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Diesel Engine Starting System | Part 2

The job of any starter motor assembly is to take a stored energy (electric, air, or hydraulic) and convert it into mechanical rotation to crank the engine fast enough to begin the engine’s ignition sequence. The most common type of starting system uses electrical energy; however, compressed air and hydraulic energy can be used as well. The following are the main comparable components of the three main types of starting systems:

We’ll first examine the different electric starter motor designs, next discuss air and hydraulic starter motors, and then look at the control circuit for starters. An electric starter will take stored electrical energy from a battery (or sometimes a capacitor) and convert it into torque at the starter’s pinion gear. The pinion then engages with the ring gear that is part of the engine’s flywheel and turns the flywheel that rotates the engine’s crankshaft. ● See Figure 2 for a cutaway of a starter and its main parts.

There are two main types of electric starter motor assemblies: Direct drive (pinion is driven directly by the armature): A direct drive electric starter has a motor that is designed to generate high torque at low speed and operate at high speed with low torque (the motor will sometimes exceed 5000 rpm) for a short length of time. It will use a solenoid actuated shift lever to push out the pinion to engage it with the ring gear before or just as the armature (rotating shaft in the motor assembly) starts turning. Gear reduction (higher speed motor output to a gear reduction and then to pinion): A gear reduction starter (planetary or pinion reduction) is designed to use a smaller higher speed electric motor to produce higher cranking torque with the same or less electrical power consumed. The heaviest and bulkiest part of a direct drive starter is the motor so by reducing motor size and weight the engineers have saved space and weight. Some direct drive starters are twice the weight as a comparable output gear reduction starter. Although this isn’t a big concern for a large machine, you will be thankful for the lighter weight whenever it comes time to change the starter. Gear reduction starter motor assemblies can have their motor offset from the output shaft or use planetary gears and have the motor shaft in-line with the output shaft. Direct Drive Starter Components ■ Starter housing: Center section that holds the pole shoes and field coils in place. ■ Nose piece: The drive end of the starter where the pinion gear is located. Holds the shift lever in place and supports the armature shaft with a bushing. ■ End cap: Opposite end of the starter from the nose piece. Supports brush holder assembly and the other armature shaft bushing. ■ Armature: The rotating part of the motor that has several windings that have each of their ends loop to a commutator bar. It will have splines to drive the starter drive. ■ Brushes: Contact the commutator bars and transfer electrical current to the armature. ■ Brush holders: Spring loaded to keep the brushes in contact with the armature. ■ Field coils: Heavy copper windings that create a strong magnetic field when current flows through them. ■ Pole shoes: Iron cores for the field coils that help to increase magnetism. ■ Solenoid: Has two windings (pull-in and hold-in) that get energized by the starter control circuit and magnetically move a plunger. The plunger is connected to a heavy contact disc that is a switch. The switch will send current from the battery terminal to the field coils. The plunger could also be connected to a shift lever that will move the pinion. ■ Pinion gear: The starter output that engages with the flywheel and cranks the engine. ■ Overrunning clutch: Drives the pinion from the armature shaft but will not allow the armature to be driven by the ring gear. ■ Shift lever: Used to push the pinion out to engage with the ring gear. (To Be Continued: For More Parts Visit Our Blog)

Diesel Engine Starting Systems | Part 1

The Importance of Starting Systems: A functional machine needs a running engine, and if the engine doesn’t crank, it doesn’t start. A properly operating and reliable starting system is a must for keeping a machine productive. For many years, diesel engines have mostly used electric motors to crank them over to start the combustion process. For some applications, an air or hydraulic motor will create the torque needed to turn the engine over. Many years ago, diesel engines were sometimes started with a smaller gas engine called a pup engine. See Figure 1 for a pup engine on an older diesel engine.

