EEP - Electrical Engineering Portal http://electrical-engineering-portal.com Electrical Engineering Portal dedicated to el. engineers Tue, 07 Jul 2015 20:54:45 +0000 en-US hourly 1 http://wordpress.org/?v=4.2.2 Let’s develop the simple PLC program for lighting control systemhttp://electrical-engineering-portal.com/lets-develop-the-simple-plc-program-for-lighting-control-system http://electrical-engineering-portal.com/lets-develop-the-simple-plc-program-for-lighting-control-system#comments Mon, 06 Jul 2015 04:05:36 +0000 http://electrical-engineering-portal.com/?p=64789

Lighting control system A lighting control system is to be developed. The system will be controlled by four switches, SWITCH1, SWITCH2, SWITCH3, and SWITCH4. These switches will control the lighting in a room based on the following criteria: Any of three of the switches SWITCH1, SWITCH2, and SWITCH3, if turned ON can turn the lighting […]

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## Lighting control system

A lighting control system is to be developed. The system will be controlled by four switches, SWITCH1, SWITCH2, SWITCH3, and SWITCH4. These switches will control the lighting in a room based on the following criteria:

1. Any of three of the switches SWITCH1, SWITCH2, and SWITCH3, if turned ON can turn the lighting on, but all three switches must be OFF before the lighting will turn OFF.
2. The fourth switch SWITCH4 is a Master Control Switch. If this switch is in the ON position, the lights will be OFF and none of the other three switches have any control.

To do: Let’s design the wiring diagram for the controller connections, assign the inputs and outputs and develop the ladder diagram which will accomplish the task.

### The wiring diagram //

The first item we may accomplish is the drawing of the controller wiring diagram. All we need do is connect all switches to inputs and the lighting to an output and note the numbers of the inputs and output associated with these connections.

The remainder of the task becomes developing the ladder diagram. The wiring diagram is shown in Figure 1.

Notice that all four switches are shown as normally open selector switches and the output is connected to a relay coil CR1. We are using the relay CR1 to operate the lights because generally the current required to operate a bank of room lights is higher than the maximum current a PLC output can carry.

Attempting to operate the room lights directly from the PLC output will most likely damage the PLC.

For this wiring configuration, the following definition list is apparent:

• INPUT IN1 = SWITCH1
• INPUT IN2 = SWITCH2
• INPUT IN3 = SWITCH3
• INPUT IN4 = SWITCH4 (Master Control Switch)
• OUTPUT OUT1 = Lights control relay coil CR1

This program requires that when SWITCH4 is ON, the lights must be OFF. In order to do this, it would appear that we need a N/C SWITCH4, not a N/O as we have in our wiring diagram. However, keep in mind that once an input signal is brought into a PLC, we may use as many contacts of the input as we need in our program, and the contacts may be either N/O or N/C.

Therefore, we may use a N/O switch for SWITCH4 and then in the program, we will logically invert it by using N/C IN4 contacts.

The ladder diagram to implement this example problem is shown in Figure 2.

The ladder was printed using graphics characters (extended ASCII characters).

Notice the normally closed contact for IN4. A normally closed contact represents an inversion of the assigned element, in this case IN4, which is defined as SWITCH 4. Remember, SWITCH 4 has to be in the OFF position before any of the other switches can take control. In the OFF position, SWITCH 4 is open.

This means that IN4 will be OFF (de-energized). So, in order for an element assigned to IN4 to be closed with the switch in the OFF position, it must be shown as a normally closed contact. When SWITCH 4 is turned ON, the input, IN4, will become active (energized). If IN4 is ON, a normally closed IN4 contact will open.

With this contact open in the ladder diagram, none of the other switches will be able to control the output.

REMEMBER: A normally closed switch will open when energized and will close when de-energized.

### PLC program to control a lighting device

How to create a PLC program to control a lighting device through a button or a SCADA system using the PLC-PROG IDE.

Reference // Programmable Logic Controllers: Programming Methods and Applications – John R. Hackworth and Frederick D. Hackworth, Jr.

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Medium Voltage Earthing Systems – Arrangements and Comparisonhttp://electrical-engineering-portal.com/medium-voltage-earthing-systems-arrangements-and-comparison http://electrical-engineering-portal.com/medium-voltage-earthing-systems-arrangements-and-comparison#comments Fri, 03 Jul 2015 04:10:39 +0000 http://electrical-engineering-portal.com/?p=64707

Neutral point connection method First, let’s define the different medium voltage earthing systems and then compare the advantages and disadvantages of each one. Earthing systems in medium voltage can be differentiated according to the neutral point connection method. The various earthing systems in medium voltage systems are different in the way they operate and each […]

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## Neutral point connection method

First, let’s define the different medium voltage earthing systems and then compare the advantages and disadvantages of each one. Earthing systems in medium voltage can be differentiated according to the neutral point connection method.

The various earthing systems in medium voltage systems are different in the way they operate and each has its advantagesand disadvantages, which we shall now consider.

### 1. Directly earthed neutral (Direct earthing)

Description: An electrical connection is made between the neutral point and earth.

Operating technique: Compulsory switching on occurrence of the first insulation fault.

• Reduces the risk of overvoltages occurring.
• Authorizes the use of equipment with a normal phase to earth insulating level.

• Compulsory tripping upon occurrence of the first fault.
• Very high fault currents leading to maximum damage and disturbance (creation of induced currents in telecommunication networks and auxiliary circuits).
• The risk for personnel is high while the fault lasts; the touch voltages which develop being high.
• Requires the use of differential protection devices so that the fault clearance time is not long. These systems are costly.

Go back to Index ↑

### 2. Unearthed neutral

Description:

There is no electrical connection between the neutral point and earth, except for measuring or protective devices. A high impedance is inserted between the neutral point and earth.

Operating technique:

No switching on occurrence of the first insulation fault – it is thus compulsory:

• To carry out permanent insulation monitoring;
• To indicate the first insulation fault;
• To locate and clear the first insulation fault;
• To switch upon occurrence of the second insulation fault (double fault).

• Provides continuity of service by only tripping upon occurrence of the second fault, subject to the network capacity not leading to a high earth fault current that would be dangerous for personnel and loads on occurrence of the first fault.

The unearthed neutral involves:

• The use of equipment whose phase-to-earth insulation level is at least equal to that of the phase-to-phase level. Indeed, when a permanent phase-earth fault occurs, the voltage of both unaffected phases in relation to earth takes on the value of the phase-to-phase voltage if tripping is not triggered on occurrence of the first fault. Cables, rotating machines, transformers and loads must therefore be chosen with this in mind;
• The risk of high internal overvoltages making it advisable to reinforce the equipment insulation;
• The compulsory insulation monitoring, with visual and audible indication of the first fault if tripping is not triggered until the second fault occurs;
• The presence of maintenance personnel to monitor and locate the first fault during use;
• Some difficulties implementing selective protection devices upon occurrence of the first fault;
• The risk of ferroresonance.

Go back to Index ↑

### 3. Resistance earthing

#### Limiting resistance earthing

A resistor is inserted between the neutral point and earth.

