Three Phase System Outline

Three Phase System Outline Single phase systems are defined by having an AC source with only one voltage waveform. Figure 1 is a simple AC circuit. Single-phase power distribution is widely used especially in rural areas, because the cost of a single-phase distribution network is low. Figure 1:- Single phase system schematic diagram Today most of the electrical power generated in the world is three-phase. Three-phase power was first conceived by Nikola Tesla. Three-phase power was the most efficient way that electricity could be produced, transmitted, and consumed. A three-phase generator has three separate but identical windings that are 1200 electrical apart from one another. 5.2 Three Phase Circuit Three-phase voltage systems are composed of three sinusoidal voltages of equal magnitude, equal frequency and separated by 120 degrees, as shown in Figure 2. It is one voltage cycle of a 3 phase system. It is labeled 0 to 360 ° (2 Ï€ radians) along the time axis. The plotted lines show the variation of instantaneous voltage (or current) over time. This power wave cycle will repeat usually  50 (50Hz), 60 (60Hz), or 400 (400Hz)  times per second, depending on the power system  frequency (Hz). The colors of the lines are in the  American Color Code for 3-phase wiring. It is black=VL1  red=VL2blue=VL3. Figure 2:- Three phase waveforms Three phase systems may or may not have a neutral wire. The neutral wire allows 3 phase systems to use a higher voltage while still supporting lower voltage 1 phase appliances. In  high voltage 3 phase distribution  situations it is common not to have a neutral wire as the loads can simply be connected between phases (phase-phase connection). 5.2.1 Advantage over Single Phase system Three phase system is better to single phase system. The reason for the advantage over single phase system is given below. The horsepower rating of three-phase motors and the KVA (kilo-volt-amp) rating of three-phase transformers is about 150% greater than for single-phase motors or transformers with a similar frame size. Figure 3:- Single-phase power falls to zero three times each cycle. Figure 4:- Three-phase power never falls to zero. The power delivered by a single-phase system pulsates, as shown in Figure 3. The power falls to zero three times during each cycle. The power delivered by a three-phase circuit pulsates also, but it never falls to zero, as shown in Figure 4. In a three-phase system, the power delivered to the load is the same at any instant. This produces superior operating characteristics for three-phase motors. In a balanced three-phase system, the conductors need be only about 75% the size of conductors for a single-phase two-wire system of the same KVA rating. This helps offset the cost of supplying the third conductor required by three-phase systems. If a magnetic field is rotate through the conductors of a stationary coil then a single phase alternating voltage can be produced. This explanation is shown in Figure 5. Figure 5:- A single-phase voltage. Since alternate polarities of the magnetic field cut through the conductors of the stationary coil, the induced voltage will change polarity at the same speed as the rotation of the magnetic field. The alternator shown in Figure 5 is single phase because it produces only one AC voltage. Figure 6:- The voltages of a three-phase system are 120 ° out of phase with each other. If three separate coils are spaced 120 ° apart, as shown in Figure 6, three voltages 120 ° out of phase with each other will be produced when the magnetic field cuts through the coils. This is the manner in which a three-phase voltage is produced. 5.2.2 Classification Three-phase supply voltages and load systems have two basic configurations: a). wye or star connection and b). delta connection. 