Definition 1
The moment of force is represented by a torque or rotational moment, being a vector physical quantity.
It is defined as the vector product of the force vector, as well as the radius vector, which is drawn from the axis of rotation to the point of application of the specified force.
The moment of force is a characteristic of the rotational effect of a force on a solid body. The concepts of “rotating” and “torque” moments will not be considered identical, since in technology the concept of “rotating” moment is considered as an external force applied to an object.
At the same time, the concept of “torque” is considered in the format of internal force that arises in an object under the influence of certain applied loads (a similar concept is used for the resistance of materials).
Concept of moment of force
The moment of force in physics can be considered in the form of the so-called “rotational force”. The SI unit of measurement is the newton meter. The moment of a force may also be called the "moment of a couple of forces", as noted in Archimedes' work on levers.
Note 1
In simple examples, when a force is applied to a lever in a perpendicular relation to it, the moment of force will be determined as the product of the magnitude of the specified force and the distance to the axis of rotation of the lever.
For example, a force of three newtons applied at a distance of two meters from the axis of rotation of the lever creates a moment equivalent to a force of one newton applied at a distance of 6 meters to the lever. More precisely, the moment of force of a particle is determined in the vector product format:
$\vec (M)=\vec(r)\vec(F)$, where:
- $\vec (F)$ represents the force acting on the particle,
- $\vec (r)$ is the radius of the particle vector.
In physics, energy should be understood as a scalar quantity, while torque would be considered a (pseudo) vector quantity. The coincidence of the dimensions of such quantities will not be accidental: a moment of force of 1 N m, which is applied through a whole revolution, performing mechanical work, imparts energy of 2 $\pi$ joules. Mathematically it looks like this:
$E = M\theta$, where:
- $E$ represents energy;
- $M$ is considered to be the torque;
- $\theta$ will be the angle in radians.
Today, the measurement of moment of force is carried out by using special load sensors of strain gauge, optical and inductive types.
Formulas for calculating moment of force
An interesting thing in physics is the calculation of the moment of force in a field, produced according to the formula:
$\vec(M) = \vec(M_1)\vec(F)$, where:
- $\vec(M_1)$ is considered the lever moment;
- $\vec(F)$ represents the magnitude of the acting force.
The disadvantage of such a representation is the fact that it does not determine the direction of the moment of force, but only its magnitude. If the force is perpendicular to the vector $\vec(r)$, the moment of the lever will be equal to the distance from the center to the point of the applied force. In this case, the moment of force will be maximum:
$\vec(T)=\vec(r)\vec(F)$
When a force performs a certain action at any distance, it will perform mechanical work. In the same way, the moment of force (when performing an action through an angular distance) will do work.
$P = \vec (M)\omega $
In the existing international measurement system, power $P$ will be measured in Watts, and the moment of force itself will be measured in Newton meters. In this case, the angular velocity is determined in radians per second.
Moment of several forces
Note 2
When a body is exposed to two equal and also oppositely directed forces, which do not lie on the same straight line, the absence of this body in a state of equilibrium is observed. This is explained by the fact that the resulting moment of the indicated forces relative to any of the axes does not have a zero value, since both represented forces have moments directed in the same direction (a pair of forces).
In a situation where the body is fixed on an axis, it will rotate under the influence of a couple of forces. If a pair of forces is applied to a free body, it will then begin to rotate around an axis passing through the center of gravity of the body.
The moment of a pair of forces is considered to be the same with respect to any axis that is perpendicular to the plane of the pair. In this case, the total moment $M$ of the pair will always be equal to the product of one of the forces $F$ and the distance $l$ between the forces (arm of the pair) regardless of the types of segments into which it divides the position of the axis.
$M=(FL_1+FL-2) = F(L_1+L_2)=FL$
In a situation where the resultant moment of several forces is equal to zero, it will be considered the same relative to all axes parallel to each other. For this reason, the effect on the body of all these forces can be replaced by the action of just one pair of forces with the same moment.
§ 92. Torque of an asynchronous motor
The torque of an asynchronous motor is created by the interaction of the rotating magnetic field of the stator with the currents in the conductors of the rotor winding. Therefore, the torque depends both on the stator magnetic flux Φ and on the current strength in the rotor winding I 2. However, only the active power consumed by the machine from the network is involved in creating torque. As a result, the torque does not depend on the current strength in the rotor winding I 2, but only from its active component, i.e. I 2 cos φ 2, where φ 2 is the phase angle between e. d.s. and current in the rotor winding.
