Que.1) Derive the Expression of inductance of two wire (1-phase) transmission line.
Ans.
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L = 2 * 10-7 ln ( D/r1' )
r1' = r1 * e-(1/4) = 0.7788 r1
How to calculate GMD of three phase line with bundled conductors? For an example see Fig-E where three phase bundles (triple conductror) are placed horizontally on transmission towers. In this case the distance between the conductors (D) is taken as distance between the centers of bundled conductors.
So, GMD = ∛(D.D.2D)
You can also calculate considering the distance from each bundled conductors of one phase to other conductors of two other phases. But the GMD calculated does not vary significantly from our simple form above. This is due to the fact that D is quite larger than d.
Post insulator is more or less similar to Pin insulator but former is suitable for higher voltage application. Post insulator has higher numbers of petticoats and has greater height. This type of insulator can be mounted on supporting structure horizontally as well as vertically. The insulator is made of one piece of porcelain but has fixing clamp arrangement are in both top and bottom end.
The main differences between pin insulator and post insulator are,
In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator because size, weight of the insulator become more. Handling and replacing bigger size single unit insulator are quite difficult task. For overcoming these difficulties, suspension insulator was developed.
In suspension insulator numbers of insulators are connected in series to form a string and the line conductor is carried by the bottom most insulator. Each insulator of a suspension string is called disc insulator because of their disc like shape.
Ans.
Inductance of a Single-phase Line
Consider two solid round conductors with radii of r1 and r2 as shown in Fig. 1.5. One conductor is the return circuit for the other. This implies that if the current in conductor 1 is I then the current in conductor 2 is -I . First let us consider conductor 1. The current flowing in the conductor will set up flux lines. However, the flux beyond a distance D + r2 from the center of the conductor links a net current of zero and therefore does not contribute to the flux linkage of the circuit. Also at a distance less than D - r2 from the center of conductor 1 the current flowing through this conductor links the flux. Moreover since D >> r2 we can make the following approximations
Fig. 1.5 A single-phase line with two conductors.
the inductance of conductor 1 due to internal and external flux as
We can rearrange L1 given in (1.18) as follows
Substituting r1¢ = r1 e-1/4 in the above expression we get
The radius r1¢ can be assumed to be that of a fictitious conductor that has no internal flux but with the same inductance as that of a conductor with radius r1 .
In a similar way the inductance due current in the conductor 2 is given by
therefore the inductance of the complete circuit is
If we assume r1¢ = r2¢ = r¢ , then the total inductance becomes
where r¢ = re-1/4.
Que.2 Define the GMD and GMR.
Ans.
It should be noted that, in all the formulas below inductance L is in Henry per unit length and not simply Henry. Here few cases are depicted.
- For a single phase line see the fig-A. The conductor inductance is
L = 2 * 10-7 ln ( D/r1' )
Here D is the distance between the centers of conductors.
r1' = r1 * e-(1/4) = 0.7788 r1
r1 is the actual radius of the conductor.
For a single phase line the return path also has inductance say L'. If the return conductor is of radius r2, then
L' = 2 * 10-7 ln ( D/r2' )
Therefore the total inductance of single phase circuit is Lt = L+L'
rearranging we get
Lt = 4 * 10 -7 ln [D / √ (r1'. r2')]
L' = 2 * 10-7 ln ( D/r2' )
Therefore the total inductance of single phase circuit is Lt = L+L'
rearranging we get
Lt = 4 * 10 -7 ln [D / √ (r1'. r2')]
- For three phase circuit whose three circular conductors are at the corners of equilateral triangle(Fig-B(i)) then the above formula for single phase case is applied here. In this case inductance per phase L is as below:
If the Denominator is renamed as Ds, then
L = 2 * 10-7 ln ( D / Ds )
Here Ds = r'
As already said r' is 0.7788 times the actual radius(r) of conductor.
L = 2 * 10-7 ln ( D / Ds )
Here Ds = r'
As already said r' is 0.7788 times the actual radius(r) of conductor.
- For three phase circuit whose three circular conductors are arbitrarily placed (Fig-B(ii)) and the conductors are transposed then,
L = 2 * 10-7 ln [ ∛(D1 . D2 . D3) / Ds ]
Beginning from the single phase line, it is observed that all the three equations for inductance of a phase conductor are similar. Remember that this formula for three phase line is not valid for non-transposed lines.
Observing the formula for single phase and three phase lines we can generalize the formula for inductance of a phase line as in the form
L = 2 * 10-7 ln ( D / Ds )
Where
D = Geometric Mean Distance (GMD)
Ds= Geometric Mean Radius (GMR)
In single phase case GMD is simply the distance between the centers of two conductors.
