Mathematical Musings

[Tideland Prius]

I've never seen anything above 3 cars. But on a test drive of the RX400h, the last 5 mins had 4 "E" squares. Weird

As a point of interest, Honda claim to be able to recapture >95% of kinetic energy via regenerative braking in the Accord hybrid.

http://www.greencarcongress.com/2004/09/un..._the_hood_.html

[DaveinOlyWA]

95%? hard to believe...

actually if you read it, it says 95.2% of the kinetic energy is available to be converted to electricity... which is good...that only means that 4.8% is lost through drag and friction losses from the tires, transmission, etc... at least that is what i take it to mean...

Honda claiming anything else is simply misleading advertising

[200Volts]

DaveinOlyWA, note how Honda says is "available" for conversion. Since most genrators-motors run at about 25-35% efficiency you are only going to "recover" 25-35% of the 95.2% "available".
Hence the basic shortcoming with electricity generation from a mechanical (motor-generator) system is that at best about 35% can be rcovered.

[chrism07924]

<div class='quotetop'>QUOTE(rwlade\";p=\"89635)</div>

I was just sitting around trying to figure out the significance of the little car icons on the consumption screen and came up with this.  
 
Assume  
 
car mass = 1,311 kilos (2890 lbs)  
velocity = 96.6 km/hr (60 mph) = 26.8 m/sec  
[/b]

Mass should be in kg, right? I though "lbs" reflects weight of a body with gravity acting upon it.

Also, I believe "velocity" is actually "instatneous velocity", a scalar value. Regular old "velocity" is a vector, with both direction & magnitude.

[Bill60546]

Dr. Bob: Very impressive musings!

<div class='quotetop'>QUOTE(200Volts\";p=\"90554)</div>

most genrators-motors run at about 25-35% efficiency you are only going to \"recover\" 25-35% of the 95.2% \"available\".
Hence the basic shortcoming with electricity generation  from a mechanical (motor-generator) system is that at best about 35% can be rcovered.[/b]

Most generators / motors are typically more than 95% efficient, and some are more than 98% efficient. If they were only 25% efficient, 75% of the electrical energy going into them would get converted into heat, and they would melt!

The bottle-neck in recapturing regenerative energy is not the motor/generator or even the power electronics (which are also very efficient) but the battery itself. The bigger the battery, or the more current it can accept, the better. The Prius battery can only handle a relatively small current input (21kW max) compared to what braking events can provide, so most braking cannot be achieved with regen alone - hence a lot of what is available is lost because the battery isn't able to absorb it all.

GMs EV1 had a big battery capable of soaking up a lot higher regen currents, and they were able to show a 44% increase in urban range by switching on regen, which I think is incredible as I wouldn't have thought that much time is spent accelerating or braking even on the urban cycle.

[dstrout]

<div class='quotetop'>QUOTE(jayman\";p=\"89765)</div>

Also - again ignoring aerodynamic drag and friction with the road surface - relatively minor increases in speed require proportionately larger energy expenditure at higher speeds.

To accelerate from 30-55 MPH takes less energy than to accelerate from 55-80 MPH.[/b]

Ack! Now you've gone & done it! :P It's really not good to ignore air resistance -- mother nature gets quite upset & steals a lot of your power if you do! Look at this:

Power required to maintain a given speed = 8.702 x 10^-6 x Cd x A x V**3

where Cd = coefficient of drag
A = square feet of frontal area
V = Velocity, in miles per hour

For the prius, Cd = .26, A = 24 sq. ft.

so, power required to maintain various speeds:

35 mph -- 2.3 hp (1.7 Kw)
55 mph -- 9 hp (6.75 Kw)
85 mph -- 33 hp (24.75 Kw)
134 mph -- 130 hp (97.5 Kw) -- Wasn't 134 mph the speed record they set in a prius?

( 1 hp = 750W = .75 Kw)

Just to beat wind drag & rolling resistance.

The 8.702 is a fudge factor that can be argued, and will change with different tires, inflation pressures, altitudes, etc, but the point carries -- power required goes as through the roof as velocity increases. It won't change by a whole lot in any event.

FWIW,

dave.

[rwlade]

<div class='quotetop'>QUOTE(chrism07924\";p=\"90714)</div>

<div class='quotetop'>QUOTE(rwlade\";p=\"89635)

I was just sitting around trying to figure out the significance of the little car icons on the consumption screen and came up with this.  
 
Assume  
 
car mass = 1,311 kilos (2890 lbs)  
velocity = 96.6 km/hr (60 mph) = 26.8 m/sec  
[/b]

Mass should be in kg, right? I though "lbs" reflects weight of a body with gravity acting upon it.

Also, I believe "velocity" is actually "instatneous velocity", a scalar value. Regular old "velocity" is a vector, with both direction & magnitude.
[/b][/quote]

Yes, the calculations were done in the MKS system. Velocity (vector) and speed (scalar) can (and usually are) functions of time. Your use of the word instantaneous implies time dependence and that is true for both the scalar and vector variables.

Don't take my musing to be a complete dynamical analyses of the system, I only was trying to get a feeling for the significance of the little car icons.

[rwlade]

<div class='quotetop'>QUOTE(dstrout\";p=\"90789)</div>

<snip>

Power required to maintain a given speed = 8.702 x 10^-6 x Cd x A x V**3

where Cd = coefficient of drag
A = square feet of frontal area
V = Velocity, in miles per hour

[/b]

Great post, Dave. Intuitively, the velocity cubed relationship makes sense. Where did that equation come from?