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I'm trying to estimate the average range reduction in mileage per hour of use of a constant electrical load within an electric car:

Assumptions (Vehicle):

$$ Usable \ vehicle \ battery \ capacity \ (B) = 84.7 \ kWh $$

$$ Vehicle \ range \ without \ additional \ load \ (R) = 220 \ miles $$

Assumptions (Constant Load):

$$ Current \ draw (A) = 45 \ amps $$

$$ Voltage \ (V) = 12 \ volts $$

$$ Running \ time \ (H) = 1 \ hour $$

This is what I thought should be the correct answer:

Car Efficiency:

$$ \frac{R_{(miles)}}{B_{(kWh)}} = E_{(miles \ per \ kWh)} $$ $$ \frac{220_{(miles)}}{84.7_{(kWh)}} = 2.6_{(miles \ per \ kWh)} $$

Constant Load:

$$ \frac{A_{(amps)} \times V_{(volts)} \times H_{(hours)}}{1000} = L_{(kWh)} $$ $$ \frac{45_{(amps)} \times 12_{(volts)} \times 1_{(hour)}}{1000} = 0.54_{(kWh)} $$

Reduction in Range:

$$L_{(kWh)} \times E_{(miles \ per \ kWh)} = mileage \ reduction \ per \ hour$$ $$0.54_{(kWh)} \times 2.6_{(miles \ per \ kWh)} = 1.4 \ miles \ per \ hour \ of \ use$$

However this must be incorrect, because as the efficiency of the vehicle drops (miles per kWh), the miles per hour of use also goes down when it should surely go up?

Andy aka
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BleedObsidian
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  • Theoretical range reduction aside, most electric vehicles have an additional 12V lead-acid battery for driving conventional 12V loads and which is not used for traction. – StarCat Apr 26 '22 at 16:13
  • @StarCat That's a fair point, however in this case I know the constant load is being powered from the car battery and the 12v auxiliary battery is isolated from this circuit. The constant load in my case is multiple high performance in-car computers, and would kill the auxiliary battery too quickly. – BleedObsidian Apr 26 '22 at 16:20
  • Your calculation is correct, your intuition wrong. If your computers use a specific amount of energy that energy would propel the inefficient car fewer miles than an efficient one. – Kevin White Apr 26 '22 at 17:05
  • _"as the efficiency of the vehicle drops (miles per kWh), the miles per hour of use also goes down"_ - The car goes slower if it is less efficient? How do you figure that? – Bruce Abbott Apr 26 '22 at 17:20
  • Please don't edit your question once an answer is given. If you screwed up in your question, please respect the existing answer(s) given by not trying to make then look silly by you altering the goalposts. That sort of action is seen as very bad form. – Andy aka Apr 26 '22 at 18:26

2 Answers2

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Calculating reduction in electric car range with an additional load

Just think about it this way

  • The usable battery capacitary is 84.7 kWh
  • The extra load draws 45 amps from 12 volts for 1 hour
  • The extra load energy is therefore 0.54 kWh
  • This leaves (84.7 - 0.54) kWh for the car = 84.16 kWh

So, the top-line number is slightly reduced having factored in the extra load. Recalculate the car's range based on this slightly degraded figure. I get 218.6 miles.

Andy aka
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  • I've had this line of thinking, 220 miles - 218.6 miles = 1.4 miles. However, when you assume the car efficiency drops (miles per kWh), the mileage reduction is less when it should surely be higher? If the powertrain of the car is now less efficient, the constant load should cause even greater mileage reduction? – BleedObsidian Apr 26 '22 at 14:50
  • I think you (@BleedObsidian) are confusing yourself by trying to throw in the extra factor of time. Your running-time constraint of one hour dictates that you are going 220 miles per hour! The extra load won't be significant if you consider wind resistance at that speed :) Conversely, the slower you drive, the longer the extra load is active. Using just kWh and distance eliminates all that. – gbarry Apr 26 '22 at 15:06
  • Okay, I agree, let's just skip any consideration of time (although I'd still like to know if it's mathematically possible to calculate without knowing speed, the constant load might not be on for the whole car journey). Throwing time aside, why is the impact on range the same as a percentage when the car's powertrain is performing LESS efficiently? The impact should surely be higher as a percentage? – BleedObsidian Apr 26 '22 at 15:17
  • This isn't a forum; it's a Q and A site. You are allotted space to make a question and I am allotted space to make an answer @BleedObsidian – Andy aka Apr 26 '22 at 15:19
  • But I'm pointing out why I think your maths is flawed? And how you haven't answered my original question, I don't care about total range, I want range reduction per unit time. – BleedObsidian Apr 26 '22 at 15:24
  • @BleedObsidian - Hi, In this situation it comes down to opinion. You [*can* use comments to request clarification of an answer](https://meta.stackexchange.com/q/19756). However e.g. If an answer's author believes clarification questions are, in their opinion, extending the question more than should be done in the SE Q&A format (e.g. like a forum) then they might decide to stop answering. Different people answer a larger or smaller range of clarification questions on their answers. It is up to each person where they stop. In the absence of further details, I think that is what's happening here. – SamGibson Apr 26 '22 at 16:06
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Well, your estimate probably won't be very good, but you can try.

First, you need to realize that there are basically 3 loss mechanisms which make up your efficiency. The first is electrical efficiency of the electronics and motor. The second is aerodynamic losses due to air friction. The third is rolling friction, caused both by axle friction and flexing of the tires as they roll.

The first should change (more or less) proportionally for small increases in motor power. You get torque increasing proportional to current, but motor losses will scale as the square of the current. The second should not change at all if the car drives at the same speed. The third will change roughly with load.

So, you need to know the relative sizes of the three terms. Then you can make a decent estimate. At high speeds, the aerodynamic drag usually dominates, for what it's worth.

WhatRoughBeast
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