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Instalaes industriais - Free Essay Example

Sample details Pages: 37 Words: 11021 Downloads: 4 Date added: 2017/06/26 Category Statistics Essay Did you like this example? Unidade 6: Instalaes industriais 6.1 Primeiras palavras Esta unidade apresenta os elementos eltricos fundamentais de uma planta industrial, trata dos transformadores de energia e dos motores eltricos. Vamos entender o funcionamento e os cuidados necessrios para prolongar a vida til desses equipamentos. 6.2 Problematizando o tema Em nossas residncias a tenso eltrica normalmente tem dois valores: 127V ou 220V. Na indstria, de um modo geral, preciso alimentar equipamentos de grandes potncias com tenses eltricas maiores para reduo correspondente da corrente eltrica. Assim, nas instalaes industriais podemos encontrar valores nominais de tenso de 380V, 440V ou at maiores. Don’t waste time! Our writers will create an original "Instalaes industriais" essay for you Create order 6.3 Texto bsico para estudos 6.3.1 Transformador de potncia um equipamento que, por meio de induo eletromagntica, transfere energia de um circuito chamado primrio para um ou mais circuitos denominados secundrio ou tercirio, respectivamente, sendo mantida a mesma frequncia, porm com tenses e correntes diferentes: Quanto ao meio isolante, os transformadores se classificam em: transformadores imersos em leo mineral isolante; transformadores a seco. Contemplaremos somente os transformadores imersos em leo, devido quase exclusividade de sua utilizao em projetos industriais. Os transformadores a seco so empregados mais especificamente em instalaes de prdios de habitao ou em locais de alto risco para a vida das pessoas e do patrimnio. So construdos, em geral, em resina epxi. Um transformador imerso em leo mineral composto basicamente de trs elementos: tanque ou carcaa; parte ativa (ncleo e enrolamentos); acessrios (terminais, ganchos, registros, etc.). O seu funcionamento est fundamentado nos fenmenos de mtua induo magntica entre os dois circuitos (primrio e secundrio) eletricamente isolados, porm magneticamente ligados. A equao fundamental de operao de um transformador : N1 nmero de espiras do enrolamento primrio; N2 nmero de espiras do enrolamento secundrio; V1 tenso aplicada nos terminais da bobina do primrio; V2 tenso de sada nos terminais da bobina do secundrio; I1 corrente que circula no enrolamento primrio; I2 corrente que circula no enrolamento secundrio. Os transformadores podem ser, quanto ao nmero de fases: monobucha (F-T); monofsico (F-N); bifsico (2F); trifsico (3F). Ao longo desta unidade s se far referncia aos transformadores trifsicos, devido sua quase total utilizao em sistemas industriais no Brasil. A 89 apresenta um transformador trifsico a leo mineral. Quanto s caractersticas eltricas, os transformadores podem ser estudados conforme os itens a seguir. 6.3.1.1 Potncia nominal a potncia que o transformador fornece continuamente a uma determinada carga, sob condies de tenso e frequncia nominais, dentro dos limites de temperatura especificados por norma. A determinao da potncia nominal do transformador em funo da carga que alimenta dada pela equao a seguir. P1= Vs tenso secundria de alimentao de carga, em V; Ic corrente da carga conectada, em A. As potncias nominais padronizadas e usuais esto discriminadas na Tabela 20. Tabela 20 Dados caractersticos de transformadores trifsicos em leo para instalao interior ou exterior classe 15kV primrio em