Poynting Vector

1

Poynting Vector and the

Flow of Power

G K SureshProfessor and Head

Department of E & C and TC ES I T Tumkur 572103

email: profgksuresh @ yahoo.com]ph: 95816-2214035 (O) 95816-2284852 (R)

2

Poynting TheoremWe have the Maxwell’s first curl equation

DH J

t

This equation can be rewritten as

D EJ H H

t t

Pre dotting the equation (2) with we get,

D E EE J E H E E H

t t

E

(1)

(2)

(3)

3

We have the vector identity

E H H E E H

But the second Maxwell’s curl equation is

0

HE

t

Using equations (4) and (5) in equation (3), we get

(4)

(5)

EE J H E E H E

t

0

H E EE J H E H

t t

(6)

(7)

4

0 0

H EE J H E E H

t t

(9)

(10)

(8)

But 2 21 1

2 2

H EH H and E E

t t t t

Substituting equation (9) in equation (8), we get

2 20 0

1 1

2 2E J H E E H

t t

Integrating both sides of equation (10) over a volume V, we get

5

2 2

2 2E J d H E d E H d

t

(11)

We apply divergence theorem to the last term on the LHS, convert the volume integral into a surface integral and get

2 2

2 2 S

E J d H E d E H d St

(12)

Let us get the physical interpretation of each term of the equation (12). We consider the equation term by term.

Term 1 Term 2 Term 3

6

Term 1 : Using generalised ohm’s law , the term on the left hand side ( E . J i.e., ) represents instantaneous ohmic power dissipated in the volume v.

.E I

E J watts per unit volumeA

J E

2 ,E E E

A conductor of cross sectional area a carrying a current I andhaving a voltage drop E per unit length will have a power loss of EI watts per unit length. The power dissipated per unit volume would be

In this case, E and J have the same direction. In general this may not be true. In such a case, the power dissipated per unit volume still is given by the product of J and the component of E along the direction of J

7

i.e., the power dissipated per unit volume is always given by

E . J

and therefore the term E J d

represents the total power dissipated in a volume v

(13)

In particular, when E represents the electric field strength required to produce the current density J in the conduction medium, the expression (13) represents power dissipated as ohmic ( I2 R ) loss.

But, if the E is an electric field strength due to a source of power, ( like due to a battery), then the power represented by (13) would be used up in driving current against the battery voltage and hence charging the battery.

8

If the direction of E is opposite to that of J, The dissipated power represented by equation (13), ie., term 1 in equation (12) would be negative. In this case, the battery would be generation power.

Term 2: In The first term on the RHS, the quantity ,in electrostatics, represents the energy density or stored electric energy per unit volume of the electric field. Also in steady magnetic fields, the quantity represents the magnetic energy density stored in the volume v. Let us assume that these quantities continue to represent the stored energy densities for time varying fields as well. Then the integral (term 2 in equation (12) ) represents the time rate at which the stored total energy inside the volume is decreasing.

21

2E

21

2H

9

Term 3. We use the conservation of energy to interpret the last term of equation (12) , the term 3.

The rate of energy dissipation in the volume v must be equal to the rate at which the stored energy inside the volume is decreasing plus the rate at which energy is entering the volume v from outside.

Thus the term of equation (12) represents the rate of flow of energy inward through the surface of the volume.

S

E H d S

10

Thus the expression without the negative sign, i.e.,

S

E H d S

represents the rate of flow of energy outward through the surface enclosed by the volume.

The interpretation of equation (12) leads to the conclusion that the integral of E X H over any closed surface gives the rate of energy flow through that surface. It is seen that the vector

P E H

has the dimension of watts per square metre.

(14)

11

It is the Poynting theorem.

P E H

Poynting theorem:

The vector product

at any point is a measure of the rate of energy flow perunit area at that point ; the direction of energy flow is perpendicular to E and H in the direction of the vector E X H .

(15)

12

Example: Power flow for a uniform plane wave.

The uniform plane travels in free space with a velocity c

80

0 0

13 10 /c m s

The total energy density due to electric and magnetic fields is given by

2 21

2E H

For a wave traveling with a velocity v0 the rate of flow of energy per unit area is

2 20

1

2P E H (16)

13

We know that for a Uniform plane wave,

E

H

Therefore equation (16) becomes

0

1

2P E H E H

00

E HP

P E H

i.e.,

14

Example 2 :Poynting vector about ac transmission lines

When a transmission line delivers ac power, the voltage and therefore the electric and magnetic fields, vary with time. Also, if it is a long line, the phases lof voltage and current (and E and H ) will vary along the length of the line.