Thus, the torque of an asynchronous motor is determined by the following expression:
M=CΦ Iφ 2 cos φ 2 , (122)
Where WITH- the design constant of the machine, depending on the number of its poles and phases, the number of turns of the stator winding, the design of the winding and the adopted system of units.
Provided that the applied voltage is constant and the motor load changes, the magnetic flux also remains almost constant.
Thus, in the expression for torque, the quantities WITH and Φ are constant and the torque is proportional only to the active component of the current in the rotor winding, i.e.
M ~ I 2 cos φ 2 . (123)
Changing the load or braking torque on the motor shaft, as is already known, changes both the rotor rotation speed and slip.
A change in slip causes a change in both the current in the rotor I 2 and its active component I 2 cos φ 2 .
The current strength in the rotor can be determined by the ratio e. d.s. to total resistance, i.e. based on Ohm's law
Where Z 2 , r 2 and x 2 - total, active and reactance of the rotor winding phase,
E 2 - e. d.s. phases of the rotating rotor winding.
Changing the slip changes the frequency of the rotor current. With a stationary rotor ( n 2 = 0 and S= 1) the rotating field crosses the conductors of the stator and rotor windings at the same speed and the frequency of the current in the rotor is equal to the frequency of the network current ( f 2 = f 1). As slip decreases, the rotor winding is crossed by a magnetic field with a lower frequency, as a result of which the frequency of the current in the rotor decreases. When the rotor rotates synchronously with the field ( n 2 = n 1 and S= 0), the conductors of the rotor winding are not crossed by the magnetic field, so the frequency of the current in the rotor is zero ( f 2 = 0). Thus, the frequency of the current in the rotor winding is proportional to slip, i.e.
f 2 = Sf 1 .
The active resistance of the rotor winding is almost independent of frequency, while e.g. d.s. and reactance are proportional to frequency, i.e. they change with slip and can be determined by the following expressions:
E 2 = S E And X 2 = S X,
Where E And X- uh. d.s. and the inductive reactance of the winding phase for a stationary rotor, respectively.
Thus we have:
and torque
Therefore, for small slips (up to approximately 20%), when the reactance X 2 = S X small compared to active r 2, an increase in slip causes an increase in torque, since this increases the active component of the current in the rotor ( I 2 cos φ 2). For large slips ( S X more than r 2) an increase in slip will cause a decrease in torque.
Thus, with increasing slip (higher values), although the current strength in the rotor increases I 2, but its active component I 2 cos φ 2 and, consequently, the torque decreases due to a significant increase in the reactance of the rotor winding.
In Fig. 115 shows the dependence of torque on slip. With some sliding S m(approximately 12 - 20%) the engine develops a maximum torque, which determines the overload capacity of the engine and is usually 2 - 3 times the rated torque.
Stable operation of the engine is possible only on the ascending branch of the torque-slip curve, i.e., when the slip changes from 0 to S m. Engine operation on the descending branch of the specified curve, i.e. when sliding S > S m, is impossible, since a stable equilibrium of moments is not ensured here.
If we assume that the torque was equal to the braking torque ( M vr = M torm) at points A And B, then if the balance of moments is accidentally disturbed, in one case it is restored, but in the other it is not restored.
Let's assume that the engine torque has decreased for some reason (for example, when the mains voltage drops), then the slip will begin to increase. If the moment equilibrium was at the point A, then an increase in slip will cause an increase in the engine torque and it will again become equal to the braking torque, i.e., the balance of moments will be restored with increased slip. If the moment equilibrium was at the point B, then an increase in slip will cause a decrease in torque, which will always remain less than the braking torque, i.e., the balance of moments will not be restored and the rotor speed will continuously decrease until the engine stops completely.
Thus, at the point A the machine will work stably, and at the point B stable operation is impossible.
If a braking torque greater than the maximum is applied to the motor shaft, the balance of moments will not be restored and the motor rotor will stop.
The torque of the motor is proportional to the square of the applied voltage, since both the magnetic flux and the current in the rotor are proportional to the voltage. Therefore, a change in network voltage causes a change in torque.
Rotation is a typical type of mechanical movement that is often found in nature and technology. Any rotation occurs as a result of the influence of some external force on the system under consideration. This force creates the so-called What it is, what it depends on, is discussed in the article.