estrela ou tringulo e secundrio em estrela 60Hz. Potncia (kVA) Tenso (V) Perdas (W) Rendimento Regulao Impedncia A vazio (no ferro) Cobre (%) (%) (%) 15 220 a 440 120 300 96,24 3,32 3,5 30 220 a 440 200 570 96,85 3,29 3,5 45 220 440 260 750 97,09 3,19 3,5 75 220 440 390 1200 97,32 3,15 3,5 112,5 220 440 520 1650 97,51 3,09 3,5 150 220 a 440 640 2050 97,68 3,02 3,5 225 380 a 440 900 2800 97,96 3,63 4,5 300 220 1120 3900 97,96 3,66 4,5 380 ou 440 1120 3700 98,04 3,61 4,5 500 220 1700 6400 98,02 3,65 4,5 380 ou 440 1700 6000 98,11 3,6 4,5 750 220 2000 10000 98,04 4,32 5,5 380 ou 440 2000 8500 98,28 4,2 5,5 1000 220 3000 12500 98,10 4,27 5,5 380 ou 440 3000 11000 98,28 4,19 5,5 1500 220 4000 18000 98,20 4,24 5,5 380 ou 440 4000 16000 98,36 4,16 5,5 6.3.1.2 Tenso nominal o valor eficaz da tenso para a qual o transformador foi projetado segundo perdas e rendimento especificado. Em geral, os transformadores so dotados de derivaes ou tapes, utilizados quase sempre para elevar a tenso de sada do secundrio, devido a uma tenso de fornecimento abaixo do valor requerido. O tape de maior valor define a tenso nominal primria do transformador, isto , a tenso para a qual foi projetado. Normalmente, o nmero mximo de derivaes fica limitado a trs, variando de 3,0 a 9,6% da tenso nominal especificada para o equipamento. Como exemplo, citando um transformador de tenso nominal de 13800V, os tapes disponveis so de 12600, 13200 e 13800V. importante lembrar que constante o produto da tenso e da corrente no primrio e no secundrio. Considerar, por exemplo, um transformador de 225kVA, tenso nominal de 13800/380V, operando numa rede com tenso nominal primria de mesmo valor. Por motivo de abaixamento da tenso de fornecimento, o transformador foi religado no tape de 12600V, logo, a corrente ser aumentada de: Vt1 It1 = Vt2 It2; Vt1 tenso no primrio no tape 1; Vt2 tenso no primrio no tape 2; It1 corrente no tape 1; It2 corrente no tape 2. 13800 It1 = 12.600 It2 It1 = A 13800 9,4 = 12600 It2 It2 = 10,29A Se a tenso de fornecimento fosse de 12400V, a tenso secundria assumiria o valor de: Vs = = 374V 6.3.1.3 Tenso nominal de curto-circuito medida curto-circuitando-se os terminais secundrios do transformador e alimentando-o no primrio com uma tenso que faa circular nesse enrolamento a corrente nominal. O valor percentual dessa tenso em relao nominal numericamente igual ao valor da impedncia em percentagem, ou seja: Zpt = Zpt tenso nominal de curto-circuito, em % ou impedncia percentual; Vnccp tenso nominal de curto-circuito aplicada aos terminais do enrolamento primrio, em V; Vnpt tenso nominal primria do transformador, em V. Caso se deseje conhecer a impedncia do transformador em valor hmico, pode-se usar a seguinte equao: ZWt = Pnt potncia nominal do transformador, em kVA; Vnt tenso nominal primria do transformador, em kV. Uma impedncia percentual de 5,5% corresponde a um transformador de 1000kVA 13800/380V e tem como impedncia hmica o valor de: Z 6.3.1.4 Componentes percentuais da tenso nominal de curto-circuito determinada a partir da composio vetorial dos componentes de tenso resistiva e reativa. O componente de tenso resistiva percentual ou resistncia percentual o componente ativo da tenso percentual, cujo valor dado pela seguinte equao: Rpt = Pcu perdas hmicas de curto-circuito, ou simplesmente perdas no cobre, em W (Tabela 20); Pnt potncia nominal do transformador, em kVA. Conhecido o valor da tenso percentual de curto-circuito do transformador, fornecido pelo fabricante, aplica-se a seguinte equao para se obter o valor da tenso reativa percentual, ou seja: Xpt = Zpt impedncia percentual de placa do transformador. Exemplo: Considerar um transformador de 225kVA, 13800-380/220V do qual se deseja saber os valores percentuais das quedas de tenso resistiva e reativa. Pcu = 2800W (Tabela 20); Zpt = 4,5% (Tabela 20). 