Current power formula. Actual and rated power. The efficiency of an electrical device. Instructions for the laboratory work "Determining the efficiency of an electric kettle" (1 course) Having made the reduction, we get the calculation

It is known that a perpetual motion machine is impossible. This is due to the fact that for any mechanism the statement is true: the total work done with the help of this mechanism (including heating the mechanism and the environment, to overcome the friction force) is always more useful work.

For example, more than half of the work of an internal combustion engine is wasted on heating the components of the engine; some heat is carried away by the exhaust gases.

It is often necessary to evaluate the effectiveness of the mechanism, the feasibility of its use. Therefore, in order to calculate what part of the work done is wasted and what part is useful, a special physical quantity is introduced that shows the efficiency of the mechanism.

This value is called the efficiency of the mechanism

The efficiency of a mechanism is equal to the ratio of useful work to total work. Obviously, the efficiency is always less than unity. This value is often expressed as a percentage. Usually it is denoted by the Greek letter η (read "this"). Efficiency is abbreviated as efficiency.

η \u003d (A_full / A_useful) * 100%,

where η efficiency, A_full full work, A_useful useful work.

Among engines, the electric motor has the highest efficiency (up to 98%). Efficiency of internal combustion engines 20% - 40%, steam turbine about 30%.

Note that for increasing the efficiency of the mechanism often try to reduce the force of friction. This can be done using various lubricants or ball bearings in which sliding friction is replaced by rolling friction.

Efficiency calculation examples

Consider an example. A cyclist with a mass of 55 kg climbs a hill with a mass of 5 kg, the height of which is 10 m, while doing 8 kJ of work. Find the efficiency of the bike. The rolling friction of the wheels on the road is not taken into account.

Solution. Find the total mass of the bicycle and the cyclist:

m = 55 kg + 5 kg = 60 kg

Let's find their total weight:

P = mg = 60 kg * 10 N/kg = 600 N

Find the work done on lifting the bike and the cyclist:

Auseful \u003d PS \u003d 600 N * 10 m \u003d 6 kJ

Let's find the efficiency of the bike:

A_full / A_useful * 100% = 6 kJ / 8 kJ * 100% = 75%

Answer: Bicycle efficiency is 75%.

Let's consider one more example. A body of mass m is suspended from the end of the lever arm. A downward force F is applied to the other arm, and its end is lowered by h. Find how much the body has risen if the efficiency of the lever is η%.

Solution. Find the work done by the force F:

η % of this work is done to lift a body of mass m. Therefore, Fhη / 100 was spent on lifting the body. Since the weight of the body is equal to mg, the body has risen to a height of Fhη / 100 / mg.

Laboratory work

Target: purposeful training of search activity, actualization of the personal meaning of students to study the topic, creation of conditions for the development of communication skills and joint activities.

Tasks:

Educational: experimental work ondetermining the efficiency of electrical appliances using the example of an electric kettle,formation of the ability to establish a connection between the elements of the content of the previously studied material and the new one.

Developing: development of skills of mental operations, improvement of the ability to formulate personally significant goals, promote the development of research and creative skills.

Educational: improving the ability to work in pairs, to form the ability to introspection.

Lesson type : lesson - workshop (2 hours)

Equipment

During the classes

Organizing time

Safety briefing when working with electrical measuring instruments.

Formulation of the problem.

Calculate the work done by the electric current

Calculate the amount of heat received by water and equal to useful work,

determine by experienceThe efficiency of electrical appliances on the example of an electric kettle;

Performing work in accordance with guidelines.

1. Consider an electric kettle. According to the passport data, determine the electrical power of the appliance P.

2. Pour water into the kettle with a volume V equal to 1 liter (1 kg)

3. Use a thermometer to measure the initial water temperature t 1 .

4. Turn on the kettle in the electrical network and heat the water to a boil.

5. Determine the boiling point of water t from the table 2 .

6. Note on the clock the time interval during which the water was heated Δŧ

Perform all measurements in SI.

7. Using the measurement data, calculate:

electric current = P∙Δt

load = cm(t 2 -t 1 )

8. Calculate the efficiency of the electric kettle using the formula

η = =

9. Enter the results of measurements and calculations in the table

P, W

V, m 3

t 1 , 0 FROM

Δt, s

t 2 , 0 FROM

A electric current ,J

Q load , J

ŋ,%

Test questions:

Why is an electric kettle spiral made of a conductor with a large cross-sectional area? Give a detailed answer.

Give examples of other electrical appliances in which the heating element is a coil. How do these devices differ from each other?