Rotation process
Before considering the concept of torque, let us characterize the systems to which this concept can be applied. A rotation system presupposes the presence of an axis around which circular motion or rotation is carried out. The distance from this axis to the material points of the system is called the radius of rotation.
From the point of view of kinematics, the process is characterized by three angular quantities:
- rotation angle θ (measured in radians);
- angular velocity ω (measured in radians per second);
- angular acceleration α (measured in radians per square second).
These quantities are related to each other by the following equalities:
Examples of rotation in nature are the movements of planets in their orbits and around their axes, and the movements of tornadoes. In everyday life and technology, the movement in question is typical for engine motors, wrenches, construction cranes, opening doors, and so on.
Determination of moment of force
Now let's move on to the immediate topic of the article. According to the physical definition, it is the vector product of the vector of application of force relative to the axis of rotation and the vector of the force itself. The corresponding mathematical expression can be written as follows:
Here the vector r¯ is directed from the axis of rotation to the point of application of the force F¯.
In this formula for the torque M¯, the force F¯ can be directed in any way relative to the direction of the axis. However, a force component parallel to the axis will not produce rotation if the axis is rigidly fixed. In most problems in physics, one has to consider forces F¯, which lie in planes perpendicular to the axis of rotation. In these cases, the absolute value of the torque can be determined using the following formula:
|M¯| = |r¯|*|F¯|*sin(β).
Where β is the angle between the vectors r¯ and F¯.
What is leverage?
The force lever plays an important role in determining the magnitude of the moment of force. To understand what we are talking about, consider the following figure.
Shown here is a rod of length L, which is fixed at the point of rotation by one of its ends. The other end is acted upon by a force F directed at an acute angle φ. According to the definition of moment of force, we can write:
M = F*L*sin(180 o -φ).
The angle (180 o -φ) appeared because the vector L¯ is directed from the fixed end to the free one. Taking into account the periodicity of the trigonometric sine function, we can rewrite this equality as follows:
Now let's turn our attention to a right triangle built on sides L, d and F. By the definition of the sine function, the product of the hypotenuse L and the sine of the angle φ gives the value of leg d. Then we come to equality:
The linear quantity d is called the lever of force. It is equal to the distance from the force vector F¯ to the axis of rotation. As can be seen from the formula, the concept of a force lever is convenient to use when calculating the moment M. The resulting formula says that the maximum torque for a certain force F will occur only when the length of the radius vector r¯ (L¯ in the figure above) is equal to lever of force, that is, r¯ and F¯ will be mutually perpendicular.
Direction of action of the quantity M¯
It was shown above that torque is a vector characteristic for a given system. Where is this vector directed? Answering this question is not particularly difficult if we remember that the result of the product of two vectors is a third vector, which lies on an axis perpendicular to the plane of location of the original vectors.
It remains to decide whether the moment of force will be directed upward or downward (toward or away from the reader) relative to the mentioned plane. This can be determined either by the gimlet rule or by the right hand rule. Here are both rules:
- Right hand rule. If you position the right hand in such a way that its four fingers move from the beginning of the vector r¯ to its end, and then from the beginning of the vector F¯ to its end, then the protruding thumb will point in the direction of the moment M¯.
- The gimlet rule. If the direction of rotation of an imaginary gimlet coincides with the direction of rotational motion of the system, then the translational motion of the gimlet will indicate the direction of the vector M¯. Remember that it only rotates clockwise.
Both rules are equal, so everyone can use the one that is more convenient for them.
When solving practical problems, different directions of torque (up - down, left - right) are taken into account using the "+" or "-" signs. It should be remembered that the positive direction of the moment M¯ is considered to be one that leads to the rotation of the system counterclockwise. Accordingly, if a certain force causes the system to rotate in the direction of the clock, then the moment it creates will have a negative value.
Physical meaning of the quantity M¯
In the physics and mechanics of rotation, the value M¯ determines the ability of a force or a sum of forces to perform rotation. Since the mathematical definition of the value M¯ includes not only the force, but also the radius vector of its application, it is the latter that largely determines the noted rotational ability. To make it clearer what kind of ability we are talking about, here are a few examples:
- Every person, at least once in his life, tried to open a door, not by grasping the handle, but by pushing it close to the hinges. In the latter case, you have to make a significant effort to achieve the desired result.
- To unscrew the nut from a bolt, use special wrenches. The longer the wrench, the easier it is to unscrew the nut.