Operating technique: Switching upon occurrence of the first fault.

• Limits fault currents (reduced damage and disturbance).
• Dampens overvoltages of internal origin in that the limiting current Il is twice as high as the capacitive current IC giving Il > 2 IC.
• Does not require the use of equipment, and in particular cables, having a special phase/earth insulation level.
• Allows the use of simple selective protection devices.

Disadvantages: Tripping on the first fault.

Go back to Index ↑

### 4. Reactance earthing

#### Limiting reactance earthing

Description: A reactor is inserted between the neutral point and earth.

Operating technique: Switching upon occurrence of the first insulation fault.

• Limits the fault currents (reduced damage and disturbance).
• Allows the implementation of simple selective protection devices if IL >> IC.
• The coil, being of low resistance, does not have to dissipate a high heat load.

• May cause high overvoltages during earth fault clearance.
• Compulsory tripping upon occurrence of the first fault.

Go back to Index ↑

### 5. Petersen coil earthing

Description:

A reactor L tuned to the network capacities is inserted between the neutral point and earth so that if an earth fault occurs, the fault current is zero.

If = IL + IC

Where:

• If – fault current
• IL – current in the neutral earthing reactor
• IC – current in the phase-earth capacitances

Operating technique: No switching upon occurrence of the first fault.

If the reactance is such that 3 L0 C0 ω0 = 1 is respected, the phase-earth fault current is zero:

• Spontaneous clearance of non-permanent earth faults;
• The installation continues to operate in spite of there being a permanent fault, with tripping necessarily occurring on the second fault;
• The first fault is indicated by the detection of the current flowing through the coil. The coil is dimensioned so that permanent operation is possible.

• Difficulties establishing the condition 3 L0 C0 ω2 = 1 due to uncertain knowledge of the network’s capacity: the result is that throughout the duration of the fault, a residual current circulates in the fault. Care must be taken to make sure this current is not dangerousfor personnel and equipment.
• The risk of overvoltages occurring is high.
• Requires the presence of monitoring personnel.
• Impossible to provide selective protection upon occurrence of the first fault if the coil has been tuned to: 3 L0 C0 ω2 = 1
If it is systematically out of tune (3 L0 C0 ω2 ≠ 1) selective protection upon occurrence of the first fault is complex and costly.
• Risk of ferro-resonance.

Go back to Index ↑

Resource: Protection of Electrical Networks – Christophe Prévé (get this book from Amazon)

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4 Main Means For The Generation Of Reactive Powerhttp://electrical-engineering-portal.com/4-main-means-for-the-generation-of-reactive-power http://electrical-engineering-portal.com/4-main-means-for-the-generation-of-reactive-power#comments Mon, 29 Jun 2015 21:11:46 +0000 http://electrical-engineering-portal.com/?p=64689

It’s all about reactive power… The four main means for the generation of reactive power are: Synchronous alternators Synchronous compensators (SC) Static var compensators (SVC) and Banks of static capacitors 1. Synchronous alternators Synchronous alternators are the main machines used for the generation of electrical energy. Besides, without going into technical details, by acting on […]

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## It’s all about reactive power…

The four main means for the generation of reactive power are:

### 1. Synchronous alternators

Synchronous alternators are the main machines used for the generation of electrical energy.

They are intended to supply electrical power to the final loads through transmission and distribution systems.

Besides, without going into technical details, by acting on the excitation of alternators, it is possible to vary the value of the generated voltage and consequently to regulate the injections of reactive power into the network, so that the voltage profiles of the system can be improved and the losses due to joule effect along the lines can be reduced.

Go back to Index ↑

### 2. Synchronous compensators

They are synchronous motors running no-load in synchronism with the network and having the only function to absorb the reactive power in excess (under-excited operation) or to supply the missing one (overexcited operation).

Where:

• E = e.m.f. induced in the stator phases
• V = Phase voltage imposed by the network to the alternator terminals I : stator current
• Xs = Stator reactance
These devices are used mainly in definite nodes of the power transmission and sub-transmission network for the regulation of voltages and of reactive power flows. The use of synchronous compensators in power distribution networks is not favourable from an economic point of view because of their high installation and maintenance costs.

Go back to Index ↑

### 3. Static var compensators

The considerable development of power electronics is encouraging the replacement of synchronous compensators with static systems for the control of the reactive power such as for example TSC (thyristor switched capacitors) and TCR (thyristor controlled reactors).

These are an electronic version of the reactive power compensation systems based on electromechanical components in which, however, the switching of the various capacitors is not carried out through the opening and closing of suitable contactors, but through the control carried out by couples of antiparallel tyristors.

TSC allow a step-by-step control of the reactive power delivered by groups of capacitors, whereas with TCR a continuous control of the reactive power drawn by the inductors is possible. By coupling a TSC with a TCR it is possible to obtain a continuous modulated regulation of the delivered/drawn reactive power.

From the point of view of applications, these devices are used above all in high and very high voltage networks.

Go back to Index ↑

### 4. Banks of static capacitors

A capacitor is a passive dipole consisting of two conducting surfaces called plates, isolated from one another by a dielectric material.

The system thus obtained is impregnated to prevent the penetration of humidity or of gas pockets which could cause electrical discharges.

The last generation capacitors are dry-type and undergo a specific treatment which improve their electrical characteristics. Using dry-type capacitors there is no risk of pollution because of the incidental leak of the impregnating substance.

According to the geometry of the metal plates, it is possible to have:

• Plane capacitors;
• Cylindrical capacitors;
• Spherical capacitors.

The 4 main parameters which characterize a capacitor are:

1. The rated capacitance C – the value obtained from the rated values of power, voltage and frequency of the
capacitor;
2. The rated power Qn – the reactive power for which the capacitor has been designed;
3. The rated voltage Un – the r.m.s. value of the alternating voltage for which the capacitor has been designed;
4. The rated frequency fn – the frequency for which the capacitor has been designed.

When an alternating voltage is applied across the plates, the capacitor is subjected to charge and discharge cycles, during which it stores reactive energy (capacitor charge) and injects such energy into the circuit to which it is connected (capacitor discharge).

Such energy is given by the following relation:

where:

• C is the capacitance;
• U is the voltage applied to the terminals of the capacitor.
Because of their capability of storing and delivering energy, capacitors are used as basic element for the realization of power factor correction banks (for all voltage levels) and of static devices for the regulation of reactive power.

In particular, the power factor correction capacitors used for low voltage applications are constituted by single- phase components of metalized polypropylene film and can be of the self-healing type.

In these capacitors, the dielectric part damaged by a discharge is capable of self-restoring; in fact, when such situations occur, the part of the polypropylene film affected by the discharge evaporates due to the thermal effect caused by the discharge itself, thus restoring the damaged part.

