I looked at the effect of acceleration and deceleration rates on energy used per acceleration-deceleration cycle a while back. To simplify the calculation I assumed the same rate for acceleration and deceleration. You do get more energy into the pack stopping faster with regen. For example, at a faster deceleration rate, 6 mph/sec, 7.7% of the vehicle K.E. is lost to work against drag and rolling resistance forces, and at a slower deceleration rate, 2 mph/sec, 23.1% is lost. If you assume a combined motor/controller loss of 20% and drive train loss of 10%, then for the faster deceleration rate 100 – (20 + 10 + 7.7) = 62% of vehicle kinetic energy goes into the battery pack, and at the slower deceleration rate 100 – (20 + 10 + 23.1) = 47% goes into the battery pack.
However, you don't get a lot of difference in energy into the pack as a percentage of total energy used when considering a trip with travel at constant speed and a number of acceleration-deceleration cycles. For Example: (1) My car accelerates at 6 mph/sec to 60 mph, drives 10 miles, then decelerates at the same rate to a stop, estimated percent of total energy used that is regained with regen is: 2.7% (2) Acceleration at 6 mph/sec to 35 mph and deceleration at same rate 10 times in 10 miles: 12.7% (3) Acceleration at 6 mph/sec to 35 mph and deceleration at same rate 20 times in 10 miles: 20.7% At 3 mph/sec, or 1.34 m/s (0 to 60 mph in 20 sec). Then the same three scenarios give 2.4%, 12%, and 20%, so a factor of 2 slower rate doesn’t change the result that much. At 2 mph/sec acceleration/deceleration rate the three scenarios give 2.2%, 11%, and 19%. Increasing total miles traveled, d, to 30 in scenario (1) gives 0.9%. Increasing stops in this scenario to 3, with 30 miles total, gives 2.7%. Increasing vehicle mass increases the percentage of energy recovered, but it’s a small effect for larger number of stop/starts. For example the first scenario goes from 2.7% to 3.8%, second goes from 12.7% to 14.7%, third goes from 20.7% to 23.2% if vehicle mass is doubled. Decreasing losses in the motor/controller and drive train of course increases the energy recovered. For example, decreasing motor/controller loss to 15% in scenario (2) increases the energy gained from 12.7% to 16.5%. -- View this message in context: http://electric-vehicle-discussion-list.413529.n4.nabble.com/Range-vs-Speed-tp4672366p4672477.html Sent from the Electric Vehicle Discussion List mailing list archive at Nabble.com. _______________________________________________ UNSUBSCRIBE: http://www.evdl.org/help/index.html#usub http://lists.evdl.org/listinfo.cgi/ev-evdl.org For EV drag racing discussion, please use NEDRA (http://groups.yahoo.com/group/NEDRA)
