AVERAGE POWER
P(t) =1/2 Vm +Im cos(angle phasor of voltage – angle phasor of current )
Where it undergoes process to come up with this equation. It is proven and tested that it undergo in correct process.
Vm = amplitude voltage in particular elements
Im = = amplitude current passing through an element
EXAMPLE PROBLEM:
Current I =20 angle of 50A flows through an impedance Z = 35 angle of -45 ohms. Find the average power delivered to impedance.
*In this problem we need to solve first the voltage to get the components in solving average power,
V = IZ
V = (20 angle of 50) (35 angle of -45)
V = 700 angle of 5v
Now we can solve the average power,
P(t) =1/2 Vm +Im cos(angle phasor of voltage – angle phasor of current )
P(t) =1/2 (700) (20) cos(5-50)
P(t) = 12250 cos(-45)
P(t) = 8662.058 Watts
I learned how to solve average power in AC analysis which is different from the process in DC analysis. It is useful in our lives especially now a days we are in the process of rising economies and having a high technology. As a college student with a course a course of Electrical Engineering, I need to know this theory because it is very important and useful especially in making projects like amplifier. Amplifier need to be match the components and materials to be compatible to each other like the speaker, transformer and etc.
EFFECTIVE or RMS VALUE
In the previous lesson/experiment I learned how to use the multimeter to measure voltages and currents in dc circuits and how to use the oscilloscope to measure the peak voltage or peak-to-peak voltage of an AC waveform. When we used to measure AC voltages or currents, the multimeter gives us something called the effective value, or rms value.
The root-mean-square (rms) value or effective value of an AC waveform is a measure of how effective the waveform is in producing heat in a resistance.
The rms value is am constant itself which depending on the shape of the function i(t).
The Effective value or rms value of an AC waveform is an equivalent DC value.
Example: If you connect a 5 Vrms source across a resistor, it will produce the same amount of heat as you would get if you connected a 5 V dc source across that same resistor. On the other hand, if you connect a 5 V peak source or a 5 V peak-to-peak source across that resistor, it will
Not produce the same amount of heat as a 5 V dc source.
That’s why rms (or effective) values are useful: they give us a way to compare ac voltages to dc voltages.
To show that a voltage or current is an rms value, we write rms after the unit: for example, Vrms = 25 V rms.
P=1/2 VmIm cos(angle of voltage – angle of current)or= Vrms I rms cos(angle of voltage – angle of current)
Resistive load only:
True power, reactive power, and apparent power for a purely resistive load.
Reactive load only:
True power, reactive power, and apparent power for a purely reactive load.
Resistive/reactive load:
True power, reactive power, and apparent power for a resistive/reactive load
I learned that RMS, which actually means Root Mean Square, is a trigonometric way to convert changing values into a more ‘average’ continuous value. So, basically, it’s more of an average performance to compare with other products’ average performances. And rms is useful in our life, example our project in electronics the amplifier, it helps to get the average power in output of the amplifier in order to check and balance the wattages of output.
APPARENT POWER and POWER FACTOR
APPARENT POWER (active + reactive)
Apparent power is a power which is easy to identify, easy to see or to know. Not literally that we see but we can measure the power.
Apparent power is a measure of alternating current (AC) power that is computed by multiplying the root-mean-square (rms) current by the root-mean-square voltage. In a direct current (DC) circuit, or in an AC circuit whose impedance is a pure resistance, the voltage and current are in phase.
Apparent power is measured in units’ volt-ampere (VA).
Apparent power symbol by/represented by “S”.
POWER FACTOR
Power factor is defined as the ratio of real power (P) to apparent power (S). This definition is often mathematically represented as kW/kVA, where the numerator is the active (real) power and the denominator is the (active+ reactive) apparent power or simply PF = P/S
or
PF = cos angle of PF= cos angle of(angle of voltage -angle of current)
.In sinusoid, PF is cosine of the phase difference between the voltage and current.
I learned that apparent power is the power which can be easily to found or to measure and apparent will divide into two power which is the reactive and real power that have their own function.