Another way to get a diesel engine started was to start it on gasoline and then switch it over to run on diesel fuel. This was a complex solution to a simple task because the engine had to have a way to vary its compression ratio, and it needed a spark ignition system and a carburetor. As 12V electrical systems became more popular and electric motor design improved, electric starters were able to get the job done. Many large diesel engines will use a 24V starting system for even greater cranking power. See Figure 2 for a typical arrangement of a heavy-duty electric starter on a diesel engine.

A diesel engine needs to rotate between 150 and 250 rpm to start. The purpose of the starting system is to provide the torque needed to achieve the necessary minimum cranking speed. As the starter motor starts to rotate the flywheel, the crankshaft is turned, which then starts piston movement. For a small four-cylinder engine, there doesn’t need to be a great deal of torque generated by a starter. But as engines get more cylinders and bigger pistons, a huge amount of torque will be needed to get the required cranking speed. Some heavy-duty 24V starters will create over 200 ft-lb of torque. This torque then gets multiplied by the gear reduction factor between the starter motor pinion gear and ring gear on the engine’s flywheel. This is usually around 20:1. See Figure 3 for how a starter assembly pinion engages with the flywheel ring gear.

Some larger engines will need two or more starters to do this. Some starters for large diesel engines will create over 15 kW or 20 hp! See Figure 4 for a double starter arrangement.

When a starter motor starts to turn the engine over, its pistons start to travel up in the cylinders on compression stroke. There needs to be between 350 and 600 psi of pressure created on top of the piston. This is the main resistance that the starter has to overcome. This pressure is what is needed to create the necessary heat in the cylinder so that when fuel is injected it will ignite. If the starting system can’t crank the engine fast enough, then the compression pressure and heat won’t be high enough to ignite the fuel. If the pistons are moving too slowly, there will be time for the compression to leak by the piston rings. Also the rings won’t get pushed against the cylinder, which again allows compression pressure to leak into the crankcase. When this happens, the engine won’t start or it starts with incomplete combustion. Incomplete combustion equals excessive emissions. This is another reason to have a properly operating starting system. The faster a starter can crank a diesel engine, the faster it starts and the quicker it runs clean. This engine cranking task is much more difficult in colder temperatures especially if the engine is directly driving other machine components such as hydraulic pumps, a torque converter, or a PTO (power take-off) drive shaft. Cold engine oil adds to the load on the starter, and this load may increase by three to four times what it would normally be in warmer weather. Engine oil that is the wrong viscosity (too thick) for the temperature will greatly increase the engine’s rolling resistance. Adding to this problem is the fact that a battery is less efficient in cold temperatures. When engineers design a cranking system, they must take into account cold weather cranking conditions and will quite often offer a cold weather starting option. This would likely include one or more of the following: bigger or more batteries, higher output starter, larger battery cables, battery blankets, oil heaters, diesel fired coolant heater, electric immersion coolant heater (block heater), and one or more starting aids like an ether injection system or an inlet heater. One more recent difficulty added to starting systems is a result of electronic controls on some engines. Some ECMs may need to see a minimum number of engine revolutions at a minimum speed before it will energize the fuel system. This equates to longer cranking times and more strain on the cranking system. Some electronic engines will crank for five seconds or longer even when the engine is warm before the ECM starts to inject fuel and the engine starts. It’s important that a machine’s starting system works properly and you should be aware of how the main components of a system work. This will give you the knowledge needed to make a proper diagnosis when you get a complaint of an engine cranking slowly or not at all. If an engine doesn’t start, then a machine isn’t working, and instead of making money, it’s costing money. The better you know how to diagnose and repair a starting system problem, the more valuable you will be as an HDET. There are lots of technicians who are good at changing starters whether the starter is faulty or not. Many times the cause of a starting complaint is something other than the starter. If a starter is used properly, it will last for well over 10,000 starts. The biggest factor in reducing the life of an electric starter is overheating from over-cranking. Never run the starter for more than a 30-second stretch, and if it does run that long, then wait at least two minutes between cranks to allow the starter to cool. For engines up to 500 hp, electric starting systems will be used for 99% of the applications. For any size engine, air and hydraulic starting systems are an option; however, they will likely only be used for special applications and usually for engines over 500 hp. (For More Parts Visit Our Blog)

Selecting, sizing Electrical Power Transformers For Buildings

While commercial building designs change, their electrical loads remain fundamentally unchanged. Properly sizing and selecting transformers ensures that these loads are accommodated.

