Total power and its components
Electrical power (P) in physics is a measure that shows how quickly the transformation or transfer of electricity occurs. The unit of measurement is watt (W, W). The value of P depends on the voltage (U) and current (I) in the closed circuit.
For direct current, the load consumed by P is the result of the product of current and voltage:
P = I*U (A*B = W).
P in a constant current circuit
Attention! In this case, the values of both electrical characteristics are constant, which means that at every second of time their values are instantaneous.
The formula changes form if there is a source of electromotive force E (EMF) in the circuit:
P = I*E.
For circuits where the current changes its values periodically along a sinusoid, this formula is not suitable. It is necessary to calculate P based on its instantaneous values in the time interval.
The total power S in its value corresponds to the expression:
S = U*I,
Where:
- U – potential difference at the terminals, (V);
- I – current, (A).
To designate S, use the non-systemic unit B*A (V*A).
Loads included in variable current circuits can be:
- active;
- reactive: capacitive or inductive.
Active load (AH)
A similar load is the elements of devices that have active resistance. The working part of such devices heats up when electricity passes through them.
The current passing through the load does work, which is spent on heating and releasing thermal energy. How is this load measured? It is measured in ohms (ohms).
Examples of AN include: iron, electric stove, hair dryer coils, lamp filament, resistive resistance.
For your information. AN behaves the same way both with constant and with time-varying current.
Example AN
Capacitive load
Devices that can store energy in an electric field and create recirculation (full or partial return) of power are called capacitive loads. A capacitive load (EC) at an alternating voltage, passing current, shifts its phase forward by 900.
The main elements related to EN are:
- capacitors;
- cable lines (capacitance between cores);
- Power lines (power lines) of ultra-high voltage;
- generators operating in overexcitation mode.
EH delivers reactive power (Q).
Circuit with capacitor
Inductive load (IN)
A load in which the current is phase-shifted back from the voltage by 900 is called inductive. She also consumes Q.
Chain with coil
When an inductor (choke), which has a low active resistance, is connected to an alternating voltage network, an emf is formed in it. Electromotive force opposes applied voltage.
Important! In the case of pure inductance L, the resistance to sinusoidal current increases with increasing frequency. The average power P released by such a load is zero.
Examples of ID are:
- asynchronous motors;
- electromagnets;
- chokes;
- reactors;
- transformers;
- rectifiers
This also includes thyristor converters.
Measuring power with a wattmeter
The power consumption of three-phase current is measured using wattmeters. This can be a special wattmeter for a 3-phase network, or a single-phase one connected according to a specific circuit. Modern electricity metering devices are often made using digital circuitry. Such designs are characterized by high measurement accuracy and greater capabilities for operating with input and output data.
Measurement options:
- Star connection with neutral conductor and symmetrical load - the measuring device is connected to one of the lines, the readings taken are multiplied by three.
- Asymmetrical current consumption in a star connection - three wattmeters in the circuit of each phase. The wattmeter readings are summed up;
- Any load and delta connection - two wattmeters connected in a circuit of any two loads. The wattmeter readings are also summed up.
In practice, they always try to make the load symmetrical. This, firstly, improves network parameters, and secondly, simplifies the accounting of electrical energy.
Negative influence of reactive load
Three-phase network power calculation
If we represent the powers in the form of vectors, then the vectors P and Q in sum will give the total power. It is equal to:
S = √ (P2 + Q2).
Total S graph
The formulas for P and Q are:
- P = U*I*sinφ (for a single-phase network) and P = √3* U*I*sinφ (for a three-phase network);
- Q = U*I* cosφ (for a 220 V network) and Q = √3*U*I* cosφ (for a 380 V line).
For your information. Calculations for a three-phase network are correct for symmetrically loaded phases. Otherwise, the powers of all phases are summed up.