COMPLEX POWER
Complex power is the product of voltage peak (Vm) and the conjugate(*) of the current divided by two.
or Vrms multiply by Irms.
symbol is bold S or “S” and the unit is VA.
S = V . I*
V is the phasor representation of voltage and I* is the conjugate of current phasor.
So if V is the reference phasor then V can be written as |V| ∠0.
(Usually one phasor is taken reference which makes zero degrees with real axis. It eliminates the necessity of introducing a non zero phase angle for voltage)
Let current lags voltage by an angle φ, so I = | I | ∠-φ
(current phasor makes -φ degrees with real axis) I*= | I | ∠φ
So,
S = |V| | I | ∠(0+φ) = |V| | I | ∠φ
(For multiplication of phasors we have considered polar form to facilitate calculation)
Writting the above formula for S in rectangular form we get
S = |V| | I | cos φ + j |V| | I | sin φ
The real part of complex power S is |V| | I | cos φ which is the real power or average power and the imaginary part |V| | I | sin φ is the reactive power.
So, S = P + j Q Where P = |V| | I | cos φ and Q = |V| | I | sin φ
example:
I learned that complex power is just like Apparent power but the different is complex power is a vector while apparent power is magnitude.
Power Factor Correction
This topic will help you to know the methods of power factor correction.
First, let me define power factor.
Power factor is the ratio between the KW (Kilo-Watts) and the KVA (Kilo-Volt Amperes) drawn by an electrical load, where the KW is the actual load power and the KVA is the apparent load power. It is written as PF= P/S.
It is a measure of how effectively the current is being converted into useful work output and more particularly is a good indicator of the effect of the load current on the efficiency of the supply system.
As what our teacher said, everytime he talks about this topic, all current flow will causes losses in the supply and distribution system. A load with a power factor of 1.0 results in the most efficient loading of the supply, and a load with a power factor of 0.5 will result inmuch higher losses in the supply system.
A poor power factor can be the result of either a significant phase difference between the voltage and current at the load terminals, or it can be due to a high harmonic content or distorted/discontinuous current waveform.
Poor load current phase angle is generally the result of an inductive load such as an induction motor and power transformer.
As what I have learned from the other subject (electronics), which is related on this topic; is that a distorted current waveform can be the result of a rectifier, switched mode power supply, discharge lighting or other electronic load.
A poor power factor due to an inductive load can be improved by the addition of power factor correction, but a poor power factor due to a distorted current waveform requires a change in equipment design or expensive harmonic filters to gain an appreciable improvement.
Many inverters are quoted as having a power factor of better than 0.95 when in reality, the true power factor is between 0.5 and 0.75. The figure of 0.95 is based on the cosine of the angle between the voltage and current, but does not take into account that the current waveform is discontinuous and therefore contributes to increased losses on the supply.
Our teacher gave us an example figure of the balanced three phase system, with balanced 3Φ load, balanced load with the apparent power of 24MVA, 0.78 pf lagging, and wye-connected capacitors. The capacitors are said to be wye-connected because of their common point, which is the neutral line.
This is the figure presented to us by our teacher:
CLICK THE IMAGE TO ENLARGE:
It is similar to single phase case. It uses capacitors to increase the power factor.
Just keep in your mind:
Keep clear about total/phase power, line/phase voltages.
To use capacitors this value should be negative.
For the solution, first we identify the given.
Line voltage = 34.5 kVrms
Frequency= 60Hz
Next, we solve for the necessary parameters, inorder to answer the question, and after we arrived to the answers, we then label the calculated values in the power triangle.
Then, we focus on the question, what value of capacitor increase the pf to 0.94 leading?
I will show you how;
Solving for angle Θ:
Θ= arccos ( 0.78 )
Θ= 38.74°
Since, the apparent power was already given, you can now use different ways on how to solve the remaining parameters.