WelderWhy, during electric welding, more heat is released precisely at the junction of the welded pieces?(when a high current is passed through metal parts that are in contact on a small plane, at the point of contact of these parts, there is maximum resistance to the passing current and, therefore, the maximum amount of heat is released).

Auto MechanicWhy are low-power devices unprofitable? Why is energy consumption inevitable when using such devices?( Due to the increase in the heating time of water, losses increase by convection, thermal conductivity, radiation)

Output.

5. Homework: prepare a report, repeat the Joule-Lenz law.

Lab report

"Determination of the efficiency of an electric kettle".

Objective: learn experimentally determine the efficiency of electrical appliances using the example of an electric kettle.

Equipment : Electric kettle, thermometer, clock with second hand.

Working process:

1. Considered an electric kettle. According to the passport data, I determined the electrical power of the electrical appliance P.

2. Pour water into the kettle with a volume V equal to 1 liter (1 kg)

3. Measured with a thermometer the initial water temperature t 1 .

4. Turned on the kettle in the electrical network and heated the water to a boil.

5. Determined from the table the boiling point of water t 2 .

6. I noticed by the clock the time interval during which the water was heated Δŧ

7. Using the measurement data, calculated:

a) the work done by the electric current, knowing the power of the kettle P and the time of heating water Δt, according to the formula A electric current = P∙Δt

b) the amount of heat received by water and equal to useful work, Q load = cm(t 2 -t 1 )

8. Calculated the efficiency of the electric kettle using the formula

η = =

9. The results of measurements and calculations entered in the table

P, W

V, m 3

t 1 , 0 FROM

Δt, s

t 2 , 0 FROM

A electric current ,J

Q load , J

ŋ,%

Answers to security questions :

Conclusion of the work done:

That is different types of energy. In this article, we will consider and study such physical concepts as the power of an electric current.

Current power formulas

Under the current power, as well as in mechanics, they understand the work that is performed per unit of time. To calculate the power, knowing the work that an electric current performs over a certain period of time, a physical formula will help.

Current, voltage, power in electrostatics are related by an equality that can be derived from the formula A=UIt. It determines the work that an electric current does:

P=A/t=UI/t=UI
Thus, the formula for direct current power in any section of the circuit is expressed as the product of the current strength and the voltage between the ends of the section.

Power units

1 W (watt) - current power of 1 A (ampere) in a conductor, between the ends of which a voltage of 1 V (volt) is maintained.

A device for measuring the power of electric current is called a wattmeter. Also, the current power formula allows you to determine the power using a voltmeter and ammeter.

An off-system unit of power is kW (kilowatt), GW (gigawatt), mW (milliwatt), etc. Some off-system units of work that are often used in everyday life are also associated with this, for example (kilowatt hour). Insofar as 1kW = 10 3 W, and 1h = 3600s, then

1kW · h \u003d 10 3 W 3600s \u003d 3.6 10 6 W s \u003d 3.6 10 6 J.

Ohm's law and power

Using Ohm's law, the current power formula P=UI is written in this form:

P \u003d UI \u003d U 2 / R \u003d I 2 / R
So, the power released on the conductors is directly proportional to the strength of the current flowing through the conductor and the voltage at its ends.

Actual and rated power

When measuring power in a consumer, the current power formula allows you to determine its actual value, that is, the one that is actually released at a given time on the consumer.

In the passports of various electrical appliances, the power value is also noted. It is called nominal. The passport of an electrical device usually indicates not only the rated power, but also the voltage for which it is designed. However, the voltage in the network may differ slightly from that indicated in the passport, for example, increase. With an increase in voltage, the current strength in the network also increases, and consequently, the current power in the consumer. That is, the value of the actual and rated power of the device may differ. The maximum actual power of the electrical device is greater than the nominal. This is done in order to prevent the failure of the device with minor changes in the voltage in the network.

If the circuit consists of several consumers, then, when calculating their actual power, it should be remembered that for any connection of consumers, the total power in the entire circuit is equal to the sum of the capacities of individual consumers.

Efficiency of an electrical appliance

As you know, ideal machines and mechanisms do not exist (that is, those that would completely convert one type of energy into another or generate energy). During operation of the device, a part of the energy expended is necessarily spent on overcoming undesirable resistance forces or is simply “dissipated” into the environment. Thus, only a part of the energy expended by us goes to perform useful work, for which the device was created.

A physical quantity that shows how much of the useful work is spent is called the efficiency factor (hereinafter referred to as efficiency).

In other words, efficiency shows how efficiently the expended work is used when it is performed, for example, by an electrical appliance.

Efficiency (denoted by the Greek letter η ("this")) is a physical quantity that characterizes the efficiency of an electrical device and shows how much of the useful work is spent.

