LP

Wouldn't that burn out your brakes?

When I was taught to drive, I was told to use a low gear when going downhill and to apply the brakes only occassionally to maintain control. Partial braking wears down your brake pads or disc and could result in the loss of your brakes. Sounds like a dangerous thing especially going downhill.

Assuming no friction, the unbraked vehicle would accelerate down the hill at 6% of 9.81m/s² losing potential energy and gaining kinetic energy in the process. The brakes are needed to dissipate the energy so that there is no increase in kinetic energy. So for a vehicle of 25000kg travelling down the slope at 40km/h, the total dissipation needed is therefore a little over 163kW. In practice the brakes will release less than this, as there will be other friction losses in the system releasing the remainder.

Engine braking, by using a higher gear would prove attractive in the case of a vehicle powered by a combustion engine as less heat would appear at the brakes.

Regenerative braking, by either recharging its batteries, by pumping electricity back into its supply wires or rails, or by speeding up its on-board flywheels, may prove an attractive energy-recovering feature of an electric or hybrid vehicle system.

I'm all with you on this one PWSlack, except for one part. I think it may just be an ambiguity in speech. It is about using engine braking to dissipate some energy, one would choose a >lower< gear to control speed. If I have a 13 speed trans in a semi I most definitely would not want to use a high gear. A rule of thumb taught around my area is that when going down a hill you use one gear lower than what you could go up the hill in.

I would like to point out how to get to the 163kW power dissipation number. I hope it is to educate the OP and not to sound like a know it all! ;-)

from Wikipedia to double check my memory

http://simple.wikipedia.org/wiki/Watt

1Watt = 1kg * m/s^2 * m / s (eq. 1)

That is that one Watt of power is needed to accelerate one kilogram of mass at one m/s^2 over a distance of meter for a time of one second.

Or to muddy the waters it can be broken down a little more.

1W (power) = 1 Joule (energy) / second (time) (eq. 1a)

1 J (energy) = 1 Newton (force) * 1 meter (distance) (eq. 1b)

1 N (force) = 1 kg (mass) * 1 m/s^2 (acceleration) (eq. 1c)

substitution will provide the equation (eq. 1) above.

Since the OP posted a very well stated question we have all the data we need:

mass = 25000kg, acceleration = 9.81m/s^2 * 6% = 0.5886, velocity = 40km/h = 11.11m/s

PowerDissipated = 25000kg * 0.5886m/s^2 * 11.1m/s = 163498 Watt ~ 163.5kW

An answer to Vulcan, if the breaks are designed to continuously dissipate a total of 164kW of heat, then theoretically the breaks could survive this grade if applied evenly. That is in theory, since a human is driving and probably doesn't have thermometers on all the break drums/rotors there is a good chance that the driver could surge the breaks and overheat them, which would most likely "glaze" the break pads and reduce their effectiveness. However, I agree with what you were taught; that when driving heavy trucks it is best to down shift a few gears at the top of a steep grade and use your engine to do most of the breaking, especially when the engine is a diesel equipped with a "Jake Break" or compression breaking valve train. There are several reasons for this, my favorite is that instead of wearing friction material off the pads, the air being pumped through the engine it getting all the abuse. And the engine is already designed to handle high temps, so using the air through the engine to carry the heat away seams a very logical thing to do. Like you said use the breaks for assisting the engine incase you underestimated the grade; or you have an old truck that jumps out of gear!

And contrary to what your physics teacher tells you, a loaded truck cannot stop in the same distance that an empty truck can. That only applies if you lock up the tires and assume constant friction even though the rubber tires get soft as they melt on the road. Not to say it's completely impractical, nevertheless, they still teach it.

Thanks. I guess if the "hill" is just a few feet then partial braking will be okay. If it's a long downhill stretch of several miles, I'd use the engine with occasional assist from the foot brake.

Thanks guys for the explanations so far, apologies for taking so long to reply, but I have been out of the office. There is one fact I think I left out, and that is that the length of this gradient is 6 km. How would this affect the formula?

Just to give you more info, this is a requirement from the Vehicle braking regulation ECE R13 that states the following:

Laden power-driven vehicles must be tested in such a manner that the energy

input is equivalent to that recorded in the same period of time with a laden

vehicle driven at an average speed of 30 km/h on a 6 per cent down-gradient

for a distance of 6 km, with the appropriate gear engaged and the endurance

braking system, if the vehicle is equipped with one, being used. The gear

engaged must be such that the speed of the engine (min-1) does not exceed the

maximum value prescribed by the manufacturer.

Sorry I said 40 km/h and it should be 30 km/h. (40 km/h is for trailers)

Quite.

The rate of loss of height is 6% of 40kph, with no acceleration. Assume earth gravity for the time being as there are no serviceable wheeled 25000kg vehicles on any other planet mankind has access to at the moment.

So, the rate of loss of potential energy = 6% x 40kph x 9.81ms^-2 x 25000kg,

or >163kW once the units are sorted out.

Some replies assume a road vehicle powered by a liquid-fuel combustion engine with an engine-braking capacity, which may be a false assumption. The 163kW could come in handy for topping up a battery or shoving up the jacksy of the power supply for an electrically-powered rail vehicle. Then some people are funny that way...

"Some replies assume a road vehicle powered by a liquid-fuel combustion engine with an engine-braking capacity, which may be a false assumption."

Yes, of course that would be a wonderful way to recover the energy stored from elevation. No doubt. I had it stuck in my head that this is a highway use straight truck with no electrical drive component. In a perfect world all rail locos would have a tender car full of batteries/caps to store power from regen breaking instead of huge banks of resistors. And the fuel inefficient LTL trucking industry would be uneconomical. I wish the day to come soon; no sarcasm.
To support your point, after looking at the speed (not used to kph) again, they do seem to support a rail car type vehicle.