5.3 Star and Delta connection The Wye is a 4-wire system. Wye configurations typically include a neutral line (N) connected to the common point (3 phase plus neutral for a total of four wires), as shown in Figure 7. Figure 7:- A wye connections is formed by joining one end of each of the windings together. The Delta, as shown in Figure 8, is a 3-wire system which is primarily used to provide power for three-phase motor loads. The system is normally ungrounded and has only one three-phase voltage available. The lack of a system ground makes it difficult to protect for ground faults. Often, a ground detection scheme, employing ground lamps, is used to provide an indication or alarm in the event of a system ground. The Delta System is sometimes corner grounded to protect for ground faults on the other two phases. Figure 8:- Three-phase delta connection 5.4 Phasor diagrams 5.4.1 Star connection The voltage measured across a single winding or phase is known as the phase voltage, as shown in Figure 9. The voltage measured between the lines is known as the line-to-line voltage or simply as the line voltage. The currents flowing in the phases are called phase currents and currents flowing in the lines are called line currents. Figure 9:- Line and phase voltages are different in a wye connection. The parallelogram method of vector addition for the voltages in a wye-connected three-phase system is shown in Figure 10. Figure 10 shows how the line voltage may be obtained using the normal parallelogram addition. Figure 10:- Phasor diagram of Star connection Voltage However, the line voltage is not equal to the phase voltage. The line voltage V1-2 is equal to the phasor difference of VA and VB. The line voltage V2-3 is equal to the phasor difference of VB and VC. The line voltage V3-1 is equal to the phasor difference of VC and VA. The line voltages are defined as: V1-2 = VA VB, V2-3 = VB-VC, and V3-1 = VC-VA. Here V1-2, V2-3, V3-1 are the line voltage (VLine) and VA, VB, VC are the phase voltage (VPhase) of Wye connection. VA, VB, VC are the reverse phase voltage of VA,VB, VC. The two phasors VA and VB are 600 apart. V1-2 = VLine = VA VB = [VPhase (-VPhase)] cos(600/2) = 2 VPhase cos300 = √3 VPhase The two phasors VB and VC are 600 apart. V2-3 = VLine = VB-VC = √3 VPhase The two phasors VC and VA are 600 apart. V3-1 = VLine = VC-VA = √3 VPhase  Ã…“ V1-2 = V2-3 = V3-1 = line voltage = VLine =√3 VPhase Current On a Wye system or star connected supply, the phase unbalance current is carried by the neutral. On a Wye system, the line current (current in the line) (ILine) is equal to the phase current (current in a phase) (IPhase) i.e. ILine = IPhase Power Total power P = 3 Power in each phase = 3 VPhase IPhase cosÃŽ ¦ = 3 (VLine/√3) ILine cosÃŽ ¦ [for Wye connection] = √3 VLine ILine cosÃŽ ¦ Where VLine and ILine are the line voltage and the line current of a star connected supply. The term cosÃŽ ¦ is called power factor of the circuit and its value is given by; cosÃŽ ¦ = R/Z Where R and Z are the resistance and impedance of a circuit. 5.4.2 DELTA CONNECTIONS In Figure 11, voltmeters have been connected across the lines and across the phase. Ammeters have been connected in the line and in the phase. Figure 11:- Voltage and current relationships in a delta connection The delta connection is similar to a parallel connection because there is always more than one path for current flow. Since these currents are 120 ° out of phase with each other, vector addition must be used when finding the sum of the currents, as shown in Figure 12. Figure 12:- Phasor Diagram of Delta connection Voltage In the delta connection, the three voltages are equal in magnitude but displaced 1200 from one another. In the delta connection, line voltage (VLine) and phase voltage (Vphase) are the same. VLine = Vphase Current In the delta connection, the line current and phase current are different. The line current is the vector sum of two individual phase currents. The line current I1 is equal to the phasor difference of IA and IC. The line current I2 is equal to the phasor difference of IB and IA. The line current I3 is equal to the phasor difference of IC and IB. The line currents are defined as: I1 = IA IC, I2 = IB IA and I3 = IC IB. Here I1, I2, I3 are the line current (ILine) and IA, IB, IC are the phase current (IPhase) of Wye connection. IA, IB, IC are the reverse phase current of IA, IB, IC. The two phasors IA and IC are 600 apart. I1 = ILine = IA IC = [IPhase (-IPhase)] cos(600/2) = 2 IPhase cos300 = √3 IPhase The two phasors IB and IA are 600 apart. I2 = ILine = IB IA = √3 IPhase The two phasors IC and IB are 600 apart. I3 = ILine = IC IB = √3 IPhase  Ã…“ I1 = I2 = I3 = ILine = line current = √3 IPhase However, the line current of a delta connection is higher than the phase current by a factor of the square root of 3 (1.732). Power Total power P = 3 Power in each phase = 3 VPhase IPhase cosÃŽ ¦ = 3 VLine- (ILine/√3) cosÃŽ ¦ [for delta connection] = √3 VLine ILine cosÃŽ ¦ Where VLine, ILine and cosÃŽ ¦ are the line voltage, the line current and power factor of a delta connected supply. 5.5 Relationship between line and phase quantities 5.5.1 Star connection On a Wye system, the line current is equal to the phase current i.e. ILine = IPhase Where ILine and IPhase are the line current and phase current of Wye connection. In a wye connected system, the line voltage is higher than the phase voltage by a factor of the square root of 3 (1.732). Two formulas used to compute the voltage in a wye connected system are: VLine = √3 VPhase = 1.732 VPhase  Ã…“ VPhase = VLine / 1.732 Where VLine and VPhase are the line voltage and phase voltage of Wye connection. 5.5.2 Delta connection In the delta connection, line voltage and phase voltage are the same. VLine = Vphase Where VLine and VPhase are the line voltage and phase voltage of delta connection. Formulas for determining the current in a delta connection are: Where ILine and IPhase are the line current and phase current of delta connection. 5.6 Power measurement by two watt meters method In two wattmeters method, current coils of the two wattmeters are connected in any two terminals of Wye system, as shown in Figure 13. The algebraic sum of two wattmeters gives the total power consumed whether the load is balanced or not i.e. Total power = W1 + W2 Figure 13:- Wye connected load Figure 14:- Phasor Diagram The power factor angle of load impedance being ÃŽ ¦ lag. The currents will lag behind their respective phase voltages by ÃŽ ¦ as shown in Fig. 14. Current through current coil of W1 = IA. Potential difference across potential coil of W1, V1-2 = VA VB. The phase angle between V1-2 and IA is (300 + ÃŽ ¦).  Ã…“ W1 = V1-2 IA cos(300 + ÃŽ ¦) Current through current coil of W2 = IB. Potential difference across potential coil of W2, V2-3 = VB-VC. The phase angle between V2-3 and IB is (300 ÃŽ ¦).  Ã…“ W2 = V2-3 IB cos(300 ÃŽ ¦) Here load is balanced, V1-2 = V2-3 = VLine = line voltage and IA = IB = ILine = line current.  Ã…“ W1 = VLine ILine cos(300 + ÃŽ ¦)  Ã…“ W2 = VLine ILine cos(300 ÃŽ ¦)  Ã…“ W1 + W2 = VLine ILine [cos(300 + ÃŽ ¦) + cos(300 ÃŽ ¦)] = VLine ILine(2cos300cosÃŽ ¦) = √3VLine ILine cosÃŽ ¦  Ã…“ W2 W1 = VLine ILine [cos(300 ÃŽ ¦) cos(300 + ÃŽ ¦)] = VLine ILine(2sin300sinÃŽ ¦) = VLine ILine sinÃŽ ¦ tanÃŽ ¦ = [√3 (W2 W1)] / (W1 + W2) Thus from the two wattmeter method, we can find ÃŽ ¦. PROBLEM 1. Three coils, each having a resistance of 20- and an inductive reactance of 15-, are connected in star to a 400V, 3-phase, 50Hz supply. Calculate (i) the line current (ii) power factor and (iii) power supplied. Solution:- VPhase = VLine / 1.732 = 400/1.732 = 231V ZPhase = √(202 + 152) = 25- (i) IPhase = VPhase/ ZPhase = 231/25 = 9.24A = ILine (ii) Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 20/25 = 0.8 lag (iii) P = √3VLine ILine cosÃŽ ¦ = √3 400 9.24 0.8 = 5121W 2. A balanced star-connected load of impedance (6 + j8)- per phase is connected to a 3-phase, 230V, 50Hz supply. Find the line current and power absorbed by each phase. Solution:- ZPhase = √(62 + 82) = 10- VPhase = VLine / 1.732 = 230/1.732 = 133V Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 6/10 = 0.6 lag IPhase = VPhase/ ZPhase = 133/10 = 13.3A = ILine P = √3VLine ILine cosÃŽ ¦ = √3 230 13.3 0.6 = 1061W 3. Three similar coils, connected in star, take a total power of 1.5kW at a power factor of 0.2 lagging from 3-phase, 400V, 50Hz supply. Calculate the resistance and inductance of each coil. Solution:- VPhase = VLine / 1.732 = 400/1.732 = 231V P = √3VLine ILine cosÃŽ ¦  Ã…“ ILine = P / (√3VLine cosÃŽ ¦) = 1500 / (1.732 