6.3.1.5 Perdas eltricas Os transformadores apresentam perdas eltricas pequenas quando comparadas com suas potncias nominais. Mas sendo uma mquina que opera, em geral, continuamente, a energia desperdiada pode ser relevante e, portanto, considerada nas avaliaes de eficincia energtica, conforme a unidade 5. As perdas dos transformadores referem-se a perdas no ncleo e a perdas nos enrolamentos. 6.3.1.6 Regulao Representa a variao de tenso no secundrio do transformador, desde o seu funcionamento em vazio at a operao a plena carga, considerando a tenso primria constante. Tambm denominada queda de tenso industrial, pode ser calculada em funo dos componentes ativo e reativo, da impedncia percentual do transformador, do fator de potncia e do fator de carga, conforme a equao a seguir. R = regulao; Fc = fator de carga; q = ngulo do fator de potncia. O valor da tenso no secundrio do transformador, correspondente s condies de carga a que est submetido, dado pela equao seguinte, ou seja: Vnst= tenso nominal do secundrio, em V. Exemplo: Considerar uma transformador de 225kVA, 13800-380/220V operando numa instalao cujo fator de carga 0,75. Deseja-se determinar o valor da regulao ou variao de tenso no secundrio, sabendo-se que o fator de potncia da carga 0,80. Os valores de Rpt e Xpt foram calculados no exemplo anterior. Logo a tenso secundria vale: 6.3.1.7 Rendimento a relao entre a potncia eltrica fornecida pelo secundrio do transformador e a potncia eltrica absorvida pelo primrio. Pode ser determinado pela equao: Pfe = perdas no ferro, em kW; q = ngulo do fator de potncia. Exemplo: Tomando como exemplo as condies previstas nos dois exemplos anteriores, determinar o rendimento do transformador de 225kVA. h = 100 1,8 = 98,2%; Pfe = 0,90kW (Tabela 20); Pcu = 2,8kW (Tabela 20). 6.3.1.8 Lquido isolante O lquido isolante nos transformadores tem a funo de transferir o calor gerado pelas partes internas do equipamento para as paredes do tanque e dos radiadores, que so resfriadas naturalmente ou por ventilao forada, fazendo com que o leo volte novamente ao interior, retirando calor e passando ao exterior, num ciclo contnuo, segundo o fenmeno de conveco. O leo mineral para transformador deve apresentar uma alta rigidez dieltrica e excelente fluidez, alm de manter as suas caractersticas naturais praticamente inalteradas perante temperaturas elevadas. Os transformadores podem conter leo mineral do tipo parafnico ou naftnico. O leo mineral inflamvel e, portanto, cuidados devem ser tomados na instalao de transformadores. No caso de projetos industriais de produtos de alto risco de incndio, usando-se transformadores a leo, estes devem ser localizados distantes e fora da rea de risco. Existe, entretanto, um tipo de lquido isolante, chamado ascarel, cujas propriedades eltricas se assemelham s do leo mineral, com a vantagem de no ser inflamvel. Devido ao seu alto poder de poluio, est proibida a sua utilizao no territrio nacional. Quando for estritamente necessria a instalao de transformadores no inflamveis, devem ser especificados transformadores a seco ou a silicone. 