- To feel the importance of the lever of force, we invite readers to do the following experiment: take a chair and try to hold it suspended with one hand, in one case lean your hand against your body, in another - perform the task with a straight arm. The latter will be an impossible task for many, although the weight of the chair remains the same.
Torque units
A few words should also be said about the SI units in which torque is measured. According to the formula written down for it, it is measured in newtons per meter (N*m). However, these units also measure work and energy in physics (1 N*m = 1 joule). The joule for the moment M¯ does not apply, since work is a scalar quantity, while M¯ is a vector.
However, the coincidence of units of moment of force with units of energy is not accidental. The work done to rotate the system, performed by the moment M, is calculated by the formula:
From this we find that M can also be expressed in joules per radian (J/rad).
Dynamics of rotation
At the beginning of the article, we wrote down the kinematic characteristics that are used to describe the rotational motion. In rotational dynamics, the main equation that uses these characteristics is the following:
The action of the moment M on a system having a moment of inertia I leads to the appearance of angular acceleration α.
This formula is used to determine the angular frequencies of rotation in technology. For example, knowing the torque of an asynchronous motor, which depends on the frequency of the current in the stator coil and on the magnitude of the changing magnetic field, as well as knowing the inertial properties of the rotating rotor, it is possible to determine to what rotation speed ω the motor rotor spins up in a known time t.
Example of problem solution
The weightless lever, which is 2 meters long, has a support in the middle. What weight should be placed on one end of the lever so that it is in a state of equilibrium, if on the other side of the support at a distance of 0.5 meters from it lies a load weighing 10 kg?
Obviously, what will happen if the moments of force created by the loads are equal in magnitude. The force creating the moment in this problem is the weight of the body. The levers of force are equal to the distances from the loads to the support. Let us write the corresponding equality:
m 1 *g*d 1 = m 2 *g*d 2 =>
P 2 = m 2 *g = m 1 *g*d 1 /d 2 .
We obtain the weight P 2 if we substitute from the problem conditions the values m 1 = 10 kg, d 1 = 0.5 m, d 2 = 1 m. The written equality gives the answer: P 2 = 49.05 newton.
Electric motor power and torque
This chapter is devoted to torque: what it is, what it is needed for, etc. We will also look at the types of loads depending on the pump models and the correspondence between the electric motor and the pump load.
Have you ever tried to turn the shaft of an empty pump by hand? Now imagine turning it while the pump is filled with water. You will feel that in this case, much more force is required to create torque.
Now imagine that you have to turn the pump shaft for several hours in a row. You would get tired faster if the pump was filled with water, and you would feel like you had expended a lot more effort in the same period of time than if you were doing the same thing with an empty pump. Your observations are absolutely correct: a lot of power is required, which is a measure of work (energy expended) per unit time. Typically, the power of a standard electric motor is expressed in kW.
Torque (T) is the product of force and force arm. In Europe it is measured in Newtons per meter (Nm).
As you can see from the formula, torque increases if force or leverage increases - or both. For example, if we apply a force of 10 N, equivalent to 1 kg, to a shaft with a lever length of 1 m, the resulting torque will be 10 Nm. When the force increases to 20 N or 2 kg, the torque will be 20 Nm. In the same way, the torque would be 20 Nm if the lever was increased to 2 m and the force was 10 N. Or with a torque of 10 Nm with a force arm of 0.5 m, the force should be 20 N.
Work and power
Now let’s dwell on the concept of “work,” which in this context has a special meaning. Work is done whenever a force - any force - causes movement. Work equals force times distance. For linear motion, power is expressed as work done at a certain point in time.
If we're talking about rotation, power is expressed as torque (T) multiplied by speed (w).
The rotational speed of an object is determined by measuring the time it takes for a certain point on a rotating object to complete a full rotation. Typically this value is expressed in revolutions per minute, i.e. min-1 or rpm. For example, if an object makes 10 complete revolutions per minute, that means its rotation speed is: 10 min-1 or 10 rpm.
So, rotation speed is measured in revolutions per minute, i.e. min-1.
Let's bring the units of measurement to a general form.
For clarity, let's take different electric motors to analyze in more detail the relationship between power, torque and speed. Although the torque and speed of electric motors vary greatly, they can have the same power.
For example, let's say we have a 2-pole motor (3000 rpm) and a 4-pole motor (1500 rpm). The power of both electric motors is 3.0 kW, but their torques are different.