400 0.2) = 10.83A = IPhase ZPhase= VPhase/ IPhase = 231 / 10.83 = 21.33- RPhase = ZPhase cosÃŽ ¦ = 21.33 0.2 = 4.27- XPhase = √(21.332 4.272) = 20.9- LPhase = XPhase/ 2Ï€f =20.9 / (2Ï€ 50) = 0.0665H 4. The load to a 3-phase supply comprises three similar coils connected in star. The line currents are 25A and kVA and kW inputs are 20 and 11 respectively. Find (i) the phase and line voltages (ii) the kVAR input and (iii) resistance and reactance of each coil. Solution:- VPhase = Apparent power / (3 IPh) = (20-103) / (3 25) = 267V VLine= √3 VPhase=1.732-267 = 462V Input kVAR = √ (kVA2 kW2) = √ (202 112) = 16.7kVAR Power factor = cosÃŽ ¦ = kW/kVA = 11/20 ZPhase= VPhase/ IPhase = 267 / 25 = 10.68- RPhase = ZPhase cosÃŽ ¦ = 10.68 11/20 = 5.87- XPhase = √(10.682 5.872) = 8.92- 5. A balanced 3-phase, delta-connected load has per phase impedance of (25+j40)-. If 400V, 3-phase supply is connected to this load, find (i) phase current (ii) line current (iii) power supplied to the load. Solution:- ZPhase = √(252 + 402) = 47.17- IPhase= VPhase/ ZPhase = 400 / 47.17 = 8.48- ILine= √3 IPhase=1.732-8.48 = 14.7A Power factor = cosÃŽ ¦ = RPhase/ ZPhase = 25/47.17 = 0.53 lag P = √3VLine ILine cosÃŽ ¦ = √3 400 14.7- 0.53 = 5397.76W 6. A balanced 3-phase load consists of three coils, each of resistance 6-, and inductive reactance of 8-. Determine the line current and power absorbed when the coils are delta-connected across 400V, 3-phase supply. Solution:- ZPhase = √(62 + 82) = 10- cosÃŽ ¦ = RPhase/ ZPhase = 6/10 = 0.6 lag VPhase = VLine = 400V IPhase= VPhase/ ZPhase = 400 / 10 = 40A ILine= √3 IPhase=1.732-40 = 69.28A P = √3VLine ILine cosÃŽ ¦ = √3 400 69.28 0.6 = 28799W 7. Two-wattmeter method is used to measure the power absorbed by a 3-phase induction motor. The wattmeter readings are 12.5kW and -4.8kW. Find (i) the power absorbed by the machine (ii) load power factor (iii) reactive power taken by the load. Solution:- W2 = 12.5kW ; W1 = -4.8kW Power absorbed = W2 + W1 = 12.5 + (-4.8) = 7.7kW tanÃŽ ¦ = [√3 (W2 W1)] / (W1 + W2) = (12.5+4.8) / 7.7 = 3.89 ÃŽ ¦ = tan-13.89 = 75.60 Power factor = cosÃŽ ¦ = cos75.60 = 0.2487lag Reactive power = √3 (W2-W1) = √3 (12.5 + 4.8) = 29.96kVAR P O I N T S TO REMEMBER 1. The voltages of a three-phase system are 120 ° out of phase with each other. 2. The two types of three-phase connections are wye and delta. 3. Wye connections are characterized by the fact that one terminal of each device is connected together. 4. In a wye connection, the phase voltage is less than the line voltage by a factor of 1.732. The phase current and line current are the same. 5. In a delta connection, the phase voltage is the same as the line voltage. The phase current is less than the line current by a factor of 1.732. IMPORTANT FORMULAE 1. On a wye system, the relation between line and phase current is: ILine = IPhase 2. On a wye system, the line voltages are defined as: V1-2 = VA VB, V2-3 = VB-VC, and V3-1 = VC-VA. 3. In the delta connection, the relation between line and phase voltage is: VLine = Vphase 4. In the delta connection, the line currents are defined as: I1 = IA IC, I2 = IB IA and I3 = IC IB 5. On a wye system, the relation between line and phase voltage is: VPhase = VLine / 1.732 6. In the delta connection, the relation between line and phase current is: OBJECTIVE QUESTIONS 1. In a two phase generator, the electrical displacement between the two phases or winding is: (a) 1200 (b) 900 (c) 1800 (d) none of these 2. The advantage of star-connected supply system is that: (a) line current is equal to phase current (b) two voltages can be used (c) phase sequence can be easily changed (d) it is a simple arranged 3. In a balanced star-connected system, line voltage are ahead of their respective phase voltages. (a) 300 (b) 600 (c) 1200 (d) none of these 4. In a star connected system, the relationship between the line voltage VL and phase voltage VPh is: (a) VL = VPh (b) VL = VPh / √3 (c) VL = √3VPh (d) none of these 5. The algebraic sum of instantaneous phase voltages in a three-phase circuit is equal to: (a) zero (b) line voltage (c) phase voltage (d) none of these 6. If one line conductor of a 3-phase line is cut, the load is then supplied by: (a) single phase voltage (b) two phase voltage (c) three phase voltage (d) none of these 7. The resistance between any two terminals of a balanced star-connected load is 12-. The resistance of each phase is: (a) 12- (b) 24- (c) 6- (d) none of these 8. A 3-phase load is balanced if all the three phases have the same (a) impedance (b) power factor (c) impedance and power factor (d) none of these REVIEW QUESTIONS 1. How many degrees out of phase with each other are the voltages of a three-phase system? 