In three phase case for conductors equidistant from each other GMD is the distance between any two phase conductors.
In the three phase case, for line conductors arbitrarily placed GMD = ∛(D1D2D3 )
In all the three cases Ds = r'.
Observing the formula for single phase and three phase lines we can generalize the formula for inductance of a phase line as in the form
L = 2 * 10-7 ln ( D / Ds )
Where
D = Geometric Mean Distance (GMD)
Ds= Geometric Mean Radius (GMR)
In single phase case GMD is simply the distance between the centers of two conductors.
In three phase case for conductors equidistant from each other GMD is the distance between any two phase conductors.
In the three phase case, for line conductors arbitrarily placed GMD = ∛(D1D2D3 )
In all the three cases Ds = r'.
- From above we can conclude that GMD is like equivalent distance between conductors. When two or more conductors per phase are used as in bundled conductors then GMD is required to be computed. Here distances from each conductor in one phase to each conductor in other phase is calculated. If for example in a single phase line there are 4 conductors in one phase and 3 conductors in other phase (Fig-C) then we will have 12 numbers of distances between the conductors. I have shown four distances only.
GMD = [D1 . D2 ........ D12] 1/12
so here GMD is the 12th root of product of 12 numbers of distances.
- GMR is calculated for each phase separately. Each of the phases may have different GMR values depending upon the conductor size and arrangement. GMR is to be calculated when each phase is comprised of more than one conductor per phase as in the example above. For GMR calculation when two or more conductors per phase are used, first product of all the groups (one group for each conductor)are found where each group is product of possible distances from one conductor to other conductors including r' of that conductor. In the above example case GMR for line with 3 conductors per phase is
GMR = [(r1'.D12.D13)(r2'.D23.D21)(r3'.D31.D32)]1/9
It should be noted that D12 = D21, D13 = D31 and D23 = D32
It should be noted that D12 = D21, D13 = D31 and D23 = D32
- Below is the formula to calculate GMR for the bundled conductors (see Fig-D)
For three conductors per phase (triple conductor)
GMR = ∛(Ds *d 2)
For four conductors per phase (quad conductor)
GMR = 1.09 ∜(Ds *d 3)
GMR = ∛(Ds *d 2)
For four conductors per phase (quad conductor)
GMR = 1.09 ∜(Ds *d 3)
How to calculate GMD of three phase line with bundled conductors? For an example see Fig-E where three phase bundles (triple conductror) are placed horizontally on transmission towers. In this case the distance between the conductors (D) is taken as distance between the centers of bundled conductors.
So, GMD = ∛(D.D.2D)
You can also calculate considering the distance from each bundled conductors of one phase to other conductors of two other phases. But the GMD calculated does not vary significantly from our simple form above. This is due to the fact that D is quite larger than d.
- For ACSR conductors GMR is specified by the manufacturer.If this GMR is called Ds. For example if two such ACSR conductors(twin conductor) are used in a bundle for each phase. The GMR of the phase conductor arrangement is calculated imagining that the supplied GMR (or Ds) as the equivalent radius of ACSR conductor.
Hence if d is the distance between the centers of the two ACSR conductors, similar to the formulas in
Fig-D,
GMR= [(Ds.d).(Ds.d)]1/4 =√(Ds . d)
Fig-D,
GMR= [(Ds.d).(Ds.d)]1/4 =√(Ds . d)
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Que.3 Explain Kelvin's law for choice of conductor.
Ans.
Kelvins Law:
The most economical area of conductor is that for which the total annual cost of transmission line is minimum.
This is called as kelvins law after Lord Kelvin who first stated in 1881.
The most economical area of conductor is that for which the total annual cost of transmission line is minimum.
This is called as kelvins law after Lord Kelvin who first stated in 1881.
The transmission line cost forms major part in the annual charges of a power system.
The cost is due to
The cost is due to
- Depreciation
- Repair and maintenance
- Loss of energy in the line due to its resistance
- The cost towards the production of the lost energy is considered
- If we decreases the area of the conductor in order to reduce the capital cost, the line losses increase.
- Similarly, if we increase the conductor cross-section to save the cost towards copper loss in the line, the weight of copper increases and hence the capital cost will be more.
Because of the above reasons, it is difficult to find the economical size of the conductor. But it becomes easy with the help of kelvin’s law.
In this post we will understand about the kelvin’s law and limitations of the kelvin’s law.
