The Lungs Research Paper Example | Topics and Well Written Essays - 750 words

The Lungs - Research Paper Example The lungs are composed of a substance which is light, porous, have a spongy texture and are highly elastic. The surface has many polyhedral areas. The lungs are heavier in a male than in a female; the left lung is smaller than the right one in order to accommodate the heart. The lungs can be examined in terms of an apex, a base, three borders (anterior and inferior) and two surfaces (Costal and mediastinal). Each lung is conical in shape and is divided into lobes by interlobular fissures. The left lung is divided into the left superior or upper lobe and the left inferior or lower lobe. The right lung is divided into the right superior lobe, the right middle lobe and the right inferior lobe. Each lobe is further divided into pyramidal bronchopulmonary segments: ten in the right and eight in the left one. The root of each lung or the hilum connects the lung to the heart and the trachea. According to Gray (1918), â€Å"the bronchus, the pulmonary artery, the pulmonary veins, the bronch ial arteries and veins, the pulmonary plexuses of nerves, lymphatic vessels, bronchial lymph glands, and areolar tissue enter and leave the lungs through the root†. The trachea, or windpipe, reaches the thorax and is divided into the right and left bronchi. In the lungs the bronchi divide into large ventral and small dorsal branches. The bronchi are divided into smaller bronchi and bronchioles. Each bronchiole further splits into two or more respiratory bronchioles connected to alveolar ducts which, in turn, are connected to a cluster of small alveoli or air sacs. The pulmonary artery, the pulmonary veins, and the pulmonary capillaries supply deoxygenated blood to the lungs and remove oxygenated blood. The bronchial arteries and the bronchial vein supply oxygenated blood for the nutrition of the lung itself. All these arteries and veins enter and leave the lung through the hilum, as do the lymphatic vessels of the lungs. Pulmonary nodes, bronchopulmonary nodes and tracheobronc heal nodes are present. The lungs are supplied with nerves from the anterior and posterior pulmonary plexuses which have small ganglia on them. (Gray, 1918). The function of the lungs is to provide a continuous gas exchange between the inhaled air and the blood in pulmonary circulation. Oxygen is supplied and carbon dioxide is eliminated by exhalation. The exchange of gases takes place through simple diffusion: oxygen diffuses from the alveoli into the blood and CO2 diffuses from the blood into the alveoli. This diffusion requires a concentration, or pressure, gradient. The concentration of oxygen in the alveoli is kept higher than in the blood and the concentration of CO2 in the alveoli is kept lower than in the blood by the breathing of fresh air into the lungs. This breathing mechanism is generated by the respiratory muscles: the exterior intercostals located between the ribs, and the diaphragm which is a sheet separating the thorax from the abdomen. During inspiration the interc ostals and the diaphragm contract, the dimension of the thoracic cavity increases, air pressure in the lungs decreases and becomes slightly negative as compared to atmospheric pressure. This causes the air to move from the atmosphere into the lungs. During exhalation the muscles relax, lung volume decreases, pressure in the lungs increases and air is exhaled. The surface tension on the walls of the alveoli caused by water molecules is reduced by the pulmonary surfactant produced by the lungs. The exchange

Vulnerability Assessment on your System at Home Essay

Vulnerability Assessment on your System at Home - Essay Example Hackers having time, resources and enthusiasm can breach into any system. Even the system with the most enhanced security architecture cannot guarantee 100% that it will not be hacked or are safe from viruses. Firewalls help to secure the network from unwanted data by packet filtering technologies. Routers can help secure Internet and network gateways. Virtual private networks transmit data after encryptions in a secure tunnel. Anti viruses software detects viruses, spywares, and malicious activities. The success of these technologies is reliant on the expertise of a professional, who is responsible for monitoring, maintaining, configuring and updating the system of regular basis. Updating the system also involve the expertise required for updating the services and kernels effectively. Continuous monitoring of the network vigilance is also an important factor. If we take an example of an enterprise network, these networks are comprised of components such as Servers, Firewalls, File servers, Intrusion detection software and hardware, routers, switches etc. As we know that security threats are always a debatable topic because they keep on disrupting networks, it will be intimidating to handle the network vigilance along with updates and maintenance of the huge enterprise network. For maintaining the enterprise networks multiple expertise are required. It is unavoidable that networks will be completely safe from threats, disruption of services and vulnerabilities. For analyzing vulnerabilities, networks must be investigated for possible weaknesses. Vulnerability assessment of the home system and network resources can reveal potential issues that can be addresses before the hacker identifies them. It is an internal audit of system security and network resources. It involves regular network vulnerability scanning of the system to ensure the co nfigurations are meeting the standards along with the checking of latest

