In the environment of auto lovers, there are often disputes about the various parameters of the engines, their power, volume, compression ratio and torque. What can the torque of the engine be, and how is it interconnected with such a parameter as engine power?
If we recall the school lessons of physics, then the power of the engine determines the product of the force per speed for the translational motion. In this case, certain coefficients were applied, depending on which units were used for the measurement. For example, if you pull a load, applying a force of 12 kg at a speed of 1m / second, then the power in this case will be 12kg / second and equal to 0.16 horsepower.
Basic concept
In Europe it is considered to be the Paris horsepower, which is equal to 75 kg / sec. In England and America, everything is much more complicated by pounds and feet - there horsepower is equal to the European standards of 1.0139 hp, which is very good. At such rates, the engines installed on spaceships develop thrust to 100 tons, while the speed is 12 km / sec, and, consequently, the power of such an engine will be equal to 16 million horsepower!
In that case, if the power is determined by the derivative of the torque, and it makes sense only during the rotation.
Calculation of the engine's torque will be equal to the product of the acting force on the arm.
If you apply a force equal to 10 kg perpendicular to the shoulder to a lever with a shoulder size of 1 meter, a torque that will be equal to 10 kilograms or 98 Nm is created - who is used to what units of measurement, on the frequency of the rotating shaft in the rotational motion. The rest is the matter of arithmetic. Let's say that if the torque is measured on the motor shaft at 6000 rpm and it will be 10 kgm, then the power of such an engine will be 83.775 horsepower or 61.6 kW - another unit of power measurement, where 1 kW equals 1 European horsepower Worldwide.
This formula for determining engine power is applied regardless of whether it is an electric, gas turbine or piston engine. For arithmetic, this does not matter. The torque will be F x R, where F is the torque and R is the torque.
Practical use
So what is more important for motorists - power or torque? Very often they can be heard from them that the pulling moment is more important, and the power is secondary.
Example
If, for example, take a small small-displacement engine developing power of 10 hp. at 6000 vol., the torque on the flywheel will be 11.7 Nm., or 1.2 kgm., In order to obtain 100 Nm. it suffices to put a reduction gearbox having a gear ratio of 8.55 and a result on the output shaft is reached. It is not worth while to remember the inevitable loss of power in the gearbox. Although the power if you take away losses will remain unchanged. There is a desire to get 1000 Nm.? Use the gearbox, which has a gear ratio of 85.5 - the whole thing is only in the selection of gear pairs.
However, it should be borne in mind that with a torque of 100 Nm at the output of the reducer, the revolutions will drop and will no longer be 6000, but slightly more than 700.
This confirms one of the basic rules of mechanics: by winning in force, we will certainly lose in speed.
To receive 1000 Nm it is possible and at 70 об. min, but it will be too slow.
And if you compare?
If the car travels on an even freeway, having a constant speed of 100 km / h, then the thrust of the engine in the places of direct contact of the driving wheels with the road surface as a result will cover the force of air resistance and rolling tires.
In this case, if we take into account the aerodynamics, the weight and pressure in the tires, for example, it will be 54 kg. Otherwise, the torque at a wheel radius of 265 mm. will be 140 Nm, with a wheel speed of about 1000 per minute and consumed power of 1500 kgm / s or 20 horsepower. Given the losses in the transmission - from the flywheel and to the point of contact of the wheel with the surface, this engine requires an engine power of about 22.5 horsepower.
And if you need to go at a speed of 200 km / h? With an increase in speed doubled. The resistance force increases fourfold - in square. In other words, this will mean that the required power will increase eight times - over the speed cube (4x2). So, the engine should be a power of 170-180 horsepower on the flywheel.
That's why not every car can develop a speed of 200 km / h. And this is with a uniform motion.
If it is necessary to give additional acceleration or when going uphill, there is a need for additional power.
Suppose that the same 22.5 horsepower at a speed of 100 km / h add about 10 hp to accelerate the physical body (II law of Newton), that is 50 hp. naturally if the latter, then the acceleration will be more vigorous.
How to increase engine power
From this it is clear that the speed of the car and its dynamics depend directly on the engine power. But how to increase its power?
If you hold a torque of 11.7 Nm with a high speed of the shaft and bring it, say in the same small-displacement engine, to 12,000 rpm, the engine's power will double and amount to 20 horsepower. In this case, the ratio is P = 1 / 716.2 M xn in which the engine power determines the value - P., at its n min-1, and the engine torque is M (kgm) at constant revolutions, and the value 1/716 , 2 this is just a coefficient of dimensionality.