Go back to Index ↑

Reference // Power factor correction and harmonic filtering in electrical plants – ABB

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12 Transformer Factory Tests Briefly Explainedhttp://electrical-engineering-portal.com/12-transformer-factory-tests-briefly-explained http://electrical-engineering-portal.com/12-transformer-factory-tests-briefly-explained#comments Mon, 29 Jun 2015 04:04:07 +0000 http://electrical-engineering-portal.com/?p=64666

Factory tests The remainder of the twelve factory tests are briefly summarized below. The details of the test set connections and formulas of some of the listed tests are already described in separatly published articles, and for the rest you are directed to ANSI/IEEE Standard C57.12.90 for these details. This list is not complete, there […]

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## Factory tests

The remainder of the twelve factory tests are briefly summarized below. The details of the test set connections and formulas of some of the listed tests are already described in separatly published articles, and for the rest you are directed to ANSI/IEEE Standard C57.12.90 for these details.

This list is not complete, there are few tests missing, not mentioned here, like Turn ratio test or Measurement of voltage ratio and check of phase displacement, but you can find them also separatly published at EEP (use Search).

### 1. No-Load Losses

The tests measures the no-load losses at specified excitation voltage and a specified frequency. Sine-wave voltages are used unless a different waveform is inherent in the operation of the transformer.

The recommended method is the average-voltage voltmeter method, employing two parallel-connected voltmeters. One voltmeter is an average-responding but RMS calibrated voltmeter and the other voltmeter is a true RMS-responding voltmeter.

The test voltage is adjusted to the specified value as read by the average-responding voltmeters. The readings of both voltmeters are used to correct the no-load losses to a sine-wave basis.

Go back to Tests ↑

### 2. No-Load Excitation Current

This current is measured in the winding used to excite the transformer with the other windings open-circuited. It is generally expressed in percent of the rated current of the winding. No-load excitation current is not sinusoidal and contains, as we have seen, odd harmonics (predominantly third harmonic current).

The ammeter used to record the no-load excitation current is an RMS meter which reads the square root of the sum of the squares of the harmonic currents.

Go back to Tests ↑

### 3. Load Losses and Impedance Voltage

The transformer must be in a specific state before the load losses and impedance voltage are measured. The temperature of the insulating liquid must be stabilized and the difference between the top and bottom oil temperatures shall be less than 5°C.

The winding temperatures must be measured (using a resistance method) before and after the test and the average taken as the true temperature. The difference in the winding temperature before and after the test must not exceed 5°C.

The two test methods for measuring load losses and impedance voltage are:

1. Wattmeter-voltmeter-ammeter method and
2. Impedance bridge method.

These tests generally apply a reduced voltage to one set of windings with the other set of windings short-circuited. For three-winding transformers, these tests are repeated for each combination of windings taken two at a time.

Go back to Tests ↑

### 4. Dielectric Tests

These tests consist of applied-voltage tests and induced-voltage tests.

Applied-voltage tests apply a high voltage to all bushings of a winding, one winding at a time, with the other windings grounded. A 60 Hz voltage is increased gradually over 15 s and held for 40 s and reduced to zero over 5 s.

Induced-voltage tests apply a high voltage across a winding with the other windings open-circuited in order to test the quality of the turn-to-turn insulation. In order to prevent core saturation at the higher excitation voltage, the frequency of the induced-voltage test is increased (typically around 120 Hz). The induced voltage is applied for 7200 cycles or 60 s, whichever is shorter.

Go back to Tests ↑

### 5. Switching Impulse Test

The switching impulse test applies a switching impulse wave between each high-voltage line terminal and ground.

The test series consists of one reduced voltage wave (50%– 70% of specified test level) followed by two full-voltage waves. Either positive or negative polarity waves, or both, may be used. A voltage oscillogram is taken for each applied wave. The test is successful if there is no sudden collapse of voltage. Successive oscillograms may differ because of the influence of core saturation.

Go back to Tests ↑

### 6. Lightning Impulse Test

The test sequence consists of one reduced full wave, two chopped waves, and two full waves. Tap connections are made with the minimum effective turns in the winding under tests and regulating transformers are set to the maximum buck position. Oscillograms are taken of each wave.

The general technique for interpreting the results is to look for differences in the shapes of the reduced full wave and the two full waves, which indicate turn-to-turn insulation failure.

Additional test criteria are found in IEEE Std. C57.98-1993. The impulse tests probably have the highest likelihood failures among all of the factory tests that are typically performed.

Go back to Tests ↑

### 7. Partial Discharge Test

This test detects radio-frequency (0.85–1.15 MHz) noise generated from partial discharges within voids in the insulation. An applied voltage is gradually increased until partial discharge starts to occur, which is the inception voltage. The voltage is then decreased until the partial discharge stops, which is the extinction voltage.

The extinction voltage must be less than the operating voltage of the transformer; otherwise, once partial discharge starts in the field (due to some voltage transient), it would continue indefinitely and possibly cause damage or failure.

Go back to Tests ↑

### 8. Insulation Power Factor

Insulation power factor is the ratio of the power dissipated in the insulation in watts to the apparent power (volt-amperes) under a sinusoidal voltage. The applied 60 Hz voltage of this test is generally lower than the operating voltage of the trans- former. The Doble Test Set is designed specifically to carry out this test.

Portable versions are used to measure the insulation power factor of transformers in the field. This test usually must be done by a trained technician. The test results are temperature-corrected to a reference temperature of 20°C.

Go back to Tests ↑

### 9. Insulation Resistance

This test applies a high-voltage DC voltage to one winding at a time with the other windings grounded. The leakage current is measured and the insulation resistance is calculated using Ohm’s law.

A Insulation resistance test set is designed specifically to carry out this test, and its meter is calibrated in megohms in order that the calculation may be avoided. The Megger as well as other manufacturers has a portable instrument that can easily carried around in the field.

Go back to Tests ↑

### 10. Noise Measurement

The noise measurement test is carried out while the transformer is energized at rated voltage with all of the cooling equipment running. Room geometry can greatly affect the measurements, so it is preferable that the transformer be inside an anechoic chamber. However, if such a chamber is not available, no acoustically reflecting surface may be within 3 m of the measuring microphone other than the floor or ground.

The recording microphones are positioned in 1 m intervals around the perimeter of the transformer, with no fewer than four (4) microphone positions for small transformers..

Sound power levels are measured over a specified band of frequencies. The sound power levels are converted into decibels (dB).

Go back to Tests ↑

### 11. Temperature Rise (Heat Run)

The transformer is energized at rated voltage in order to generate core losses. The windings are connected to a loading transformer that simultaneously circulates rated currents through all of the windings in order to develop load losses.

Naturally, the excitation voltage and the applied circulating currents are electrically 90° apart to minimize the KW requirements for this test. Nonetheless, a large power transformer can consume up to 1 MW of total losses and the heat run test is an expensive test to perform.

Therefore, in order to reduce the total expense, heat run tests are normally performed on only one transformer on a purchase order for multiple transformers, unless the customer chooses to pay for testing additional units.

Go back to Tests ↑

### 12. Short-Circuit Test

The short-circuit test is generally reserved for a sample transformer to verify the design of a core and coil assembly unless the customer specifies that a short-circuit test be performed on transformers that are purchased.