S= 24MVA
Solving for real power:
P= ScosΘ
P= 18.72 MW
Now, to complete the power triangle, solve for the reactive power:
Q=SsinΘ
Q= 15.01 MVAR
Power Triangle:
CLICK THE IMAGE TO VIEW IT CLEARER:
OLD:
NEW:
Using the 0.94 power factor leading:
Solving for angle Θ:
Θ= arccos ( 0.94 )
Θ=19.94°
Since, the real power is the same with the real power at the power triangle, you can now use different ways on how to solve the remaining parameters.
P= 18.72MW
Solving for real power:
S=P/PF ,
S= 19.91 MVA
Now, to complete the power triangle, solve for the reactive power:
Q=SsinΘ
Q= 6.79 MVAR
Power Triangle:
CLICK THE IMAGE TO VIEW IT CLEARER:
It is inverted because the PF is leading.
SUPER-IMPOSED:
Qcapacitor= Qnew- Qold
Solving for the 3Φ capacitor:
C= Q/2πf V^2rms
C= -21.79×10^6/ 2π(60) (34.5×10^3)^2
C= |-48| microFarad
C= + 48 microFarad
Solving for the per phase capacitor:
jQc= -jωCV^2rms
You will come up with the same answer, which is
C= + 48 microFarad
CLICK THE IMAGE TO ENLARGE:
LESSON LEARNED:
In the power factor correction, you should know how to solve the angles, apparent, real, reactive power, and the other parameters involved to solve the capacitor needed to correct the power factor. You also consider the Power triangle , to analyze the problem better. Most loads on an electrical distribution system fall into one of three categories; resistive, inductive or capacitive. In a maunfacturing plant, the most common is likely to be inductive. Typical examples of this include transformers, fluorescent lighting and AC induction motors. Most inductive loads use a conductive coil winding to produce an electromagnetic field, allowing the motor to function. All inductive loads require two kinds of power to operate: Active power (kwatts) – to produce the motive force Reactive power (kvar) – to energise the magnetic field. Almost all loads are inductive.In order to cancel the reactive component of power, we must add reactance of the opposite type. This is called power factor correction.
Why is Power Factor Important?
For the load with Fp = 0.6, the generator had to supply 133 more amperes in order to do the same work (P)!
Larger current means larger equipment (wires, transformers, generators) which cost more.
Larger current also means larger transmission losses (think I^2R).
Because of the wide variation in possible current requirements due to power factor, most large electrical equipment is rated using apparent power (S) in volt-amperes (VA) instead of real power (P) in watts (W).
For more details, just click on these links:
http://www.youtube.com/watch?v=R2l0qSZPLbM
http://www.youtube.com/watch?v=dPFKcUxbNuQ
http://www.youtube.com/watch?v=9pWOm77KDJM
http://www.youtube.com/watch?v=6Le9_zyWULE
ELECTRICITY CONSUMPTION COST
First, let me define what cost, consumption, and electricity mean:
Cost is an amount paid or required in payment for a purchase; a price.
Consumption is the act or process of consuming ;the state of being consumed; an amount consumed.
Electricity is the most common form of energy used in the home. Different characteristics of electricity are measured using different quantities and units,
Now, you already know their definitions, we must also consider the necessary terms about this topic, including:
VOLTAGE (symbol V, measured in Volts), which is a measure of the force of flow of electricity (and its “potential strength/ potential difference‟ when electricity is not actually flowing).CURRENT (symbol I, measured in Amperes), which is a measure of the volume of flow of electricity.
However, neither voltage nor current is particularly
useful to the householder, because neither is
a measure of the rate of consumption of electricity,
which is what we really want to know…
POWER (symbol P, measured in Watts) is a
measure of the rate at which electricity flows
through an appliance, effectively the rate of
electricity consumption. For example, use your imagination, just imagine the pipe
analogy, if you multiply the water pressure in a pipe
by its diameter, you will determine how much water
is flowing through that pipe every second. So, to
calculate the amount of power used by an appliance,
you simply multiply the force of the flow (in volts)
by the volume of flow (in amps) to get the rate at
which electricity is flowing through it (in watts):
P= V x I
If you look at an appliance’s compliance plate or
specifications, just look for “W‟ or “kW‟!
