Electric motors have a high coefficient of performance (efficiency), but it is still far from the ideal indicators that designers continue to strive for. The thing is that during the operation of the power unit, the conversion of one type of energy into another takes place with the release of heat and inevitable losses. The dissipation of thermal energy can be recorded in different components of any type of engine. Power losses in electric motors are a consequence of local losses in the winding, in steel parts and at mechanical work. Additional losses contribute, albeit insignificantly.
Magnetic power loss
When magnetization reversal occurs in the magnetic field of the armature core of an electric motor, magnetic losses. Their value, consisting of the total losses of eddy currents and those that arise during magnetization reversal, depends on the frequency of magnetization reversal, the values of the magnetic induction of the back and armature teeth. A significant role is played by the thickness of the sheets of electrical steel used and the quality of its insulation.
Mechanical and electrical losses
Mechanical losses during operation of an electric motor, like magnetic ones, are permanent. They consist of losses due to bearing friction, brush friction, and engine ventilation. The use of modern materials, the performance characteristics of which are improving from year to year, allows minimizing mechanical losses. In contrast, electrical losses are not constant and depend on the load level of the electric motor. Most often they arise due to heating of brushes and brush contact. The efficiency decreases due to losses in the armature winding and excitation circuit. Mechanical and electrical losses are the main contributors to changes in engine efficiency.
Additional losses
Additional power losses in electric motors consist of losses arising in equalizing connections and losses due to uneven induction in the armature steel at high loads. Eddy currents, as well as losses in pole pieces, contribute to the total amount of additional losses. It is quite difficult to accurately determine all these values, so their sum is usually taken to be within the range of 0.5-1%. These numbers are used when calculating total losses to determine the efficiency of the electric motor.
Efficiency and its dependence on load
The coefficient of performance (COP) of an electric motor is the ratio of the useful power of the power unit to the power consumed. This indicator for engines with a power of up to 100 kW ranges from 0.75 to 0.9. for more powerful power units, the efficiency is significantly higher: 0.9-0.97. By determining the total power losses in electric motors, the efficiency of any power unit can be calculated quite accurately. This method of determining efficiency is called indirect and it can be used for machines of various capacities. For low-power power units, the direct load method is often used, which consists of measuring the power consumed by the engine.
The efficiency of an electric motor is not a constant value; it reaches its maximum at loads of about 80% of power. It reaches its peak value quickly and confidently, but after its maximum it begins to slowly decrease. This is associated with an increase in electrical losses at loads exceeding 80% of the rated power. The drop in efficiency is not large, which allows us to speak of high efficiency indicators of electric motors in a wide power range.
Example. The average engine thrust is 882 N. For 100 km of travel, it consumes 7 kg of gasoline. Determine the efficiency of its engine. Find a rewarding job first. It is equal to the product of force F and the distance S covered by the body under its influence Аn=F∙S. Determine the amount of heat that will be released when burning 7 kg of gasoline, this will be the work expended Az = Q = q∙m, where q is the specific heat of combustion of the fuel, for gasoline it is equal to 42∙10^6 J/kg, and m is the mass this fuel. The engine efficiency will be equal to efficiency=(F∙S)/(q∙m)∙100%= (882∙100000)/(42∙10^6∙7)∙100%=30%.
In general, to find the efficiency of any heat engine (internal combustion engine, steam engine, turbine, etc.), where work is performed by gas, has an efficiency coefficient equal to the difference in the heat given off by the heater Q1 and received by the refrigerator Q2, find the difference in the heat of the heater and refrigerator, and divide by the heat of the heater efficiency = (Q1-Q2)/Q1. Here, efficiency is measured in submultiple units from 0 to 1; to convert the result to a percentage, multiply it by 100.
To obtain the efficiency of an ideal heat engine (Carnot machine), find the ratio of the temperature difference between the heater T1 and the refrigerator T2 to the heater temperature efficiency = (T1-T2)/T1. This is the maximum possible efficiency for a specific type of heat engine with given temperatures of the heater and refrigerator.
For an electric motor, find the work expended as the product of power and the time it takes to complete it. For example, if a crane electric motor with a power of 3.2 kW lifts a load weighing 800 kg to a height of 3.6 m in 10 s, then its efficiency equal to the ratio useful work Аp=m∙g∙h, where m is the mass of the load, g≈10 m/s² acceleration free fall, h – the height to which the load was lifted, and the work expended Az=P∙t, where P is the engine power, t is the time it operates. Get the formula for determining the efficiency=Ap/Az∙100%=(m∙g∙h)/(P∙t) ∙100%=%=(800∙10∙3.6)/(3200∙10) ∙100% =90%.