Unfortunately, increasing the rotational speed of the piston engine shaft is not so simple. All parts are subjected to heavy loads, such as inertia force, friction.
If you unscrew the motor shaft from 6000 rpm to 12000 rpm, then in this case the inertia forces that load the parts will increase fourfold. In the eight-cylinder engines of Formula 1, the volume of only 2.4 liters when reaching a maximum capacity of 19,500 vol. min. the inertia forces are significantly higher than the 6000 rpm values. min. and not at all 3.25 times. This value must be multiplied by it (3.25x3.25 = 10.5). As a result, the force of inertia at such speeds will increase by 10.5 times.
The friction of moving parts is increasing even more rapidly (using the same values from 6,000 to 19,500 rpm), it will increase 35 times. A significant reduction in the required amount of fuel-air mixture entering the cylinders of the engine inevitably leads to a drop in the torque.
For each engine there is a point indicating the moment of inflection on the power curve determined by the speed of the crankshaft.
And after this point the power will no longer rise, but on the contrary will fall. Not to mention the danger, such as the possibility of twisting the engine with increasing forces of inertia with subsequent destruction.
Moments affecting engine power
You can go and by the way of increasing the torque. Here the main thing is to ensure forced air supply or boost. If two times more fuel-air mixture is pumped through the engine, then the engine torque and power will increase by about 2 times at the same speed. But in this case, the heat loads increase significantly, and new tasks appear that need to be solved.
In order to fully understand what affects the power of the engine, you can take as an example the calculation of the power of a four-stroke engine with one cylinder. According to the initial data, the piston diameter is 80 mm, and the piston stroke is 100 mm. The shaft rotation is 3600 rpm. The average indicator value of the gases is 8 kg / cm². The force of the gas pressure on the piston can be determined by multiplying the value of the bottom area of the piston by the value of the indicator value of the gases.
Power calculation
The bottom of the piston is represented as a circle with an area equal to a constant value - 3.14 (Pi) and multiplied by the radius in the square R2. In turn, the radius is half the diameter. In this case, 40 mm. accordingly, the area of the piston head will be approximately 50 cm2. (3.14x4 = 50.24). pressing on the piston is calculated as follows: 8 kg / cm² x 50 cm² = 400 kg. It follows that the work done by the piston for each of its cycles at a stroke of 100 mm, or 0.1 m, is 400 kg x 0.1 m = 40 kgm. Due to the fact that the working cycle of four-stroke engines is performed in two turns of the shaft, at a speed of 3600 rpm the number of cycles will be 3600: 2 = 1800 per minute. In a second it will be 1800: 60 = 30 full cycles. Engine power will be equal to 40kg x 30 = 1200 kgm / sec or 1200: 75 = 16 horsepower.
The power that develops due to the gases inside the engine is called the indicator power and is determined by the corresponding diagram giving its values to a special device - the indicator. A certain part of this power is spent on overcoming the details of the crank mechanism in the friction engine.
Design features of the engine
As a result, the engine power developed on the shaft will be less than the indicator power by about 20-30 percent. When considering this case, the effective engine power will be on the order of 12-14 horsepower. It follows that the power directly depends on the diameter of the piston, the magnitude of its stroke, the average values of the test gas pressure and the crankshaft speed per unit time.
The power increases with increasing engine speed only to certain values, which depend on the engine design.
This can be explained by the fact that in connection with the increase in turnover, mechanical losses such as friction, a decrease in the indicator pressure of gases, and the filling of the cylinders with a fuel mixture are significantly increased. The fillability of the mixture is reduced by increasing resistance to passage in the valves and shortening the opening time of the intake valves.
In production, the torque of the engines is determined by means of special stands, on which the operation of the engine is characterized for one torque. Knowing these values, you can easily determine the useful power of the engine. The values of the shaft speed corresponding to the maximum torque and engine power do not match. In the event that the power values develop at 2800-3600 rpm, the maximum engine torque will be reached at 1400-2100 rpm. In the case of full opening of the shutter, the fuel mixture is supplied in the largest volume into the cylinders and the average pressure of the indicator gases reaches its maximum values, providing the greatest torque.
Economy of engines
Engine Elasticity Parameters
If you look again at the torque curve, you can see that it gives the main characteristic of the engine - it's its elasticity. In principle, this curve is not favorable for all ICEs - significantly worse than for gas turbine engines or electric motors. They give the highest torque at low rpm, even when the shaft is stopped. Approximately like a horse, stopped, tensed and pulled out the cart. So with the car engine will not work. He immediately stall.