A 600W hairdryer uses 10 times more power
than a 60W lightbulb. Similarly, a 6kW
(6000W) air-conditioning unit uses 10 times
more power than a 600W hairdryer.
Our teacher gave us a photocopy, which is an electric bill (statement of account) in the year 2012 and 2014, that we need to analyze. The said electric bill is categorized in four groups, generation and transmission, second is distribution, the third one includes S.C discount subsidy and lifeline rate (others), and the fourth one is the government. Here in our place, electricity is solely provided by SOCOTECO II or South Cotabato II Electric Cooperative, Inc. It covers the SOCCSKSARGEN areas of South Cotabato, Cotabato, Sultan Kudarat, Sarangani and General Santos City
SOCOTECO II provides Electricity to all the barangays of General Santos City and ensures that every household and business establishment enjoy the full benefits of a reliable electrical power, with as little power interruptions as possible.
The cooperative has likewise been instrumental in keeping the city abreast with the latest technology in communications by providing efficient Electricity to power communications infrastructure in the ICT field.
Our teacher let us calculate the amount to be paid relating the current charges total amount and the KWH used.
Based on the electric bill given to us students, we calculated this ff. values:
2012:
KWH used: 237
Current bill: P 1620.62
Calculation: P 1620. 62 divided by 237
P 1620.62 / 237 = P 6.838
2014:
KWH used: 293
Current bill: P 2218.03
Calculation: P 2218.03 divided by 293
P 2218.03 / 293 = P 7.570
Upon getting these values, our teacher gave us another activity about this topic. That was a triad activity, wherein we listed different appliances/ other things we used at our homes relating to electricity consumption, and we enjoyed it. In computing, I ended up to a P 2000 plus value of electricity bill in our own households. The value I had calculated in the classroom is almost the same in our actual bill at home, which is 2000 plus also.
Example:
The electricity consumption cost per household depends on family size, living habits, number and age of electrical appliances and hours of usage. Customer can calculate the estimate electricity cost for different appliances if he knows the following:
Power rating of the electrical appliance and its efficiency
Number of hours appliances being used
The domestic tariff rate per kilowatt – hour (kWh)
Electricity consumption usually increases due to the following reasons:
Additional electrical appliances as the family member grow
Electrical loading or size of the appliances
Modern life style leads to using more electrical appliances
Longer usage of appliances
Capacity of appliances which can be adjusted at maximum, result on high load factor, e.g air condition, fan, water heater, etc.
Replacement of smaller appliances to bigger capacity
Faulty appliance will result in appliance operating longer hour and wasting electricity, e.g. refrigerator with faulty thermostat, shortage of refrigerant, or defective door gasket
To calculate: Electricity consumption (kWh) = Power (watts) x Hour of Operation x 30 days ÷ 1000. ( depending on how many days in a month you used the appliance/s)
CLICK THE IMAGES TO ENLARGE:
1000W x 5hr= 5 000 Wh
5 000 Wh x 15 days= 75 000 Wh
75 000/ 1000= 75 kWh
750W x 7 hr= 5 250 Wh
5 250 Wh x 30 days= 157 500 Wh
157 500/ 1000= 157.5 Kwh
150 W x 5 hr= 750 Wh
750 Wh x 30 days= 22 500 Wh
22 500/ 1000= 22.5 Kwh
75W x 7 hr= 525 Wh
525 Wh x 30 days= 15 750Wh
15 750/ 1000= 15.75 Kwh
36 W x 7 hr= 252 Wh
252 Wh x 30 days= 7 560 Wh
7 560/ 1000= 7.56 Kwh
850 W x .5 hr= 425 Wh
425 Wh x 20 days= 8 500 Wh
8 500/ 1000= 8.5 Kwh
850 W x .5 hr= 425 Wh
425 Wh x 15 days= 6 375 Wh
6 375/ 1000= 6. 375 Kwh
730 W x .75 hr= 547.5 Wh
547.5 Wh x 30 days= 16 425 Wh
16 425/ 1000= 16.425 Kwh
We make a table with Quantity, Equipment, Hours/day, Wattage (W), Wattage Hour (WHr), Estimated kWH monthly.