Transformers, along with other power distribution apparatus, remain a fundamental component in electrical systems distribution for commercial buildings. This article presents several useful design concepts for selecting and sizing transformers in the design of electrical systems for commercial buildings. Transformers change voltage levels to supply electrical loads with the voltages they require. They supply the required incoming electrical service to the buildings. Transformer primary and secondary voltages can be 2,400; 4,160; 7,200; 12,470; and 13,200 for 15-kV Class, and 120, 208, 240, 277, and 480 for 600-V Class. Transformers are located either outdoors or inside buildings in an electrical room or other areas as permitted by code. The electrical phase characteristics associated with the transformer’s primary side is 3-phase, 3-wire or Delta connected. The secondary is 3-phase, 4-wire or Wye connected.

Construction types There are different construction types for transformers used in commercial buildings. Our understanding of their general characteristics will allow the designer and end user to make the proper selection for the electrical system application. Following are some of the transformer types available in the industry along with a few of their characteristics: Ventilated dry-type transformers are ventilated by air, use larger space for clearance, and use different insulating materials to augment the dielectric strength of the air. They contain an enclosure surrounding the windings for their mechanical protection and the safety of personnel. This type is the most common to be used in the building indoor electrical system distribution. See Table 1 for typical dry-type transformer ratings, dimensions, and weights. Sealed dry-type transformers are similar to dry type in most of their characteristics. The difference is they contain an enclosed tank with nitrogen or other dielectric gas to protect the windings. They can be installed outdoors or indoors. They are useful in areas with a corrosive or dirty atmosphere. Cast-coil transformers are constructed with the primary and secondary windings encapsulated in reinforced resin. They can be installed where moisture or airborne contaminants exist. Nonventilated dry-type transformers are similar to the ventilated type but are totally enclosed. This type can be installed in areas with corrosive or dirty atmospheric conditions where it would be impossible to use a ventilated-type transformer. Oil-filled transformers are constructed with the windings encased in an oil-tight tank filled with insulating mineral oil. It is good practice to regularly test this type of transformer in order to determine dielectric breakdown, which affects its useful life. Application types There are different ways in which transformers are installed and used as part of a commercial building electrical system. These application types include: Indoor distribution transformers are used with panelboards and are separately mounted to supply the specific electrical load requirements in a system-specific application within the system distribution. Several transformer types rated higher than 600 V for oil insulated type, higher than 35,000 V for dry type, and other transformers rated higher than 600 V are required to be located in vault rooms, which must be built with fire-rated enclosures depending on the transformer type and applicable local authority requirements, when indoors. Transformers that are not over 600 V and are part of the indoor building electrical system distribution have both primary and secondary voltages below 600 V with the most common voltage level change from 480 V to 208 Y/120 V. Pad-mounted transformers are installed outside and are considered the first option for supplying service entrance voltage to the building electrical system based on the project size and requirements. They typically have primary voltages higher than 600 V and secondary voltages lower than 600 V with compartments for the associated protective devices assembled in an integral tamper-resistant and weatherproof unit. In addition, the size of the commercial facility will determine the appropriate approach for designing the electrical distribution system for the specific application. In this electrical system design, the transformer can be used as part of a substation, primary unit substation, secondary unit substation, or network configuration. Sizing The electrical size of the transformer load is rated in kVA. This rating provides the associated power output delivered for a specific period by the loads connected to the transformer on the secondary side of the equipment. The loads, which are calculated as part of the building electrical system design phase, are shown in the construction documents’ respective equipment schedules in VA or kVA. A general approach to determining transformer capacity and selecting