AC Theory Which Connection Gives Term Paper

  • Length: 9 pages
  • Subject: Physics
  • Type: Term Paper
  • Paper: #9880861

Excerpt from Term Paper :

No voltage is produced in the auxiliary rotor winding throughout synchronous operation. The capability of a synchronous motor to work at leading power factor makes it appropriate to be used for power factor improvement. When a synchronous motor is used solely for power factor improvement and not for driving any mechanical load and it is called a synchronous condenser. (Electricity and Motors)

9) Detail a specific application for a Capacitor Start Induction Run Motor.

A relay and a start capacitor are present in the control box in a capacitor-start/induction-run or CSIR system. The start capacitor is linked to the start winding in the motor. The motor starts by means of both windings; but as the motor in the CSIR system reaches to speed, the relay takes away the start winding and the start capacitor from the circuit. This occurs in about one-third of a second, and the motor then runs on the run winding only without capacitor. This is the reason for the current in the red lead of a CSIR motor to be zero after the motor has started. (CSIR vs. CSCR: What's the Difference?)

The CSIR motor has a very wide range of uses. The starting mechanism in the capacitor start motor is either a mechanical or solid-state electronic switch. When the motor reaches about 75% of rated speed, the starting mechanism disconnects both the start winding and the capacitor. As the capacitor is in line with the start circuit, it generates more starting torque, typically 200 to 400% of rated load. A specific application for this type of motor is belt-drive devices. This is inclusive of small conveyors, large blowers and pumps, as well as many direct-drive or geared systems. These are the workhorses of widely used and reasonably available single-phase industrial motors. (Single-phase Electric Motors Characteristics & Applications)

10) What is the basic condition for transformer maximum efficiency?

Transformer efficiency is related to its power losses. All these losses can be explained by two factors. The first factor is winding copper loss. Since there are two sets of windings, there exist two components to copper loss, namely, primary and secondary winding copper loss. The second factor representing transformer power losses is core loss. Hysteresis is the reason for the core losses and the hysteresis is a function of many features of the core steel, all decided by the manufacturing process. If supply frequency is unvarying, the core losses for any given transformer remain invariable. Maximum efficiency of transformer occurs if winding copper loss becomes equal to core loss. (Basics of Transformer Voltage Efficiency)

11) How do the transformer copper losses and iron losses vary with load current?

Transformers have two main constituents that influence losses, namely the core and the coils. The usual core is an assembly of laminated steel. Core losses are typically linked to magnetizing or energizing the core. These losses, also called as no-load losses exist for the whole time the transformer is powered on, irrespective of whether there's any load or not. Core losses are approximately constant from no-load to full-load when supplying linear loads. They signify an incessant cost, for the 25- to 40-year life of the transformer. The coil losses, normally referred to as load losses, are connected with supplying power to the connected load. For linear loads, these losses are principally I2R losses. Putting it otherwise, load losses enlarge by the square of current from no-load to full-load, driven by the resistance of the coil. (Overcoming Transformer Losses)

12) What effects does the load's p.f.lag or lead, have on the transformer regulation?

Power factor is the relation between the KW and the KVA which is drawn by an electrical load where the KW denotes the real load power and the KVA is the load power. It is a gauge of how efficiently the current is being changed into constructive work output and more specifically is a fine pointer of the effect of the load current on the efficiency of the supply system. All current will lead to losses in the supply and distribution system. A load with a power factor of 1.0 lead to the most effective loading of the supply and a load with a power factor of 0.5 which lead to enhanced higher losses in the supply system. A poor power factor can be the consequence of either a major phase difference between the voltage and current at the load terminals, or it can be because of a high harmonic content or distorted/discontinuous current waveform. Weak load current phase angle is generally the due to an inductive load like the induction motor, power transformer, lighting ballasts, and welder or induction furnace. (Power Factor)

13) on the open circuit test what loss is being ignored?

If a transformer having the secondary open-circuited being energized at levels of primary voltage, then the input power shows the core loss of the transformer, which means that Input power equals core loss. As a result of this in open circuit test, load or copper loss is being ignored at all levels. (Electrical Machine Applications)

14) on the Short circuit test, what loss is being ignored?

If a transformer with its secondary short-circuited is energized at a diminished primary voltage which leads rated secondary current to flow through the short circuit; then the input power stands for the load loss of the transformer: that is, Input power = Primary copper loss + Secondary copper loss + Stray loss. It is to be noted that the temperature rise should be permitted to even out as conductor resistance change with temperature. Hence, in Short circuit test, core or iron loss is ignored. (Electrical Machine Applications)

15) Explain how to determine the copper loss at the various load factors.

Copper loss is power which is being vanished in the primary and secondary windings of a transformer. This occurs due to the ohmic resistance of the windings. We could hence prove that copper loss, in watts, could be found by means of an Equation: Copper Loss = I2P RP + I2S RS where IP is the primary current; IS the secondary current; and RP is the primary winding resistance. Finally RS -- denotes the secondary winding resistance. (Transformer Losses and Efficiency)

16) Why can we ignore the copper loss in the open circuit test and the iron loss in the short circuit test?

In the open circuit, since it could be understood that the input power denotes the total core loss, copper loss is being ignored. In short circuit test, it could further be noted that since the input power is copper loss, iron loss is being ignored. (Electrical Machine Applications)

17) When does a transformer work at maximum efficiency?

Efficiency is related to transformer's power losses, and two factors give an explanation for almost all of these losses. One is winding copper loss. Since there are two sets of windings, there exists two parts to copper loss: primary and secondary winding copper loss. The second factor making up for transformer power losses is core loss. Core losses are due to hysteresis, which is a function of many features of the core steel or iron, all decided by the manufacturing process. Luckily, the core losses for any given transformer remain unchanged if supply frequency is constant. To get maximum efficiency, the condition is: winding copper loss becomes equal to core loss. (Basics of Transformer Voltage Efficiency)

18) What advantages has this method of determining the efficiency of a transformer over an on -load input -output test?

To calculate transformer efficiency in other than no-load situation, we must first compute the equivalent resistance or ER of both the primary and secondary together with the load. The attempt required to reach the ER, which changes with facility reconfigurations, is difficult to validate for usual applications. Nevertheless, we can effortlessly compute no-load efficiency. The first step is to multiply the output voltage by the output current. Replicate this step for the input. Then, the output results are to be divided by the input results. No-load efficiency provides a foundation for evaluating transformers or testing a transformer against a specification. It will not tell how efficient a transformer is when in use. That is why we still compute transformer efficiency in conditions other than no-load. (Basics of Transformer Voltage Efficiency)

Yet, the no-load computation is often worthwhile as it is so easy. For instance, firstly, one has to set up a baseline for each transformer. When one go through troubles that point to a transformer fault, re-computing no-load efficiency once more can cut down troubleshooting time noticeably. Is there any value for the transformer efficiency under load conditions? The answer is Yes, if one can compute the ER of everything on the load side, something few installations merit paying for. Thus, we move back to the fundamental rule of thumb in the real world for maximizing transformer efficiency namely to load the…

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