Putting all the necessary information and calculated values in the table we made. Then, to complete the table, we summed up all the electricity consumption (kWh):
Electricity consumption (kWh)= 75 kWh + 157.5 Kwh +22.5 Kwh + 15.75 Kwh + 7.56 Kwh + 8.5 Kwh + 6. 375 Kwh + 16.425 Kwh
Electricity consumption (kWh)= 309.61 kWh
After we had solved the total Kwh, we calculate the amount to be paid based on the values we obtained upon solving it. Different values in the year 2012 and 2014, which are P 6.838 and P 7.570 respectively.
2012: 309.61 kWh *P 6.838 = P 2 117.113
2014: 309.61 kWh *P 7.570 =P 2343.7477
Thus, in the year 2014, the payment increases to 226. 6347 pesos..
The image above relates to the percentage of home energy usage.Electricity is more than numbers on a utility bill; it is the
foundation of everything we do. All of us use electricity every
day—for cooking, heating and cooling rooms,
manufacturing, lighting, and entertainment. We rely on electricity
to make our lives comfortable, productive, and enjoyable.
There are many things we can do to use less electricity and use
it more wisely. These things involve electricity conservation and
electricity efficiency. Electricity conservation is any behavior that
results in the use of less electricity. Electricity efficiency is the use
of technology that requires less electricity to perform the same
function.
TIPS TO SAVE ELECTRICITY TODAY:
Easy low-cost and no-cost ways to save energy
Set a proper schedule in ironing all your clothes.
Use less hot water ,for example shower instead of bath, and take shorter showers. Only fill the kettle as much as you need it. Wash a full load of dishes, rather than one dish at a time. Use cold water where possible for laundry washing.
Reduce/ lower the refrigerator’s thermostat to lower utility bills and manage your heating and cooling systems efficiently.
Air dry dishes instead of using your dishwasher’s drying cycle.
Turn things off when you are not in the room such as lights, TVs, entertainment systems, and your computer and monitor.
Switch off equipment when not in use, turn appliances off at the wall plug, rather than leaving them on standby as this can still draw about 20% or more of normal electricity use. (Examples are TVs, music systems, computers, phone chargers etc.)
Use efficient lighting, Compact Fluorescent Lamps (CFLs) use 75% less power than old incandescent bulbs, and last much longer.
Air dry clothes.
Check to see that windows and doors are closed when heating or cooling your home.
The activity gave us more additional details about electricity consumption and how the electricity conserved at home.
For more information, just click the links below:
http://www.youtube.com/watch?v=ZxN5Hx64jmk
http://www.youtube.com/watch?v=mTYwsSjWowU
http://www.youtube.com/watch?v=9qB9_z6AKLY
Lessons I’ve learned:
Electricity is beneficial because it is clean, cheap, safe and a convenient source of energy. Most of technology is based or needs electricity. One advantage of electricity is that we have a more convenient way of life – light, power to our house, it can give you lighting, heat, and computers wouldn’t work without it, etc. Electricity is called an energy carrier because it is a safe way to move energy from one place to another, and it can be used for so many tasks. As we use more technology, the demand for electricity grows. Learning how to conserve energy and use it efficiently are important goals for everyone. Some important tips I want to advice to save energy for lesser amount/ payment are:
Measure and monitor your home electricity consumption and costs.
Educate everyone in the home, including children and domestic helpers.
Remember that saving requires both behaviour and equipment. Example, it’s no use installing an efficient shower head, if you then shower for twice as long.
In calculating the total wattage:Total wattage = wattage of equipment x number of equipment. For Wattage hour, it is equal to wattage of equipment x hour/day and for Estimated kWh use monthly = (wattage hour x days use in a month)/1000 = kwH.
Therefore, it is better to have a right use of electricity inorder to save money. It’s hard to imagine life without electricity, but you must be responsible enough not to waste electricity, but to use it properly. There, you can minimize your bill. Make sure you’re paying a low price per kilowatt. Use as few kilowatts as possible. With this in mind, we have developed a simple way to help you lessen your bill.