Increasing Torque and Speed

Hmmm, sounds a bit homeworky to me, but I appreciate your confusion.
On no account take anything a cat says as necessarilly true.

[1] It surely depends on the characteristics of the engine. I'm not sure your statement the torque is inversely proportional to speed is true. It may be a good practical generalisation, but I'm not sure it's a necessity.
[2]Explain both output as well as consumed power.
Oh c'mon, this has got to be about electric motors, surely it's self evident? There will be some losses, heat, friction etc resistive losses in the windings, magnetic losses. Obviously the electrical watts put into the motor will be more than the mechanical watts you get out (although there are some over unity nutters about who think they can get over 100% efficiency). Basically efficiency is power out/power in expressed as a percentage..
I'm affraid cats don't do basic arithmetic, they are too lazy...and as for formulae...puhlease.

Maybe one of the mech guys will set you straight, but at least I've set the ball rolling.
Del

"the torque is inversely proportional to speed" actually this sentence is for a gearbox.. sorry for not mentioning it.

Well, this is what a gearbox does. It increases one of those two parameters by reducing the other.

Hello,

You can simply increase or decrease both speed and torque simultaneously just by increasing or decreasing input, of course up to the point of saturation and below burn out limits,

For example one can increase RPM and torque of a DC motor and an engine, just by putting more voltage in its terminal and by injecting more fuel respectively.

The law you are talking about is for a motor and engine of a given output power. For example one can design a 1 HP motor or engine with torque T & speed S, toque T/2 &speed 2S, Torque 2T & speed S/2 etc.

This means for given case

1Hp = K x T x S

1Hp = K x (T/2) x 2S

1Hp = K x 2T x (S/2) etc

Is it like turbocharger/supercharger/Nitrous oxide purged/installed to the engine....!!????

What does 'K' mean in 1Hp = K x T x S????

Is it like turbocharger/supercharger/Nitrous oxide purged/installed to the engine....!!????

All these offer changed environmental conditions, so I would say No. Its similar to running a motor at lower temp to reduce its ohmic losses.

What does 'K' mean in 1Hp = K x T x S????

Ans is Constant

Hp = (2 pie x N x T)/746

N =torque in N-m and N =speed in RPS

Sounds Good All to Me..... Brilliant. Now some good similitudes exist some how as well for example in a fan set. As we increasing motor speed and momentum the torque are figthing air resistance until it got an steady (RPM) balancing coasting, now if we approaching a piece of cardboard in front of it slowly we will hear and feel a good amount of air pressure trying to escape around and also the fan will stalling due air resistance increment, the torque are draining out at this point but motor are trying to sustain it speed settings regardless resistance due it's HP designed which are trying to keep-up with loading (Resistance to Motion) impossed by the air flow restriction (Blockage) in front of it.

If we make use of a transmission (Gears) then we'll be on better position to counteract (Load) resistance situation conditions, absolutely... And motor or engine will be only in charge of keep-up speed (RPM's) neccesary to sustain torque requirements when needs its, definetetly!

Hp-Speed-Torque-Load-Inertia-GearRatio- all play few factors on that "Puppy Out Put" I'll Bet. And remember 'you have the technology' I knew you can do it!!!! Jaaa HeeeHeeeiiiii Good Move Buddys....Awesome and keep it cool...Lol.......!!Where I left my 'Safety Glass' ??? Gooosshhh....??? OHhhJoouuhhh!....Allset Now Buddies.....

Crank That Puppy-Up,

MC

Respected Sir,

1. As per my knowledge, it is quit impossible to increase Torque and speed simultaneously by means of external devices like gear box etc.. but if we increase the CC of engine then it is possible.

2. Further, the bore diameter to height ratio helps to make high speed or high torque engine.

3. I think, the rating of electrical motor (5 HP) is given for full load conditions. i.e. It will consume 5 HP electricity at full load condition..The efficiency of electrical motor is quit high (92% to 98%) hence almost all power is transmitted to shaft.

Formulae:-- 1 HP= 0.746 KW

Electricity consumed by motor= V*I .......(single phase)

= V*I*cos(phi) ...(Three phase)

where V= voltage, I= Current, phi= phase angle

"As per my knowledge, it is quit impossible to increase Torque and speed simultaneously by means of external devices "

Between the limits- more the input more the output,

Had you ever taken your bike and car to a mechanic with complaining about pic-up or it don't catches peak speed? if yes, How does he corrected it.and what does that correction means?

