Work, Energy and Powerwebstaff.kmutt.ac.th/~werapon.chi/ENE240/ENE240_2... · Single-Phase...

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1 1 Work, Energy and Power Work is an activity of force and movement in th di ti ff (J l ) the direction of force (Joules) Energy is the capacity for doing work (Joules) Power is the rate of using energy (Watt) P = W / t , Work done in the time E/t E t f di th ti 2 = E / t , Energy transferred in the time

Transcript of Work, Energy and Powerwebstaff.kmutt.ac.th/~werapon.chi/ENE240/ENE240_2... · Single-Phase...

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Work, Energy and Power

• Work is an activity of force and movement in th di ti f f (J l )the direction of force (Joules)

• Energy is the capacity for doing work (Joules)

• Power is the rate of using energy (Watt)

P = W / t , Work done in the time E / t E t f d i th ti

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= E / t , Energy transferred in the time

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Electrical Power

• DC: Instantaneous power at time t

P(t) = I(t) V(t)

= I2(t) R

= V2(t) / R

• AC: Mean power per one cycle

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Pavg = 1/T t=0T P(t) dt

= Irms2 R

= Vrms2 / R

Apparent and True Power• True power is the power used by the load

in Watt unit (W), P = I2R = V2/R

• Reactive power in Volt-Amps-Reactive unit (VAR),

Q = I2X = V2/X• Apparent power is the power that is

applied to the load in Volt-Amps unit (VA)

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applied to the load in Volt Amps unit (VA),

S = I2Z = V2/Z

where impedance Z = R + jX

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Power Factor

cos = True Power / Apparent Power

• Pure resistance, cos(0) = 1 perfect

• Pure inductance, cos(90) = 0

• Pure capacitance, cos(–90) = 0

Im

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True Power

Reactive Power

Re

Im

0

Apparent Power

Power Factor (Cont’d)

P = I V cos

U t ki d f l d b i d• Up to a kind of loads being powered

• In phase or = 0 (resistor)

• Lagging or current lags voltage, > 0 (inductor)

• Leading or current leads voltage, < 0 (capacitor)

• Lower power factor (higher | | ), higher losses

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p ( g | | ), g

• Need to improve power factor

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Power Factor (Cont’d)I = sin(t+t)

= sin(t+t)

V = sinI = sin(+)

= sin(+)

= 0 I = sin() , resistor

= 90 I = sin(+90)

t t

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= 90 I = sin(+90) = cos() , inductor

= –90 I = sin(–90) = –cos() , capacitor

Power Measurement in DC Circuit

• Ammeter measures current which flow into th lt t d l d h i ( )the voltmeter and load, shown in (a).

• Voltmeter measures voltage drop across the ammeter in addition to that dropping across the load, shown in (b).

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Wattmeter

• To measure the true power

T fi d il i d tScale

P i t• Two fixed coil is used to provide a magnetic field that is proportional to a current passing through it, instead of using a permanent magnet

Spring

Pointer

Fixed Coil

Moving Coil(Rotatable)

I1I2

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permanent magnet.

IB

Right Hand’s RuleElectrodynamometeras an ammeter (I2)

(Rotatable)

Meter Terminal

Wattmeter (Cont’d)

• Fixed coils, B I1• Moving coil, Torgue B I2Moving coil, Torgue B I2

= K I1 I2= K I2 , I = I1 = I2 I2 ammeter (non-linear scale)

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Fixed Fixed

to AC sourceto load circuit

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Wattmeter (Cont’d)

• Full-scale deflection current 5-100 mA

• For larger FSD up to 20 A, it is shunted by a low resistance.

• Electrodynamometer can be used for both DC and AC (rms) measurement.

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Fixed Fixed

Shunt

Wattmeter (Cont’d)

• Electrodynamometer as a voltmeter V2 V

Fixed Fixed

to AC sourceto load circuit

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Wattmeter (Cont’d)• Electrodynamometer as a wattmeter

IV • Loading error!• Loading error!

