Power Measurement and Energy Consumption Cost

Power Measurement

Power is symbolize as (P). It is defined as the amount of energy consumed per unit time. The unit of power is known as the Watt (W). The average power that is absorbed by the load is measured by a wattmeter. Loads consume electric power, converting it to other forms such as mechanical work, heat, light, etc. Examples of loads are electrical appliances, such as light bulbs, electric motors, and electric heaters. When we are using AC, power is determined not only by the r.m.s. values of the voltage and current, but also by the phase angle which will determine the power factor.

Energy Consumption Cost

Every appliance we have in our house has its own corresponding power. This power really matters on how much we pay in our electric bills that is why it is important to know how much power present in our appliance. Since we are paying for the electric energy over a period of time, we have to consider how long we use our appliances in our house.

TYPICAL WATTAGES OF VARIOUS APPLIANCES

·        Aquarium = 50–1210 Watts

·        Clothes washer = 350–500

·        Clothes dryer = 1800–5000

·        Fans Ceiling = 65–175

·        Hair dryer = 1200–1875

·        Heater (portable) = 750–1500

·        Clothes iron = 1000–1800

·        Microwave oven = 750–1100

·        Personal computer

·        CPU - awake / asleep = 120 / 30 or less

·        Monitor - awake / asleep = 150 / 30 or less

·        Laptop = 50

·        Radio (stereo) = 70–400

·        Refrigerator (frost-free, 16 cubic feet) = 725

·        Televisions (color)

·        19" = 65–110

·        27" = 113

·        36" = 133

·        53" - 61" Projection = 170

·        Flat screen = 120

·        Toaster = 800–1400

·        VCR/DVD = 17–21 / 20–25

·        Vacuum cleaner = 1000–1440

Formulas :

Power = Energy / Time

Energy = Power x Duration of Usage (Time)

Energy consumption per day = Power Consumption  x Hours used / Number of Days

TIPS TO CONSERVE ENERGY

·  Clean or replace furnace and air-conditioner filters regularly, following  manufacturer's instructions.
·  Set the thermostats of the refrigerator at the appropriate temperature.
·  Leave thermostat’s fan switch on “auto”.
·  Lessen the hours of using the appliances.

·  Unplug electronic appliances and gadgets when not in use.

·  When buying new appliances, be sure to purchase energy-efficient 

·  Replace light bulbs with CFL’s.

I've learned that the more power, wattage and energy used per unit of time, The higher electrical bill we will have. To conserve energy and to decrease our bill, we must use house efficient appliances. Those appliances which generates heat, contributes to high consumption of electricity such as Rice Cooker, Electric Stove, Flat Iron and Etc. 

Balanced Three-Phase Circuits

Balanced Three-Phase Circuits 

The voltages in the three-phase power system are produced by a
synchronous generator. In a balanced system, each of the three
instantaneous voltages have equal amplitudes but are separated from the
other voltages by a phase angle of 120. The three voltages (or phases) are
typically labeled a, b and c. The common reference point for the three
phase voltages is designated as the neutral connection and is labeled as n.
We may define either a positive phase sequence (abc) or a negative phase
sequence (acb) as shown below. The three sources Van, Vbn and Vcn are
designated as the line-to-neutral voltages in the three-phase system.

LINE-TO-LINE VOLTAGES

An alternative way of defining the voltages in a balanced three-phase
system is to define the voltage differences between the phases. These
voltages are designated as line-to-line voltages. The line-to-line voltages
can be expressed in terms of the line-to-neutral voltages by applying
Kirchoff’s voltage law to the generator circuit, which yields

Inserting the line-to-neutral voltages for a positive phase sequence into the
line-to-line equations yields

If we compare the line-to-neutral voltages with the line-to-line voltages, we find the following relationships,

THREE-PHASE CONNECTIONS

The sources and loads in a three-phase system can each be connected in either a wye (Y) or delta ()) configuration. Note that the wye connections are line-to-neutral while the delta connections are line-to-line with no neutral. Also note the convention on the node designations (lowercase letters at the source connections and uppercase letters at the load

connections).

BALANCED WYE-WYE CONNECTION

The balanced three-phase wye-wye connection is shown below. Note that the line impedance for each of the individual phases in included in the circuit. The line impedances are assumed to be equal for all three phases. The line currents (IaA, IbB and IcC) are designated according to the source/load node naming convention. The source current, line current, and load current are all one in the same current for a given phase in a wye-wye connection.

Note that the line current magnitudes are equal and each line current lags the respective line-to-neutral voltage by the impedance phase angle 2Z. Thus, the balanced voltages yield balanced currents. The phasor diagram for the line currents and the line-to-neutral voltages is shown below. If we lay the line-to-neutral voltage phasors end to end, they form a closed triangle (the same property is true for the line currents). The closed triangle shows that the sum of these phasors is zero.

BALANCED DELTA - WYE CONNECTION

A balanced delta - wye system consists of a balanced delta connected source and a balanced connected load. There are also the line voltages as well as the phase voltages. We can generate an equation through getting a loop from the circuit and this may help us solved for the line currents. Like in other connections, We may also transform delta connected souce to wye connected source.

THREE-PHASE CONNECTIONS INVOLVING DELTA SOURCES OR LOADS
In addition to the wye-wye three-phase connection, there are three

other possible configurations of wye and delta sources and loads.

The most efficient way to handle three-phase circuits containing delta sources and/or loads is to transform all delta connections into wye connections.