2. What are the two main types of three-phase connections? 3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage drop across each phase? 4. A wye-connected load has a phase current of 25 A. How much current is flowing through the lines supplying the load? 5. A delta connection has a voltage of 560 V connected to it. How much voltage is dropped across each phase? 6. A delta connection has 30 A of current flowing through each phase winding. How much current is flowing through each of the lines supplying power to the load? 7. A three-phase resistive load has a phase voltage of 240 V and a phase current of 18 A. What is the power of this load? 8. If the load in question 7 is connected in a wye, what would be the line voltage and line current supplying the load? 9. An alternator with a line voltage of 2400 V supplies a delta-connected load. The line current supplied to the load is 40 A. Assume the load is a balanced three-phase load, what is the impedance of each phase? 10. If the load is pure resistive, what is the power of the circuit in question 9? PRACTICE PROBLEMS 1. Three similar coils are star connected to a 3-phase, 400V, and 50Hz supply. If the inductance and resistance of each coil are 38.2mH and 16- respectively, determine (i) line current (ii) power factor (iii) power consumed. 2. Three 50- resistors are connected in star across 400V, 3-phase supply. (i) Find phase current, line current and power taken from the main. (ii) What would be the above value if one of the resistors were disconnected? 3. Calculate the active and reactive components of current in each phase of a star-connected 10,000 volts, 3-phase generator supplying 5,000kW at a lagging power factor 0.8. Find the new output if the current is maintained at the same value but the power factor is raised to 0.9 lagging. 4. Three 20 µF capacitors are star-connected across 420V, 50Hz, 3-phase, three wire supplies. (i) Calculate the current in each line. (ii) If one of the capacitors is short-circuited, calculate the line currents. (iii) If one of the capacitors is open-circuited, calculate the line currents and potential difference across each of the other two capacitors. 5. If the phase voltage of a 3-phase star connected alternator be 231V, what will be the line voltages (i) when the phases are correctly connected (ii) when the connections of one of the phases are reversed? 6. Calculate the phase and line currents in a balanced delta connected load taking 75kw at a power factor 0.8 from a 3-phase 440V supply. 7. Three identical resistances, each of 18-, are connected in delta across 400V, 3-phase supply. What value of resistance in each leg of balanced star connected load would take the same line current? 8. Three similar resistors are connected in star across a 415V, 3-phase supply. The line current is 10A. Calculate (i) the value of each resistance (ii) the line voltage required to give the same line current if the resistors were delta-connected. 9. Two wattmeters are used to measure power in a 3-phase balanced load. The wattmeter readings are 8.2kW and 7.2kW. Calculate (i) total power (ii) power factor and (iii) total reactive power. 10. A balanced 3-phase load takes 10kW at a power factor of 0.9 lagging. Calculate the readings on each of the two wattmeters connected to read the input power. 11. Three identical coils, each having a resistance of 20- and a reactance of 20- are connected in (i) star (ii) delta across 440v, 3-phase lines. Calculate for each method of connection the line current and readings on each of the two wattmeters connected to measure the power.