For the simple case of a lossless line terminated in its characteristic impedance which is a pure resistance, the variation in time and along the line of both voltage and current will be given by

cosz

tv

15

For any value of z and t there will be a certain distribution of the Poynting vector over a plane parallel to the x and y axes. At every point in this plane, P will be parallel to the z axis. The Poynting vector is given by

2cos ( , )z

P E H A t f x yv

The function f( x ,y) will not vary with z or t. for a fixed valueof time, the total power passing through a plane will vary with the position of the plane, that is with z, whereas for a fixed value of z the power through the plane will vary with time. It will be noted that the power flow past a given plane is pulses of double frequency, a fact readily appreciated when observing the flicker of a 25 Hz bulb.

16

Instantaneous, Average and Complex Poynting vector

In ac circuits, the Instantaneous power is always given by the product of the instantaneous voltage and the instantaneous current as

WV

I

W V I

The quantities may be expressed in terms of phasors V and I as follows:

V and I

Re{ } Re{| | } | | cos( )

Re{ } Re{| | } | | cos( )

jj t j t

j t j i j ti

V V e V e e V t

I I e I e e V t

17

| || | cos( ) cos( )

| || |cos( ) cos(2 )

2

i

i i

W V I t t

V It

We observe that this consists of a dc (average) part and an oscillatory part. Therefore the average power is

dc (average) part oscillatory part

| | | |cos

2av

V IW

where θ = θv - θi

VreIre

V

I

Iim

Vimθ v- θ i

θ v

θ i

(17)

18

Reactive Power:

Wav and Wreact are the in-phase and out – of – phase components of the volt ampere product. The presence of an out-of-phase or reactive component indicates that the instantaneous power changes sign (reversed\s direction of flow ) over the part of an ac cycle.

| | | |sin

2react

V IW

Another useful quantity, reactive power,also called VAR, Reactive Volt- Ampere

(18)

19

Consider now the complex power, W defined as one-half of the product of V and the complex conjugate of I

Complex power, W

1 1

2 2v ij j

reactW V I V e I e

1

2jV I e

av reactW jW

Similar relations exist between Poynting Vector, the electric field strength and the magnetic field strength:

(19)

(20)

20

P E H

In phasor form, we may write the above relation as

1

2P E H

This gives the expressions for the average and reactive parts of power flow per square meter as

1Re{ }2avP E H

1Im{ }2reactP E H

(21)

(22)

(23)

(24)

21

Only mutually perpendicular components of E and H contribute to power flow; the direction of the flow is normal to the plane containing E and H.

Thus in rectangular coordinates, the complex flow of power per unit area normal to y-z plane is

1

2x y z z yP E H E H

In other directions, corresponding expressions can be obtained.

(25)

22

In Spherical coordinates the outward flow of complex power per unit area is

1

2xP E H E H

Poynting Vector in Complex Form:

The Maxwell’s Curl Equations in Phasor form are

( )H j E J and E j H

Here J represents nonohmic currents like convection currents or specified source current.

(26)

23

Real and Imaginary parts of the complex Poynting Vector :

( )E H H E E H

Using the Maxwell’s curl equations in the above we get

( ) ( )E H H j H E J j E E

( )E H j H H j E E E J

We have the Vector identity

Integrating over the volume V surrounded by the surface S we get

24

Time average Stored energyDensities, electric and magnetic

( )S v v v

E H dS j H H E E dv E E dv E J dv

1 1Re Re

2 2S v v

P dS E E dv E J dv

The Poynting theorem next is separated into real and imaginary parts as

1Im 2 ( ) Im

2S v v

P dS H H E E dv E J dv

(27)

(28)

25

For the uniform plane wave propagating in +z direction, we have associated Ex and Hy components. For this we obtain

ˆ ˆ ˆx x y y z zE a H a P a

In perfect dielectrics, the E and H field amplitudes are given by

where η is real. The power density amplitude is therefore

0 cos ( )x xE E t z

0 cos ( )xy

EH t z

220 cos ( )x

z

EP t z

(29)

26

In a lossy dielectric Ex and Hy are not in time phase. Also

0 cos ( )zx xE E e t z

Letting n

We get the magnetic field as

0 cos ( )xy n

EH t z

27

From this we get the Power density as

220 cos ( )cos ( )zx

z n

EP e t z t z

220 cos (2 2 ) cos ( )

2zx n nE t z

e

Therefore the average power is

220

0

cos (2 2 ) cos ( )1

2

Tzx n n

av

E t zP e dt

T

28

2201cos

2zx

av n

EP e

Note that the power density drops off as whereas Ex and Hy fall off as

2 ze

ze

The phasor form of Poynting vector may be written also as

1

2 S SP E H

(29)

(30)

29

In the present case

and

0 ˆj zS x xE E e a

0 0ˆ ˆjj z j zx xS y y

E EH e a e e a

Ex0 is assumed real.

Work out D 12.6