Fixed Fixed

to AC source to load circuit

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Fixed Fixed

to AC source to load circuit

Wattmeter (Cont’d)

• Electrodynamometer

Index Needle

Fixed Coli

Spiral SpringIndex Needle

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Moving Coli

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Advantages• Used on AC as well as DC measurements

• Free from eddy current and hysteresis error (the difference in two measurements of the(the difference in two measurements of the same quantity when the measurement is approached from opposite directions)

• Accurate measurement of RMS values of voltages irrespective of waveforms.

• Because of precision grade accuracy and same calibration for AC and DC measurements, it is useful as transfer type and calibration instruments. 15

Disadvantages

• As the instrument has square law response, the scale is non-uniform.

• These instruments have small torque/weight ratio, so the frictional error is considerable.

• More costly than PMMC instruments.

• Adequate screening of the movements against stray magnetic fields is essential.

• Power consumption is comparably high because of their construction.

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Single-Phase Wattmeter

• Based on the principles of the dynamometer

• The current passing through moving coil and load is proportional to the potential difference across the load.

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Single-Phase Wattmeter (Cont’d)

= K I1 I2

I

V

LoadCurrent Coil

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= K I1 I2= K I (V/R)

= (K/R) (IV)

P

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• Low-current high-voltage load

Single-Phase Wattmeter (Cont’d)

Preading = Iload2(Rfixed+Rload)

Pload = Iload2Rload

Fixed

Supply Load

Iload

• High-current low-voltage loadP V 2 / (R || R )

Wattmeter reads high due to the potential drop across the fixed coil (if neglect the reactance).

MovingSupply Load

Fixed

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Preading = Vload2 / (Rmoving || Rload)

= Vload2 (Rmoving+Rload)

RmovingRload

Pload = Vload2 / Rload

Wattmeter reads high due to the current through the movable coil (if neglect the reactance).

MovingSupply LoadVload

+

• Compensated wattmeter

Single-Phase Wattmeter (Cont’d)

To reduce the systematic error, the compensating

Compensating

Fixed

Supply LoadMoving

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y , p gcoil current is, however, in the opposite direction to the load current and cancels out the proportion of the magnetic flux due to the moving coil current (the same number of turns).

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Crossed-Coil Power Factor Meter• Two moving coils are

crossed and connected in series to an inductor and

1 0.90.90.80.8series to an inductor and

resistor.

• For PF = 1, the coil with R will be in phase with the line current (moving), while the coil with L will be

FixedCoil Load

Lag Lead

0.80.8

Supply

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while the coil with L will be out of phase by 90 (no moving).

• For PF = 0, coil L moves and coil R does not move.

L R

Three-Phase Voltages

• Balanced three-phase voltage

VA = 2 V cos(t) A ( )VB = 2 V cos(t – 120) VC = 2 V cos(t – 240)

= 2 V cos(t + 120)

V V VV

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VA VB VC

0t

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Three-Phase Power DistributionThe three-phase system uses less conductor material

to transmit electric power that the others.

Y-Type, VAB = VAN – VBN

|VAB| = |VAN| cos30 + |VBN| cos30= 3 |VAN| , |VAN| = |VBN|

VLine = 3 VPhase

A VAB

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N120

120BC

3030

VAN

VBN

–VBN

Three-Phase Loads

3-Phase Generator

3-Phase Transmission Line

3-Phase Loads

ILine

Neutral IPhase

ILineVPhase

VLine

IPhase

Star-type balanced-loads,

P1 = P2 = P3 = P = IP VP cos

Star-Type,VLine = 3 VPhase

ILine = IPhase

Delta-Type,VL = VP

IL = 3 IP

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Ptotal = P3 = 3 IP VP cos= 3 (IL) (VL / 3 ) cos= 3 IL VL cos