Assume
A = Cross section of conductor
A = Cross section of conductor
C = total initial cost towards conductor
C is directly proportional to A
C ∝ A
C = PA
where P is a constant.
Let r be the annual rate of interest and depreciation.
The annual fixed cost C1 = Cr = PAr
Since line losses are inversely proportional to the area of the conductor
The annual cost on lost energy,
The annual fixed cost C1 = Cr = PAr
Since line losses are inversely proportional to the area of the conductor
The annual cost on lost energy,
C2 = Q/A where Q is a constant.
Total annual cost C = C1 + C2
= PAr + Q/A
For C to be minimum,
C/dA = 0
Pr – Q/A2 =0
Pr = Q/A2
Pr.A2 = Q
A2 = Q/Pr
A = √ (Q/Pr)
The equation shows that
“The economical cross-section of the conductor is that for which the annual charge on the conductor equals the annual charge for the loss of energy in the conductor”.
This is known as Kelvin’s law.
“The economical cross-section of the conductor is that for which the annual charge on the conductor equals the annual charge for the loss of energy in the conductor”.
This is known as Kelvin’s law.
Limitations of Kelvin’s Law
This law has many problems and limits as we are selecting the cross-section from an economical point of view.We did not consider the electrical behaviour of the line.
This law has many problems and limits as we are selecting the cross-section from an economical point of view.We did not consider the electrical behaviour of the line.
- It is not easy to estimate the energy loss in the line without actual load curves, which are not available at the time of estimation.
- Kelvin’s law did not consider many physical factors like voltage regulation, corona loss, temperature rise etc.
- The assumption that annual cost on account of interest and depreciation on the capital outlay is not 100% true.
- The conductor size determined by this law may not be always practicable one.
- The rates of interest and depreciation may vary from time to time.
- The diameter of the conductor may be so small as to cause high corona loss.
- The conductor may be too weak to stamp from mechanical point of view.
- Cost of insulation in cables is assumed to be independent of the cross-section of the conductor which is only an approx. assumption.
Remember that
Kelvin’s law should not be applied to underground cables and high voltage overhead lines. Kelvin’s law may be successfully used for overhead lines of voltage below 30KV.
Kelvin’s law should not be applied to underground cables and high voltage overhead lines. Kelvin’s law may be successfully used for overhead lines of voltage below 30KV.
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Que.4 Explain types of insulator.
Ans.
There are mainly three types of insulator used as overhead insulator likewise
1. Pin Insulator 2. Suspension Insulator 3. Strain Insulator In addition to that there are other two types of electrical insulator available mainly for low voltage application, e.i. Stay Insulator and Shackle Insulator.
1. Pin Insulator 2. Suspension Insulator 3. Strain Insulator In addition to that there are other two types of electrical insulator available mainly for low voltage application, e.i. Stay Insulator and Shackle Insulator.
Pin Insulator
Pin Insulator is earliest developed overhead insulator, but still popularly used in power network up to 33KV system. Pin type insulator can be one part, two parts or three parts type, depending upon application voltage. In 11KV system we generally use one part type insulator where whole pin insulator is one piece of properly shaped porcelain or glass. As the leakage path of insulator is through its surface, it is desirable to increase the vertical length of the insulator surface area for lengthening leakage path. In order to obtain lengthy leakage path, one, tow or more rain sheds or petticoats are provided on the insulator body. In addition to that rain shed or petticoats on an insulator serve another purpose. These rain sheds or petticoats are so designed, that during raining the outer surface of the rain shed becomes wet but the inner surface remains dry and non-conductive. So there will be discontinuations of conducting path through the wet pin insulator surface.
In higher voltage like 33KV and 66KV manufacturing of one part porcelain pin insulator becomes difficult. Because in higher voltage, the thickness of the insulator become more and a quite thick single piece porcelain insulator can not manufactured practically. In this case we use multiple part pin insulator, where a number of properly designed porcelain shells are fixed together by Portland cement to form one complete insulator unit. For 33KV tow parts and for 66KV three parts pin insulator are generally used.
Designing Consideration of Electrical Insulator
The live conductor attached to the top of the pin insulator is at a potential and bottom of the insulator fixed to supporting structure of earth potential. The insulator has to withstand the potential stresses between conductor and earth. The shortest distance between conductor and earth, surrounding the insulator body, along which electrical discharge may take place through air, is known as flash over distance.