The smaller the angle φ between vectors S and P, the higher the power factor cosφ. Complete coincidence of vectors is prevented by Q. The load on power lines and power supply system equipment increases with a large angle value. Overheating of wires and wear of power system equipment occurs.
In practice, the main consumers of manufacturing enterprises are transformers, electric motors and long-distance cables. Therefore, IN is in the lead there, consuming Q. There is an overconsumption of consumed energy, for which enterprises are punished with fines.
Reactive power (RP) carries with it the following disadvantages:
- does not perform useful work;
- causes unnecessary energy consumption and unexpected overload of electrical equipment;
- may cause an emergency.
To compensate for PM, it is necessary to include capacitive elements in parallel with such consumers. For this purpose, Q compensation devices are built. They can be capacitor or inductive, depending on what type of reactive load predominates. Capacitor units, including capacitor batteries, are placed both at power substations and in separate blocks. Such compensation replenishes the reactive component of the energy consumed from the supplier.
Complete condenser unit KKU
Calculation of power by current and voltage
Electric current power formula
You can calculate consumption P by knowing these two parameters I and U of the network. Before choosing cables or wires for wiring in an apartment, you need to decide on the P consumers that can be connected to them. The calculation is made after the measuring instruments record the current readings of the current strength I (A), as well as the voltage U (V).
Single-phase 220 volt network
When connecting an active load to the circuit, use the formula: P = U*I. If there is a phase shift between U and I, use the formula: P = U*I* cosφ.
Three-phase network voltage 380 V
In a three-phase AC network with a phase shift, the result of the last formula is multiplied by √3. The value of the angle cosφ can be clarified in the reference book.
Cosφ table for household devices
When choosing a wire cross-section, the total power of future consumers and the network voltage are usually known.
All you need is the current strength, the formula through power and voltage is:
I = P / (U *cosφ).
The formula for calculating current using power and voltage has the following components:
- P – known power of the device, (W);
- U – supply voltage, (220/380 V);
- cosφ – phase shift angle.
Current calculation can be done using an online calculator.
Online calculator – general view of the interface
Relationship between basic quantities
The most common task faced by ordinary consumers is calculating the actual current strength. So how to correctly calculate amperage based on known voltage and power values? It is necessary to solve it by justifying the cross-sectional values of the cores and the rating of the machine, having technical information about the devices that will be powered into this circuit.
After calculating the current, cables with the smallest permissible cross-section are often selected. However, this is not always correct, since such a solution leads to significant restrictions when it is necessary to add new electrical appliances to the network.
Sometimes it is necessary to carry out reverse calculations and determine what total power can be connected to devices at a known voltage and maximum permissible current, which is limited by existing wiring.
You can solve these two problems for a single-phase circuit using a simple formula:
I = S/U;
S = U * I,
where S is the total total power of all electrical consumers.
A pie chart reflecting Ohm's law and expressing the dependence of power, current, voltage and resistance is suitable for calculating the parameters of a single-phase circuit
To solve the problem of calculating current using known or calculated values of power and voltage in a three-phase circuit, you need to know the total load imposed on each phase.
Both the required cross-section of the cable cores and the minimum permissible rating of the machine are selected according to the busiest line, considering that:
S = 3 * max{S1, S2, S3}.
I = S / (U * 1.73).
The permissible power for each phase can be calculated using the following formula:
S1,2,3 < S / 3 = I * U / 1.73,
where I is the maximum permissible current for existing wiring.
Power at currents: direct and alternating
Calculation of voltage drop in a cable
When it becomes necessary to calculate how much installed equipment will consume, you need to remember that there is a difference between the P value when applying DC and AC voltage.
The formula P for constant current shows P as the product of the instantaneous values of I and U. In this case, the moment in time can be absolutely any.
The expression P under conditions of sinusoidal electron motion takes into account the angle by which the phases of the current and voltage are shifted. The cosine of this angle is multiplied by the product of current and voltage over a period of time T. This is the period of time during which the current changes its value from positive to negative:
T = 1/f, where f is the frequency of 50 Hz.