I've learned that in a connected balanced source, Line currents and phase  currents are equal,while in a delta - connected balanced source, line voltage and phase voltage is equal also. It is easy to solve for the parameters if the system is in wye-wye connection. so for us to have an easy solving, we have to convert it into wye connection. 

Power Triangle

What is Power Triangle?

The relationship between real power, reactive power and apparent power. Which can be expressed by representing the quantities as vectors. Real power is represented as a horizontal vector and reactive power is represented as a vertical vector. The apparent power vector is the hypotenuse of a right triangle formed by connecting the real and reactive power vectors. This representation is often called the power triangle. Using the Pythagorean Theorem, the relationship among real, reactive and apparent power is: 

(apparent power)^2 = (real power)^2 + (reactive power)^2 

Real and reactive powers can also be calculated directly from the apparent power, when the current and voltage are both sinusoids with a known phase angle between them.

Below is the figure of a Power triangle.

The power triangle graphically shows the relationship between real (P), reactive (Q) and apparent power (S).

The power triangle also shows that we can find real (P) and reactive (Q) power, given S and the impedance angle θ.

Sample Problem :

Calculate the complex power for the circuit and correct the power factor to
unity by connecting a parallel reactance to the load.

Solution
Known Quantities: Source voltage; load impedance.
Find:

1. S = Pav + jQ for the complex load.

2. Value of parallel reactance required for power factor correction resulting in pf = 1.

Schematics, Diagrams, Circuits, and Given Data: 

VS = 117∠0 V; RL = 50 ohms ; jXL = j86.7 ohms .

Assumptions: Use rms values for all phasor quantities in the problem.

Analysis:

First, we compute the load impedance:

ZL = R + jXL = 50 + j86.7 = 100∠1.047

Next, we compute the load current:

and the complex power, as defined in equation
S = VI = 117∠0 × 1.17∠1.047 = 137∠1.047 = 68.4 + j118.5 W


Therefore
Pav = 68.4 W Q = 118.5 VAR

To compute the reactance needed for the power factor correction, we observe that we need
to contribute a negative reactive power equal to −118.5 VAR. This requires a negative
reactance and therefore a capacitor with QC = −118.5 VAR. The reactance of such a

capacitor is given by

In this lesson, I've learned that this topic is really essential and "MUST LEARN" topic because you cannot proceed to other topics for a reason that this topic is the base line of them all. You cannot find the values of the apparent power, real power and complex power if you don't know how to draw the power triangle and how it is used. In this particular topic, We students must know also the different formula to be use so that we can compute for the different power.

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

In this topic, I've learned that  resistors absorbs the real power and dissipates in the form of heat. Capacitors absorbs the reactive power and dissipates in the form of electric. Inductor absorbs the reactive power and dissipates in the form of magnetic field. Using this topic complex power, we can determine if it is inductive, resistive or capacitive. In complex power, we have to consider the phase network which voltage and current represented in complex form.

Apparent 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.

In  this topic, I find apparent power easy to solve for a reason that it is easy to found/measure. Apparent will be divided into two power which is the reactive and real power that has their own function.

Maximum Average Power Transfer

Maximum Average Power Transfer

The maximum power transfer theorem states that, to obtain maximum external power from a source with a finite
 internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

RL= Re{ZTh} and XL = - Im{ZTh}

Where V2Th and I2N represent the square of the sinusoidal peak values.

We’ll next illustrate the theorem with some examples.

Example 1

R1 = 5 kohm, L = 2 H, vS(t) = 100V cos wt, w = 1 krad/s.

a) Find C and R2 so that the average power of the R2-C two-pole will be maximum

b) Find the maximum average power and the reactive power in this case.

c) Find v(t) in this case.

The solution by the theorem using V, mA, mW, kohm, mS, krad/s, ms, H, m F units:v

a.) The network is already in Thévenin form, so we can use the conjugate form and determine the real and imaginary components of Z_Th:

R₂ = R₁ = 5 kohm; wL = 1/w C = 2 ® C = 1/w²L = 0.5 mF = 500 nF. 

b.) The average power: 

P_max = V²/(4*R₁) = 100²/(2*4*5) = 250 mW

 The reactive power: first the current: 

I = V / (R₁ + R₂ + j(wL – 1/wC)) = 100/10 = 10 mA

Q = - I²/2 * X_C = - 50*2 = - 100 mvar

c.) The load voltage in the case of maximum power transfer: 

V_L = I*(R₂ + 1/ (j w C ) = 10*(5-j/(1*0.5)) =50 – j 20 = 53.852 e ^{-j 21.8°} V

and the time function: v(t) = 53.853 cos (wt – 21.8°) V

In this topic, I noticed that the way we solve DC Circuits compared to AC Circuits is the same but there are some that is not.  In solving problems with regards to Maximum average Power Transfer is that we really have to inspect carefully the loads so that we can get correctly the needed unknown and Maximum Power transfer will occur.

EFFECTIVE or RMS VALUE

EFFECTIVE or RMS VALUE

• We have a few different ways to specify the size of an ac current or voltage.
• We can give either
• the peak value, or
• the peak-to-peak value, or
• Something called the effective value (also called rms value).

• These distinctions apply only to ac, not to dc.

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.                                                                                                                        

In this topic, I've learned that RMS (Root Mean Square) is used to measure varying signals effective value and the mathematical relationship to peak voltage do varies depending on the type of waveform.