1. When insulator is wet, its outer surface becomes almost conducting. Hence the flash over distance of insulator is decreased. The design of an electrical insulator should be such that the decrease of flash over distance is minimum when the insulator is wet. That is why the upper most petticoat of a pin insulator has umbrella type designed so that it can protect, the rest lower part of the insulator from rain. The upper surface of top most petticoat is inclined as less as possible to maintain maximum flash over voltage during raining.
2. To keep the inner side of the insulator dry, the rain sheds are made in order that these rain sheds should not disturb the voltage distribution they are so designed that their subsurface at right angle to the electromagnetic lines of force.
Post Insulator
Post insulator is more or less similar to Pin insulator but former is suitable for higher voltage application. Post insulator has higher numbers of petticoats and has greater height. This type of insulator can be mounted on supporting structure horizontally as well as vertically. The insulator is made of one piece of porcelain but has fixing clamp arrangement are in both top and bottom end.
The main differences between pin insulator and post insulator are,
| SL | Pin Insulator | Post Insulator |
| 1 | It is generally used up to 33KV system | It is suitable for lower voltage and also for higher voltage |
| 2 | It is single stag | It can be single stag as well as multiple stags |
| 3 | Conductor is fixed on the top of the insulator by binding | Conductor is fixed on the top of the insulator with help of connector clamp |
| 4 | Two insulators cannot be fixed together for higher voltage application | Two or more insulators can be fixed together one above other for higher voltage application |
| 4 | Metallic fixing arrangement provided only on bottom end of the insulator | Metallic fixing arrangement provided on both top and bottom ends of the insulator |
Suspension Insulator
In higher voltage, beyond 33KV, it becomes uneconomical to use pin insulator because size, weight of the insulator become more. Handling and replacing bigger size single unit insulator are quite difficult task. For overcoming these difficulties, suspension insulator was developed.
In suspension insulator numbers of insulators are connected in series to form a string and the line conductor is carried by the bottom most insulator. Each insulator of a suspension string is called disc insulator because of their disc like shape.
Advantages of Suspension Insulator
1. Each suspension disc is designed for normal voltage rating 11KV(Higher voltage rating 15KV), so by using different numbers of discs, a suspension string can be made suitable for any voltage level.
2. If any one of the disc insulators in a suspension string is damaged, it can be replaced much easily.
3. Mechanical stresses on the suspension insulator is less since the line hanged on a flexible suspension string.
4. As the current carrying conductors are suspended from supporting structure by suspension string, the height of the conductor position is always less than the total height of the supporting structure. Therefore, the conductors may be safe from lightening.
2. If any one of the disc insulators in a suspension string is damaged, it can be replaced much easily.
3. Mechanical stresses on the suspension insulator is less since the line hanged on a flexible suspension string.
4. As the current carrying conductors are suspended from supporting structure by suspension string, the height of the conductor position is always less than the total height of the supporting structure. Therefore, the conductors may be safe from lightening.Disadvantages of Suspension Insulator
1. Suspension insulator string costlier than pin and post type insulator.
2. Suspension string requires more height of supporting structure than that for pin or post insulator to maintain same ground clearance of current conductor.
3. The amplitude of free swing of conductors is larger in suspension insulator system, hence, more spacing between conductors should be provided.
2. Suspension string requires more height of supporting structure than that for pin or post insulator to maintain same ground clearance of current conductor.
3. The amplitude of free swing of conductors is larger in suspension insulator system, hence, more spacing between conductors should be provided.
Strain Insulator
When suspension string is used to sustain extraordinary tensile load of conductor it is referred as string insulator. When there is a dead end or there is a sharp corner in transmission line, the line has to sustain a great tensile load of conductor or strain. A strain insulator must have considerable mechanical strength as well as the necessary electrical insulating properties.
| Rated System Voltage | Number of disc insulator used in strain type tension insulator string | Number of disc insulator used in suspension insulator string |
| 33KV | 3 | 3 |
| 66KV | 5 | 4 |
| 132KV | 9 | 8 |
| 220KV | 15 | 14 |
Stay Insulator
For low voltage lines, the stays are to be insulated from ground at a height. The insulator used in the stay wire is called as the stay insulator and is usually of porcelain and is so designed that in case of breakage of the insulator the guy-wire will not fall to the ground.
Shackle Insulator or Spool Insulator
The shackle insulator or spool insulator is usually used in low voltage distribution network. It can be used both in horizontal and vertical position. The use of such insulator has decreased recently after increasing the using of underground cable for distribution purpose. The tapered hole of the spool insulator distributes the load more evenly and minimizes the possibility of breakage when heavily loaded. The conductor in the groove of shackle insulator is fixed with the help of soft binding wire.