Thus, the torque of a 4-pole electric motor is twice the torque of a two-pole electric motor with the same power.
How are torque and speed generated?
Now that we've covered the basics of torque and speed, we need to look at how they are created.
In AC motors, torque and speed are created by the interaction between the rotor and the rotating magnetic field. The magnetic field around the rotor windings will tend to the magnetic field of the stator. In real operating conditions, the rotor speed always lags behind the magnetic field. Thus, the magnetic field of the rotor crosses the magnetic field of the stator and lags behind it and creates a torque. The difference in the rotational speed of the rotor and stator, which is measured in %, is called the sliding speed.
Slip is the main parameter of the electric motor, characterizing its operating mode and load. The greater the load the electric motor must handle, the greater the slip.
Keeping in mind what was said above, let’s look at a few more formulas. The torque of an induction motor depends on the strength of the magnetic fields of the rotor and stator, as well as on the phase relationship between these fields. This relationship is shown in the following formula:
The strength of the magnetic field primarily depends on the design of the stator and the materials from which the stator is made. However, voltage and frequency also play an important role. The torque ratio is proportional to the square of the stress ratio, i.e. if the supplied voltage drops by 2%, the torque therefore decreases by 4%.
The rotor current is induced through the power supply to which the electric motor is connected, and the magnetic field is partially created by the voltage. The input power can be calculated if we know the motor power supply data, i.e. voltage, power factor, current consumption and efficiency.
In Europe, shaft power is usually measured in kilowatts. In the US, shaft horsepower is measured in horsepower (hp).
If you need to convert horsepower to kilowatts, simply multiply the corresponding value (in horsepower) by 0.746. For example, 20 hp. equals (20 0.746) = 14.92 kW.
Conversely, kilowatts can be converted to horsepower by multiplying the kilowatt value by 1.341. This means that 15 kW equals 20.11 hp.
Motor torque
Power [kW or hp] relates torque to speed to determine the total amount of work that must be done in a given period of time.
Let's look at the interaction between torque, power and speed, and their relationship with electrical voltage, using Grundfos electric motors as an example. Electric motors have the same power rating at both 50 Hz and 60 Hz.
This entails a sharp reduction in torque at 60 Hz: 60 Hz causes a 20% increase in speed, which leads to a 20% decrease in torque. Most manufacturers prefer to specify motor power at 60 Hz, so as line frequency drops to 50 Hz, motors will produce less shaft power and torque. Electric motors provide the same power at 50 and 60 Hz.
A graphical representation of the torque of the electric motor is shown in the figure.
The illustration represents a typical torque/speed characteristic. The following are terms used to describe the torque of an AC motor.
Starting torque(Mp): Mechanical torque developed by an electric motor on the shaft during startup, i.e. when current is passed through an electric motor at full voltage while the shaft is locked.
Minimum starting torque(Mmin): This term is used to refer to the lowest point on the torque/speed curve of an electric motor whose load is increased to full speed. For most Grundfos electric motors, the minimum starting torque is not specified separately, since the lowest point is at the locked rotor point. As a result, for most Grundfos motors the minimum starting torque is the same as the starting torque.
Locking torque(Mblock): Maximum torque is the torque produced by an AC motor at rated voltage, supplied at rated frequency, without sudden changes in rotational speed. It is called the ultimate overload torque or maximum torque.
Torque at full load(MP): Torque required to produce rated power at full load.
Pump Load and Motor Load Types
The following types of loads are distinguished:
Constant power
The term "constant power" is used for certain types of loads that require less torque as the rotation speed increases, and vice versa. Constant power loads are typically used in metalworking applications such as drilling, rolling, etc.
Constant torque
As the name implies - “constant torque” - it is implied that the amount of torque required to operate a mechanism is constant, regardless of the speed of rotation. An example of such an operating mode is conveyors.
Variable torque and power
“Variable torque” - this category is of greatest interest to us. This torque is relevant for loads that require low torque at low speed and require higher torque as the speed increases. A typical example is centrifugal pumps.
The rest of this section will focus solely on variable torque and power.
Having determined that variable torque is typical for centrifugal pumps, we must analyze and evaluate some of the characteristics of a centrifugal pump. The use of variable speed drives is subject to special laws of physics. In this case it is laws of similarity , which describe the relationship between pressure differences and flow rates.