The efficiency is determined (as in mechanics) by the formula:

η = A P / A Z 100%

If the power of the electric current is known, the formulas for determining the CCD will look like this:

η \u003d P P / P Z 100%

Before determining the efficiency of a device, it is necessary to determine what is useful work (what the device is created for), and what is work expended (work is done or what energy is expended to perform useful work).

A task

An ordinary electric lamp has a power of 60 W and an operating voltage of 220 V. What work does the electric current in the lamp do, and how much do you pay for electricity during the month, at a tariff of T = 28 rubles, using the lamp for 3 hours every day?
What is the current in the lamp and the resistance of its spiral in working condition?

Solution:

1. To solve this problem:
a) calculate the lamp operating time for a month;
b) we calculate the work of the current strength in the lamp;
c) we calculate the monthly fee at the rate of 28 rubles;
d) calculate the current strength in the lamp;
e) we calculate the resistance of the lamp coil in working condition.

2. The work of the current strength is calculated by the formula:

A = P t

The current strength in the lamp will help to calculate the current power formula:

P=UI;
I = P/U.

The resistance of the lamp coil in working condition from Ohm's law is:

[A] = Wh;

[I] \u003d 1V 1A / 1B \u003d 1A;

[R] = 1V/1A = 1Ω.

4. Calculations:

t = 30 days 3 h = 90 h;
A \u003d 60 90 \u003d 5400 W h \u003d 5.4 kW h;
I \u003d 60/220 \u003d 0.3 A;
R \u003d 220 / 0.3 \u003d 733 Ohms;
V \u003d 5.4 kWh 28 k / kWh \u003d 151 rubles.

Answer: A \u003d 5.4 kWh; I \u003d 0.3 A; R = 733 Ohm; B = 151 rubles.

Content:

In the process of moving charges within a closed circuit, a certain work is performed by the current source. It can be useful and complete. In the first case, the current source moves charges in the external circuit, while doing work, and in the second case, the charges move in the entire circuit. In this process, the efficiency of the current source, defined as the ratio of the external and total resistance of the circuit, is of great importance. If the internal resistance of the source and the external resistance of the load are equal, half of all power will be lost in the source itself, and the other half will be released at the load. In this case, the efficiency will be 0.5 or 50%.

Electric circuit efficiency

The considered efficiency is primarily associated with physical quantities characterizing the rate of conversion or transmission of electricity. Among them, in the first place is the power, measured in watts. There are several formulas for its definition: P = U x I = U2/R = I2 x R.

In electrical circuits, there can be a different voltage value and charge value, respectively, and the work performed is also different in each case. Very often there is a need to estimate the speed with which electricity is transmitted or converted. This speed is the electrical power corresponding to the work performed in a certain unit of time. In the form of a formula, this parameter will look like this: P=A/∆t. Therefore, the work is displayed as the product of power and time: A=P∙∆t. The unit of measure for work is .

In order to determine how efficient a device, machine, electrical circuit or other similar system is, in terms of power and work, efficiency is used - efficiency. This value is defined as the ratio of useful energy spent to the total amount of energy supplied to the system. The efficiency is denoted by the symbol η, and mathematically defined as the formula: η \u003d A / Q x 100% \u003d [J] / [J] x 100% \u003d [%], in which A is the work done by the consumer, Q is the energy given by the source . In accordance with the law of conservation of energy, the efficiency value is always equal to or below unity. This means that useful work cannot exceed the amount of energy spent on its completion.

Thus, the power losses in any system or device are determined, as well as the degree of their usefulness. For example, in conductors, power losses are formed when an electric current is partially converted into thermal energy. The amount of these losses depends on the resistance of the conductor, they are not an integral part of the useful work.

There is a difference, expressed by the formula ∆Q=A-Q, which clearly shows the power loss. Here, the relationship between the growth of power losses and the resistance of the conductor is very clearly visible. The most striking example is an incandescent lamp, the efficiency of which does not exceed 15%. The remaining 85% of the power is converted into thermal, that is, into infrared radiation.

What is the efficiency of the current source

The considered efficiency of the entire electrical circuit makes it possible to better understand the physical essence of the efficiency of the current source, the formula of which also consists of various quantities.

In the process of moving electric charges along a closed electrical circuit, a certain work is performed by the current source, which differs as useful and complete. During the performance of useful work, the current source moves charges in the external circuit. At full work, the charges, under the influence of a current source, move already throughout the circuit.

In the form of formulas, they are displayed as follows:

Useful work - Apolesis = qU = IUt = I2Rt.
Complete work - Afull = qε = Iεt = I2(R +r)t.