The graph of the torque of a conventional ICE shifted to the left of 1000 rpm, as a rule, is not drawn, since the engine simply can not operate at revolutions below the idle speed. At the time, these parameters are much higher at the electric motor, and as the load increases, the electric motor loses its revolutions, increases the torque, giving resistance to the end.
So there was an idea of creation of hybrid engines combining in itself two elements of power units - internal combustion and the electric motor.
In this case, the load is distributed according to the needs that have arisen. The electromotor takes on the load exactly there, where the engine's capabilities are limited.
Conclusion!
So what is most important in the engine - power or torque? Certainly, the torque of the engine is needed in a wide range of shaft speeds. Even at the highest speed, and this can mean only one thing - power is more important.
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For each asynchronous motor, a nominal mode, i.e., a continuous operation mode, in which the engine does not overheat above the set temperature, can be determined. The moment Mnom corresponding to the nominal regime is called. rated torque. The corresponding nominal slip is for medium-power asynchronous motors s H0M = 0.02 ... 0.06, i.e. The nominal speed n and is within
n nom = n 0 (1 - s 0) = (0.94 ... 0.98) n 0
The ratio of the maximum torque to the rated torque to M = Mmax / Mn is called the overload capacity of an induction motor. Usually to m = 1.8 ... .2.5.
When starting up, that is, when starting from and during acceleration, the asynchronous motor is in conditions substantially different from the conditions of normal operation. The torque developed by the motor must exceed the moment of load resistance, otherwise the engine will not be able to accelerate. Thus, from the point of view of starting the engine, its starting point plays an important role.
The ratio of the starting torque M n developed by the engine in a stationary state, that is, for n = 0, to the nominal torque k n = M n / M nom is called the multiplicity of the starting torque.
The maximum moment M max is called the critical moment of the asynchronous machine. The operation of the machine with a torque exceeding the rated speed is only possible for a short time, otherwise the service life of the machine is reduced due to its overheating.
As a result of the interaction of the rotating magnetic flux with the currents induced by it in the conductors of the rotor winding, forces acting on these conductors in the tangential direction arise. Find the value of the torque created by these forces on the shaft of the machine.
Electromagnetic power transmitted to the rotor by a rotating magnetic field is equal:
where M em is the electromagnetic moment acting on the rotor.
In accordance with the scheme of substitution of one phase of the machine:
From these expressions we find: 
Taking into account the current of the rotor, EMF, inductive resistance, we get:
We introduce the constant 
and neglecting the moment of friction, we represent the expression of the moment on the shaft in the form:
If the magnetic flux Φ is expressed in webs, the current I 2 - in amperes, the torque will be obtained in Newton meters (Nm).
The rotational moment of the machine depends on the changes in the load, t, t 2
and, but it can be represented as a one-variable function. As such a variable for an asynchronous motor, it is most convenient to choose a slip s.
According to the previously studied formulas:
Assuming that the frequency of the network is unchanged we introduce
36. Methods for regulating the speed of hells with short circuits. rotor
37. Start and adjustment of the speed of rotation of AD with f.r.
Adjustment by changing the slip is performed by changing the resistance R p of the adjusting rheostat in the rotor circuit.
The introduction of the rheostat into the rotor circuit changes the dependence of the torque M on the sliding of s, without affecting the magnitude of the maximum moment. Three characteristics of M (s): the natural (nonrebreather) characteristic 1 corresponds to the short-circuited rotor winding (rheostat resistance = 0), rheostat (artificial) characteristics 2 and 3 to the one and two stages of the rheostat introduced.
The introduction of a rheostat into the rotor circuit has a positive effect on the starting current, reducing it approximately 2 times compared to a short-circuited blood pressure.
Disadvantages of this method: 1) low efficiency due to losses in the rheostat R p; 2) reduction of rigidity of mechanical characteristics; 3) the speed can only be adjusted downwards.
Start of an asynchronous motor with a phase rotor.The start-up of asynchronous motors differs significantly from the conditions of normal operation. The torque of the motor at start-up must exceed the moment of load resistance, the starting torque plays a role. The second important starting characteristic is the starting current. The multiplicity of the starting current for squirrel-cage motors is 5-7, which may be unacceptable for the motor or for the network and may have a smooth start value. Start-up of the motor with a phase rotor is carried out through a 3-phase rheostat, each phase of which is connected through brushes and rings to one of the phases of the rotor. At the start of the start-up, the rheostat is fully inserted, at the end of the start it is output and all three phases of the rotor are short-circuited. The number of stages of the rheostat is taken more than two and the switching process at start-up is usually automated. The introduction of active resistances into the phase rotor circuit increases the torque and makes the start smooth and limits the starting current. This starting method has a number of advantages, but it is only applicable to motors with a phase rotor.