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- how to determine efficiency
Efficiency (coefficient of efficiency) is a dimensionless quantity that characterizes operating efficiency. Work is a force that influences a process over a period of time. The action of force requires energy. Energy is invested in strength, strength is invested in work, work is characterized by effectiveness.
Instructions
Calculation of efficiency by determining the energy spent directly to achieve the result. It can be expressed in units necessary to achieve the result of energy, strength, power.
To avoid mistakes, it is useful to keep the following diagram in mind. The simplest includes the elements: “worker”, energy source, controls, paths and elements for conducting and converting energy. The energy spent on achieving a result is the energy spent only by the “working tool”.
Next, you determine the energy actually spent by the entire system in the process of achieving the result. That is, not only the “working tool”, but also the controls, energy converters, and also the costs should include the energy dissipated in the energy conduction paths.
And then you calculate the efficiency using the formula:
Efficiency = (A / B)*100%, where
A – energy required to achieve results
B is the energy actually spent by the system to achieve results. For example: 100 kW was spent on power tool work, while the entire power system of the workshop consumed 120 kW during this time. The efficiency of the system (workshop power system) in this case will be equal to 100 kW / 120 kW = 0.83*100% = 83%.
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Please note
The concept of efficiency is often used to evaluate the ratio of planned energy expenditures to those actually spent. For example, the ratio of the planned amount of work (or the time required to complete the work) to the actual work performed and time spent. You should be extremely careful here. For example, we planned to spend 200 kW on work, but spent 100 kW. Or they planned to complete the work in 1 hour, but spent 0.5 hours; in both cases the efficiency is 200%, which is impossible. In fact, in such cases, what economists call “Stakhanov syndrome” occurs, that is, a deliberate underestimation of the plan in relation to the actually necessary costs.
1. You must evaluate energy costs in the same units.
2. The energy expended by the entire system cannot be less than that spent directly on achieving the result, that is, the efficiency cannot be more than 100%.
Sources:
- how to calculate energy
Tip 3: How to calculate the efficiency of a tank in game World of Tanks
The efficiency rating of a tank or its efficiency is one of the comprehensive indicators of gaming skill. It is taken into account when admitting to top clans, e-sports teams, and companies. The calculation formula is quite complex, so players use various online calculators.
Calculation formula
One of the first calculation formulas looked like this:R=K x (350 – 20 x L) + Ddmg x (0.2 + 1.5 / L) + S x 200 + Ddef x 150 + C x 150
The formula itself is shown in the picture. This formula contains the following variables:
- R – player’s combat effectiveness;
- K – average number of tanks destroyed (total number of frags divided by total number of battles):
- L – intermediate level tank;
- S – average number of detected tanks;
- Ddmg – average amount of damage dealt per battle;
- Ddef – average number of base defense points;
- C – average number of base capture points.
The meaning of the received numbers:
- less than 600 – bad player; About 6% of all players have such efficiency;
- from 600 to 900 – below average player; 25% of all players have such efficiency;
- from 900 to 1200 – average player; 43% of players have such efficiency;
- from 1200 and above – a strong player; there are about 25% of such players;
- over 1800 – a unique player; there are no more than 1% of them.
American players use their WN6 formula, which looks like this:
wn6=(1240 – 1040 / (MIN (TIER,6)) ^ 0.164) x FRAGS + DAMAGE x 530 / (184 x e ^ (0.24 x TIER) + 130) + SPOT x 125 + MIN(DEF,2.2) x 100 + ((185 / (0.17+ e^((WINRATE - 35) x 0.134))) - 500) x 0.45 + (6-MIN(TIER,6)) x 60
In this formula:
MIN (TIER,6) – the average level of the player’s tank, if it is greater than 6, the value 6 is used
FRAGS – average number of tanks destroyed
TIER – average level of the player’s tanks
DAMAGE – average damage in battle
MIN (DEF,2,2) – average number of base capture points shot down, if the value is greater than 2.2, use 2.2
WINRATE – overall winning percentage
As you can see, this formula does not take into account base capture points, the number of frags on low-level vehicles, the percentage of wins and the impact of the initial exposure on the rating does not have a very strong effect.
Wargeiming has introduced in the update an indicator of a player’s personal performance rating, which is calculated using a more complex formula that takes into account all possible statistical indicators.
