Understanding the Power of an Engine

[fog37]

Hello,

In mechanics, power is defined as the rate at which energy is either produced or consumer: $\frac {E}{t}$. A different but equivalent definition is $P= F v$. In the rotational world, we talk about rotational power and the expression becomes $$P= \tau {\omega}$$
where $\tau$ is torque and $\omega$ is the rotational speed in RPM.

So two engines having the same power can have different torques and speeds. In fact at each different torque value corresponds a different speed value.

Can an engine produce a minimum torque and a maximum torque? How is torque measured? I know the definition of torque as $\tau= F r$, as the agent that changes the rotational state (i.e. accelerates) of a body but I am still not able to grasp it in the context of an engine. What is the lever arm in the case of an engine? Is an object attached to the shaft of the motor to determine torque? I know an engine has a torque and speed curve describing it.

Rotational speed is easy to grasp: it is just how fast the motor's axis (shaft) is spinning.

Thanks!

 

[jrmichler]

First look up Prony brake, then look up dynamometer. That should answer your questions.

 

[Dr.D]

In the rotational world, we talk about rotational power and the expression becomes

P=τω​

P= \tau {\omega}
where τ\tau is torque and ω\omega is the rotational speed in RPM.

You were doing OK until you got to rpm. That is not correct. The equation is correct only for omega in rad/s.

 

[xxChrisxx]

An engine power thread that might not descend into a pile of nonsense.

What is the lever arm in the case of an engine?

Is an object attached to the shaft of the motor to determine torque?

The question is how do these components produce any torque/power measured?

You were doing OK until you got to rpm. That is not correct. The equation is correct only for omega in rad/s.

I think OP is being conceptual making ∝ more appropriate. One might pedantically argue that he never stated the units of torque or power he wanted to use.

 

[Dr.D]

The OP did state that omega was in rpm. That does not work in any system of units I know about.

 

[xxChrisxx]

I think you are missing the point of the OP's question. It's one of concepts, not systems of units. He's not confused because he's getting the wrong numerical answer because he's using rpm and not rad/s. He's asking where that torque comes from.

He should have used P ∝ Tw or P = k.Tw

My point was frivolous really. P = Tw works for any units as long as they are dimensionally correct. The units will just be unconventional.
More torque; MOAR POWAHHH.
More speed; MOAR POWAHHH.

edit: proof reading fail

 

[russ_watters]

Can an engine produce a minimum torque and a maximum torque?

That feels like a trivial yes...did you mean something else?

How is torque measured?... What is the lever arm in the case of an engine? Is an object attached to the shaft of the motor to determine torque?

For convenience it is measured at the wheels, so the lever arm is the radius of the car tire. Torque at the engine is calculated by multiplying by gear ratio.

This also translates directly to force for propelling the car forward.

 

[Dr.D]

For convenience it is measured at the wheels, so the lever arm is the radius of the car tire. Torque at the engine is calculated by multiplying by gear ratio.

This is only true for a wheeled vehicle; it does not apply to a stationary engine driving a load such as a pump or a generator. Actually, there is another problem with the idea of measuring it at the wheels, and that is the matter of multiple inefficiencies that each reduce power.

There are two power figures that are really meaningful for an IC engine. One is the indicated power as calculated from the P-V diagram. The other is the shaft power, as measured at the crank shaft output. Indicated power is the power rekeased by the combustion process. Shaft power is the indicated power reduced by the various losses in the engine itself. The equation
P = omega*T
applies to the shaft power. It cannot be applied directly to the indicated power.

 

[fog37]

Great. Thanks.

I have read about an absorption dynamometer to measure power, torque and speed. This is my understanding: fist the engine's shaft turns at a certain angular speed with no side break, i.e. no load, applied to it. The break exerts a force that generates a torque if applied to the side of the shaft. The lever arm is the radius of the shaft. This is when the engine has maximum speed but exerts zero torque.

The side break is then applied. A torque is now generated and tries to slow down the spinning wheel. The angular velocity is reduced but the shaft still spins. The torque engine at that moment is equivalent to the torque applied by the brake. As we continue to increase the side break force, hence the external torque, the speed decreases up to a point when the torque is so large that the shaft does not spin anymore (zero angular speed).

Different motors, even with the same maximum power, will have different pairs of speed/torque values. The product speed times torque represents power. I guess, for the same motor, the speed x torque products are not all equal and maximum power is the product with the largest value. This means the motor is capable of different powers up to a maximum power...

Is that correct?

 

[russ_watters]

This is only true for a wheeled vehicle; it does not apply to a stationary engine driving a load such as a pump or a generator.

Good point - I got it in my head this was about cars, but re-reading the OP, it isn't that specific.

Actually, there is another problem with the idea of measuring it at the wheels, and that is the matter of multiple inefficiencies that each reduce power.

I guess I always assumed that power delivered to the wheels (or pump, winch, whatever) is what is most relevant, so including the losses would be preferred. But it depends on what the purpose is for doing the calc/measurement.