Drug Abuse Essay Example for Free

Drug Abuse Essay 1.0 Topic and Chosen aspect Drug abuse among teenagers is the chosen aspect for this report. Drug abuse is defined as the excessive use of drugs without medicine justification. It is reported that 50% of teenagers have abused a drug of some kind (Teendrugrehabs n.d.). 1.1 Critical evaluation of findings One of the main factors of drug abuse among teenagers is due to peer pressure. Based on statistics 80% of teenagers are affected by peer pressure (Wallace 2011). Wanting to be accepted by their peers, they are willing to do anything. Their desire to be socially accepted will cause them to be influence by their peers. Thus, this will lead them to the abuse of drugs to make things more exciting and not left out. Besides that, it is very hard to say â€Å"NO!† to their peers. Furthermore, another reason for teenagers to be easily involved in drug abuse is due to the easy access to drugs. 54% of the students of private high schools reported that drugs are accessible in their schools while 61% of the public high school students state that their schools are â€Å"drug-infested† (Pahuriray 2012). As drugs can be easily obtained everywhere. 2.0 Processes Various steps were undertaken to attain the sources needed for the report. However, a lot of the sources from the internet is too vague and the statistics are already outdated.

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.

Sexual Assault and Child Abuse Essay -- Compare Contrast

There are many different types of victims we have discussed over the course of this class, but we’re only going to talk about two types in the following paper. These two types of victims are common just as any another victim across America. These include sex assault victims and child abuse victims, which are both primary victims in cases. The two share a tie together, both are a victim of abuse and can cause lifelong consequences, but they also pose many differences as well. Many questions arise when talking about victims, for example why is a child or adult being abused and what are the life altering affects to these actions. Throughout this paper we discuss both sexual assault victims and child abuse victims and compare and contrast between the two. We’ll start with child abuse victims and the affects and reasons of this abuse. There are four types of child abuse and I will list them in order from least to greatest, neglect, physical, sexual and emotional abuse. Neglect takes first with the NCVS having 54% reports of child neglect in 2007. Neglect is a very serious form of abuse it is the failure for a parent or guardian to provide for a child’s basic needs, including physical and educational needs. We grow and development drastically in the first twelve years of our life so when parental guidance and love is absent it affects a child’s developmental skills along with learning right from wrong. Many forms of neglect occur in larger households and with households with domestic violence. Many parents with multiple kids become too busy focusing on the older children they tend to forget the youngest one. So it’s common for a three year old to walk out of the front door and on to the street when no one is there to tell him or... ...formation about Domestic Violence and Services: http://www.domesticviolenceservices.com/rape.html †¢ †¢ Karmen, A. (2013). Crime Victims: An Introduction to Victimology. Belmont, CA: Wadsworth/Thompson Learning. †¢ †¢ Kilpatrick, D. G. (2000). The Mental Health Impact of Rape. Retrieved from National Violenc Against Women Prevention Research Center: http://www.musc.edu/vawprevention/research/mentalimpact.shtml †¢ †¢ Rainn. (2009). Effects of Sexual Assault. Retrieved from Rape, Abuse and Incest National Netwrok: http://www.rainn.org/get-information/effects-of-sexual-assault †¢ †¢ Shannon. (2007). Was it my fault? Self-Blame and Survivors. Retrieved from Pandora's Project: http://www.pandys.org/articles/selfblameandsurvivors.pdf> †¢ †¢ RAINN.org. (2009). Types of sexual violence. Retrieved from: http://www.rainn.org/get-information/types-of-sexual-assault.