How to increase efficiency
From the formula Kx(350-20xL) it is clear that the higher the level of the tank, the fewer efficiency points are obtained for destroying tanks, but the more for causing damage. Therefore, when playing low-level vehicles, try to take more frags. At high level – deal more damage (damage). The number of points received or knocked down for capturing a base does not affect the rating much, and more efficiency points are awarded for knocked down capture points than for captured base capture points.
Therefore, most players improve their statistics by playing at lower levels, in the so-called sandbox. Firstly, most players at the lower levels are beginners who have no skills, do not use a pumped-up crew with skills and abilities, do not use additional equipment, and do not know the advantages and disadvantages of a particular tank.
Regardless of what vehicle you play on, try to knock down as many base capture points as possible. Platoon battles greatly increase the effectiveness rating, as players in a platoon act in a coordinated manner and achieve victory more often.
The term "efficiency" is an abbreviation derived from the phrase "coefficient of efficiency". In the very general view it represents the ratio of the resources expended and the result of the work performed using them.
Efficiency
The concept of coefficient of performance (COP) can be applied to the most various types devices and mechanisms whose operation is based on the use of any resources. So, if we consider the energy used to operate the system as such a resource, then the result of this should be considered the amount of useful work performed on this energy.In general, the efficiency formula can be written as follows: n = A*100%/Q. In this formula, the symbol n is used to denote efficiency, the symbol A represents the amount of work done, and Q is the amount of energy expended. It is worth emphasizing that the unit of measurement for efficiency is percentage. Theoretically, the maximum value of this coefficient is 100%, but in practice it is almost impossible to achieve such an indicator, since in the operation of each mechanism there are certain energy losses.
Engine efficiency
The internal combustion engine (ICE), which is one of the key components of the mechanism of a modern car, is also a variant of a system based on the use of a resource - gasoline or diesel fuel. Therefore, the efficiency value can be calculated for it.
Despite all the technical achievements of the automotive industry, the standard efficiency of internal combustion engines remains quite low: depending on the technologies used to design the engine, it can range from 25% to 60%. This is due to the fact that the operation of such an engine is associated with significant energy losses.
Thus, the greatest loss in the efficiency of the internal combustion engine occurs in the operation of the cooling system, which takes up to 40% of the energy generated by the engine. A significant part of the energy - up to 25% - is lost in the process of exhaust gas removal, that is, it is simply carried away into the atmosphere. Finally, approximately 10% of the energy produced by the engine is spent on overcoming friction between the various parts of the internal combustion engine.
Therefore, technologists and engineers involved in the automotive industry are making significant efforts to increase the efficiency of engines by reducing losses in all of the listed items. Thus, the main direction of design developments aimed at reducing losses related to the operation of the cooling system is associated with attempts to reduce the size of the surfaces through which heat transfer occurs. Reducing losses in the gas exchange process is carried out mainly using a turbocharging system, and reducing losses associated with friction is done through the use of more technologically advanced and modern materials when designing the engine. According to experts, the use of these and other technologies can raise the efficiency of internal combustion engines to 80% and higher.
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- About the internal combustion engine, its reserves and development prospects through the eyes of a specialist
Work done by a constant force on a straight section
Let's consider a material point M to which a force F is applied. Let the point move from position M 0 to position M 1, passing the path s (Fig. 1).
To establish a quantitative measure of the influence of force F on the path s, let us decompose this force into components N and R, directed respectively perpendicular to the direction of movement and along it. Since the component N (perpendicular to the displacement) cannot move the point or resist its movement in the direction s, the action of the force F on the path s can be determined by the product Rs.
This quantity is called work and is denoted W.
Hence,
W = Rs = Fs cos α,
that is, the work of a force is equal to the product of its modulus by the path and the cosine of the angle between the direction of the force vector and the direction of movement of the material point.
Thus, work is a measure of the force applied to a material point during some movement.
Work is a scalar quantity.
Considering the work of force, we can distinguish three special cases: the force is directed along the displacement (α = 0˚), the force is directed in the direction opposite to the displacement (α = 180˚), and the force is perpendicular to the displacement (α = 90˚).
Based on the value of the cosine of the angle α, we can conclude that in the first case the work will be positive, in the second – negative, and in the third case (cos 90˚ = 0) the work of the force is zero.
So, for example, when a body moves downward, the work of gravity will be positive (the force vector coincides with the displacement), when the body rises up, the work of gravity will be negative, and when the body moves horizontally relative to the surface of the Earth, the work of gravity will be zero.