 

[russ_watters]

Different motors, even with the same maximum power, will have different pairs of speed/torque values. The product speed times torque represents power. I guess, for the same motor, the speed x torque products are not all equal and maximum power is the product with the largest value. This means the motor is capable of different powers up to a maximum power...

Is that correct?

Yes. Note that power has to start at zero, but torque doesn't. The torque curve is much flatter, so the power curve steadily rises:

 

[fog37]

Hello russ_watter,

I am not clear on your last comment: power has to start at zero, but torque doesn't. Assuming the motor is either fully ON or fully OFF, when the motor is ON and no load is attached to the motor's shaft, the shaft spins at the highest possible angular speed $\omega$. I would say the torque $\tau$ is zero at that moment and power $P=\tau \omega = 0$ as well since there is not load the product is zero.

As an external retarding force is applied to the shaft causing an external torque, the speed $\omega$ decreases but the motor's torque increases to match and counteract the external retarding torque trying to slow the shaft down.

When we compare two motors, say two small electric motors having the same power, the rated power must be the maximum power since the two motors can exhibit a range of output powers up to a maximum. Suppose we turn on both motors and trace their graph $\tau$ vs $\omega$. These curves will look different. Also, to be able to make a fair comparison, the shaft radius should be the same. While the angular speed $\omega$ is not affected by the shaft radius, the torque is. That seems to imply that for the same motor, increasing the shaft gives a larger torque hence a larger output power $P$...Is that true? I wouldn't think so.

 

[Randy Beikmann]

Hello russ_watter,

I am not clear on your last comment: power has to start at zero, but torque doesn't. Assuming the motor is either fully ON or fully OFF, when the motor is ON and no load is attached to the motor's shaft, the shaft spins at the highest possible angular speed $\omega$. I would say the torque $\tau$ is zero at that moment and power $P=\tau \omega = 0$ as well since there is not load the product is zero.

As an external retarding force is applied to the shaft causing an external torque, the speed $\omega$ decreases but the motor's torque increases to match and counteract the external retarding torque trying to slow the shaft down.

When we compare two motors, say two small electric motors having the same power, the rated power must be the maximum power since the two motors can exhibit a range of output powers up to a maximum. Suppose we turn on both motors and trace their graph $\tau$ vs $\omega$. These curves will look different. Also, to be able to make a fair comparison, the shaft radius should be the same. While the angular speed $\omega$ is not affected by the shaft radius, the torque is. That seems to imply that for the same motor, increasing the shaft gives a larger torque hence a larger output power $P$...Is that true? I wouldn't think so.

The torque in the shaft is not affected by shaft radius. The stress in the shaft will be different for the same torque though, if you have a different radius.

If you mean the shaft torque capacity before it fails, that’s a different issue.

 

[Dr.D]

I think the gist of the point about "power starts at zero" is simply that, with zero speed, there is no work being done and hence no rate of work. A prime example is a large steam engine, stalled against a load. It may be exerting tremendous force to move the load, but if there is no motion, there is no work done and no power transfer.

 

[fog37]

Dr.D, I see it now: at stall, power is zero and $\omega=0$ but torque is not zero since there is a large load that the shaft is trying, unsuccessfully, to rotate. It is a "static" torque that cannot cause acceleration.
So, in general, the engine in car has the some maximum power, say $500 W$, but different torques $\tau$ at different rotational speeds $\omega$ of the shaft. The transmission in a car is a group of rotating gears that allows the wheels to rotate at a speed larger or smaller than the current rotational speed of the engine. Without transmission, the wheels of the car would only be able to rotate at the same speed as the crankshaft (between zero and a maximum value $\omega_{max}$), correct? When the car is not moving, a lot of torque is needed to win the inertia and the angular speed is small in low gears. The shaft speed is higher than the wheel's speed but more torque is applied to the wheel. The large output torque is due to the larger diameter of the output shaft. Randy mentions that the torque in the shaft is not affected by the shaft radius because the stress will be different for the same torque but changing ratio. I am still not sure about that point.

Torque seems to be important at the very beginning when the car is still moving slowing and not when the car has already reached a higher speed.

 

[Randy Beikmann]

This can all be simplified. With a manual transmission, the propelling force at the tires (total) equals the engine torque multiplied by the overall gear ratio (engine to axle), divided by the radius of the driving tires. I’m ignoring friction and driveline inertia, which is a good first approximation. This is independent of speed.

 

[Randy Beikmann]

This can all be simplified. With a manual transmission, the propelling force at the tires (total) equals the engine torque multiplied by the overall gear ratio (engine to axle), divided by the radius of the driving tires. I’m ignoring friction and driveline inertia, which is a good first approximation. This is independent of speed.

I guess I should add that power is then the propelling force times vehicle speed, or engine torque times engine speed. If done in the correct units, they come out the same.