the proper rating for the design application is to obtain the calculated design load from the respective electrical schedule and add 20% spare capacity for future load growth to be shown in the equipment schedule, unless otherwise directed by the facility based on design parameters. For example, the code-based demand load of a 208 Y/120 V, 3-phase, 4-wire panelboard is 42 kVA, which does do not include spare capacity for future growth. Therefore, the transformer size required for converting the system voltage from 480 V, 3-phase, 3-wire to 208 Y/120 V, 3-phase, 4-wire is: Transformer size in kVA = 42 kVA x 1.25 = 52.5 kVA Therefore, a 75 kVA transformer would be selected for this application out of the available standard ratings for a 480 V primary to 208 Y/120 V secondary. The most common building industry standard ratings are 3, 6, 9, 15, 30, 37.5, 45, 75, 112.5, 150, 225, 300, 500, 750, and 1,000 kVA. The above simple calculation meets the intent to achieve the normal life expectancy of a transformer, which is based on the following basic conditions: The transformer is equal to or less than its rated kVA and rated voltage. The average temperature of the cooling air during a 24-hour period is 86 F. The temperature of the cooling air at no time exceeds 104 F. Selecting Transformer selection starts with the kVA rating required to supply the loads connected in the electrical system. Another consideration for indoor distribution transformers is the type of load: linear or nonlinear. Linear loads include resistive heating and induction motors; nonlinear loads are produced by electronic equipment that contributes to the distortion of the electrical power signals by generating harmonics. The harmonics resulting from nonsinusoidal currents generate additional losses and heating of the transformer coils, which reduce the transformer life expectancy. Indoor transformers for nonlinear loads can be selected with a K rating, which allows the transformer to withstand nonlinear conditions in the electrical system. K-rated transformers do not mitigate or eliminate harmonics. However, they do protect the transformer itself from damage caused by harmonics. For harmonic mitigation, K-rated transformers can be combined with harmonic filters or chokes. For linear load applications, transformers should be selected with lower core losses. Other factors that should be considered in selecting transformers are voltage ratings for both primary and secondary, voltage taps, efficiency, impedance value, type of cooling and temperature rise, voltage insulation class, basic impulse level, and sound level. Practical applications In the past two years, two large projects in Miami Dade County have been built: the Florida International University football stadium and Miami International Airport South Terminal. Both projects included dry-type 480 V, 3-phase to 208 Y/120V V step-down transformers (in NEMA 2 enclosures), ranging from 15 kVA to 112.5 kVA in the electrical system distribution design. The 18,688-seat FIU football stadium was designed with about 12 transformers as part of the electrical system distribution in order to supply general-use receptacles, small motors, and other loads in the stadium building structure and the attached field house building. The MIA South Terminal expansion was designed with about 50 transformers with similar intent as the stadium’s but a more diverse group of loads for the 208 Y/120 V 3-phase, 4-wire system, which also included lighting loads, signage, telecommunication, security systems, and other loads part of this building project (Figure 1). Installation The installation of power transformers and transformer vaults must comply with the requirements of National Electrical Code (NFPA 70) article 450 and specific local authority having jurisdiction requirements. Some principles to consider for transformer installation include locating them in isolated rooms with proper ventilation, clearances, and accessibility. Otherwise, they can be installed on open walls or steel columns or above suspended ceilings. In addition, there are other specific requirements based on the transformer type, such as weatherproof enclosures for dry-type transformers installed outdoors or a transformer vault room for oil-insulated transformers installed indoors. In addition, a good design and installation require the proper transformer feeder and overcurrent protection device size based on NEC articles 240, 250, 450, and applicable sections of Article 310 (Figure 2). Looking ahead Transformers remain a fundamental component of electrical distribution systems. Equipment operation characteristics will continue to change. However, their operating principles will remain with the same. The industry trend is to continue building transformers with less core losses, and that comply with Energy Star efficiency requirements.

What Does Power Factor Mean?