The customer should be cognizant of the ever-present risk of damaging the transformer during short-circuit tests.

A low-voltage impulse (LVI) current waveform is applied to the transformer before and after the applications of short-circuit test.  The ‘‘before’’ and ‘‘after’’ oscillograms of the LVI currents are compared for significant changes in waveshape that could indicate mechanical damage to the windings.

Go back to Tests ↑

Reference // Power Transformers Principles and Applications – John J. Winders, Jr.
(Purchase from Amazon)

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Take Into Account the Losses When Purchasing Power Transformerhttp://electrical-engineering-portal.com/take-into-account-the-losses-when-purchasing-power-transformer http://electrical-engineering-portal.com/take-into-account-the-losses-when-purchasing-power-transformer#comments Fri, 26 Jun 2015 04:01:41 +0000 http://electrical-engineering-portal.com/?p=64659

Transformer Price and Losses The Connection Losses and purchase price should be considered when deciding which transformer to purchase. The purpose of this technical article is to present a uniform approach that can be used to determine the dollar value of these losses over the life of the transformer. Below is typical wording of a transformer loss evaluation clause […]

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## Transformer Price and Losses

### The Connection

Losses and purchase price should be considered when deciding which transformer to purchase. The purpose of this technical article is to present a uniform approach that can be used to determine the dollar value of these losses over the life of the transformer.

Below is typical wording of a transformer loss evaluation clause for insertion into bidding documents that specifies how losses will be evaluated.

Load, no-load and auxiliary losses at 50 MVA for the 30/40/50 MVA transformer will be evaluated as follows:

• No-Load Losses \$/kW 2450
• Load Losses \$/kW 1304
• Auxiliary Losses \$/kW 756

The cost of losses for each transformer will be calculated by multiplying the appropriate dollars/kW values above by the guaranteed load losses at 55°C rating and no-load losses at 100% voltages. This cost will be added to the bid price for evaluation.”

### Let’s see an example //

Using the loss evaluation factors given above, determine which manufacturer’s transformer has the lowest evaluated cost including losses.

#### 161/34.5 kV, 30/40/50 MVA Transformer

 Manufacturer A’s Transformer Manufacturer B’s Transformer Bid price \$424,500 \$436,000 No -load losses 59 kW 53 kW Load losses at 50 MVA (at 55°C temperature rise) 224 kW 218 kW Auxiliary losses (at 50 MVA 55°C temperature rise) 2.0 kW 2.5 kW

### … and play with numbers //

 Manufacturer A’s Transformer Manufacturer B’s Transformer Bid Price = \$424,500 = \$436,000 Total cost of No-load Losses 59 kW (2450 \$/kW) = \$144,550 53 kW (2450 \$/kW) = \$129,850 Total cost of Load Losses 224 kW (1304 \$/kW) = \$292,096 218 kW (1304 \$/kW) = \$284,272 Total cost of Auxiliary Losses 2.0 kW ( 756 \$/kW) = \$ 1,512 2.5 kW (756 \$/kW) = \$1,890 TOTAL COST // = \$862,658 = \$852,012

Although the transformer from Manufacturer A has the lowest bid price, it is obvious that the transformer from Manufacturer B has the lowest evaluated total cost.

### Bidding and Penalty Values

In addition to giving loss evaluation values, the bid documents should also have penalty values that the manufacturer is to be charged for every kilowatt by which the actual tested transformer losses exceed the guaranteed losses upon which the bids are evaluated.

It is important to have such penalty values in order to give an incentive to the manufacturers to provide the most accurate guaranteed loss values possible. The penalty values should be expressed in the same dollars per kW manner as the bid evaluation values but should be somewhat higher.

An increment of approximately 20 percent is recommended.

### Few words about above mentioned losses //

The three different types of transformer losses that should be evaluated separately are:

1. Load losses (sometimes called copper or coil losses);
2. No-load losses (sometimes called core or iron losses); and
3. Auxiliary losses (electric fan losses, other such equipment losses).

Load losses are primarily from the I2R losses in the transformer windings and eddy current losses. If a value of load losses is not directly given, load losses can be determined by subtracting no-load losses from total losses (If the total losses at full load are 100 kW and the no-load losses are 10 kW, then the load (or copper or coil) losses are 90 kW).

No-load losses consist of the hysteresis and the eddy current losses in the iron core of the transformer and the I2R losses in the windings due to the excitation current.

Auxiliary losses consist of the power necessary to drive the auxiliary cooling pumps and fans.

The formulae below yield the total costs of the losses that should be added to the purchase price of the transformer as shown in the example above:

Where:

• G = peak ratio
• K = peak responsibility factor
• SI = the system capital investment in dollars per kilowatt required to supply the power losses of the transformer;
• 8760 = the number of hours in a year
• EC = the cost of energy in dollars per kilowatt-hour;
• FCR = fixed charge rate for capital investment expressed as a decimal in dollars per dollar of investment;
• LFA = the loss factor for auxiliary equipment;
• LFT = the transformer loss factor which is the ratio of average transformer losses to peak transformer losses;
• TNLL = the transformer’s guaranteed no-load losses in kilowatts;
• TLL = the transformer’s guaranteed load losses in kilowatts;
• TAL = the losses due to transformer auxiliary equipment in kilowatts

### Values for formulae

#### SI – System Investment Charge

The System Investment (SI) charge is the cost of generation and transmission facilities per kilowatt necessary to supply the additional demand resulting from the transformer losses at the system peak.

Since a transformer located directly at a generating station does not require an investment in transmission facilities, the SI value used to evaluate the losses in the generating station transformer should be less than the SI of a transformer to be located at the receiving end of a transmission line.

One method for determining the SI value involves adding the construction cost (dollars per kilowatt) of a recently completed or soon to be completed generating station to the cost of the transmission facilities (dollars per kilowatt) required to connect the transformer to the plant.

If power is purchased rather than self-generated, the SI value can be determined by dividing the demand charge in dollars per kW per year by the fixed charge rate (FCR). Since there is more than one method of evaluating the SI value, the method that is judged to yield the most realistic results should be used.

#### FCR – Fixed Charge Rate

The fixed charge rate (FCR) represents the yearly income necessary to pay for a capital investment. FCR is expressed as a percentage of capital investment. The rate covers all costs that are fixed and do not vary with the amount of energy produced.

The rate includes interest, depreciation, taxes, insurance, and those operations and maintenance expenses that do not depend on system kilowatt-hours sold. The interest rate used should be the same as the interest rate of the loan acquired to purchase the transformers. If loan funds are not used, a blended rate of the interest earned on deposited funds should be used.

The practice of including some operations and maintenance expenses in the fixed charge rate is a matter of judgment. Some typical values for the components of the carrying charge rate are as follows:

 Interest 7.5% Depreciation 2.75% Insurance 0.60% Taxes 1.00% Operations and Maintenance Carrying 2.76% Carrying Charge Rate 14.61%

#### EC – Energy Charge

The energy charge (EC) is the cost per kilowatt-hour for fuel and other expenses that are directly related to the production of electrical energy.

