liveforphysics said:
Myself, I'm just going to make sure you know ALL your thermal calcs are wrong, for each and every example. Many wrong by a factor of 10x or more.
LFP must have woken up on the wrong side today. Auraslip did more than 90% of the students (and faculty) at our university (and most other universities) would be capable of: Looking up various theory, trying to apply it. Just didn't get it quite right.
So maybe write a thread called, "is this how to calculate a battery temp rise" rather than putting more incorrect stuff people inevitably will think is true on the intarwibs?
auraslip said:
Lets say we want to draw 40 amps,
Heat produced is (I**2)R
Yes. Now we only need the internal resistance Ir of the batteries. Lets go back to your Ping and Headway example. The Ir
It's still R sub i, even for Jag. Ri. Saying Ir means current through some resistive path, we are using our unit to represent a psudo resistance internal to the battery.of your current 48V20Ah Ping pack happens to be listed on ebikes.ca simulator as 0.20Ohm. Hence
P = 40^2*.2 = 320W
Now for the Headway pack I had some notes that a 72V10Ah pack would have Ir = 0.24Ohm. Warning: not sure how correct this is. I probably found it on the Internet, and hopefully computed it from discharge graphs at some similar current, but my notes are a couple of years old (and convinced me not to buy Headways at the time, so I never tried in reality) Someonw with actual headways could measure the Vdrop at 40A and tell us.
for a 72v pack made from 22 cells, resistance should be in that ball-park if you have perfect interconnects, as the cells are around 10mohm each.
P=40^2*.24 = 384W
Both of these battery packs are waisting a lot of the chemical energy jsut heating themselves instead of powering your bike.
Now suppose you could be satisfied with just 30A, then things look much better (thanks to the nonlinear I^2 term being much smaller)
Ping
P = 30^2*.2 = 180W
Headways
P = 30^2*.24 = 216W
auraslip said:
How much does that heat the pack?
The amount of heat energy (q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature)
q = m x Cg x (Tf - Ti)
Lipo has a specific heat capacity of 1010 j/kg/Celsius. Weight is 5780 gm.
No idea where you got the heat capacity from.
It came from my thermal calcs on other pack designs, and I got it from a white paper on thermal modeling for LiPo pack cell cooling. Water is about 4kJ/C/kg. Anyhow, heating would be
Ping (10kg) @40A
\Delta T = 15min*60s*P/Cg/m = 15*60*320/1010/10 = 28C temperature rise above ambient
Headways (7kg) @40A
\Delta T = 15*60*608/1010/7 = 49 degrees above ambient
The problem here is ion movement becomes much less restricted as the cells start to heat, and resistance drops. This is why you see the voltage very often RAISE on low-C rate cells as they start to discharge, the Ri can be cut in half as the cells start to warm, so the modeling using room-temp Ri numbers doesn't work in practice as the cells start to warm. With LiPo it works because the delta-T is so small the Ri value stays close to constant. TexasPyro did some good Ri testing showing an A123 M1 cell having something like a 3-4x change in Ri depending on temps.
These are both a bit on the high side for temp rises.
Now in practice a pack gets cooled from the airflow, so heat rise would be slowed down. Still 30A would be more reasonable for continuous use than 40A for both these cell types. 40A peak at acceleration is no problem.