Privacy In Demand Essay -- essays research papers

  Ã‚  Ã‚  Ã‚  Ã‚  Like most countries and especially the United States their inhabitants enjoy a certain level of privacy. People don’t generally want intimate information to be accessible to the public eye. In fact many people go to great lengths to hide everything about themselves. What exactly is the definition of privacy? Well, privacy is the expectation that confidential personal information disclosed in a private place will not be disclosed to third parties, when that disclosure would cause either embarassment or emotional distress to a person of reasonable sensitivities. This information includes facts, images (ex: photographs and videotapes), and disparaging opinions. When over zealous law enforcement officials demand access to telephone conversations, e-mail or other electronic communication they are violating the unwritten code of privacy. When organizations from the private sector purchase intimate information about medical records either for commercial purposes, or to challenge your insurance eligibility or employment suitability. Unfortunatly this is a common practice in the United States and it is wrong.   Ã‚  Ã‚  Ã‚  Ã‚  First of all, what does the government do to secure this private information? The answer is very little. There are bascially two different laws that effect privacy. These two laws are the Privacy Act of 1974 and the Freedom of Information act. At a first inspection the two laws seem to work against each other. In short the Privacy Act of 1974 keeps information in government records concerning individuals discreet. The Privacy Act of 1974 gives the individual the rights to see and copy files that the federal government maintains on him or her. It also gives the right to know who else has access to that information, and to request a change to any information that is not accurate. The most important part of this law is the fact that the government is not allowed to use any information for any purpose other than the one for which it was initially collected. This is important and will be addressed later on. The Freedom of Information Act is used mostly to pry open government files. It was designed to help individuals obtain information about the actions of government. The law proclaims that any citizen is to be given access to government records unless the disclosure involves litigation, the CIA, personal m... ... license from every state.† Basically this states that a mugshot data base will be created by virtually all non-criminals. This is a violation of privacy. When the DMV issued the driver’s license there was never any intent to create a mugshot from the information on the card. In 1992 The DPPA(Federal Drivers Privacy Protection Act) was created to make a nation mugshot database. This act authorized the sale of driver’s names, addresses, birthdates, social security numbers, driver’s license numbers, digital signatures, and digital photographs to private companies for the purpose of making a registry of identifying information. Fortunatly, this act was ruled unconstitutional for it was in violation of the tenth amendment. However, before this act was ruled unconstitutional the state of South Carolina sold the complete contents of it driver’s license information for a mere five thousand dollars.   Ã‚  Ã‚  Ã‚  Ã‚  Now with the introduction of the internet it is becoming increasingly difficult to control the publication of personal and private information. Any information that is collected should not be used for any other purpose except for what it was originally accepted.