The low power factor reduces the distribution capacity of the electrical system by increasing the current flow. Therefore, having a low-power feature does not work and is expensive. But what is electrical energy and what is the effect? The standard distribution system is limited to the current manageable value; the energy factor, expressed as a percentage, is an indication of the total current value that can be used to create a function (active energy). The proximity of 1.00 (100%), decreases the amount of current required to perform the specified function. For example, a power load of 0.80 means that only 80% of the energy is successfully used to do the job. In a perfect world, all the power taken from the energy system will be transformed into a useful function, but this is not the case in the real world. To fully define power, complex calculations are required. Easy to understand, however, the U.S. Department of Energy. You use a simple simulation of the force required for the horse to pull the trolley down the track. Appropriately, the horse would be placed in front of a train car to provide the most efficient pulling force; however, that does not always happen. The gravitational angle represents a change in electrical energy - the smaller the angle, the better the factor, the greater the angle, the lower the factor

1. Angles affect useful function. The analogy shown here provides visual aids to help us understand the power factor. An element of energy is defined as the measure of the actual (active) energy in the visible (total) energy. When a horse is led close to the center of the track, the side-pull angle decreases, and the actual force is closer to the visible force. Source: U.S. Department of Energy The absolute power needed to pull the train car is the visible power. The actual power that drives the train car is real power. Unused power from the pulling part of the horse is active force. In other words, real power, also called active force (kW), performs the actual function of movement, heat, and light. Active power, or inactive power (kVar), supports the magnetic field of the active load (frequency intake). Currently used to create active energy is not used to create work; however, this currently places a burden on the distribution system, electricity supplier, and local electricity bill. Total working capacity of working capacity and non-working capacity total power (physical energy): Emerging Power = √ (Real Power2 + Active Power2) used to calculate the power factor: Power factor = Real strength / visible strength = angle cosine (ϕ) Power and Current Foundations To understand the element of energy, we must first understand a certain basic dynamic current (AC) and related wave forms. The voltage in the AC system alternates between positive and negative (in the sinusoidal state) and forces the current to behave in the same way. This happens 60 times per second (in the 60-Hz system), from 0 to 360 degrees. Unlike AC systems, the electric current in the direct current (DC) mode does not change. Because the rapid value of AC power continues to change, science has defined a different value for AC values, i.e. the value of RMS (root means square). The RMS value of the AC waveform produces the same thermal effect as the DC waveform of the same value. RMS is the square root of the square root definition of a group of instant values (cycle). Where the current voltage is sinusoidal, RMS voltage and current can be found in peak (pk) voltage and current: VRMS = Vpk / √2 119.5 VRMS = 169 Vpk / 1.414 Similarly, IRMS = Ipk / √2 75 ARMS = 106 Apk / 1.414 You may be wondering, what does this have to do with power? AC power calculation requires knowledge of RMS voltage, current RMS, and sinusoidal phase relationships. Therefore, in summary, RMS is a measure of the effect of temperature, calculated from the waveform, which allows the comparison of AC and DC. Any phase shift from the state of pure sinusoidal radio indicates a power factor. The following is a comparison of how the power factor affects the release of kVA into two different loads of the same phase. With a 9-kW electric heater space (120 VAC, 75 A) with 1.0 power factor (PF): P = -1ϕ x 120 VAC x 75 A x 1.0 PF = 9 kW kVA = -1ϕ x 120 VAC x 75 A = 9 kVA 9-kW (120 VAC, 75 A) battery charger with 0.866 PF: P = -1ϕ x 120 VAC x 86.6 A x 0.866 PF = 9 kW kVA = -1ϕ x 120 VAC x 86.6 A = 10.392 kVA Although each load uses 9 kW of power, the power factor for charging the battery charger is 0.866. The low power feature requires an additional 11.6 A to operate, which is ultimately supplied by the power company. Not only should the previous active add-ons be purchased, but also the size of the distribution system should be increased to handle additional extras. What Affects the Power of the Feature? An element of energy refers to the relationship between active energy (useful energy) and physical (total) energy. This relationship is a measure of how well electricity is used. Linear Resistive Loads. In the AC system, the loads are separated by the current drawing. The resistive load line is an opposing load that is free of energy-intensive or energy-resistant materials, such as space heater and incandescent lamps. Current voltages and currents meet zero simultaneous connections. The force curve (P) in Figure 2 is calculated by the force (V) and the current (I), shown as the positive area of the graph. In this example, the current voltage is 120 VRMS and 75 ARMS, respectively. The output of these two is 9 kVA or 9 kW. Electric current and current are "in the category," and 100% energy (working capacity) is used effectively to do useful work. The power factor for this type of load is 1.0.