6.3.1.9 Especificao O pedido de compra de um transformador deve conter, no mnimo, os seguintes elementos: potncia nominal; tenso nominal primria; tenso nominal secundria; derivaes desejadas (tapes); perdas mximas no ferro e no cobre; ligao dos enrolamentos; tenso suportvel de impulso; impedncia percentual; acessrios desejados (especificar). Exemplo: Transformador trifsico de 225kVA, tenso nominal primria 13,8kV, tenso nominal secundria 380/220V, com derivaes 13,8/13,2/12,6kV, dispondo de ligao dos enrolamentos em tringulo primrio e em estrela secundria com neutro acessvel, impedncia nominal percentual de 4,5%, frequncia de 60Hz, perdas mximas no cobre de 2800W, perdas mximas no ferro de 900W e tenso suportvel de impulso 95kV. 6.3.2 Motores eltricos de corrente alternada O motor eltrico uma mquina que transforma energia eltrica em energia mecnica de utilizao. Os motores eltricos de induo trifsicos so utilizados na maioria das aplicaes industriais. 6.3.2.1 Motores trifsicos So aqueles alimentados por um sistema trifsico a trs fios, em que as tenses esto defasadas de 120 eltricos. Representam a grande maioria dos motores empregados nas instalaes industriais. A 90 mostra os seus principais componentes. Podem ser do tipo induo ou sncrono. 6.3.2.1.1 Motores de induo So constitudos de duas partes bsicas: estator e rotor. O estator formado por trs elementos: Carcaa: constituda de uma estrutura de construo robusta, fabricada em ferro fundido, ao ou alumnio injetado, resistente corroso e com superfcie aletada, que tem como principal funo suportar todas as partes fixas e mveis do motor. Ncleo de chapas: constitudo de chapas magnticas adequadamente fixadas ao estator. Enrolamentos: dimensionados em material condutor isolado, dispostos sobre o ncleo e ligados rede de energia eltrica de alimentao. O rotor constitudo por quatro elementos bsicos: Eixo: responsvel pela transmisso da potncia mecnica gerada pelo motor. Ncleo de chapas: constitudo de chapas magnticas adequadamente fixadas sobre o eixo. Barras e anis de curto-circuito (motor de gaiola): constitudo de alumnio injetado sobre-presso. Enrolamentos (motor com rotor bobinados): constitudos de material condutor e dispostos sobre o ncleo. Demais componentes: Ventilador: responsvel pela remoo do calor acumulado na carcaa. Tampa defletora: componente mecnico provido de aberturas instaladas na parte traseira do motor sobre o ventilador. Terminais: conectores metlicos que recebem os condutores de alimentao do motor. Rolamentos: componentes mecnicos sobre os quais est fixado o eixo. Tampa: componente metlico de fechamento lateral. Caixa de ligao: local onde esto fixados os terminais de ligao do motor. As correntes rotricas so geradas eletromagneticamente pelo estator, nico elemento do motor ligado linha de alimentao. O comportamento de um motor eltrico de induo relativo ao rotor comparado ao secundrio de um transformador. O rotor pode ser constitudo de duas maneiras: a) Rotor bobinado Constitudo de bobinas, cujos terminais so ligados a anis coletores fixados ao eixo do motor e isolados deste. So de emprego frequente nos projetos industriais, principalmente quando se necessita de controle adequado movimentao de carga, ou se deseja acionar uma determinada carga atravs de reostato de partida. Esses motores so construdos com o rotor envolvido por um conjunto de bobinas normalmente interligadas, em cono estrela, com os terminais conectados a trs anis, presos mecanicamente ao eixo do motor, porm isolados eletricamente, e ligados por meio de escovas condutoras a uma resistncia