Diagnosis of Psychological Disorders and Treatment Essay -- Female Sex

Introduction In this paper we shall discuss two (2) fictional clients and their symptoms, and attempt to diagnose the clients using the DSM-IV TR. We then shall discuss specific treatment strategies that are based on the four (4) approaches discussed during the Introduction to Psychology course and briefly discuss ethical obligations in regard to the client during their treatment. Disorders Amanda is a 25 year old Caucasian female whom has a history of sexual abuse as a child and is in therapy to deal with emotional issues in regards to the abuse. She has been married for three (3) years and enjoys sexual activity with her husband. Her concern lies in the fact that although she is enjoying sexual intercourse and other sexual activities with her husband, she has noticed since becoming sexually active at age 17 that there have been reoccurring issues in regards to maintaining adequate lubrication. This has been frustrating for both Amanda and her husband and caused some strain on their marriage. Regina is a 28 year old Caucasian female. She has been married for nine months to her husband, they have two children together and each have one child from a previous relationship. Regina and her husband are seeking family counseling in regards to issues pertaining on ways better integrate their blended family. During the initial intake Regina is forthcoming but her husband brings up her â€Å"immense need† to clean and maintain their home. He gives specifics to what he describes as a â€Å"compulsiveness†, after talking with Regina she explains she feels an overpowering need to clean and pick up after the children. She goes on to explain her mother was not one for keeping a tidy house and would be embarrassed that it was â€Å" dirty â€Å"when frien... ...ue. Another important one is patient confidentiality and maintaining that confidentiality throughout treatment. If a patient does not trust you they are unlikely to get the most out of the therapy strategy you are pursuing. Lastly, it is very unbecoming to have an inappropriate relationship with a client during their treatment and can cause more harm to the patient. References Female Sexual Arousal Disorder in Sexual Disorders and Dysfunctions at ALLPSYCH Online. (n.d.). Retrieved from http://allpsych.com/disorders/sexual/sexualarousal.html Obsessive-Compulsive Disorder in Anxiety Disorders at ALLPSYCH Online. (n.d.). Retrieved from http://allpsych.com/disorders/anxiety/ocd.html Stangor, C. (2010). Introduction to Psychology. Irvington, NY: Flat World Knowledge, Inc. Weiten, W. (2005). Psychology: Themes & Variations (6th ed.) Thompson/ Wadsworth