2. Linear resistive loads. The current Voltage is in the power category equal to 1.0 of the resistance loads. Sincerely: Ametek Solidstate Controls Linear Non-Resistive / Reactive Loads. It is not uncommon to find only contradictory loads; many loads have an additional functionality. These inactive / inactive loads make up a large percentage of all loads. The current state of energy has shifted from electric current to “out of phase.” When the load generates power, the current lags set the voltage; if the load is strong, the current track. Industrial facilities often have residual energy loads (incoming loads). These types of loads can be induction motors, choke, and transformers. Lead power loads (capacitive loads) are less common and are usually underground cables or power switch switching modes. In Figure 3, the same load from Figure 2 now has a voltage and a waveform of the current coming out of the phase by 30 degrees. Because this is a dynamic wave system, the current is left behind(lagging).

3. Import duties. Voltage and current are not in the category of direct non-combat / active loads. In this powerful loading example, the current lags are 30 degrees with a power of 0.866. Sincerely: Ametek Solidstate Controls Indistinguishable Loads - Harmonics. Modern industrial settings not only carry powerful, powerful, and powerful objects, but many also include state-of-the-art equipment, such as power converters, DC drives, frequency-frequency drives (VFDs), electronic ballast, arc welding, and heat- in a controlled oven. All of these are indirect loads, or loads where the current is not sinusoidal, even if the voltage is sinusoidal. The non-sinusoidal nature of these waveforms is expressed using harmonics. Harmonics is a type of wave of varying frequencies at recurring frequencies of the basic frequency of electrical energy (50 Hz or 60 Hz). They are placed on top of the current sinusoidal form to create a complete current formula. Figure 4 is an example of such a current waveform.

4. Non-linear (Offline) loads. This graph shows the voltage and current power supply for non-harmonics. Shown without modification of the current 30-degree section for clarity. Sincerely: Ametek Solidstate Controls The RMS value of all current is obtained by summarizing the current RMS value of each harmonic. Given the 60-Hz wave form, this means that the second frequency of harmonic frequency is 120 Hz (60 Hz x 2 = 120 Hz) and the 3, 4, and 5 harmonic frequencies are approximately 180 Hz , 240 Hz, and 300 Hz, respectively. As a repetition of basic frequency, harmonics can be expressed as 2f, 3f, 4f, etc. Total harmonic current distortion (THD) is the sum of all the harmonic components of the current waveform compared to the basic part of the current wave. As shown below, the RMS value for current harmonics is higher than the current RMS value. ITHD = RMS for current harmonics / RMS current base = √ (I22 + I32 + I42 +…) / I1 x 100% For sinusoidal waveforms only, the variability of the phase between power and electricity is sufficient to measure the strength of the power factor (PF). In non-sinusoidal waveforms, the term displacement power factor (DpPF) is used to measure phase transitions between the bases of two waveforms (50-Hz or 60-Hz). In the same non-sinusoidal waveforms, the term is defined to measure the effect of harmonics over the PF. This term is called the distension power factor (DF). DF = 1 / √ (1 + THD2) To find the total energy (PFT), the following equation is used: PFT = DF x DpPF Power Factor integration In direct loads, the power triangle is the right triangle that shows the relationship between active, active, and tangible force. The relationship between active energy and perceived energy is PF. The value can range from 0.0 to 1.0. Operating power, also called true power, real power, or operating power, performs real movement / heat / light function etc and is measured in watts (W). Active power supports a magnetic or electrical field in devices, such as solenoid coils, motor windings, transformer windings, capacitors and ballasts, without performing the actual function. This extra energy is measured in volt-amperes reactive (VAR) and is sometimes called “wattless”. Visual power combines working capacity and operating power, and is measured in volt-amperes (VA). The phase angle (ϕ) in degrees, represents the "inefficiency" of the load and corresponds to the total impedance value (Z) in the current flow of the load. The larger the phase angle, the greater the effective power. Non-linear loads add an extra element to the full (visible) power without adding to the active force, further further reducing the power factor. https://www.youtube.com/channel/UCFVmeTU84v-ymkDGeyC0B4g https://www.youtube.com/channel/UCB2AfhtTsUstDZ7FE7rpWlw