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Que.5 Explain different type of DC distribution connection scheme.
Ans.
Distribution of electric power is done by distribution networks. Distribution networks consist of following main parts
- Distribution substation,
- Primary distribution feeder,
- Distribution Transformer,
- Distributors,
- Service mains.
The transmitted electric power is stepped down is substations, for primary distribution purpose. Now these stepped down electric power is fed to the distribution transformer through primary distribution feeders. Over head primary distribution feeders are supported by mainly supporting iron pole (preferably rail pole). The conductors are strand aluminum conductors and they are mounted on the arms of the pole by means of pin insulators. Some times in congested places, underground cables may also be used for primary distribution purposes.
Distribution transformers are mainly 3 phase pole mounted type. The secondary of the transformer is connected to distributors. Different consumers are fed electric power by means of the service main. These service mains are tapped from different points of distributors. The distributors can also be re-categorized by distributors and sub distributors. Distributors are directly connected to the secondary of distribution transformers whereas sub distributors are tapped from distributors. Service main of the consumers may be either connected to distributors or sub distributors depending upon the position and agreement of consumers. In this discussion of electrical power distribution system, we have already mentioned about both feeders and distributors. Both feeder and distributor carry the electrical load, but they have one basic difference. Feeder feeds power from one point to another without being tapped from any intermediate point. As because there is no tapping point in between, the current at sending end is equal to that of receiving end of the conductor. The distributors are tapped at different points for feeding different consumers; and hence the current varies along their entire length.
Distribution transformers are mainly 3 phase pole mounted type. The secondary of the transformer is connected to distributors. Different consumers are fed electric power by means of the service main. These service mains are tapped from different points of distributors. The distributors can also be re-categorized by distributors and sub distributors. Distributors are directly connected to the secondary of distribution transformers whereas sub distributors are tapped from distributors. Service main of the consumers may be either connected to distributors or sub distributors depending upon the position and agreement of consumers. In this discussion of electrical power distribution system, we have already mentioned about both feeders and distributors. Both feeder and distributor carry the electrical load, but they have one basic difference. Feeder feeds power from one point to another without being tapped from any intermediate point. As because there is no tapping point in between, the current at sending end is equal to that of receiving end of the conductor. The distributors are tapped at different points for feeding different consumers; and hence the current varies along their entire length.
Radial Electrical Power Distribution System
In early days of electrical power distribution system, different feeders were radially come out from the substation and connected to the primary of distribution transformer directly.
But radial electrical power distribution system has one major drawback that in case of any feeder failure, the associated consumers would not get any power as there was no alternative path to feed the transformer. In case of transformer failure also, the power supply is interrupted. In other words the consumer in the radial electrical distribution system would be in darkness until the feeder or transformer was rectified.
But radial electrical power distribution system has one major drawback that in case of any feeder failure, the associated consumers would not get any power as there was no alternative path to feed the transformer. In case of transformer failure also, the power supply is interrupted. In other words the consumer in the radial electrical distribution system would be in darkness until the feeder or transformer was rectified.
Ring Main Electrical Power Distribution System
The drawback of radial electrical power distribution system can be overcome by introducing a ring main electrical power distribution system. Here one ring network of distributors is fed by more than one feeder. In this case if one feeder is under fault or maintenance, the ring distributor is still energized by other feeders connected to it. In this way the supply to the consumers is not affected even when any feeder becomes out of service. In addition to that the ring main system is also provided with different section isolates at different suitable points. If any fault occurs on any section, of the ring, this section can easily be isolated by opening the associated section isolators on both sides of the faulty zone.
In this way, supply to the consumers connected to the healthy zone of the ring, can easily be maintained even when one section of the ring is under shutdown. The number of feeders connected to the ring main electrical power distribution system depends upon the following factors.
In this way, supply to the consumers connected to the healthy zone of the ring, can easily be maintained even when one section of the ring is under shutdown. The number of feeders connected to the ring main electrical power distribution system depends upon the following factors.
- Maximum demand of the system : If it is more, then more numbers of feeders feed the ring.
- Total length of the ring main distributors : It length is more, to compensate the voltage drop in the line, more feeders to be connected to the ring system.
- Required voltage regulation : The number of feeders connected to the ring also depends upon the permissible allowable, voltage drop of the line.
The sub distributors and service mains are taken off may be via distribution transformer at different suitable points on the ring depending upon the location of the consumers. Sometimes, instead of connecting service main directly to the ring, sub distributors are also used to feed a group of service mains where direct access of ring distributor is not possible.
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