Electric vehicle
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| 193 Figure 8.3 Electric scooter of the type simulated at various points in this chapter. The photograph was taken in a Berlin car park 8.3.2 Modelling the Acceleration of an Electric Scooter For our first example we will take an electric scooter. No particular model is being considered, but the vehicle is similar to the electric scooters made by Peugeot and EVS, an example of which is shown in Figure 8.3. • The electric scooter has amass of 115 kg, with atypical passenger of mass 70 kg, so total mass m = 185 kg The moment of inertia of the motor is not known, so we will adopt the expedient suggested at the end of Section 8.2.5, and increase m by 5% in the linear acceleration term only. A value of 194 kg will thus be used from m in the final term of Equation The drag coefficient C d is estimated as 0.75, a reasonable value fora small scooter, with a fairly ‘sit-up’ riding style The frontal area of vehicle and rider ism The tyres and wheel bearings give a coefficient of rolling resistance, μ rr = 0.007. • The motor is connected to the rear wheel using a 2:1 ratio belt system, and the wheel diameter is 42 cm. Thus G = 2 and r = 0.21 m The motor is an 18 V ‘Lynch’-type motor, of the type discussed in the previous chapter, Section 7.1.2. Equation (7.8) has been recalculated for 18 V, giving T = 153 − 1.16 ω (8.14) • As in Section 7.1.2, the maximum current is controlled by the maximum safe current, in this case 250 A, so, as shown in Equation (7.9), the maximum torque T max is 34 N m. 194 Electric Vehicle Technology Explained, Second Edition The critical motor speed, ω c , after which the torque falls according to Equation (occurs when 153 − 1.16 ω ∴ ω = 153 − 34 1 .16 = 103 rads The gear system is very simple, and of low ratio, so we can assume a good efficiency. A value of η g of 0.98 is estimated. An effect of this will be to reduce the torque, and so this factor will be applied to the torque. When the torque is constant, as in Equation (8.13), we have 0 .21 × 0.98 × 34 = 0.007 × 185 × 9.8 + 0.625 × 0.6 × 0.75v 2 + 194 dv dt 317 = 12.7 + 0.281v 2 + 194 dv dt thus 194 dv dt = 304 − 0.281v 2 and so dv dt = 1.57 − 0.00145v 2 (8.15) This equation holds until the torque begins to fall when ω = ω c = 103 rads, which corresponds to 103 × 0.21/2 = 10.8 ms. After this point the torque is governed by Equation (8.14). If we substitute this, and the other constants, into Equation (8.12) we obtain 0 .21 × 0.98 × 153 − 1.16 2 0 .21 v = 0.007 × 185 × 9.8 + 0.625 × 0.6 × 0.75v 2 + 194 dv dt 1428 − 103v = 12.7 + 0.281v 2 + 194 dv dt and so dv dt = 7.30 − 0.53v − 0.00145v 2 (8.16) There are many practical and simple ways of solving these differential equations. Many modern calculators will solve such equations, remembering that there is a simple initial condition that v = 0 when t = 0. However, the most versatile next step is to derive a simple numerical solution, which can then easily be used in Excel or MATLAB®. The derivative of v is simply the difference between consecutive values of v divided by the time step. Applying this to Equation (8.15) gives us v n +1 − v n δt = 1.57 − 0.00145v 2 n For a program such as Excel or MATLAB® we need to rearrange this equation to obtain the value of the next velocity from the current velocity. This is done as follows: v n +1 = v v + δt × 1 .57 − 0.00145v 2 n (8.17) Electric Vehicle Modelling 195 This equation holds for velocities up to the critical velocity of 10.8 ms, after which we have to use Equation (8.16), approximated inexactly the same way as we have just done for Equation (8.15), which gives v n +1 = v v + δt × 7 .30 − 0.53v − 0.00145v 2 n (8.18) The MATLAB® script file below shows how to solve these equations using this program. Figure 8.4 is a plot of the solution using a time step δt of 0.1 seconds. Exactly the same result can be obtained with almost equal ease using Excel. It is left as an exercise for the reader to produce an Excel spreadsheet or MATLAB® script file where many of the machine parameters, such as the gear ratio G, are left as easily altered variables, so that the effect of changing it on the vehicle’s performance can be noted ScootA - electric scooter acceleration. t=linspace(0,50,501); % 0 to 50 sins steps vel=zeros(1,501); % 501 readings of velocity d=zeros(1,501);% Array for storing distance travelled dT=0.1; % 0.1 second time step for n Now follow equations (8.17) & (if vel(n)<10.8 % Torque constant till this point vel(n+1)= vel(n) + dT*(1.57 - (0.00145*(vel(n)^2))); elseif vel(n)>=10.8 vel(n+1)=vel(n)+dT*(7.30-(0.53*vel(n))-(0.00145*(vel(n)^2))); end; d(n+1)=d(n) + 0.1*vel(n); % Compute distance travelled. end; vel=vel.*3.6; % Multiply by 3.6 to convert ms to kph plot(t,vel); axis 30 0 50]); xlabel('Time / seconds'); ylabel('Velocity / kph'); title('Full power (WOT) acceleration of electric scooter'); The result of this simulation is shown in Figure 8.4, and shows that the performance is somewhat as might be expected from a fairly low-power motor. The acceleration is unspectacular, and the top speed is about 30 mph or 48 kph, on level ground. However, this is reasonably compatible with safe city riding. The acceleration of such vehicles is sometimes given in terms of the standing start 100 m times, and the power of such MATLAB® script files is that they can very easily be changed to produce such information. If the plot line in the file above is changed as follows, then Figure 8.5 is obtained. plot(t,d); axis 15 0 While we are not claiming that our model exactly represents any particular commercial designs, it is worth noting the following points from the specification of the Peugeot Scoot’Elec performance specification: 196 Electric Vehicle Technology Explained, Second Edition 4 6 8 30 mph 10 Time / s Full power (WOT) acceleration of electric scooter V elocity/kph 12 14 16 18 20 0 5 10 15 20 25 30 35 40 45 50 0 Download 3.49 Mb. Share with your friends:
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