Forces that do positive work are called moving forces, forces, and those performing negative work – resistance forces.
The unit of work is the joule (J):
1 J = force×length = newton×meter = 1 Nm.
A joule is the work done by a force of one newton over a path of one meter.
Work of force on a curved section of track
In an infinitely small area ds, the curvilinear path can be conventionally considered rectilinear, and the force can be considered constant.
Then the elementary work dW of the force along the path ds is equal to
dW = F ds cos (F ,v) .
The work on the final displacement is equal to the sum of the elementary works:
W = ∫ F cos (F ,v) ds .
Figure 2a shows a graph of the relationship between the distance traveled and F cos (F ,v). The area of the shaded strip, which can be taken as a rectangle with an infinitesimal displacement ds, is equal to the elementary work on the path ds:
dW = F cos (F ,v) ds ,
F on the final path s is graphically expressed by the area of the figure OABC, limited by the abscissa axis, two ordinates and the curve AB, which is called the force curve.
If the work coincides with the direction of movement and increases from zero in proportion to the path, then the work is graphically expressed by the area of the triangle OAB (Fig. 2 b), which, as is known, can be determined by half the product of the base and the height, i.e., half the product of the force and the path :
W = Fs/2.
Theorem on the work of the resultant
Theorem: the work of the resultant system of forces on a certain section of the path is equal to the algebraic sum of the work of the component forces on the same section of the path.
Let a system of forces (F 1, F 2, F 3,...F n) be applied to the material point M, the resultant of which is equal to F Σ (Fig. 3).
The system of forces applied to a material point is a system of converging forces, therefore,
F Σ = F 1 + F 2 + F 3 + .... + F n.
Let us project this vector equality onto the tangent to the trajectory along which it moves material point, Then:
F Σ cos γ = F 1 cos α 1 + F 2 cos α 2 + F 3 cos α 3 + .... + F n cos α n.
Let's multiply both sides of the equality by an infinitesimal displacement ds and integrate the resulting equality within the limits of some finite displacement s:
∫ F Σ cos γ ds = ∫ F 1 cos α 1 ds + ∫ F 2 cos α 2 ds + ∫ F 3 cos α 3 ds + .... + ∫ F n cos α n ds,
which corresponds to the equality:
W Σ = W 1 + W 2 + W 3 + ... + W n
or abbreviated:
W Σ = ΣW Fi
The theorem is proven.
Theorem about the work of gravity
Theorem: the work done by gravity does not depend on the type of trajectory and is equal to the product of the force modulus and the vertical displacement of the point of its application.
Let a material point M move under the influence of gravity G and, over a certain period of time, move from position M 1 to position M 2, passing the path s (Fig. 4). 
On the trajectory of point M, we select an infinitesimal section ds, which can be considered rectilinear, and from its ends we draw straight lines parallel to the coordinate axes, one of which is vertical and the other horizontal.
From the shaded triangle we get that
dy = ds cos α .
The elementary work of force G on the path ds is equal to:
dW = F ds cos α .
The total work of gravity G on the path s is equal to
W = ∫ Gds cos α = ∫ Gdy = G ∫ dy = Gh.
So, the work done by gravity is equal to the product of the force and the vertical displacement of the point of its application:
W=Gh;
The theorem is proven.
An example of solving the problem of determining the work of gravity
Problem: A homogeneous rectangular array ABCD with a mass m = 4080 kg has the dimensions shown in Fig. 5. 
Determine the work required to flip the array around edge D.
Solution.
Obviously, the required work will be equal to the work of resistance performed by the force of gravity of the array, while the vertical movement of the center of gravity of the array when tipping over edge D is the path that determines the magnitude of the work of gravity.
First, let's determine the gravity of the array: G = mg = 4080×9.81 = 40,000 N = 40 kN.
To determine the vertical displacement h of the center of gravity of a rectangular homogeneous array (it is located at the intersection point of the diagonals of the rectangle), we use the Pythagorean theorem, based on which:
KO 1 = ОD – КD = √(ОК 2 + КD 2) – КD = √(3 2 +4 2) - 4 = 1 m.
Based on the theorem on the work of gravity, we determine the required work required to overturn the massif:
W = G×KO 1 = 40,000×1 = 40,000 J = 40 kJ.
The problem is solved.
Work done by a constant force applied to a rotating body
Let's imagine a disk rotating around a fixed axis under the influence of constant force F (Fig. 6), the point of application of which moves along with the disk. Let us decompose the force F into three mutually perpendicular components: F 1 – circumferential force, F 2 – axial force, F 3 – radial force.
