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)

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