Calculating power at the wheels (newbie question)

[AlmostClever]

I'm trying out some ideas for improving drive train efficiencies. For this, I'm trying to model the power necessary at the wheels of a truck.
To validate this, I've entered the specs of a real truck, and the Heavy-vehicle version of the Urban Dynamometer Drive cycle.
My calculation is that the power necessary is dv/dt * the forces to move the truck.
Problem is, I'm getting numbers which are about three times the max output of the engine on the actual truck.

I've considered dividing by pi , but I would like to know what I'm doing wrong.

Please forgive me if this post is inappropriate. I'm just a big-diesel mechanic toying with some new ideas for hybrid powertrains.
Let me know if I shouldn't have posted here, and I'll ask for the post to be removed.
Thanks

 

[dgallup]

Did you take into account the torque multiplication of the gear box and differential?

 

[AlmostClever]

I'm trying to work out the power required at the wheel. I'm planning to try to specify my powertrain from the power and torque necessary at the wheels.
I'm actually calculating forces to move the truck from drag, rolling resistance, inertia and grade. I'm trying to calculate the power from that.
Power at the wheels and at the engine should be the same, though of course torque is multiplied.
Unless I'm missing something, which I obviously am.

 

[BobM3]

Power is velocity x force, not dV/dt x force.

 

[dgallup]

I think you have mixed your units.


In terms of mechanical energy, one watt is the rate at which work is done when an object is moved at a speed of one meter per second against a force of one newton.

1W = 1Js-1 = 1kgm2s-3 = 1Nms-1

 

[AlmostClever]

Thanks for the help BobM3 and dgallup.

I'm going to have another look at what I'm doing to make sure I've got all of my units in the right place and such.
At first glance it seems OK, but something must be wrong.
I just can't see how 790 kW can be right for a 33 ton truck.

 

[NormPeterson]

Time for a sanity check by the method of alternate calculation.

Try working backwards from the traction requirement using torques, gearing, and a factor to account for having less than 100% efficiency to arrive at the necessary engine torque. See what the required HP is at the various engine rpms involved.


Norm

 

[AlmostClever]

Thank you everybody.
I think I shall take a step back and start again at first principles.
I'll try your suggestions and be very careful with any assumptions I'm making.

Thanks for the help.

 

[black2003cobra]

BobM3 actually already gave you the answer. You just need to "expand" it a little.

Power = F*v, but F = m*a, and a = dv/dt. So Power = m*v*a = m*v*(dv/dt).

If you convert units, and express acceleration in G's, velocity in mph, and use vehicle weight in lbs, you should get

HP = w*v*G/375

(I assume you know how to calculate acceleration in G's.)

 

[black2003cobra]

P.S. Probably should have added that the above is the power needed to accelerate the vehicle, after taking off the various losses. (That due to rolling resistance, aero drag, driveline losses, etc.)

 

[YvesLLewelyn]

Almost Clever - Power at the engine and at the wheels is not the same - up to about one third of the engine power (and torque) can disappear on its way to the wheels.

You can roughly calculate the power needed to maintain a steady speed on a flat road - that is, the power needed to overcome air and rolling resistance etc.
Take the vehicle (truck or car or whatever) up to over 60mph, put it out of gear and time how long it takes to slow from 60mph to 50mph. For example assume it takes 15 seconds to slow. This is a change in speed of 15 feet per second (the calculation has to be in fps). If it took 15 secs - this a deceleration rate of 1ft/sec/sec. The power needed to cause this deceleration can be calculated from F=ma (this is also the power needed to maintain speed). The weight of the truck 66,000lbs (assumimg US tons) has to be converted to mass by dividing by 32 (gravitational constant). F then equals 66000 divided by 32 times 1 (rate of deceleration) this works out to be about 2063ft lbs (which is the required force). To convert this into power you multiply the force by the average speed of the truck as it decelerates - or about 80fps. This gives about 165000ftlbs/sec. To convert to horsepower divide by 550 (which is the definition of one HP) - this gives an answer of about 300HP - which seems a bit high but is of the correct order (maybe the deceration time would be more than 15secs). This calculation is very approximate but if you use the same method when testing it should show up any changes you make to the truck.
The power needed to climb a grade is a similar calculation. If the road rises one foot after 88ft travelled at 60mph (88fps) the truck has a vertical velocity of 1ft/sec. This is work being done against gravity so this is just the weight of the truck (the gravitational constants cancel out), so the power needed is 66000 divided by 550 which is 120HP - which looks about right. This would have to be added to the power needed to maintain a steady speed - so about 420HP would needed to charge up the grade at about 55mph.
There are also other road test methods that can be used to calculate power required to maintain steady speed.