Although the costs per kilowatt-hour will vary with the level of demand, a single energy charge representing an average cost per kilowatt-hour throughout the load cycle should be used for the sake of simplicity.

Equations 1 and 2 are based on the assumption that the energy charge remains constant throughout the life of the transformer. If power is purchased, EC will be the kWh (or energy) cost of power.

#### K – Responsibility Factor

The peak responsibility factor (K) is intended to compensate for the transformer peak load losses not occurring at the system peak losses. This means that only a fraction of the peak transformer losses will contribute to the system peak demand.

The value of K can be determined from:

NOTE: It should be pointed out that K is squared in Equations 2 and 3 because K is a ratio of loads while losses are proportional to the load squared. Any value of K that seems appropriate can be used.

The following are recommended values that appear to be reasonable.

 Transformer Type K K2 Generator Step-up 1.00 1.00 Transmission Substation 0.90 0.81 Disrtribution Substation 0.80 0.64

#### LFT

The transformer loss factor is defined as the ratio of the average transformer losses to the peak transformer losses during a specific period of time. For the sake of simplicity, the equations assume that the transformer loss factor is a constant and that it does not change significantly over the life of the transformer.

The transformer loss factor can be determined directly using the equation:

LFT can also be approximated from the load factor (the average load divided by the peak load for a specified time period) using the empirical equation below:

Where:

Load factor is the ratio of the average load over a period of time to the peak load occurring in that period. The load factor is a commonly available system parameter. The one-hour integrated peak value should be used.

#### G – Peak Ratio

The peak ratio is defined by the equation:

For the peak annual transformer load, the one hour integrated peak value should be used.  The purpose of the peak ratio is to relate the value of Equation 2 to the full rated transformer load and not to the peak transformer load that would otherwise result if G were not in the equation.

If the total kVA of all transformers is known for your system and the peak kW (or kVA) load is known, then the average peak ratio for your system would be:

If the peak kW is known, but the peak kVA is unknown, assume a reasonable power factor on peak and calculate peak kVA as follows:

kVA = kW / power factor

If the transformer being purchased has a peak ratio different from the average, use that value. If the transformer will be installed at a known substation, use the billing data and assumed load growth for that substation.

The equations above are based on the assumption that the peak annual transformer load remains the same throughout the life of the transformer. If the load on the transformer is expected to increase annually, then use a reasonable equivalent level yearly peak load value based on experience even though the expected peak loading value will increase every year.

#### LFA – Auxiliary Loss Factor

The auxiliary loss factor compensates for the transformer auxiliary equipment that operates during only part of the transformer’s load cycle. For a transformer with two stages of cooling:

LFA = (0.5)• (probability first stage of cooling will be on at any given time) + (Eq. 8) (0.5)• (probability second stage of cooling will be on at any given time)

The choice of the proper probabilities in the above equation is a matter of judgment based on historical system loading patterns. It is expected that the above probabilities under normal loading patterns will be extremely low.

Since energy use and losses associated with transformer auxiliaries are extremely small over the life of the transformer, they could be ignored. The capital cost associated with auxiliaries are significant and should be considered.

### Simplified Losses Calculation Tool //

Reference // Guide for the Evaluation of Large Power Transformer Losses – United States Department of Agriculture – Rural Utility Service

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Example for Coordination of Cascaded Circuit Breakershttp://electrical-engineering-portal.com/example-for-coordination-of-cascaded-circuit-breakers http://electrical-engineering-portal.com/example-for-coordination-of-cascaded-circuit-breakers#comments Wed, 24 Jun 2015 04:21:38 +0000 http://electrical-engineering-portal.com/?p=64633

Inputs for Coordination Calculation A 440 V 60 Hz switchboard feeds a 4-wire distribution board for small loads such as socket outlets. The switchboard has a fault making capacity of 100kA rms. After applying diversity factors to the loads the total load current is 90 A. Moulded case circuit breakers (MCCBs) rated at 16 A […]

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## Inputs for Coordination Calculation

A 440 V 60 Hz switchboard feeds a 4-wire distribution board for small loads such as socket outlets. The switchboard has a fault making capacity of 100kA rms. After applying diversity factors to the loads the total load current is 90 A. Moulded case circuit breakers (MCCBs) rated at 16 A and 32 A are to be used for the loads.

The installation will use cables having copper conductors and XLPE insulation. The cable from the switchboard to the distribution board is 20 metres in length.

A typical load cable is 15 metres in length and will carry a current of 29 A at a power factor of 0.85 lagging.

Ignore the presence of induction motors at the switchboard and find the following:

1. Rating of the incoming circuit breaker.
2. Size of the incoming cable.
3. Size of the load cable.
4. Check that the MCCB coordination is complete.

The following sequence will be used to calculate the results:

### Let’s dive into solution!

#### 1. Choose the upstream MCCB at the switchboard and its settings

From a manufacturer’s data sheet a 125 A MCCB with an adjustable 100 A thermal release is chosen. The thermal release is set to 90 A to match the total load.

#### 2. Choose the incoming feeder cable

From a manufacturer’s data sheet several cables can be compared for the same ambient conditions and laying arrangements. Their details are:

• 50 mm2 cable, maximum current 124 A, R = 0.492, X = 0.110 ohms/km.
• 70 mm2 cable, maximum current 159 A, R = 0.340, X = 0.106 ohms/km.
• 95 mm2 cable, maximum current 193 A, R = 0.247, X = 0.093 ohms/km.
The 70 mm2 cable is chosen since the rating of the 50 mm2 cable is just too low.

Go back to Index ↑

#### 3. Choose the downstream load MCCB and its settings

From a manufacturer’s data sheet a 32 A MCCB with an adjustable 32 A thermal release is chosen. The thermal release is set to 29 A to match its load.

#### 4. Find the upstream fault source impedance

For a prospective symmetrical fault current of 100 kA rms the upstream fault source impedance Zup is:

Go back to Index ↑

#### 5. Find the cut-off, or let-through, current from the switchboard

From a manufacturer’s data sheet a 125 A MCCB has a let-through current Ip of 25 kA peak for a prospective fault current Is of 100 kArms.

#### 6.  Find the impedance of the incoming cable

The impedance Zc1 of the incoming cable is:

Go back to Index ↑

#### 7. Find the impedance of the load cable

The impedance Zc2 of the incoming cable is:

From a manufacturer’s data sheet several cables can be compared for the same ambient conditions and laying arrangements. Their details are:

• 6 mm2 cable, maximum current 33.8 A, R = 3.91, X = 0.130 ohms/km.
• 10 mm2 cable, maximum current 46.7 A, R = 2.31, X = 0.126 ohms/km.
The 6 mm2 cable is chosen provisionally, since its rating is above the 32 A rating of the MCCB that feeds it.

