Path: ...!fu-berlin.de!uni-berlin.de!news.dfncis.de!not-for-mail
From: "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
Newsgroups: sci.physics.research
Subject: electric-car energy model
Date: 22 Dec 2024 08:57:10 GMT
Lines: 134
Approved: hees@itp.uni-frankfurt.de (sci.physics.research)
Message-ID: <lsq2j6F2p92U1@mid.dfncis.de>
X-Trace: news.dfncis.de ih1AGlf0YosxOznYWm7mIAHIcdUCAzrNeo+qmIWc3oUzfRaOHsUlhoifsR
Cancel-Lock: sha1:dpSOFB6VO4UVF8w3eiTi9FjPRQM= sha256:2yt43z15hrUxaiE+Cqe07C+UQ95bwqDbj9XF4kfOdAc=
Bytes: 7208

What determines the energy consumption and driving range of a modern
electric car?  Applying some elementary physics, it's quite easy to
construct an energy model and make rough estimates for the various
parameters.

Let's start with the usable battery capacity (available energy) E_total.
This is usually quoted in kilowatt-hours (kWh) (= 3.6e6 Joules).  Let's
consider a 100%-down-to-0% full discharge of the battery, even though
practical usage normally only uses a fraction of the battery capacity.
And let's assume driving at a steady speed on paved, level roads, with
no headwinds/tailwinds.

When driving at any but the slowest of speeds, most of the energy goes
into turning the car's wheels, doing work against aerodynamic drag and
tire friction.  This process involves battery power (typically starting
out as high-voltage DC from many battery cells in series) being converted
into whatever voltage and current the drive motor(s) use (modulated by
the accelerator-pedal position), the driving electric motor(s) converting
this to mechanical rotation, and the drivetrain converting this rotation
into mechanical rotation of the driving wheels (loosing some energy to
bearing friction and other mechanical losses).  None of these processes
are 100% efficient; let's call the net efficiency from battery energy to
rotation of the driving wheels eta_drivetrain.

If the car is 2-wheel-drive there's also some energy lost in wheel-bearing
friction in the non-driving wheels; for simplicity we'll ignore this, or
equivalently lump it into eta_drivetrain.

This web page
  https://www.reddit.com/r/teslamotors/comments/ja89w1/tesla_model_3_lr_performance_consumption/
suggests that for one popular model of electric car, eta_drivetrain is
on the order of 0.8, so we'll use that as our estimate.

How much power P_at_wheels is needed at the wheels to move the car?
If the car is moving at a (constant) speed v, the work-energy theorem
gives
  P_at_wheels = F_drag v
where F_drag is the overall drag force.

There are two main energy-dissipation mechanisms: air resistance and
dissipation in flexing the tires,
so we take
  F_drag = F_drag,air + F_drag,tires.

For the high--Reynolds-number regime appropriate to cars, the air
resistance drag is well-approximated by
  F_drag,air = 1/2 C_d rho A v^2
where rho is the density of air (1.2 kg/m^3 at sea level & standard
temperature, falling in well-known ways with increasing altitude and/or
temperature), C_d is a dimensionless drag coefficient depending on the
car's shape (C_d is typically around 0.25 to 0.4 for modern cars), A
is the car's frontal area (on the order of a few m^2), and v is again
the driving speed.

According to the web page
  https://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html
tire drag (on a smooth paved road) can be approximated by
  F_drag,tires = C_rolling mg
where mg is the car's weight (empty weight + people + any cargo) and
C_rolling is a dimensionless drag coefficient given approximately by
  C_rolling = 0.005 + (0.01 + 0.0095 v_100^2)/p_bar
where p_bar is the tire pressure (I presume above atmospheric pressure)
in units of bars (1 bar = 1e5 Pascal = about 15 lb/inch^2) and v_100
is the car's speed in units of 100 km/hr.

(Because F_drag,tires is (in this approximation) linear in mg, we don't
need to worry about the car's weight distribution.)

There is also some energy used for non-propulsion purposes.  This
may include cabin heating and windshield defrosters, air conditioning,
battery heating or cooling, windshield wipers, headlights, power
steering/brakes, power windows, collision-avoidance radar, and dashboard
and other miscellaneous onboard electrical/electronic equipment.  This
"overhead" varies widely depending on ambient conditions; notably, cabin
and battery heating/cooling may vary from zero to many kilowatts.

As a (very) crude approximation let's model this as a constant power
draw from the battery, converted to whatever voltage/current the various
equipment uses at an average efficiency of eta_overhead to provide a
(constant) usable power of P_overhead all the time the car is turned
on.  For moderate temperatures where none of cabin/battery heating,
windshield defrosters, or air conditioning are operating, let's estimate
P_overhead = 200 Watts and eta_overhead = 0.7.  (This is < eta_drivetrain
because I suspect the main drivetrain is heavily optimized for maximum
efficiency, whereas the smaller motors and power supplies are probably
optimized more for minimum cost.)

Putting all the pieces together, the drivetrain power drawn from the
battery is
  F_drag v/eta_drivetrain
while the overhead power drawn from the battery is
  P_overhead/eta_overhead
so that the total power drawn from the battery is the sum of these,
  F_drag v/eta_drivetrain + P_overhead/eta_overhead

Thus the time to completely deplete an initially-fully-charged battery
is
  E_total / (F_drag v/eta_drivetrain + P_overhead/eta_overhead)
and the distance driven (i.e., car's range) d is
  d = v E_total / (F_drag v/eta_drivetrain + P_overhead/eta_overhead)
    = E_total / (F_drag/eta_drivetrain + P_overhead/(v eta_overhead))

Inserting some numbers for my car,
  E_total = 64 kWh = 230e6 Joules
  p_bar = 2.25
  A = 2.5 m^2
  C_d = 0.35
  m = 2000 kg
and driving speeds of
  v_100 = 0.5 (for driving at 50 km/h), 0.8 (80 km/hr), or 1.2 (120 km/h)
              (i.e., v = 13.9, 22.2, or 33.3 m/s respectively)
I get
  C_rolling = 0.0105, 0.0121, and 0.0155 respectively
  C_drag = 0.35
  F_drag,tires = 206, 238, and 304 Newtons respectively
  F_drag,air   = 101, 259, and 582 Newtons respectively
with the final result
  d = 569, 363, and 206 km respectively

These numbers are within 20% or so of reality (that's closer than I was
expecting before doing the calculation!), suggesting that at least the
major components of our model (drivetrain efficiency, tire drag at low
speeds, and air resistance at high speeds) are roughly correct.

The sharp drop-off in range with increasing speed is particularly notable
(and quite realistic).

-- 
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
   currently on the west coast of Canada
   "The programmers outside looked from Web 2.0 firm to AI company, and from
    AI company to Web 2.0 firm, and from Web 2.0 firm to AI company again;
    but already it was impossible to say which was which."
                   -- /Ars Technica/ comment by /ubercurmudgeon/, 2024-05-09