 

[robinfisichella]

Hi Everyone

Im doing practically the same thing, im designing a vehicle and trying to calculate the power required by its motor.

To do this i am using the following formula;

Tractive Force = Resistive Forces + Inertial Forces + gravitational forces

Resitive forces = Aero drag + Mechanical losses + Rolling resistance

Inertial forces = Me(dv/dt) (zero for constant velocity)

Me=vehicle effective mass = M + 4.Iw/Rr^2 + Ig.Gfd/Rr^2
M=vehicle mass,
I=inertia
Gfd=final drive ratio
Rr = rolling RADIUS


What i did was find the required torque at the wheel by

Torque_max = Tractive force x wheel radius
then
Power = rotational speed x Torque_max*

rotational speed is in rad/s (NOT SURE ABOUT THIS)

*torque can be scaled down by efficiency coefficents and other transmission losses if you like.

 

[black2003cobra]

robin - Yes, you have that right. In that form, (P = ? * ?), the angular velocity ? is in rad/sec. Watch your other units, too, of course.

 

[robinfisichella]

Thanks Blackcobra,

Almost Clever:

I think the main issue for this type of calc is the unknown values for efficiencies, this data is extrememly difficult to get hold of and the only realisitic way to obtain it is by having a chassis dyno.

Even with all this though, at high speeds and with bluff body shapes the overiding resistive force is aero drag.

If anyone has any rough figures for transmission efficiencies i would be interested to see them?

 

[GregLocock]

diff 85%
gearbox 90-98%


Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[robinfisichella]

Im having a big issue with my calcs,

In order to accelerate my vehicle from 0-60mpg (0-26m/s) I use F=m(dv/dt), this is giving me a force of around 1000 N at the tyres, which translates as a torque of around 440 Nm which is way high!

My stats are;

M=410
Wheel radius=330mm (13 inch)
0-26m/s (0-60mph)=10 s

I did F=ma to get force at contact patch (im modeling the vehicle as just a wheel at the moment, i just want to know the torque required by the electric motor).

 

[YvesLLewelyn]

Maybe the torque seems high because there is normally torque multiplication through the gearbox and diff when accelerating - maybe 3.5:1 in the gearbox (average of 1st. and 2nd. gears) and 3.5:1 in the diff - very roughly approx. 12:1 overall. Dividing 410 by 12 gives a more reasonable answer. I think most electric cars use gearboxes these days.
410Kg is a very light car.

 

[robinfisichella]

Yes true, I did think all these things, though the car being light should reduce the required torque.

I wasnt aware electric cars had much in the way of gearing?

My overall aim is to spec an electric motor and its torque curve. In that sense i just need to know the maximum torque and power required and then pick a motor which can provide it and pick and choose between the most preferable torque curve....

 

[AlmostClever]

Thanks for all the input everyone.

I am now convinced that I've got the torque numbers right.
They add up, and they make sense in relation to what the actual machine can do.

However, I'm still mis-calculating the power requirement somehow.
Using the same formula as Robin, I'm trying to calculate power at the wheels. I'm getting answers at about 750 kW, for a truck whose engine can only produce about 300 kW.

I'm not taking any mechanical/efficiency losses into account yet, because I'm still at the wheel. Once I progress back towards the power plant, I'll start doing efficiency calculations as well.

It has just occurred to me that perhaps my drive-cycle data is a bit off.
It seems to have the truck accelerating from 27 mph to 32.5 mph in the space of two seconds. I'm not sure if this is reasonable or not, but it seems rather quick for a 33 ton truck.

 

[black2003cobra]

If you are calculating torque at the wheels, make sure you are also using angular velocity at the wheels in the expression, P = ? * ?. (Not sure that's it, but it's something else to be careful about.)