trifsica provida de cursor rotativo. Assim, as resistncias so colocadas em srie com o circuito do enrolamento do rotor, e a quantidade utilizada depende do nmero de estgios de partida adotado, que, por sua vez, dimensionado em funo, exclusivamente, do valor da mxima corrente admissvel para acionamento da carga. A 91 mostra esquematicamente a ligao dos anis acoplados ao reostato de partida, com a barra de curto-circuito medianamente inserida. J a 92 mostra tambm a ligao de um motor com reostato de partida ajustado para acionamento em trs tempos. Na 92, pode-se observar que, quando acionado o contator geral C1 ligado aos terminais 1-2-3, o motor parte sob o efeito das duas resistncias inseridas em cada bobina rotrica. Aps um certo perodo de tempo previamente ajustado, o contator C3 curto-circuita o primeiro grupo de resistncia do reostato, o que equivale ao segundo estgio. Decorrido outro determinado perodo de tempo, o contator C2 opera mantendo em curto-circuito o ltimo grupo de resistncias do reostato, o que equivale ao terceiro estgio. Nessa condio, o motor entra em regime normal de funcionamento. Os motores de anis so particularmente empregados na frenagem eltrica, controlando adequadamente a movimentao de cargas verticais, em baixas velocidades. Para isso, usam um sistema combinado de frenagem sobressncrona ou subsncrona com inverso das fases de alimentao. Na etapa de levantamento, o motor acionado com a ligao normal, sendo que tanto a fora necessria para vencer a carga resistente quanto a velocidade de levantamento so ajustadas pela insero ou retirada dos resistores do circuito do rotor. Para o abaixamento da carga, basta inverter duas fases de alimentao e o motor comporta-se como gerador, em regime sobressncrono, fornecendo energia rede de alimentao, girando, portanto no sentido contrrio ao funcionamento anterior. Esses motores so empregados no acionamento de guindastes, correias transportadoras, compressores a pisto, etc. b) Rotor em gaiola Esse tipo de rotor constitudo de um conjunto de barras no isoladas atravs de anis condutores curto-circuitados. O motor de induo opera normalmente a uma velocidade constante, variando ligeiramente com a aplicao da carga mecnica no eixo. O funcionamento de um motor de induo baseia-se no princpio da formao de campo magntico rotativo produzido no estator pela passagem da corrente alternada em suas bobinas, cujo fluxo, por efeito de sua variao, se desloca em volta do rotor, gerando, neste, correntes induzidas que tendem a se opor ao campo rotativo, sendo, no entanto, arrastado por ele. O rotor em nenhuma hiptese atinge a velocidade do campo rotativo, pois, do contrrio, no haveria gerao de correntes induzidas, eliminando-se o fenmeno magntico rotrico, responsvel pelo trabalho mecnico do rotor. Quando o motor est girando sem a presena de carga mecnica no eixo, comumente chamado motor a vazio, o rotor desenvolve uma velocidade angular de valor praticamente igual velocidade sncrona do campo girante do estator. Adicionando-se carga mecnica ao eixo, o rotor diminui a sua velocidade. A diferena existente entre as velocidades sncronas e a do rotor denominada escorregamento, em termos percentuais, dado pela seguinte equao: Ws= velocidade sncrona; W = velocidade angular do rotor. 