))
= 2.453e+04 N
P_trac = F_trac_total * V_vec
= 2.453e+04 * 13.89
= 3.407e+05 kW
T = 2.453e+04*0.388*sin 7
= 6253 Nm
C). Acceleration of 0.2g, no grade, headwind of 5 kmph in vehicle direction, no trailer
F_r = 0.013*2400*9.81*cos 0
= 306.1 N
F_w = 0.5*1.225*2.4*0.35*((13.89+7.0632)+5)^2
= 256.8 N
F_g = 2400 * 9.81* sin 0
= 0
Ft_max = 2400*(d(20.95))/dt
= 2400*0
= 0 N
F_trac = F_r + F_w + F_g + Ft_max
= 306.1 + 256.8 + 0 + 0
= 562.8 N
P_trac = F_trac * V_veh
= 7818 kW
T = 562.8*0.388*sin 0
= 0 Nm