ESL Essay on Belonging (China Coin and Rabbit Proof Fence) Essay

A sense of belonging or not belonging can emerge from the connections made with people, places, groups, communities and the larger world. How does this apply to â€Å"The China Coin†? Through a study of the novel â€Å"the China Coin† by Allan Baillie, it can be seen that a sense of belonging or not belonging can emerge from the connections made with people, places, groups, communities and the larger world. This essay will explore how a sense of belonging or not belonging develops from the main characters’ (Joan and Leah’s) connections with each other and with places, specifically Good Field Village. It will also examine how a character’s connection with the nation of China at large gives rise to their sense of belonging or not belonging. In the novel â€Å"The China Coin†, the author uses various language techniques to illustrate Leah and Joan’s sense of belonging or not belonging, which emerges from their connection with each other. The novel begins with Leah feeling disconnected with Joan even though they are the only two members of their family left. This can be seen by the way Baillie uses metaphor to compare Joan to â€Å"an evil aunt, who flies a broom on full moon†¦Ã¢â‚¬ . This conveys Leah’s sense of not belonging to the relationship. A sense of belonging unfolds later in the novel after Leah and Joan both go through much together, hence leaving them with a more intimate connection with each other. After hearing grandfather implying that he wants them to stay so that he can trick them into paying for a ‘Hong Kong house’, Joan and Leah’s strong connection is expressed by their ability to communicate without even using words. Baillie uses polysyndeton to emphasize this in the sentence â€Å"And both mother and daughter stopped and grinned at each other†. As can be seen, both these examples clearly show that Leah and Joan’s sense of not belonging or belonging to each other has emerged from their connection with each other. Apart from that, Allan Baillie has also used language features in the novel to draw attention to the sense of belonging or not belonging that stem from connections with a place, specifically, Go od Field Village in â€Å"The China Coin†. When the main characters first arrive in Good Field village, Joan felt accepted immediately as she spoke Cantonese fluently and quickly formed a connection with Jade. As a result, a sense of belonging was generated in Joan. This is reinforced by the author’s use of simile to describe them as women who â€Å"had been neighbours for years†. Contrastingly, Leah, who was not as fluent in  the language, could not converse with Jade and Joan. As a result, she did not feel a sense of belonging to Good Field village. This is portrayed by Baillie’s use of the third person narrative voice, which tells the audience â€Å"Leah felt suddenly alone†. All this suggests that the concept of belonging or not belonging develops from one’s connection with a place. Lastly, language features used in the novel â€Å"The China Coin† has supported the fact that a sense of belonging or not belonging can emerge from connections with the larger world. This can be seen in the character of the young boy who puts up political posters at the restaurant where Joan and Leah are eating. He does not feel connected with the state as he does not agree with the current political situations and wants democracy instead. This generates a sense of not belonging in the character, which is further reinforced as he uses hyperbole to describe other protesters and himself as â€Å"Enemies of the State†. Similarly, Ke, who is disconnected with the principles of China’s political agendas, feels like he does not belong to China at large. This is evidenced in the use of dialogue where Ke tells Leah about what he wants changed in the political system. He tells her that he wants â€Å"Democracy! No more guanxi! No more influence, no more back-door deals!†. From this, it can be inferred that one’s sense of not belonging can rise up from one’s relationship with the world at large. In conclusion, Leah and Joan’s connection with each other and with places such as Good Field Village give rise to their impression of belonging or not belonging. Similarly, a sense of not belonging can be seen to emerge from connections that other characters have with the larger nation of China.

Didnt Get into Grad School Can You Reapply Next Year

Question: I was rejected from a grad school and now Im confused. I have a pretty decent GPA and research experience, so I dont get it. Im wondering about my future and am considering my options. Can I reapply to the same school? Does this sound familiar? Did you receive a rejection letter in response to your graduate school application? Most applicants receive at least one rejection letter. Youre not alone. Of course, that doesnt make rejection any easier to take. Why are Graduate School Applicants Rejected? No one wants to receive a rejection letter. Its easy to spend a lot of time wondering what happened. Applicants are rejected by grad programs for a variety of reasons.   GRE scores that are below the cut-off is one reason. Many grad programs use GRE scores to weed out applicants easily without viewing their application. Likewise, a low GPA might be to blame. Poor recommendation letters can be devastating to a grad school application. Asking the wrong faculty to write on your behalf or not paying attention to signs of reluctance can lead to neutral (that is, poor) references. Remember, all reference letters describe applicants in glowingly positive terms. A neutral letter is therefore interpreted negatively. Reconsider your references. Poorly written admissions essays can also the culprit. A large part of whether you get accepted to a program is fit - whether your interests and skills match the programs training and needs. But sometimes there isnt a good reason for rejection. Sometimes its just about the numbers: too many students for too few slots.There are multiple variables at play and its likely that youll never know the specific reason(s) you were rejected. You Can Apply to the Same Graduate Program After Being Rejected Does it match your academic interests?Does it offer preparation for the career you desire?Do your credentials match the requirements?Is there faculty with whom you would like to work?Do those faculty have slots open in their labs? Are they accepting students? If you decide to reapply, carefully analyze the application you submitted this year to determine whether it represented you well and whether it was the best application that you could assemble. Consider all of the parts listed above. Ask for feedback and advice from your professors - especially those who wrote your reference letters. Look for ways to improve your application. Good luck!