Similarly, Q = 3 IL VL sin

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Three-Phase Wattmeter• For three-phase system with balanced

load, just using single-phase wattmeter, P3

Voltage Coil

Current Coil

R di

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ReadingP = 3 /3 ILVLcos

Unbalanced-Load 3-Phase Wattmeter

P1

Using 3 Meters

P2

P3

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P3

By using 3 wattmeters,

Ptotal = P1 + P2 + P3

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Balanced-Load 3-Phase WattmeterBrondel’s theory,P = I1V1 + I2V2 + I3V3

P1

Kirchoff’s law,V1 = V3 + V1V2 = V3 + V2

1

P2

S

I1

I2I3

V1

V2

V3

Common

V1

0

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Therefore,P = (I1+I2+I3)V3

+ I1V1 + I2V2= P1 + P2

I3V20

Balanced-Load 3-Phase Wattmeter (Cont’d)

By using 2 wattmeters,

P1 = IR VRB cos1

P1R

IR

VRB = VRS – VBS

Balanced, |VRS| = |VBS|

lagging

P2

B

Y

S

Using 2 Meters

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VBS

–VBS

VRS

VRB

120IR

lagging

60 301 = 30 –

1

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Balanced-Load 3-Phase Wattmeter (Cont’d)

P1 = ILVLcos(30 – )= ILVL ( 3 /2 cos + 1/2 sin )

Similarly, P2 = ILVL ( 3 /2 cos – 1/2 sin )

VBS

V

120

30

2 = 30 +

29Note: cos( ) = cos cos sin sin

–VBSVYS

VYB

IY

Balanced-Load 3-Phase Wattmeter (Cont’d)

Total Power,

Pt t l = P1 + P2Ptotal P1 + P2

= 3 ILVL cos

The same as using 1 meter for balanced load system

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load system

and P1 – P2 = ILVL sin

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Balanced-Load 3-Phase Wattmeter (Cont’d)

Reactive Power,

Q = 3 ILVL sin = 3 (P1 – P2)

(P1–P2)/(P1+P2) = ILVLsin / 3 ILVLcos= tan / 3

tan = 3 (P1 – P2) / (P1 + P2)

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Balanced-Load 3-Phase Wattmeter (Cont’d)

sin2 + cos2 = 1

tan2 + 1 = 1/cos23(P P )2/(P +P )2 + 1 = 1/cos23(P1–P2)2/(P1+P2)2 + 1 = 1/cos2

Power factor,

cos = 1 / [ 1 + 3(P1–P2)2/(P1+P2)2 ]

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Watt-Hour-Meter• To measure the electrical energy supplied to

industrial and domestic consumers in Joules.

Meters for smaller services such as small• Meters for smaller services, such as small residential customers, can be connected directly in-line between source and customer.

• For larger loads, more than about 200 ampere of load, current transformers are used, so that the meter can be located other than in line with the service conductors.

• The meters fall into two basic categories, electromechanical and electronic.

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Watt-Hour-Meter (Cont’d)Electricity meters operate by continuously measuring

the instantaneous voltage and current to give energy used in Unit (kW-hr).

Generated Torque by Eddy Current = kg I V cosMagnetic Breaking Torque = kb N

where N is number of revolution per unit time

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p

(gearing to a mechanical counter to count watt-hours)

kb N = kg I V cosN = (kg/kb) IVcos

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Watt-Hour-Meter (Cont’d)

Voltage Coil

NGear Wheels for Dial Rotation

Rotating Aluminium Disk

Voltage Coil

Current Coil

S

S

NEddy Currrent

Magnetic break

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Current Coil

Therefore, number of revolution refers to the energy used by loads.

References

• P. Purkait, B. Biswas, S. Das and C. K l “El t i l d El t iKoley, “Electrical and Electronics Measurements and Instrumentation”

• https://en.wikipedia.org/wiki/Electricity_meter#Types_of_meters

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