Based on this, it is possible to derive formulas for the useful and total power of the current source:

Useful power - Рpolez = Apolez / t = IU = I2R.
Apparent power - Рfull = Apfull/t = Iε = I2(R + r).

As a result, the formula for the efficiency of the current source takes the following form:

η = Ause/ Atot = Ruse/ Ptot = U/ε = R/(R + r).

The maximum useful power is achieved at a certain value of the resistance of the external circuit, depending on the characteristics of the current source and load. However, attention should be paid to the incompatibility between maximum net power and maximum efficiency.

Investigation of the power and efficiency of the current source

The efficiency of a current source depends on many factors, which should be considered in a certain sequence.

To determine, in accordance with Ohm's law, there is the following equation: i \u003d E / (R + r), in which E is the electromotive force of the current source, and r is its internal resistance. These are constant values that do not depend on the variable resistance R. With their help, you can determine the useful power consumed by the electrical circuit:

W1 \u003d i x U \u003d i2 x R. Here R is the resistance of the consumer of electricity, i is the current in the circuit, determined by the previous equation.

Thus, the power value using finite variables will be displayed as follows: W1 = (E2 x R)/(R + r).

Since it is an intermediate variable, in this case the function W1(R) can be analyzed for an extremum. To this end, it is necessary to determine the value of R, at which the value of the first derivative of the useful power associated with the variable resistance (R) will be equal to zero: dW1/dR = E2 x [(R + r)2 - 2 x R x (R + r) ] = E2 x (Ri + r) x (R + r - 2 x R) = E2(r - R) = 0 (R + r)4 (R + r)4 (R + r)3

From this formula, we can conclude that the value of the derivative can be zero only under one condition: the resistance of the power receiver (R) from the current source must reach the value of the internal resistance of the source itself (R => r). Under these conditions, the value of the efficiency η will be determined as the ratio of the useful and total power of the current source - W1/W2. Since at the maximum point of useful power the resistance of the energy consumer of the current source will be the same as the internal resistance of the current source itself, in this case the efficiency will be 0.5 or 50%.

Tasks for current power and efficiency

In reality, the work done with the help of any device is always more useful work, since part of the work is done against the friction forces that act inside the mechanism and when moving its individual parts. So, using a movable block, they perform additional work, lifting the block itself and the rope and, overcoming the friction forces in the block.

We introduce the following notation: we denote useful work by $A_p$, and complete work by $A_(poln)$. In doing so, we have:

Definition

Coefficient of performance (COP) called the ratio of useful work to full. We 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 mechanisms with an efficiency equal to one (and even more than one) do not exist.

And so, the coefficient of efficiency is a physical quantity that shows the share that useful work is from all the work done. With the help of efficiency, the efficiency of a device (mechanism, system) that converts or transmits energy that performs work is evaluated.

To increase the efficiency of mechanisms, you can try to reduce the friction in their axes, their mass. If friction can be neglected, the mass of the mechanism is significantly less than the mass, for example, of the load that the mechanism lifts, 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 a gain in 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 by the path traveled under the influence of this force, then we transform formula (3) into the form:

Expression (4) shows that using a simple mechanism, we gain in strength as much as we lose on the way. 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 to overcome friction forces, therefore it is approximate.

Efficiency in power transmission

The efficiency factor 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, the following formula is used:

\[\eta =\frac(Q_n-Q_(ch))(Q_n)\left(6\right),\]

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:

\[\eta =\frac(T_n-T_(ch))(T_n)\left(7\right),\]

where $T_n$ - heater temperature; $T_(ch)$ - refrigerator temperature.

Examples of tasks for efficiency

Example 1

The task. The crane engine has a power of $N$. For a time interval equal to $\Delta t$, he lifted a load of mass $m$ to a height $h$. What is the efficiency of the crane?\textit()

Solution. Useful work in the problem under consideration is equal to the work of lifting the 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:

The total work that is done when lifting a load can be found using the definition of power:

Let's use the definition of the efficiency factor 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

The task. An ideal gas performs a Carnot cycle, while the efficiency of the cycle is equal to $\eta $. What is the work in a gas compression cycle at constant temperature? The work done by the gas during expansion is $A_0$

Solution. The efficiency of the cycle is defined as:

\[\eta =\frac(A_p)(Q)\left(2.1\right).\]

Consider the Carnot cycle, determine in which processes heat is supplied (it will be $Q$).

Since the Carnot cycle consists of two isotherms and two adiabats, we can immediately say that there is no heat transfer in adiabatic processes (processes 2-3 and 4-1). 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 completely to perform 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:

We transform the 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,\ $finally we get:

Answer.$A_(34)=\left(\eta -1\right)A_0$