The impedance Zc2 of the load cable is:

Go back to Index ↑

#### 8. Find the fault current at the distribution board, point B

From a manufacturer’s data sheet the contact impedance data for low voltage MCCBs are:

 MCCB (Rating in Amps) Resistance (in Ohms) Reactanse (in Ohms at 60Hz) 16 0.01 neglect 20 0.008 neglect 25 0.0065 neglect 32 0.005 0.000009 50 0.0027 0.000016 63 0.002 0.000025 80 0.0014 0.000042 100 0.0011 0.00007 125 0.0008 0.0001 160 0.00055 0.00015 200 0.0004 0.0002 250 0.00029 0.00027 320 0.0002 0.0004

Hence the upstream MCCB impedance Zm1 is 0.0008 + j 0.0001 ohms. Therefore the fault impedance Zfb is:

The fault making current Ifb is:

Where Vp is the line-to-neutral voltage. Locate the point R for 26,195 A on the prospective curve in Figure 1.

Go back to Index ↑

#### 9. Find the fault current at the beginning of the load cable, point C

Hence the downstream MCCB impedance Zm2 is 0.005+j0.000009 ohms. Add this to Zfb to give the fault impedance Zfc as:

The fault making current Ifc is:

Locate the point S for 17,443 A on the prospective curve in Figure 1.

Go back to Index ↑

#### 10. Find the fault current at the end of the load cable, point D

Add Zc2 to Zfc to give the fault impedance Zfd as:

The fault making current Ifd is:

Locate the point U for 3473 A on the prospective curve in Figure 1.

Go back to Index ↑

#### 11. Check the peak making capacity and peak let-through capacity of the MCCBs chosen above

The following manufacturer’s data are typical for 125 A and 32 A MCCBs:

 MCCB Rating Making capacity Let-through capacity kApeak (cut-off) kArms kApeak 32 A 95 209 *** 6.0 125 A 132 290 *** 25.0

*** Approximate values of the doubling factor taken to be 2.2

Hence the peak making capacity of the 32 A MCCB is well in excess of the let-through peak current of the 125 A MCCB.

Go back to Index ↑

#### 12. Find the highest I2t value for the upstream MCCB

Locate two points P and Q on the curve of the upstream MCCB as follows,

 Point Current in p.u. Current in Amps Time in seconds I2t P 14 406 6 989016.0 Q 602 17,450 0.0016 487204.0

Hence I2t at P exceeds that at Q.

#### 13. Calculate a suitable size for the load cable to satisfy the I2t duty

For XLPE cables the ‘k factor’ for the I2t is 143. The cross-sectional area A is:

The next standard cross-sectional area is 10 mm2.

Go back to Index ↑

#### 14. Calculate the volt-drop in the load cable

The usual limit to volt-drop in three-phase cables feeding static loads is 2.5% at full load.

Where, Iflc = 29 A, L = 15 m and φ = 54.5495 degrees. For a 6 mm2 cable the volt-drop is found to be:

which is well within the limit of 2.5%.

Go back to Index ↑

#### 15. Select the largest conductor size from the above calculations

Comparing the conductor sizes found in 13. and 14. gives the larger as 10 mm2, and this size should be used. Revise the calculation of the fault current IfdThe impedance Zc2 of the load cable is:

Add Zc2 to Zfc to give the fault impedance Zfd as:

The fault making current Ifd is:

Go back to Index ↑

#### 16. Plot the results

The results are plotted in Figure 1.

Refrence // Switchgear and Motor Control Centres – Handbook of Electrical Engineering: For Practitioners in the Oil, Gas and Petrochemical Industry by Alan L. Sheldrake (Download here)

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Calculation of motor starting time as first approximationhttp://electrical-engineering-portal.com/calculation-of-motor-starting-time-as-first-approximation http://electrical-engineering-portal.com/calculation-of-motor-starting-time-as-first-approximation#comments Mon, 22 Jun 2015 03:48:22 +0000 http://electrical-engineering-portal.com/?p=64619

Motor starting operations The problems connected to motor starting operations are fundamentally linked to the type of motor which a determined motor operational torque “CM” offers, to the starting modality and to the connected load which has a determined load torque “C ”. A generic curve of the above mentioned quantities is shown in the Figure […]

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## Motor starting operations

The problems connected to motor starting operations are fundamentally linked to the type of motor which a determined motor operational torque “CM offers, to the starting modality and to the connected load which has a determined load torque “C ”.

The necessary starting torque “Ca can be expressed as:

Ca = CM – CL

and shall be well calibrated to prevent it from being either too low, so as starting is not too long and heavy – which causes risks of temperature rise for the motor – or from being too high on the joints or on the operating machines.

A generic curve of the above mentioned quantities is shown in the Figure 1 below.

The concept of motor starting time “ta” can be associated to this concept of properly calibrated starting and can be evaluated making reference to concepts linked to the motion dynamics, but also by introducing simplifying hypotheses which allows, however, an evaluation with a good approximation.

It is possible to relate the acceleration torque, expressed as a difference between the motor operational torque and the load torque, to the moment of inertia of the motor “JM, of the load “JL and to the motor angular speed, to obtain the following formula:

where the expression of “dω” assumes the following form:

and it is obtained by differentiating the well known expression for the motor angular speed:

Through simple mathematical operations and applying the method of integral calculus, it is possible to make the unknown quantity “ta” explicit by the following expression:

To express the value of the acceleration torque, it is necessary to introduce some simplifications:

The first one consists in considering an average value for the motor operational torque to be expressed as:

CM = 0.45 x (Cs + Cmax)

where CS represents the inrush torque and Cmax the maximum torque;

The second one concerns the torque due to the load and which can be correct by applying the multiplying factor KL linked to the load typology as in Table 1 below.

Table 1 – Values of factor KL

 Type of comparable loads Load Coefficient Lift Fans Piston Pumps Flywheel KL 1 0.33 0.5 0

In order to better understand the significance of the coefficient KL we associate to the type of load indicated in the table the torque characterizing the starting phase of the load by means of the following assumptions:

• Lift = load torque constant during acceleration
• Fans = load torque with square law increase during acceleration
• Piston pumps = load torque with linear increase during acceleration
• Flywheel = zero load torque.

With these assumptions, the acceleration torque can be expressed as:

These hypotheses allow to obtain the motor starting time with the aid of the following formula

The starting time allows to define whether a normal or a heavy duty start must be realized and to choose correctly the protection and switching devices. The above mentioned parameters relevant to the motor are given by the manufacturer of the motor.

As an example, Table 2 below shows the values that these parameters can take for three-phase asynchronous motors of common use and typically present on the market. Obviously the parameters relevant to the load characterize each single application and must be known by the designer.

Table 2 – Typical values of some electrical and mechanical parameters of a three-phase asynchronous motor

## Calculation of the starting time of a motor

Making reference to the data of the above table, here is an example of calculation of the starting time of a motor, according to the theoretical treatment previously developed.