39.40. The device, the principle of the DC motor. Methods of stimulation. EMF winding of the armature and electromagnetic moment The device and principle of the DC motor The DC motor consists of a fixed part of the stator and a rotating armature part separated by an air gap. To the inner surface of the stator, the main ones are attached to the additional poles. The main poles with excitation windings serve to create the main magnetic flux Φ in the machine, and additional poles to reduce sparking.
The anchor consists of a shaft, a core, a winding and a collector. The collector contains copper plates isolated from each other, which are connected to the sections of the armature winding. Fixed brushes are applied to the collector; Anchoring the armature winding with an external electrical circuit. As a result of the interaction of the armature current I and of the magnetic flux Ф, a torque is created, M = SMP, where CM is the moment constant, which depends on the design data of the machine. The torque M, of the engine is balanced by the moment of resistance Ms of the working machine. When the armature rotates with a frequency n, its winding crosses the magnetic flux Ф and in it, according to the law of electromagnetic induction, counter-emf E = CeFn is induced, where Ge is the constructive constant.
The voltage on the armature of the armature U is equal to the sum of the EMF and the voltage drop at the anchor circuit resistance U = E + RdIa = CeFn, whence the armature current is Ia = (U-CeFn) / Rn, and the rotational speed n = (U-RdIa) / CeF /
Depending on the method of supplying the field winding, the DC generators are:
| a B C D) |
Fig. 50. Excitation of the generator: a - independent, b - parallel, c - consecutive, r - mixed.
With independent excitation, the OM is fed from an extraneous source. It is used in cases where it is necessary to regulate the excitation current I V and the voltage U at the terminals of the machine within a wide range. The armature current is equal to the load current I = I n (Figure 50, a)
Generators with self-excitation have OB, fed from the generator itself.
When the OB is switched on in parallel with the armature winding, we have a generator with parallel excitation (Fig. 50, b), for which I n = I n + I v. Powerful machines of normal performance I in usually make 1-3%, and for small machines - up to several tens of% of armature current. In a generator with series excitation (Fig. 50, c), ORP is connected in series with the armature, i.e.
I n = I n = I c.
Mixed excitation generators have two excitation windings, the OB is connected in parallel to the armature, and the other ORP is sequential (Figure 50, d). The main one is usually OB. ORP magnetizes the machine with an increase in the load current, which compensates for the voltage drop U in the armature winding and the demagnetizing effect of the armature response.
The torque T 1 of the drive shaft is the moment of the driving forces, its direction coincides with the direction of rotation of the shaft. The moment T 2 of the driven shaft is the moment of the resistance forces, so its direction is opposite to the direction of rotation of the shaft.
If the power is known Rand the angular velocity ω, the torque
T= P / ω, Nm
From here on the drive shaft , on the slave ,
that is, P 1 = T 1 ω 1, P 2 = T 2 ω 2 .
Substituting the values P 1 and P 2in the formula for determining the efficiency, we obtain
T 2 ω 2 = T 1 ω 1 η,
Control questions:
1. What is the reason for the need to introduce a transmission as an intermediate link between the engine and the executive element of the machine?
2. List the gearing gear.
3. List the transmissions using flexible communication.
4. What characterizes the gear ratio? How is it determined (including for multistage transmission)?
5. What is a reducer?
6. How do transmission characteristics, such as power, torque, speed, vary from the one-stage gearbox leading to the output shaft?
7. What programs are called variators?
8. Which transmissions serve to convert rotary motion into translational motion?
ü Answer test questions
TASKS
1 Determine the moment on the drive shaft of the gear shown, if the output power from the transmission is 6.6 kW; the input and output speed is 60 and 15 rad / s, respectively; efficiency = 0.96
2 Determine the speed of rotation of the third gearbox shaft, if known:
Rpm,:
n 4, z 4 n 3, z 3
 |
n 2, z 2 n 1, z 1
a) rpm;
b) rpm;
c) rpm;
d) rpm;
e) rpm
3 Determine the torque on the second shaft of the gearbox (see figure to task 2), if known: Nm,: Efficiency of tooth = 0.97, efficiency of push = 0.99
4 
Determine the required power of the electric motor, if the output power of the transmission is 12.5 kW; The efficiency of the belt drive is 0.96; Worm gear reducer efficiency 0,82
5 Determine the required power of the winch motor if the speed of lifting the load is 4 m / s; weight of cargo 1000 N; The drum efficiency is 0.9; Efficiency of cylindrical gear 0.98
a) 3.53 kW b) 2.15 kW