Firstly, the pump flow is directly proportional to the rotation speed. This means that if the pump runs at 25% higher speed, the flow will increase by 25%.
Secondly, the pump pressure will change in proportion to the square of the change in rotation speed. If the rotation speed increases by 25%, the pressure increases by 56%.
Thirdly, what is especially interesting is that power is proportional to the cube of the change in rotation speed. This means that if the required speed is reduced by 50%, this equates to an 87.5% reduction in power consumption.
In summary, the laws of similarity explain why the use of variable speed drives is more appropriate in applications where variable flow and pressure are required. Grundfos offers a range of electric motors with an integrated frequency converter that regulates the speed to achieve exactly this purpose.
Just like feed, pressure and power, the amount of torque required depends on the rotation speed.
The figure shows a cross-section of a centrifugal pump. The torque requirements for this type of load are almost the opposite of those required for "constant power". For variable torque loads, the torque requirement at low speed is low and the torque requirement at high speed is high. In mathematical expression, torque is proportional to the square of the rotation speed, and power is proportional to the cube of the rotation speed.
This can be illustrated using the torque/speed characteristic we used earlier when talking about motor torque:
As the motor accelerates from zero to rated speed, the torque can vary significantly. The amount of torque required at a given load also varies with speed. In order for an electric motor to be suitable for a particular load, it is necessary that the torque of the electric motor always exceeds the torque required for a given load.
In the example, the centrifugal pump at rated load has a torque of 70 Nm, which corresponds to 22 kW at a rated speed of 3000 rpm. In this case, the pump requires 20% torque at rated load when starting, i.e. approximately 14 Nm. After starting, the torque drops slightly and then increases to full load as the pump picks up speed.
Obviously, we need a pump that will provide the required flow/pressure (Q/H) values. This means that the electric motor must not be allowed to stop, in addition, the electric motor must constantly accelerate until it reaches its rated speed. Therefore, it is necessary that the torque characteristic matches or exceeds the load characteristic over the entire range from 0% to 100% rotation speed. Any “excess” moment, i.e. The difference between the load curve and the motor curve is used as the rotation acceleration.
Matching the electric motor to the load
If you need to determine whether the torque of a particular motor meets the load requirements, you can compare the speed/torque characteristics of the motor with the speed/torque characteristics of the load. The torque produced by the motor must exceed the torque required by the load, including periods of acceleration and full speed.
Characteristics of the dependence of torque on the rotation speed of a standard electric motor and centrifugal pump.
If we look at the characteristic, we will see that when accelerating the electric motor, it starts at a current corresponding to 550% of the full load current.
As the motor approaches its rated speed, the current decreases. As would be expected, during the initial start-up period losses on the motor are high, so this period should not be long to prevent overheating.
It is very important that the maximum rotation speed is achieved as accurately as possible. This is related to power consumption: for example, a 1% increase in rotation speed over the standard maximum results in a 3% increase in power consumption.
Power consumption is proportional to the diameter of the pump impeller to the fourth power.
Reducing the diameter of the pump impeller by 10% leads to a decrease in power consumption by (1- (0.9 * 0.9 * 0.9 * 0.9)) * 100 = 34%, which is equal to 66% of the rated power. This dependence is determined solely in practice, as it depends on the type of pump, the design of the impeller and how much you reduce the diameter of the impeller.
Motor start time
If we need to size an electric motor for a specific load, for example for centrifugal pumps, our main task is to provide the appropriate torque and power at the rated operating point, because the starting torque for centrifugal pumps is quite low. The starting time is quite limited, since the torque is quite high.
It is not uncommon for complex motor protection and control systems to take some time to start up so that they can measure the motor's starting current. The starting time of the electric motor and pump is calculated using the following formula:
tstart = time required for the pump motor to reach full load speed
n = motor speed at full load
Itotal = inertia, which requires acceleration, i.e. inertia of the electric motor shaft, rotor, pump shaft and impellers.
The moment of inertia for pumps and motors can be found in the relevant technical data.
Misb = excess torque accelerating rotation. The excess torque is equal to the motor torque minus the pump torque at various speeds.
As can be seen from the above calculations performed for this example with a 4 kW electric motor of a CR pump, the start time is 0.11 seconds.
Number of motor starts per hour
Today's sophisticated motor control systems can control the number of starts per hour for each specific pump and motor. The need to control this parameter is that every time the electric motor is started and then accelerated, a high starting current consumption is noted. The starting current heats the electric motor. If the motor does not cool down, the continuous load from the inrush current will significantly heat the motor stator windings, resulting in motor failure or shortened insulation life.