6.3.2.2 Rendimento de uma mquina Exemplo: Um motor consome 5kW de potncia eltrica e fornece 4kW de potncia mecnica em seu eixo. Qual o seu rendimento? Exemplo: Um motor trifsico de 5kW (o valor que aparece na plaqueta do motor a sua potncia nominal de sada) e 380V tem um rendimento de 0,80 e um FP = 0,85. Qual a corrente recebida pelo motor na carga nominal? Exemplo: Na placa de identificao de um motor trifsico esto as seguintes informaes: 380V; 60A; cosq = 0,85; 30kW. Qual o rendimento desse motor? Pe = 1,73 x 380 x 60 x 0,85 Pe = 33567W ou Pe = 33,57kW h = 30/33,57 h = 0,89 ou 89% Exemplo: Um motor trifsico consome 11,8cv, tem um FP 0,85 e alimentado por 220V. Calcule a corrente de linha do circuito, a potncia reativa e a aparente. Considerar h = 1. 1cv = 735W (1hp = 746W) Pe = 11,8cv = 8,67W Potncia de entrada UL = 220V cosq = 0,85 Tringulo de potncias: 6.3.2.3 Motores monofsicos de induo Os motores monofsicos so, em relao aos motores trifsicos, de pequeno uso em instalaes industriais. So construdos normalmente para pequenas potncias (at 15 cv, em geral). Os motores monofsicos so providos de um segundo enrolamento colocado no estator e defasado de 90 eltricos do enrolamento principal, e que tem a finalidade de tornar rotativo o campo estatrico monofsico. Isso o que permite a partida do motor monofsico. O torque de partida produzido pelo defasamento de 90 entre as correntes do circuito principal e do circuito de partida. Para se obter essa defasagem, liga-se ao circuito de partida um condensador, de acordo com o esquema da 93 (a). O campo rotativo assim produzido orienta o sentido de rotao do motor. A fim de que o circuito de partida no fique ligado desnecessariamente aps o acionamento do motor, um dispositivo automtico desliga o enrolamento de partida, passando o motor a funcionar normalmente em regime monofsico. Esse dispositivo pode ser acionado por um sistema de fora centrfuga, conforme a 93 (a). A bobina que liga o circuito de partida desenergizada pelo decrscimo do valor da corrente no circuito principal aps o motor entrar em regime normal de funcionamento. O condensador de partida do tipo eletroltico e tem a caracterstica de funcionar somente quando solicitado por tenses com polaridade estabelecida. montado, normalmente, sobre a carcaa do estator por meio de um suporte que tambm tem a finalidade de proteg-lo mecanicamente. A Tabela 21 fornece as caractersticas bsicas dos motores monofsicos. Os motores monofsicos podem ser do tipo induo ou sncrono, cujas caractersticas bsicas so idnticas s que foram estabelecidas para os motores trifsicos correspondentes. (a) tipo fora centrfuga (b)tipo decrscimo da corrente Tabela 21 Caractersticas dos motores eltricos monofsicos. Potncia Nominal Corrente 220(V) Veloci dade Fator de Potncia Rela- o Rela o Conjugado Rendimento Momento de Inrcia Nomi- nal Cm/Cn cv kW A rpm % Inp/In Cp/Cn Kgf . m % % Kg . m2 2 polos 1,5 1,1 7,5 3,535 75 7,8 2,9 0,31 2,3 75 0,0020 2 1,5 9,5 3,530 76 7,2 2,9 0,61 2,3 76 0,0024 3 2,2 13,0 3,460 77 7,6 30, 0,81 2,2 77 0,0064 4 3,0 18,0 3,515 79 8,7 2,8 0,61 2,6 79 0,0093 5 3,7 23,0 3,515 81 7,9 2,8 1,00 2,6 81 0,0104 7,5 5,5 34,0 3,495 78 6,2 2,1 1,50 2,1 78 0,0210 10 7,5 42,0 3,495 82 7 2,1 2,00 2,6 82 0,0295 4 polos 1 0,75 5,8 1,760 71 8,2 3,0 0,41 2,5 71 0,0039 1,5 1,1 7,5 1,760 75 8,7 2,8 0,61 2,9 75 0,0052 2 1,5 9,5 1,750 77 8,7 3,0 0,81 2,8 77 0,0084 3 2,2 14,0 1,755 79 8,5 3,0 1,20 2,8 79 0,0163 4 3,0 19,0 1,745 80 7,1 2,90 1,60 2,6 80 0,0183 5 3,7 25,0 1,750 81 7,5 3,0 2,00 2,6 81 0,0336 7,5 5,5 34,0 1,745 84 7,4 3,0 3,10 2,6 84 0,0378 10 7,5 46,0 1,745 85 7,6 3,0 4,10 2,5 85 0,0434 6.3.2.4 Motores tipo universal So aqueles capazes de operar tanto em corrente contnua como em corrente alternada. So amplamente utilizados em aparelhos eletrodomsticos, tais como enceradeiras, liquidificadores, batedeiras, etc. So constitudos de uma bobina de campo em srie com a bobina da armadura, e de uma bobina de compensao, que pode estar ligada em srie ou em paralelo com a bobina de campo, cuja compensao denominada respectivamente de condutiva ou indutiva. 