Oilpan4 said:
Because being able to move that much heat in an hour isn't something that is going to fit in a car. The car could heat it up its own battery that fast with resistance heaters but you would need an air conditioner almost the size of a car to cool it down in 10 minutes.
The Model 3 can reportedly maintain a speed of ~130 mph until the pack is empty. If its efficiency at 130 mph is 700 Wh/mile, its continuous power consumption is about 91 kW, and it thus has a heat rejection capacity (from resistive heating in the pack, not including motor) of at least 4600 W. Add in the heat generated by the motor (around 85-90% efficiency at 91 kW?) and you get around 15 kW total heat rejection. Taking that value from a linear point of view (obviously conductive temperature change isn't linear with time) it should only take around 20 minutes to take 3 kWh out of a hot pack with today's Model 3 TMS.
700 Wh/mile * 130 mph = 91 kWh/h = 91 kW.
I = 91 kW / 355V = 256 A.
M3LR pack R = 0.03 ohms / 42p * 96s = 0.07 ohms.
Heat power = (256 A)^2 * 0.07 ohms = 4.6 kW
Moreover, the Model 3 can reject that much heat while maintaining a battery temperature less than 45°C, regardless of ambient. Since conductive heat exchange is roughly linear with the difference in temperature, it should be substantially easier to cool off a pack at 60°C to 30°C, and in fact the Penn State research suggests cooling is 8-12x as efficient at higher temperatures.
So, in my view, achieving "hot charging" without long cooldown periods is more about tweaking coolant flow rates and radiator fan speeds on e.g. a Model 3 rather than a complete reinvention of the battery.