Understanding The Basics Of Electricity By Thinking Of It As Water.

Understanding the basics of electricity by thinking of it as water “We believe that electricity exists because the electric company keeps sending us bills for it. But we cannot figure out how it travels inside wires.” — Dave Barry The fundamental laws of electricity are mathematically complex. But using water as an analogy offers an easy way to gain a basic understanding. Electricity (I)– Voltage, Current, and Resistance The three most basic components of electricity are voltage, current, and resistance. VOLTAGE is like the pressure that pushes water through the hose. It is measured in volts (V). CURRENT is like the diameter of the hose. The wider it is, the more water will flow through. It is measured in amps (I or A). RESISTANCE is like sand in the hose that slows down the water flow. It is measured in ohms (R or Ω).

Voltage, current, and resistance are all related. If you change one of them in a circuit, the others will change, too. Specifically, voltage is equal to current multiplied by resistance (V = I x R). Thinking about water, if you add sand into the hose and keep the pressure the same, it’s like reducing the diameter of the hose… less water will flow.

Electricity (II) – DC, AC, Batteries, and Transformers How does electricity work in electronics and the grid? DIRECT CURRENT or DC is similar to the normal flow of water in a hose – it flows in one direction, from the source to the end. Historically, DC was originally championed by Thomas Edison in the famous Current Wars of the late 1800s. DC lost the war for the grid but it has found an even more exciting role in modern electronics like computers, phones, and televisions. ALTERNATING CURRENT or AC is like the water flowing back and forth within the hose many times per second. The water analogy breaks down a little here but AC is easily created by electric generators (also called alternators). Nikola Tesla and George Westinghouse championed AC over DC and they eventually prevailed. AC is now the global standard for delivering electricity to homes and buildings via the grid.

BATTERIES can be thought of as water pumps that circulate water through a hose that travels in a closed loop back to the battery. There are many metrics used for the capacity of batteries and not all are immediately logical. They include amp-hours and kilowatt hours. Batteries can only generate DC power. TRANSFORMERS are like holding your thumb partially over the end of the hose to get the water to spray farther. The water volume (power) remains the same but the pressure (voltage) increases as the diameter (current) decreases. This is exactly what transformers do for overhead powerlines. The electricity can travel farther with fewer losses because the resistance (sand) doesn’t impede the electricity (water) when the current is lower (smaller diameter hose). Transformers only work with AC. The ability to move electricity over long distances is the main reason AC beat out DC a century ago. Electricity (III) – Power and Energy Now, let’s keep using the hose analogy to dive into the murkier waters of circuits (pun intended, sorry). POWER is like the volume of water that is flowing from the hose, given a specific pressure and diameter. Electric power is measured in watts (W). And larger systems are measured in kilowatts (1 KW = 1000 watts) or megawatts (1 MW = 1,000,000 watts). ENERGY is like measuring the volume of water that has flowed through the hose over a period of time, like filling a 5 gallon bucket in a minute. Electric energy is often confused with electric power but they are two different things – power measures capacity and energy measures delivery. Electric energy is measured in watt hours (wh) but most people are more familiar with the measurement on their electric bills, kilowatt hours (1 kWh = 1,000 watt hours). Electric utilities work at a larger scale and will commonly use megawatt hours (1 MWh = 1,000 kWh).