Xacc 280 Final

Financial Analysis Coca-Cola and PepsiCo XACC/280 Financial Analysis An official financial analysis for a specific company needs two years of financial data from the company and from a competitor in the same industry. This financial analysis is between PepsiCo, Inc. and Coca-Cola. Pepsi and Coca-Cola dominate the beverage market worldwide. In addition to sodas, they also distribute a variety of water and energy drinks. Based on the analysis, the investor will be able to make a better investment choice. Liquidity, solvency, and profitability are the three characteristics that will be used to see a company’s success.A simple financial statement will not demonstrate the company’s power because it is a general idea of the company’s position and does not display business developments. The company’s business developments are vital for potential investors because they determine vertical and horizontal analysis. These characteristics are also used to define the ra tio analysis. Ratio analysis is dividing two numbers to get a number of percentages that can be used to compare companies in the same industry. Examining the entire company’s financial trends for a set period of time, an investor will see a factual description of the company’s financial condition.This is the financial analysis an investor desires to review prior to spending money. Liquidity measures a company’s ability to pay their debts when they are due. It is identified as a ratio or percentage of the current liabilities and calculated by dividing the current cash by the current liabilities. It is a fast way to understand if the company’s future is appealing to the investor. If the company is not turning a profit quick enough, it may be a sign of liquidity problems. This is the primary reason why an investor should compare two competitors while looking at the liquidity ratio.To define the current liquidity we use the formula: current ratio = current as sets divided by current liabilities. The most vital measure of this situation is that the right information is used from the balance sheet. PepsiCo, Inc. ’s Liquidity Ratio: 2005 Current Ration = $10,454 (current assets) divided by $9. 406 (liabilities)=1. 11% 2004 Current Ration = $8,639 (current assets) divided by $6,752 (liabilities) =1. 28% Coca-Cola Liquidity Ratio: 2005 Current Ratio= $10,250 (Current Assets) divided by $9,836(Current Liabilities) = 1. 4 % 2004 Current Ratio= $12,281 (Current Assets) divided by $11,133(Current Liabilities) = 1. 10% The evaluation of current assets compared to the current liabilities is solvency. Divide the current assets by current liabilities to determine the solvency. It will show the company’s long-term responsibilities. When the ratio is higher the company is more likely to meet their obligations and have growth in their industry by expenditures. The lower the ratio the company is less likely to meet their obligations. The st andard ratio for solvency is about 20% dependent upon the industry.The measure a company’s ability to generate assets versus expenses in an allotted period of time is profitability. If the ratio is higher or equal to their competitors’ preceding time period the company is in good standing. Periods of time are used to determine profitability and it is common to see the profitability increase and decrease all through the year. This is reason profitability is examined on a full fiscal year. The examination of the trends is horizontal analysis. An income statement, retained earnings statement and balance sheet are ways to implement the horizontal analysis.It will show the company’s financial data for a set period of time and a great instrument to know if the company is worth investing in. PepsiCo, Inc. ’s Horizontal Analysis: 2005 total current assets $10,454 divided by $8,639 total current assets 2005 = 21% 2004 total current liabilities $9,404 divided by $6 ,752 total current liabilities 2004 = 39%. PepsiCo, Inc. had an increase in total assets by making loans. Pepsi increased their debt over the previous years and this shows an unstable business. Coca-Colas Horizontal Analysis: 005 total current assets $10,250 / $12,281 = 83. 5% or a decrease from 2004 to 2005 2004 total current assets $9,836 / $11,133 = 88. 4% of an 11. 6% decrease. Coca-Cola had a decrease in their debt. This indicates that Coca-Cola has a more solid business plan during this time period. Coca-Cola is also in a better monetarily stable place to pay back their debt. Vertical Analysis or the common size analysis is calculated by dividing the Balance Sheet items by the company’s total assets. This number is then turned into a percentage for easier comparison.This percentage represents the growth within the company. Positive retained earnings and debt retention are shown through positive and negative percentages. PepsiCo, Inc. ’s Vertical Analysis: 2005 â €“ $1,716 (cash and cash equivalent) divided by $31,727 (total assets) =5. 4% 2004 – $1,280 (Cash and cash equivalent) divided by $27,987 (total assets) = 4. 6% The proper way to do vertical analysis for the PepsiCo. Inc. is to use the different line items from the balance sheet and divide each one by the company’s total assets. Both of these items are taken from the balance sheet.The total cash percentage and the cash equivalent percentages are related to the total assets. Coca-Cola Vertical Analysis: 2005 – $4,701(Cash, and Cash Equivalent) divided by $29,427(Total Assets) = 1. 6 % 2004 – 6,707(Cash and Cash Equivalent) divided by $31,441(Total Assets) = 2. 1 % Observing all the facts, it is clear to see that Coca-Cola has lower assets. Lowering their assets means that Coca-Cola used their assets to pay down or payoff their debt. This is a fact that most investors would strongly look at while determining where to invest.At this point Coca-Cola is ab le to spend more that will allow Coca-Cola to grow financially. After all of the factual numbers are observed, it is the time to decide which company is better to invest in. The only other elements that require examination is the personal choice and media influence. To look at which company has better advertising or taste is not the best way to decide but is a factor that is shared among un-educated investors. A financial investor would not encourage the investor to invest from his or her gut feelings.They would try to persuade the investor to use facts and figures as well as the reputation of the company. Although Coca-Cola posted stronger numbers in the Vertical Analysis, PepsiCo, Inc. posted stronger numbers in the Liquidity category. The Horizontal Analysis was also not the strongest showing for the PepsiCo, Inc. were lower even though they weren’t decreasing at the rate of Coca-Cola. This is harder to pick a better investment. Although Coca-Cola is decreasing their perce ntage of liabilities their total percentage of liabilities was higher.These facts and figures determine that Coca-Cola and PepsiCo. Inc, are both strong companies. They are also strong competitors. However, Coca-Cola seems to be handling their monies and financial investments in a more effective way. It seems that Coca-Cola is a stouter and more sensible investment. References: Principal of Financial Accounting 6th Edition. Weygandt,Kiesco, Kimmel www. pepsico. com/index. html#/flash/pepsico_slide. swf PepsiCo. com The Coca Cola Company. www. thecoca-colacompany. com/