6.3.2.5 Motores assncronos trifsicos com rotor em gaiola Os motores de induo trifsicos, com rotor em gaiola, so usados na maioria das instalaes industriais, principalmente em mquinas no suscetveis a variaes de velocidade. Para obteno de velocidade constante, devem-se usar motores sncronos normalmente construdos para potncias elevadas, devido a seu alto custo relativo, quando fabricados em potncias menores. A seguir, sero estudadas as principais caractersticas dos motores de induo trifsicos com rotor em gaiola. 6.3.2.5.1 Potncia nominal a potncia que o motor pode fornecer no eixo, em regime contnuo, sem que os limites de temperatura dos enrolamentos sejam excedidos aos valores mximos permitidos por norma dentro de sua classe de isolamento. Sempre que so aplicadas aos motores cargas de valor muito superior ao da potncia para a qual foram projetados, os seus enrolamentos sofrem um aquecimento anormal, diminuindo a vida til da mquina, podendo danificar o isolamento at se estabelecer um curto-circuito interno que caracteriza a sua queima. A potncia desenvolvida por um motor representa a rapidez com que a energia aplicada para mover a carga. Por definio, potncia a relao entre a energia gasta para realizar um determinado trabalho e o tempo em que o mesmo foi executado. Isso pode ser facilmente entendido quando se considera a potncia necessria para levantar um objeto pesando 50kgf do fundo de um poo de 40m de profundidade, durante um perodo de tempo de 27s. A energia gasta foi de 50kgf x 40m = 2000kgf m. Como o tempo para realizar esse trabalho foi de 27s, a potncia exigida pelo motor foi de Pm1 = 2000/27kgf.m/s = 74kgf m/s. Se o mesmo trabalho tivesse que ser realizado em 17s, a potncia do motor teria que ser incrementada para Pm2 = 2000/17kgf m/s = 117kgf.m/s. Considerando que 1 cv equivale a 75 kgf m/s, ento as potncias dos motores seriam: Pm1 = cv Pm2 = 1.1/2 cv A potncia nominal normalmente fornecida em cv, sendo que 1cv equivale a 0,736kW. A potncia nominal de um motor depende da elevao de temperatura dos enrolamentos durante o ciclo de carga. Assim, um motor pode acionar uma carga com potncia superior sua potncia nominal at atingir um conjugado um pouco inferior a seu conjugado mximo. Essa sobrecarga, no entanto, no pode resultar em temperatura dos enrolamentos superiores sua classe de temperatura. Do contrrio, a vida til do motor ser sensivelmente afetada. Quando o motor opera com cargas de regimes intermitentes, a potncia nominal do motor deve ser calculada levando em considerao o tipo de regime. A Tabela 22 fornece as principais caractersticas dos motores de induo de rotor em curto-circuito. Vale ressaltar que estes so valores mdios e podem variar, em faixas estreitas, para cada fabricante, dependendo de sua tecnologia e projeto construtivo. Tabela 22 Motores assncronos trifsicos com rotor em curto-circuito. Potncia Nominal Potncia Ativa Corrente Nominal Velocidade (rpm) Fator de Potncia Relao Inp/In Rendimento Conjugado Nominal cv kW 220V 380V % Kgf.m 2 polos 1 0,7 3,3 1,9 3440 0,76 6,2 0,81 0,208 3 2,2 9,2 5,3 3490 0,76 8,3 0,82 0,619 5 4 13,7 7,9 3490 0,83 9,0 0,83 1,020 7,5 5,5 19,2 