Hopefully, this offers a helpful introduction to the basics of electricity. We’d love to hear your feedback and suggestions, so please leave suggestions and comments below.

Dancing LED Circuit

Dancing LED Circuit.
Sharing it by demand of one of our YouTube subscriber.
Components explanation.
D = Diode having part number 1N4007.
R = Resistor(Value of each one is specified).
C= Capacitor(Value of each one is specified).
LED = Light Emitting Diode is a small size light source.
555 Timer = A small size timer IC available in market with name "Triple Five Timer IC" or "555 IC".
5V = A DC power source having 5 Volt output. It may be a battery or may be a power supply. Both are available in market. Different batteries and different power supplies have different output voltages so before purchasing ask about its output voltage. You can make a power supply by your own having 5 volts output. In this circuit upper side is connected to positive terminal of DC power supply and lower side with ground symbol is connected to negative terminal of DC power supply.
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Three Phase Induction Motor Star-Delta Starter Complete Circuit Diagram(Power And Control Wiring)

Three Phase Induction Motor Star-Delta Starter Complete Circuit Diagram(Power And Control Wiring)
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Electric load calculation and selection of transformer according to that load(Diversity Method).
Video link is given below.
https://youtu.be/PV9rxjhO2sU

How to find size of the electrical cable/wire for a given load?

How to find size of the electrical cable/wire for a given load?
1) Type of load : For feeder,control panel,distribution board (DB) or motor
2) Power Source : AC /DC power for Three phase or single phase
3) Full load current (Ampere)
4) Circuit Breaker size
5) Types of cable ( PVC,XLPE,Fire proof)
6) Installation Condition ( Underground, trunking,tray or conduit)
7) Distance / length of cable run from source to load
Calculation for sizing electrical cable/wire
Ib ≤ In ≤ Iz
Ib = Amperage load
In = Circuit Breaker size
Iz = Current rating for cable/wire

Example :

We plan to install 1 unit induction motor 75 hp for running a water pump.Power factor value is 0.8 and motor efficiency is 0.85.Power supply for this motor is 415 volt 3 phase 50Hz.Distance from panel to motor is 75 meter.This panel and pump is located at outside of factory.Please size the suitable circuit breaker,electrical cable and also type of cable for this application.

Calculation

Ib ≤ In ≤ Iz
Ib = ( hp x 746) / (√3 x Volt x PF x Efficiency)
Ib = (75 x 746) / (√3 x 415 x 0.8 x 0.85)
Ib = 114.5 Ampere
In = Ib ≤ breaker sizing
In = 114 ≤ 150 ampere
In = 150 ampere (we decide to used MCCB 3 pole with 150 ampere rating)
Iz = In ≤ Iz
Iz = 150 ≤ Iz
Iz = 187 ampere ( carried capacity for 50mm² cable size)
So we decide to used cable size = 50 mm² with 4 core cable and clipped direct to cable tray.To confirm this cable size,we need to calculate the voltage drop.
Voltage Drop Calculation
Below is a voltage drop calculation that I made for cable power supply from panel to induction motor.
From Panel to Induction motor = 75 meter
mV = 0.87 (Refer Table)
Formula voltage drop : Vd = Ib x length(meter) x mV / 1000
Vd = (114.5 x 75 x 0.87) / 1000
Vd = 7.47 ( Acceptable refer to IEE 17th edition )

* Voltage drop acceptable value is 4% x 415 volt = 16.6 Volt
Conclusion

From this calculation, we are fully confident to laying cable size 50mm² with 150 ampere of MCCB, using multi-core PVC Armored cable (due to outdoor wiring application) and also cable laying with cable tray.
That is why all Electrician and Electrical Engineer need to know about cable sizing calculation to make sure wiring installation project successful and meet the requirement of Electrical Safety and Regulation without any disaster.
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