When the disk is rotated through an infinitesimal angle dφ, the force F will perform elementary work, which, based on the resultant work theorem, will be equal to the sum of the work of the components.
It is obvious that the work of the components F 2 and F 3 will be equal to zero, since the vectors of these forces are perpendicular to the infinitesimal displacement ds of the point of application M, therefore the elementary work of force F is equal to the work of its component F 1:
dW = F 1 ds = F 1 Rdφ .
When the disk is rotated through a final angle φ F is equal to
W = ∫ F 1 Rdφ = F 1 R ∫ dφ = F 1 Rφ,
where the angle φ is expressed in radians.
Since the moments of the components F 2 and F 3 relative to the z axis are equal to zero, then, based on Varignon’s theorem, the moment of force F relative to the z axis is equal to:
M z (F) = F 1 R .
The moment of force applied to the disk relative to the axis of rotation is called torque, and, according to the standard ISO, denoted by the letter T:
T = M z (F), therefore, W = Tφ.
The work done by a constant force applied to a rotating body is equal to the product of torque and angular displacement.
Example of problem solution
Task: a worker rotates the winch handle with a force F = 200 N, perpendicular to the radius of rotation.
Find the work expended during time t = 25 seconds, if the length of the handle r = 0.4 m, and its angular velocityω = π/3 rad/s.
Solution.
First of all, we determine the angular movement φ of the winch handle in 25 seconds:
φ = ωt = (π/3)×25 = 26.18 rad.
W = Tφ = Frφ = 200×0.4×26.18 ≈ 2100 J ≈ 2.1 kJ.
Power
The work done by any force can be done over different periods of time, that is, at different speeds. To characterize how quickly work is done, in mechanics there is the concept of power, which is usually denoted by the letter P.
Power is the work done per unit of time.
If the work is done uniformly, then the power is determined by the formula
P = W/t.
If the direction of the force and the direction of displacement coincide, this formula can be written in another form:
P = W/t = Fs/t or P = Fv.
The power of the force is equal to the product of the modulus of the force and the speed of the point of its application.
If work is done by a force applied to a uniformly rotating body, then the power in this case can be determined by the formula:
P = W/t = Tφ/t or P = Tω.
The power of the force applied to a uniformly rotating body is equal to the product of the torque and the angular velocity.
The unit of power is watt (W):
Watt = work/time = joule per second.
Concept of energy and efficiency
The ability of a body to do work when transitioning from one state to another is called energy. Energy is a general measure various forms movement of matter.
In mechanics, various mechanisms and machines are used to transmit and convert energy, the purpose of which is to perform useful functions specified by man. In this case, the energy transmitted by the mechanisms is called mechanical energy, which is fundamentally different from thermal, electrical, electromagnetic, nuclear and other known types of energy. We will look at the types of mechanical energy of the body on the next page, but here we will only define the basic concepts and definitions.
When transmitting or converting energy, as well as when performing work, energy losses occur, since mechanisms and machines used to transfer or convert energy overcome various resistance forces (friction, environmental resistance, etc.). For this reason, part of the energy during transmission is irretrievably lost and cannot be used to perform useful work.
Efficiency
The part of the energy lost during its transfer to overcome resistance forces is taken into account using efficiency mechanism (machine) that transmits this energy.
Efficiency (efficiency) denoted by the letter η and is defined as the ratio of useful work (or power) to expended:
η = W 2 /W 1 = P 2 /P 1.
If the efficiency takes into account only mechanical losses, then it is called mechanical Efficiency.
It's obvious that Efficiency– is always a proper fraction (sometimes expressed as a percentage) and its value cannot be greater than one. The closer the value Efficiency to one (100%), the more economically the machine operates.
If energy or power is transmitted by a number of sequential mechanisms, then the total Efficiency can be defined as a product Efficiency all mechanisms:
η = η 1 η 2 η 3 ....η n ,
where: η 1, η 2, η 3, .... η n – Efficiency each mechanism separately.
As is known, on at the moment mechanisms have not yet been created that would completely convert one type of energy into another. During operation, any man-made device spends part of the energy on resisting forces or wastes it in vain. environment. The same thing happens in a closed electrical circuit. When charges flow through conductors, the full and useful load of electrical work is resisted. To compare their ratios, you will need to calculate the coefficient of performance (efficiency).
Why do you need to calculate efficiency?
The efficiency of an electrical circuit is the ratio of useful heat to total heat.
