0
$\begingroup$

So let's say I have 2.5KW motor that runs at 48V, drawing peak around 55A. Let's also say that I have an 18650 battery bank, which consists of 13 batteries in series, to give me 48V, and 4 parallel rows of those batteries. The batteries are rated at constant max current draw of 4.8A each. Now, the question, which in itself is really simple and basic, do those 55A get distributed between the whole bank (13S4P, 52 Batteries, drawing 1A approx. out of each battery), or do I have to deal with a 14A peak load on each battery?

$\endgroup$
  • 1
    $\begingroup$ I hope this is a theoretical question, in practice a 1.5Kw motor can be quite dangerous. $\endgroup$ – PhysicsDave Oct 15 '18 at 23:39
0
$\begingroup$

The later but the batteries can't supply the peak load, the best they'll do is 4 rows times 4 amps per row which is 16amps peak. Could be tough on the motor if it can't get moving, all the energy will turn to heat in the windings, not good.

$\endgroup$
  • $\begingroup$ You are also liable to explode the batteries. $\endgroup$ – user253751 Oct 15 '18 at 23:45
0
$\begingroup$

I believe you're approaching the calculation incorrectly.

Each 'row' has 13 series connected batteries and so the current through one of those batteries is through all the others in the row. That is, the maximum continuous current through a row is $4.8\,\mathrm{A}$.

Thus, the maximum continuous power delivered by each row (assuming no voltage droop at maximum continuous current)

$$P_{max} = 48\,\mathrm{V}\cdot 4.8\,\mathrm{A} = 230.4\,\mathrm{W}$$

So, 4 rows cannot supply $2.5\,\mathrm{kW}$ of continuous power to the motor.

$\endgroup$
0
$\begingroup$

Consider a 2500 watt motor driven at full power by a 48 volt battery pack. The proposed power supply is composed of 4 parallel branches of 13 batteries in series (3.68 volts/cell). What is the current through each of the batteries in series?

Background

  • The battery used in this battery pack, the 18650, is the battery used in the Tesla Model S. The energy capacity of the 18650 is 2000-3500 mAh. The 18650 is a Lithium Ion battery with dimensions of 18mm x 65 mm (slightly larger than the AA battery, which is 14mm x 50mm).
  • The current rating of the 18650 battery is typically between 15-30amps. The amperage rating is dependent upon the temperature chosen (i.e., the higher the current drawn, the higher the temperature of the battery).

Analysis:

  • The current drawn by a 2500 watt motor powered a 48 volt DC source is: $I = P/V= 2500 watts/48 volts = 52 amps$
  • 52 amps supplied by four battery banks in parallel carries $52amps/4 \ branches= 13 amps/branch$.
  • 13 amps flows in each parallel branch, and 13 amps flows through each of the batteries in series in that branch.
  • Limiting the current through each battery to 4.8 amps (i.e., through each parallel branch) requires connecting a total of eleven parallel branches
    ($52 amps/(4.8 amps/branch) = 10.8 branches$).
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.