 Three-phase asynchronous motor – 4 poles Frequency 160 kW Frequency 50 Hz Rated speed 1500 rpm Speed at full load 1487 rpm Moment of inertia of the motor JM = 2.9 Kgm2 Moment of inertia of the load JL = 60 Kgm2 Load torque CL = 1600 Nm Rated torque of the motor CN = 1028 Nm Inrush torque Cs = 2467 Nm (Cs = 2.4 x 1028) Max. torque Cmax = 2981 Nm (Cmax = 2.9 x 1028) Load with constant torque KL = 1

Cacc = 0.45 · ( CS + Cmax) – KL· CL = 0.45 · (2467 + 2981) – (1 · 1600) = 851.6 Nm

from which
ta = (2 · π · 1500 · (2.9 + 60)) / 60 · 851.6 = 11.6 s

Load with quadratic rising torque KL = 0.33

Cacc = 0.45 · ( CS + Cmax) – K· CL = 0.45 · (2467 + 2981) – (0.33 · 1600) = 1923.6 Nm

from which
ta = (2 · π · 1500 · (2.9 + 60)) / 60 · 1923.6 = 5.14 s

For both typologies of load, the esteemed motor starting time results to comply with the instruction given by the manufacturer regarding the maximum time admitted for DOL starting. This indication can be also taken as a cue for a correct evaluation of the thermal protection device to be chosen.

Reference // Three-phase asynchronous motors: generalities and proposals for the coordination of protective devices – ABB

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5 Components of Current Drawn by the DC Voltage Testing of Insulationhttp://electrical-engineering-portal.com/5-components-of-current-drawn-by-the-dc-voltage-testing-of-insulation http://electrical-engineering-portal.com/5-components-of-current-drawn-by-the-dc-voltage-testing-of-insulation#comments Fri, 19 Jun 2015 04:17:16 +0000 http://electrical-engineering-portal.com/?p=64555

Current drawn by the insulation When DC voltage is applied to an insulation, the electric field stress gives rise to current conduction and electrical polarization. Consider an elementary circuit as shown in Figure 1 below, which shows a DC voltage source, a switch, and an insulation specimen. However, this current immediately drops in value, and […]

The post 5 Components of Current Drawn by the DC Voltage Testing of Insulation appeared first on EEP - Electrical Engineering Portal.

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## Current drawn by the insulation

When DC voltage is applied to an insulation, the electric field stress gives rise to current conduction and electrical polarization. Consider an elementary circuit as shown in Figure 1 below, which shows a DC voltage source, a switch, and an insulation specimen.

When the switch is closed, the insulation becomes electrified and a very high current flows at the instant the switch is closed.

However, this current immediately drops in value, and then decreases at a slower rate until it reaches a nearly constant value.

The current drawn by the insulation may be analyzed into several components as follows:

### 1. Capacitance charging current

The capacitance charging current is high as the DC voltage is applied and can be calculated by the formula:

• C represents charging current
• RA represents absorption current
• RL represents volumetric leakage current (dielectric loss)

where:

• ie is the capacitance charging current
• E is the voltage in kilovolts
• R is the resistance in megohms
• C is the capacitance in microfarads
• t is the time in seconds
• e is Napierian logarithmic base
The charging current is a function of time and will decrease as the time of the application of voltage increases. It is the initial charging current when voltage is applied and therefore not of any value for test evaluation.

Test readings should not be taken until this current has decreased to a sufficiently low value.

Go back to Components ↑

### 2. Dielectric absorption current

The dielectric absorption current is also high as the test voltage is applied and decreases as the voltage application time increases, but at a slower rate than the capacitance charging current. This current is not as high as the capacitance charging current.

The absorption current can be divided into two currents called reversible and irreversible charging currents. This reversible charging current can be calculated by the formula:

ia = VCDT−n

where:

• ia is the dielectric absorption current
• V is the test voltage in kilovolts
• C is the capacitance in microfarads
• D is the proportionately constant
• T is the time in seconds
• n is a constant
The irreversible charging current is of the same general form as the reversible charging current, but is much smaller in magnitude. The irreversible charging current is lost in the insulation and thus is not recoverable.

Again, sufficient time should be allowed before recording test data so that the revers- ible absorption current has decreased to a low value.

Go back to Components ↑

### 3. Surface leakage

The surface leakage current is due to the conduction on the surface of the insulation where the conductor emerges and points of ground potential.

This current is not desired in the test results and should therefore be eliminated by carefully cleaning the surface of the conductor to eliminate the leakage paths, or should be captured and guarded out of the meter reading.

Go back to Components ↑

### 4. Partial discharge current

The partial discharge current, also known as corona current, is caused by overstressing of air at sharp corners of the conductor due to high test voltage. This current is not desirable and should be eliminated by the use of stress control shielding at such points during tests.

This current does not occur at lower voltages (below 4000 volts), such as insulation resistance test voltages.

Go back to Components ↑

### 5. Volumetric leakage current

The volumetric leakage current that flows through the insulation volume itself is of primary importance. This is the current that is used to evaluate the conditions of the insulation system under test. Sufficient time should be allowed for the volumetric current to stabilize before test readings are recorded.

The total current, consisting of various leakage currents as described above, is shown in Figure 2.

Go back to Components ↑

Reference // Electrical Power Equipment Maintenance and Testing – Paul Gill
(Purchase from Amazon)

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5 Copper Busbar Jointing Methodshttp://electrical-engineering-portal.com/5-copper-busbar-jointing-methods http://electrical-engineering-portal.com/5-copper-busbar-jointing-methods#comments Wed, 17 Jun 2015 04:30:20 +0000 http://electrical-engineering-portal.com/?p=64536

Efficient joints in copper busbar conductors Efficient joints in copper busbar conductors can be made very simply by: Bolting Clamping Riveting Soldering Welding Bolting and clamping are used extensively on-site. Shaped busbars may be prefabricated by using friction stir welding. 1. Bolted joints (most common) Bolted joints are formed by overlapping the bars and bolting […]

The post 5 Copper Busbar Jointing Methods appeared first on EEP - Electrical Engineering Portal.

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## Efficient joints in copper busbar conductors

Efficient joints in copper busbar conductors can be made very simply by:

Bolting and clamping are used extensively on-site. Shaped busbars may be prefabricated by using friction stir welding.

### 1. Bolted joints (most common)

Bolted joints are formed by overlapping the bars and bolting through the overlap area. They are compact, reliable and versatile but have the disadvantage that holes must be drilled or punched through the conductors, causing some distortion of the current flow in the bar.

Bolted joints also tend to have a less uniform contact pressure than those made by clamping but, despite these issues, bolted joints are very commonly used and have proven to be reliable.

They can be assembled on-site without difficulty.

Go back to Methods ↑

### 2. Clamped joints (most common)

Clamped joints are formed by overlapping the bars and applying an external clamp around the overlap. Since there are no bolt holes, the current flow is not disturbed resulting in lower joint resistance. The extra mass at the joint helps to reduce temperature excursions under cyclic loads.

Well-designed clamps give an even contact pressure and are easy to assemble, but take up more space than a bolted joint and are more expensive to manufacture.

Go back to Methods ↑

### 3. Riveted joints (difficult, but…)

Riveted joints are similar to bolted joints. They can be efficient if well made. It is difficult to control the contact pressure. They cannot easily be dismantled or tightened in service and they are difficult to install.