For clarity, let's give an example. By finding the efficiency of a motor, it is possible to determine whether its primary operating function justifies the cost of electricity consumed. That is, its calculation will give a clear picture of how well the device converts the received energy.
Pay attention! As a rule, efficiency does not have a value, but is a percentage or a numerical equivalent from 0 to 1.
Efficiency is found using a general calculation formula for all devices as a whole. But to get its result in an electrical circuit, you first need to find the force of electricity.
Finding the current in a complete circuit
It is known from physics that any current generator has its own resistance, which is also called internal power. Apart from this meaning, the source of electricity also has its own power.
Let's give values to each element of the chain:
- resistance – r;
- current strength – E;
So, to find the current strength, the designation of which will be - I, and the voltage across the resistor - U, it will take time - t, with the passage of charge q = lt.
Due to the fact that the power of electricity is constant, the work of the generator is entirely converted into heat released to R and r. This amount can be calculated using the Joule-Lenz law:
Q = I2 + I2 rt = I2 (R + r) t.
Then the right sides of the formula are equated:
EIt = I2 (R + r) t.
Having carried out the reduction, the calculation is obtained:
By rearranging the formula, the result is:
This final value will be the electrical force in this device.
Having made a preliminary calculation in this way, the efficiency can now be determined.
Calculation of electrical circuit efficiency
The power received from the current source is called consumed, its definition is written - P1. If this physical quantity passes from the generator into the complete circuit, it is considered useful and is written down - P2.
To determine the efficiency of a circuit, it is necessary to recall the law of conservation of energy. In accordance with it, the power of the receiver P2 will always be less than the power consumption of P1. This is explained by the fact that during operation in the receiver there is always an inevitable waste of converted energy, which is spent on heating the wires, their sheath, eddy currents, etc.
To find an assessment of the properties of energy conversion, an efficiency is required, which will be equal to the ratio of the powers P2 and P1.
So, knowing all the values of the indicators that make up the electrical circuit, we find its useful and complete operation:
- And useful. = qU = IUt =I2Rt;
- And total = qE = IEt = I2(R+r)t.
In accordance with these values, we find the power of the current source:
- P2 = A useful /t = IU = I2 R;
- P1 = A total /t = IE = I2 (R + r).
Having performed all the steps, we obtain the efficiency formula:
n = A useful / A total = P2 / P1 =U / E = R / (R +r).
This formula turns out that R is above infinity, and n is above 1, but with all this, the current in the circuit remains in a low position, and its useful power is small.
Everyone wants to find increased efficiency. To do this, it is necessary to find conditions under which P2 will be maximum. The optimal values will be:
- P2 = I2 R = (E / R + r)2 R;
- dP2 / dR = (E2 (R + r)2 - 2 (r + R) E2 R) / (R + r)4 = 0;
- E2 ((R + r) -2R) = 0.
In this expression, E and (R + r) are not equal to 0, therefore, the expression in brackets is equal to it, that is, (r = R). Then it turns out that the power has a maximum value, and the efficiency = 50%.
As you can see, you can find the efficiency of an electrical circuit yourself, without resorting to the services of a specialist. The main thing is to maintain consistency in the calculations and not go beyond the given formulas.
Video
Efficiency is a characteristic of the operating efficiency of a device or machine. Efficiency is defined as the ratio of the useful energy at the output of the system to the total amount of energy supplied to the system. Efficiency is a dimensionless value and is often determined as a percentage.
Formula 1 - efficiency
Where- A useful work
—Q total work that was spent
Any system that does any work must receive energy from outside, with the help of which the work will be done. Take, for example, a voltage transformer. A mains voltage of 220 volts is supplied to the input, and 12 volts are removed from the output to power, for example, an incandescent lamp. So the transformer converts the energy at the input to the required value at which the lamp will operate.
But not all the energy taken from the network will reach the lamp, since there are losses in the transformer. For example, the loss of magnetic energy in the core of a transformer. Or losses in the active resistance of the windings. Where electrical energy will turn into heat before reaching the consumer. This thermal energy is useless in this system.
Since power losses cannot be avoided in any system, the efficiency is always below unity.
Efficiency can be considered for the entire system, consisting of many individual parts. So, if you determine the efficiency for each part separately, then the total efficiency will be equal to the product of the efficiency coefficients of all its elements.
In conclusion, we can say that efficiency determines the level of perfection of any device in the sense of transmitting or converting energy. It also indicates how much energy supplied to the system is spent on useful work.
