Work force power efficiency. Efficiency. Formula, definition. Test questions and assignments
In reality, the work done with the help of any device is always more useful work, since part of the work is performed against the frictional forces that act inside the mechanism and when moving its individual parts. Thus, using a movable block, they perform additional work by lifting the block itself and the rope and overcoming the frictional forces in the block.
Let us introduce the following notation: useful work will be denoted by $A_p$, total work by $A_(poln)$. In this case we have:
Definition
Efficiency factor (efficiency) called the ratio of useful work to complete work. Let us denote the efficiency by the letter $\eta $, then:
\[\eta =\frac(A_p)(A_(poln))\ \left(2\right).\]
Most often, the efficiency is expressed as a percentage, then its definition is the formula:
\[\eta =\frac(A_p)(A_(poln))\cdot 100\%\ \left(2\right).\]
When creating mechanisms, they try to increase their efficiency, but there are no mechanisms with an efficiency equal to one (let alone more than one).
And so, efficiency is a physical quantity that shows the proportion that useful work makes up of all the work produced. Using efficiency, the efficiency of a device (mechanism, system) that converts or transmits energy and performs work is assessed.
To increase the efficiency of mechanisms, you can try to reduce friction in their axes and their mass. If friction can be neglected, the mass of the mechanism is significantly less than the mass, for example, of the load that lifts the mechanism, then the efficiency is slightly less than unity. Then the work done is approximately equal to the useful work:
The golden rule of mechanics
It must be remembered that winning at work cannot be achieved using a simple mechanism.
Let us express each of the works in formula (3) as the product of the corresponding force and the path traveled under the influence of this force, then we transform formula (3) to the form:
Expression (4) shows that using a simple mechanism, we gain in power as much as we lose in travel. This law is called the “golden rule” of mechanics. This rule was formulated in ancient Greece by Heron of Alexandria.
This rule does not take into account the work of overcoming friction forces, therefore it is approximate.
Energy transfer efficiency
Efficiency can be defined as the ratio of useful work to the energy expended on its implementation ($Q$):
\[\eta =\frac(A_p)(Q)\cdot 100\%\ \left(5\right).\]
To calculate the efficiency of a heat engine, use the following formula:
\[\eta =\frac(Q_n-Q_(ch))(Q_n)\left(6\right),\]
where $Q_n$ is the amount of heat received from the heater; $Q_(ch)$ - the amount of heat transferred to the refrigerator.
The efficiency of an ideal heat engine that operates according to the Carnot cycle is equal to:
\[\eta =\frac(T_n-T_(ch))(T_n)\left(7\right),\]
where $T_n$ is the heater temperature; $T_(ch)$ - refrigerator temperature.
Examples of efficiency problems
Example 1
Exercise. The crane engine has a power of $N$. In a time interval equal to $\Delta t$, he lifted a load of mass $m$ to a height $h$. What is the efficiency of a crane?\textit()
Solution. Useful work in the problem under consideration is equal to the work of lifting a body to a height $h$ of a load of mass $m$; this is the work of overcoming the force of gravity. It is equal to:
We find the total work done when lifting a load using the definition of power:
Let's use the definition of efficiency to find it:
\[\eta =\frac(A_p)(A_(poln))\cdot 100\%\left(1.3\right).\]
We transform formula (1.3) using expressions (1.1) and (1.2):
\[\eta =\frac(mgh)(N\Delta t)\cdot 100\%.\]
Answer.$\eta =\frac(mgh)(N\Delta t)\cdot 100\%$
Example 2
Exercise. An ideal gas performs a Carnot cycle, with the efficiency of the cycle being $\eta$. What is the work done in a gas compression cycle at constant temperature? The work done by the gas during expansion is $A_0$
Solution. We define the efficiency of the cycle as:
\[\eta =\frac(A_p)(Q)\left(2.1\right).\]
Let's consider the Carnot cycle and determine in which processes heat is supplied (this will be $Q$).
Since the Carnot cycle consists of two isotherms and two adiabats, we can immediately say that in adiabatic processes (processes 2-3 and 4-1) there is no heat transfer. In isothermal process 1-2, heat is supplied (Fig. 1 $Q_1$), in isothermal process 3-4 heat is removed ($Q_2$). It turns out that in expression (2.1) $Q=Q_1$. We know that the amount of heat (the first law of thermodynamics) supplied to the system during an isothermal process goes entirely to doing work by the gas, which means:
The gas performs useful work, which is equal to:
The amount of heat that is removed in the isothermal process 3-4 is equal to the work of compression (the work is negative) (since T=const, then $Q_2=-A_(34)$). As a result we have:
Let us transform formula (2.1) taking into account the results (2.2) - (2.4):
\[\eta =\frac(A_(12)+A_(34))(A_(12))\to A_(12)\eta =A_(12)+A_(34)\to A_(34)=( \eta -1)A_(12)\left(2.4\right).\]
Since by condition $A_(12)=A_0,\ $we finally get:
Answer.$A_(34)=\left(\eta -1\right)A_0$
This is the power that it can provide for a long time without overheating above the permissible temperature. The normal service life of a power transformer should be at least 20 years. Since the heating of the windings depends on the amount of current flowing through them, the transformer passport always indicates the total power S nom in volt-amperes or kilovolt-amperes.