Go back to Methods ↑

### 4. Soldered or brazed joints (rare)

Soldered or brazed joints are rarely used for busbars unless they are reinforced with bolts or clamps since heating under short-circuit conditions can make them both mechanically and electrically unsound.

Go back to Methods ↑

### 5. Welded joints (not very safe…)

Welded joints are made by butting the ends of the bars and welding. They are compact and have the advantage that the current-carrying capacity is unimpaired, as the joint is effectively a continuous copper conductor. However, it may not be safe or desirable to make welded joints in situ.

Welding of copper is discussed in Copper Development Association Publication 98, Cost-Effective Manufacturing: Joining of Copper and Copper Alloys (Download here).

Go back to Methods ↑

## Joint Resistance Calculation

In principle, a clamped or bolted joint is made by bringing together two flat surfaces under controlled (and maintained) pressure, as shown in Figure 6.

The resistance of a joint is mainly dependent on two factors:

1. The streamline effect or spreading resistance, Rs, due to the diversion of the current flow through the joint
2. The contact resistance or interface resistance of the joint, Ri.

The total joint resistance, Rj, is given by:

Rj = Rs+ Ri

This applies specifically to direct current applications. Where alternating currents are flowing, the changes in resistance due to skin and proximity effects in the joint zone must also be taken into account.

Go back to Methods ↑

Reference // Copper for Busbars – Guidance for Design and Installation – Copper Development Association (Download guidance)

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Basics of DC Motors For Electrical Engineers – Beginnershttp://electrical-engineering-portal.com/basics-of-dc-motors-for-electrical-engineers-beginners http://electrical-engineering-portal.com/basics-of-dc-motors-for-electrical-engineers-beginners#comments Mon, 15 Jun 2015 04:14:50 +0000 http://electrical-engineering-portal.com/?p=64527

General about DC motors Separate field excitation DC motors are still sometimes used for driving machines at variable speed. These motors are very easy to miniaturize, and essential for very low powers and low voltages. They are also particularly suitable, up to high power levels (several megawatts), for speed variation with simple, uncomplicated electronic technologies […]

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## General about DC motors

Separate field excitation DC motors are still sometimes used for driving machines at variable speed. These motors are very easy to miniaturize, and essential for very low powers and low voltages.

They are also particularly suitable, up to high power levels (several megawatts), for speed variation with simple, uncomplicated electronic technologies for high performance levels (variation range commonly used from 1 to 100).

Their characteristics also enable accurate torque regulation, when operating as a motor or as a generator. Their nominal rotation speed, which is independent of the line supply frequency, is easy to adapt by design to suit all applications.

They are however less rugged than asynchronous motors and much more expensive, in terms of both hardware and maintenance costs, as they require regular servicing of the commutator and the brushes.

### Construction of DC motor //

A DC motor is composed of the following main parts:

#### Field coil or stator

This is a non-moving part of the magnetic circuit on which a winding is wound in order to produce a magnetic field. The electro-magnet that is created has a cylindrical cavity between its poles.

#### Armature or rotor

This is a cylinder of magnetic laminations that are insulated from one another and perpendicular to the axis of the cylinder. The armature is a moving part that rotates round its axis, and is separated from the field coil by an air gap. Conductors are evenly distributed around its outer surface.

#### Commutator and brushes

The commutator is integral with the armature. The brushes are fixed. They rub against the commutator and thus supply power to the armature conductors.

### Operating principle

When the field coil is energized, it creates a magnetic field (excitation flux) in the air gap, in the direction of the radii of the armature. This magnetic field “enters” the armature from the North pole side of the field coil and “exits” the armature from the South pole side of the field coil.

When the armature is energized, currents pass through the conductors located under one field coil pole (on the same side of the brushes) in the same direction and are thus, according to Laplace’s law, subject to a force.

The conductors located under the other pole are subject to a force of the same intensity in the opposite direction. The two forces create a torque which causes the motor armature to rotate (see Figure 1).

When the motor armature is powered by a DC or rectified voltage supply U, it produces back emf E whose value is:

E = U – RI

where RI represents the ohmic voltage drop in the armature.

The back emf E is linked to the speed and the excitation by the equation:

E = k ω Φ

Where:

• k is a constant specific to the motor
• ω is the angular speed
• Φ is the flux

This equation shows that at constant excitation the back emf E (proportional to ω) is an image of the speed.

The torque is linked to the field coil flux and the current in the armature by the equation:

T = k Φ I

If the flux is reduced, the torque decreases.

There are two methods for increasing the speed //

1. Either increase the back emf E, and thus the supply voltage at constant excitation: this is known as “constant torque” operation.

2. Or decrease the excitation flux, and thus the excitation current, while keeping the supply voltage constant: this is known as “reduced flux” or “constant power” operation. This operation requires the torque to decrease as the speed increases (see Figure 2 below). However, for high reduced flux ratios this operation requires specially adapted motors (mechanically and electrically) to overcome switching problems.

The operation of this type of device (DC motor) is reversible //

If the load opposes the rotation movement (the load is said to be resistive), the device provides a torque and operates as a motor.

If the load is such that it tends to make the device rotate (the load is said to be driving) or it opposes the slow-down (stopping phase of a load with a certain inertia), the device provides electrical energy and operates as a generator.

### Various types of DC motor

#### Parallel excitation (separate or shunt)

The coils, armature and field coil are connected in parallel or supplied via two sources with different voltages in order to adapt to the characteristics of the machine (e.g.: armature voltage 400 volts and field coil voltage 180 volts).

The direction of rotation is reversed by inverting one or other of the windings, generally by inverting the armature voltage due to the much lower time constants. Most bidirectional speed drives for DC motors operate in this way.

#### Series wound

The design of this motor is similar to that of the separate field excitation motor. The field coil is connected in series to the armature coil, hence its name. The direction of rotation can be reversed by inverting the polarities of the armature or the field coil.

This motor is mainly used for traction, in particular on trucks supplied by battery packs. In railway traction the old TGV (French high-speed train) motor coaches used this type of motor. More recent coaches use asynchronous motors.

#### Compound wound (series-parallel excitation)

This technology combines the qualities of the series wound motor and the shunt wound motor. This motor has two windings per field coil pole. One is connected in parallel with the armature. A low current (low in relation to the working current) flows through it. The other is connected in series.

It is an added flux motor if the ampere turns of the two windings add their effects. Otherwise it is a negative flux motor. But this particular mounting method is very rarely used as it leads to unstable operation with high loads.

### Open DC motor – School project (VIDEO)

Here we have an open DC motor with permanent magnet for the field. An old armature from a faulty hedge trimmer was used. As it was for a school project The following topics had to be covered.

• Chemical energy – We used a lead acid battery
• Kinetic energy – We got the armature rotating
• Noise – We got audio from the speaker
• Heating – We got some heating in the armature, and commutator bars
• Light – We made a special LED driven via coil
• Magnetic – We established that the armature won’t rotate without the magnet underneath.

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Reference // Cahier technique no. 207 – Electric motors – Schneider Electric

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