Features of calculations in alternating electricity circuits
To perform calculations of P, the load consumed in circuits of varying electricity, it is necessary to clearly separate the switching circuits. They can be single-phase or three-phase.
In single-phase circuits, P is found by multiplying the current value by the voltage value (220 V). In this case, the presence of a phase shift between them is taken into account.
In three-phase networks with a voltage of 380 V, two cases are considered:
- symmetrical load per phase;
- asymmetric phase load.
In the first case, P is found using the formula:
P = √3*U*I* cosφ.
In the second case, it is necessary to calculate P for each phase (A, B, C). P value is the result of the summation:
P total = PfA + PfB + PfS.
Attention! When it is necessary to find the total power of a three-phase circuit, the values of reactive Q are found using the same principle.
It is also possible to calculate the current by power, knowing what voltage: phase (220 V) or linear (380 V), by expressing it from the general formula P.
Three-phase network power calculation
Hello readers of my site. Today, using a real example, we will look at a formula with which you can calculate the power (load) of a three-phase network.
But first you need to decide what power you have, since it comes in two types:
1. uniform (symmetrical)
2. uneven (asymmetrical)
An example of a uniform load is when you have an electric motor running. That is, the same current flows through all phases. Not large differences; they can be neglected here. And in the neutral wire the current is zero. In this case, the formula looks like this:
P = √3*Uф*I* cos (φ) = 1.73Uл*I* cos (φ)
Where Uph is the phase voltage
Ul is the linear voltage
I is the current that flows in the conductor. It can be measured with a clamp meter.
cos (φ) – power factor. Usually they take 0.76
An uneven load is when the current in all phases is different. For example, the lighting of a room is powered from a three-phase network. One row of lamps was turned on, and all the lamps there were burning. In the second row, 7 lamps do not light, and in the third, 12. In this case, you need to take a pliers and measure the current in all phases. And the formula will look like this:
Calculation of power consumption
The need arises in calculating electrical power when it is necessary to determine how much energy a particular consumer consumes. Or to calculate the load that the room wiring wires must withstand. To select the diameter of the conductor that will be used for wiring, you need to calculate the total power consumption Ppotr of all household appliances simultaneously plugged into the outlet.
Pin = Pnom*T,
Where:
- Pnom – rated power of the device, (W);
- T – operating time of the device, (h).
If an incandescent lamp with Pnom = 60 W operates for four hours during the day, then Pin = Pnom * T = 100*4 = 400 W.
Table for determining Pnom of some household appliances
Calculation of electrical circuits
All formulas used to calculate electrical circuits follow from one another.
Electrical characteristics relationships
So, for example, using the formula for calculating power, you can calculate the current strength if P and U are known.
To find out how much current an iron (1100 W) connected to a 220 V network will consume, you need to express the current strength from the power formula:
I = P/U = 1100/220 = 5 A.
Knowing the calculated resistance of the electric stove spiral, you can find P of the device. Power through resistance is determined by the formula:
P = U2/R.
There are several methods that allow you to solve the problems of calculating various parameters of a given circuit.
Methods for calculating electrical circuits
Calculating power for various types of current circuits helps to correctly assess the condition of power lines. Household and industrial devices selected in accordance with the specified parameters Pnom and S will work reliably and withstand maximum loads for years.
Calculation of engine power: methods and necessary formulas
Someone needs to calculate the power of the engine unit in order to calculate the car tax.
For some, it is important to independently calculate the power of the compressor engine. For some, it is important to know exactly the power of the machine in order to compare it with the one that was declared. In general, power calculation and engine selection are two inseparable processes. These are not the only reasons why car enthusiasts try to independently calculate the engine power of their cars. This is quite difficult to do without the necessary formulas for calculation. They will be presented in this article so that every motorist can calculate for himself how much the real engine power of his car is.