Depending on the power factor cosφ 2 at which consumers operate, more or less useful power can be obtained from the transformer. When cosφ 2 = l, the power of consumers connected to it can be equal to its rated power S nom. At cosφ 2.
Power factor.
The power factor cosφ of a transformer is determined by the nature of the load connected to its secondary circuit. As the load decreases, the inductive reactance of the transformer windings begins to have a strong effect and its power factor decreases. When there is no load (at no load), the transformer has a very low power factor, which worsens the performance of AC sources and electrical networks. In this case, the transformer must be disconnected from the AC mains.
Power losses and efficiency.
When transferring power from the primary winding of a transformer to the secondary, power losses occur both in the wires of the primary and secondary windings themselves (electrical losses and or copper losses) and in the steel of the magnetic core (steel losses).
When idling, the transformer does not transmit electrical energy to the consumer. The power it consumes is spent mainly on compensating for power losses in the magnetic circuit due to the action of eddy currents and hysteresis. These losses are called steel losses or no-load losses. The smaller the cross-section of the magnetic circuit, the greater the induction in it, and therefore the no-load losses. They also increase significantly when the voltage supplied to the primary winding increases above the nominal value. When operating powerful transformers, no-load losses amount to 0.3-0.5% of its rated power. Nevertheless, they strive to reduce them as much as possible. This is explained by the fact that steel losses do not depend on whether the transformer is running idle or under load. And since the total operating time of the transformer is usually quite long, the total annual energy losses during no-load operation are significant.
When under load, electrical losses in the winding wires (copper losses) are added to the no-load losses, proportional to the square of the load current. These losses at rated current are approximately equal to the power consumed by the transformer during a short circuit when voltage is applied to its primary winding U k. For powerful transformers they are usually 0.5- 2 % rated power. Reducing the total losses is achieved by appropriate selection of the cross-section of the wires of the transformer windings (reduction of electrical losses in the wires), the use of electrical steel for the manufacture of the magnetic core (reduction of losses from magnetization reversal) and delamination of the magnetic core into a number of sheets isolated from each other (reduction of losses from eddy currents).
The efficiency of the transformer is equal to
The efficiency of the transformer is relatively high and reaches 98-99% in high-power transformers. In low-power transformers, efficiency can decrease to 50-70%. When the load changes, the efficiency of the transformer changes, as the useful power and electrical losses change. However, it remains of great importance in a fairly wide range of load changes (Fig. 119.6). With significant underloads, the efficiency decreases, since the useful power decreases, and the losses in steel remain unchanged. A decrease in efficiency is also caused by overloads, since electrical losses increase sharply (they are proportional to the square of the load current, while useful power is only to the current to the first power). The efficiency has its maximum value at a load when the electrical losses are equal to the losses in steel.
When designing transformers, they strive to ensure that the maximum efficiency value is achieved at a load of 50-75% of the rated load; this corresponds to the most probable average load of the operating transformer. This kind of load is called economic.
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 during mechanical operation. 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 occur. 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 figures are used to calculate the 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 powers. 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 suggests high efficiency indicators of electric motors over a wide power range.
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 is 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 be converted into heat without 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.
In practice, it is important to know how quickly a machine or mechanism does work.
The speed at which work is done is characterized by power.
Average power is numerically equal to the ratio of work to the period of time during which work is performed.
If Dt ® 0, then, going to the limit, we obtain the instantaneous power:
.
(8)
,
(9)
N = Fvcos.
In SI, power is measured in watts(Wt).
In practice, it is important to know the performance of mechanisms and machines or other industrial and agricultural equipment.
For this purpose, the coefficient of performance (efficiency) is used.
The efficiency factor is the ratio of useful work to all expended.
.
(10)
.
1.5. Kinetic energy
The energy possessed by moving bodies is called kinetic energy(W k).
Let's find the total work done by the force when moving the m.t. (body) along the path section 1–2. Under the influence of force, the m.t. can change its speed, for example, it increases (decreases) from v 1 to v 2.
We write the equation of motion of m.T. in the form
Full work 
or 
.
After integration 
,
Where 
called kinetic energy. (eleven)
Therefore,
.
(12)
Conclusion: The work done by a force when moving a material point is equal to the change in its kinetic energy.
The obtained result can be generalized to the case of an arbitrary m.t. system: 
.
Consequently, the total kinetic energy is an additive quantity. Another form of writing the kinetic energy formula is widely used: 
.
(13)
Comment: kinetic energy is a function of the state of the system, depends on the choice of reference system and is a relative quantity.
In the formula A 12 = W k, A 12 must be understood as the work of all external and internal forces. But the sum of all internal forces is zero (based on Newton's third law) and the total momentum is zero.
But this is not the case in the case of the kinetic energy of an isolated system of m.t. or bodies. It turns out that the work done by all internal forces is not zero.
As can be seen from Fig. 6, the work done by force f 12 to move a m.t. with mass m 1 is positive
A 12 = (– f 12) (– r 12) > 0
and the work of force f 21 to move m.t. (body) with mass m 2 is also positive:
A 21 = (+ f 21) (+ r 21) > 0.
Consequently, the total work of the internal forces of an isolated m.t. system is not equal to zero:
A = A 12 + A 21 0.
Thus, the total work of all internal and external forces goes to change kinetic energy.