Hello Guest

You said

3. I think, the rating of electrical motor (5 HP) is given for full load conditions. i.e. It will consume 5 HP electricity at full load condition..The efficiency of electrical motor is quit high (92% to 98%) hence almost all power is transmitted to shaft.

This is wrong. The rated power of a motor (the nameplate figure) is the shaft output mechanical power. The electrical input power is rated power/efficiency. Also efficiency of a 5HP motor is more like 85% than 92 - 98%.

Cheers.......Codey

Hmm, yes, it does look homework. Just work from first principles (think I got the right spelling ? !!). Power is the product of Force and Velocity. For a rotary machine like an engine this becomes

Power = Torque x Rotational Speed.

Horsepower has its origins in imperial units

the imperial unit for Torque is ft-lbf

the unit (imperial and metric) for Rotational speed is radians/sec

- definition of rad is the angle whose circumference equals radius

so you can see that there are 2 pi radians in 360 degrees

so that with (say) Revs/Minute as an input, rotational speed = RPM x 60 x 2 x pi rad/sec

1 HP is defined as = 550 ft-lbf/sec

so any combination of Torque in ft-lbf and RPM x 60 x 2 x pi that multiplies to 550 is 1 HP

Its a similar deal in metric except that the units are Nm for Torque and remembering that 1 Watt is a Nm/sec. If you convert lbf to N and ft to m you find that

1HP = 550 ft-lbf = 746 W.

Now think about the above in terms of a machine.

"Engine" usually means "internal combustion engine" and Torque is produced by combusting gases pushing on a piston and turning a crank. At full throttle this is roughly independent of speed in the mid range so Torque is somewhere near flat in the middle, with a drop-off at high and low speed. The exact curve depends on the design of the engine and one would not normally hear phrases like "torque changes inversely as speed" in discussion about this sort of engine. To get torque increasing at the same time as speed one would simply have to start with low revs and a small amount of throttle and then increase the throttle as speed increased. Very strange and I don't think this is what you want.

"Motor" is the word that is more commonly used when with electricity is involved and it is the case that "torque varies inversely with speed" in certain designs of motor - the series universal motor being a classic case.

The mechanics of electrical force are that a wire with a current in it is forced sideways when it is located in a transverse magnetic field. More current or more turns (i.e. wire) and more force or torque is generated. However, you don't get noth'n for noth'n. When the conductor moves through the magnetic field it cuts the field and a counter voltage is generated that tries to stop the current flowing in the wire that generates the torque. This is called "back emf" or "back electro motive force".

Put the two effects above together in a series motor - where the field and rotor windings are connected in series - and you can see that when the motor starts out and the speed is very low, the current is roughly equal to the applied voltage divided by the resistance of the field and rotor circuit. As the speed increases the back emf increases proportionally and this subtracts proportionally from the net voltage driving the current through the "field" and the "rotor" SO the current reduces inversely and one finishes up with torque reducing inversely as the square motor speed - because both the current and magnetic field generated by the current are reduced as speed increases.

If the motor had a field generated by permanent magnets and a wound rotor, then the torque would increase more near proportionality with decrease in speed.

You can get similar but more limited results in other sorts of motor, for example in an induction motor where as the load increases, the slip increases and the torque increases, the induced rotor current increases and the stator current in turn increases.

I cannot immediately think of any conventional electric motor whose torque increases with speed, unusual end effects excluded like when an induction motor is starting, as a function of fundamental construction. All things can of course be arranged if electronics are introduced.

As for your 5 HP question. This would be the maximum output power. The real input power would be probably 20% more than that if the motor were around 83% efficient. One additional matter for new players is that simply multiplying the rated voltage and consumed current is likely to give an even higher figure than the real power figure above because real power is the product of the vector component of current that is in phase with the applied voltage. In AC machinery the current is usually not exactly in phase with the voltage because of inductive and capacative effects. This is called "power factor".

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in bike u can measure torque at two points,

1) just at out put shaft of engine where is torque is proportional to speed,
2) at out put of gear box (at chain sprocket) where torque at low gear will be high and at top gear will be less at constant engine RPM.

In an internal combustion engine one would normally choose an engine that had the required power for the designed application at full demand. Any lesser power demands would usually require restricting the air, and thus fuel fed to the engine. Or in the case of a diesel engine restricting the fuel supply per combustion cycle.

It should also be possible to lower power at a given engine speed by retarding ignition timing, or retarding valve timing, if your gasoline engine has these provisions.

But why?