Typically, the number of starts a motor can make per hour is the responsibility of the motor supplier. For example, Grundfos specifies the maximum number of starts per hour in the technical data for the pump, since the maximum number of starts depends on the moment of inertia of the pump.
Power and efficiency (eta) of the electric motor
There is a direct relationship between the power consumed by the electric motor from the network, the power on the electric motor shaft and the hydraulic power developed by the pump.
In the manufacture of pumps, the following designations are used for these three different power types.
P1 (kW) The electrical input power of the pumps is the power that the pump motor receives from the electrical power source. Power P! is equal to the power P2 divided by the efficiency of the electric motor.
P2 (kW) Motor shaft power is the power that the electric motor transmits to the pump shaft.
P3 (kW) Pump input power = P2, assuming that the coupling between the pump and motor shafts does not dissipate energy.
P4 (kW) Hydraulic power of the pump.
So, for the equilibrium of a body fixed on an axis, it is not the force modulus itself that is important, but the product of the force modulus and the distance from the axis to the line along which the force acts (Fig. 115; it is assumed that the force lies in a plane perpendicular to the axis of rotation). This product is called the moment of force about the axis or simply the moment of force. The distance is called the leverage. Denoting the moment of force by the letter , we get
Let us agree to consider the moment of force positive if this force, acting separately, would rotate the body clockwise, and negative otherwise (in this case, we must agree in advance from which side we will look at the body). For example, forces and in Fig. 116 should be assigned a positive moment, and force a negative one.
Rice. 115. The moment of force is equal to the product of its modulus and the arm
Rice. 116. Moments of forces and are positive, moments of force are negative
Rice. 117. The moment of force is equal to the product of the modulus of the force component and the modulus of the radius vector
The moment of force can be given another definition. Let us draw a directed segment from a point lying on the axis in the same plane as the force to the point of application of the force (Fig. 117). This segment is called the radius vector of the point of application of the force. The vector modulus is equal to the distance from the axis to the point of application of the force. Now let's construct the force component perpendicular to the radius vector. Let us denote this component by . It is clear from the figure that , a . Multiplying both expressions, we get that .
Thus, the moment of force can be represented as
where is the modulus of the force component perpendicular to the radius vector of the point of application of the force, is the modulus of the radius vector. Note that the product is numerically equal to the area of the parallelogram constructed on the vectors and (Fig. 117). In Fig. 118 shows forces whose moments about the axis are the same. From Fig. 119 it is clear that moving the point of application of the force along its direction does not change its moment. If the direction of the force passes through the axis of rotation, then the leverage of the force is zero; therefore, the moment of force is also equal to zero. We have seen that in this case the force does not cause rotation of the body: a force whose moment about a given axis is equal to zero does not cause rotation around this axis.
Rice. 118. Forces and have the same moments about the axis
Rice. 119. Equal forces with the same shoulder have equal moments about the axis
Using the concept of moment of force, we can formulate in a new way the conditions of equilibrium of a body fixed on an axis and under the influence of two forces. In the equilibrium condition expressed by formula (76.1), there is nothing more than the shoulders of the corresponding forces. Consequently, this condition consists in the equality of the absolute values of the moments of both forces. In addition, to prevent rotation from occurring, the directions of the moments must be opposite, that is, the moments must differ in sign. Thus, for the equilibrium of a body fixed on an axis, the algebraic sum of the moments of the forces acting on it must be equal to zero.
Since the moment of force is determined by the product of the modulus of force and the shoulder, we obtain the unit of moment of force by taking a force equal to one, the shoulder of which is also equal to one. Therefore, the SI unit of moment of force is the moment of force equal to one newton and acting on an arm of one meter. It is called a newton meter (Nm).
If a body fixed on an axis is acted upon by many forces, then, as experience shows, the equilibrium condition remains the same as in the case of two forces: for the equilibrium of a body fixed on an axis, the algebraic sum of the moments of all forces acting on the body must be equal to zero. The resulting moment of several moments acting on a body (component moments) is called the algebraic sum of the component moments. Under the action of the resulting moment, the body will rotate around the axis in the same way as it would rotate under the simultaneous action of all component moments. In particular, if the resulting moment is zero, then the body fixed to the axis is either at rest or rotates uniformly.