11,5 3480 0,83 7,4 0,83 1,540 10 7,5 28,6 16,2 3475 0,85 6,7 0,83 2,050 15 11 40,7 23,5 3500 0,82 7,0 0,83 3,070 20 15 64,0 35,5 3540 0,73 6,8 0,83 3,970 25 18,5 69,0 38,3 3540 0,82 6,8 0,86 4,960 30 22 73,0 40,5 3535 0,88 6,3 0,89 5,960 40 30 98,0 54,4 3525 0,89 6,8 0,90 7,970 50 37 120,0 66,6 3540 0,89 6,8 0,91 9,920 60 45 146 81,0 3545 0,89 6,5 0,91 11,880 75 55 178 98,8 3550 0,89 6,9 0,92 14,840 100 75 240,0 133,2 3560 0,90 6,8 0,93 19,720 125 90 284,0 158,7 3570 0,90 6,5 0,93 24,590 150 110 344,0 190,9 3575 0,90 6,8 0,93 29,460 Potncia Nominal Potncia Ativa Corrente Nominal Velocidade (rpm) Fator de Potncia Relao Inp/In Rendimento Conjugado Nominal cv kW 220V 380V % Kgf.m 4 polos 1 0,7 3,8 2,2 1715 0,65 5,7 0,81 0,420 3 2,2 9,5 5,5 1720 0,73 6,6 0,82 1,230 5 4 13,7 7,9 1720 0,83 7,0 0,83 2,070 7,5 5,5 20,6 11,9 1735 0,81 7,0 0,84 3,100 10 7,5 26,6 15,4 1740 0,85 6,6 0,86 4,110 15 11 45,0 26,0 1760 0,75 7,8 0,86 6,120 20 15 52,0 28,8 1760 0,86 6,8 0,88 7,980 25 18,5 64,0 35,5 1760 0,84 6,7 0,90 9,970 30 22 78,0 43,3 1760 0,83 6,8 0,90 11,970 40 30 102,0 56,6 1760 0,85 6,7 0,91 15,960 50 37 124,0 68,8 1760 0,86 6,4 0,92 19,950 60 45 150,0 83,3 1765 0,86 6,7 0,92 23,870 75 55 182,0 101,1 1770 0,86 6,8 0,92 29,750 100 75 244,0 135,4 1770 0,87 6,7 0,92 39,670 125 90 290,0 160,9 1780 0,87 6,5 0,94 49,310 150 110 350,0 194,2 1780 0,87 6,8 0,95 59,170 180 132 420,0 233,1 1785 0,87 6,5 0,95 70,810 200 150 470,0 271,2 1785 0,87 6,9 0,95 80,000 220 160 510,0 283,0 1785 0,87 6,5 0,95 86,550 250 185 590,0 327,4 1785 0,87 6,8 0,95 95,350 300 220 694,0 385,2 1785 0,88 6,8 0,96 118,020 380 280 864,0 479,5 1785 0,89 6,9 0,96 149,090 475 355 1.100,0 610,5 1788 0,89 7,6 0,96 186,550 600 450 1384,0 768,1 1790 0,89 7,8 0,96 265,370 Como informao adicional, a seguir so dadas as expresses que permitem determinar a potncia de um motor para as atividades de maior uso industrial: a) Bombas Pb = Pb potncia requerida pela bomba, em kW; Q quantidade do lquido, em m3/s; peso especfico do lquido, em kg/dm3; = 1 kg /dm3 para a gua; H altura manomtrica e perdas nas tubulaes, em m; eficincia da bomba; 0,87 para bombas a pisto; 0,40 para bombas centrfugas. Exemplo de Aplicao Calcular a potncia nominal de um motor que ser acoplado a uma bomba centrfuga, cuja vazo de 0,50m3/s. A altura de recalque mais a de elevao de 15m e destina-se captao de gua potvel. Pm = b) Elevadores de carga Pe = Pe potncia requerida pelo motor do guindaste, em kW; 0,70; C carga a ser levantada, em kg; V velocidade, em m/s; 0,50 V 1,50 m/s elevadores para pessoas; 0,40 V 0,60 m/s elevadores para carga. Exemplo de Aplicao Determinar a potncia nominal de um motor de um elevador de carga destinado a levantar uma carga mxima de 400kg. Pe = 3,36kW Pm = 5 cv (Tabela 22) c) Ventiladores Pv = Pv potncia requerida pelo ventilador, em kW; Q vazo, em m3/s; P presso, em N/m2; rendimento; 0,50 0,80 para ventiladores com P 400mmHg; 0,35 0,50 para ventiladores com 100 P 400mmHg; 0,20 0,35 para ventiladores com P 100mmHg. Obs.: 1 mmHg = 9,81 N/m2; 1 N/m2 = 1,02 10-3 kgf/m2. Exemplo de Aplicao Qual a potncia nominal de um motor para ser acoplado ao eixo de um ventilador com vazo de 2m3/s e presso de 200mmHg? Sabe-se que o rendimento do ventilador de aproximadamente 40%. d) Compressores Pc = Pc potncia requerida pelo compressor, em kW; Wc velocidade nominal do compressor, em rps; Cnc conjugado nominal do compressor, em Nm; ac rendimento de acoplamento:0,95. Exemplo de Aplicao Determinar a potncia de um compressor, sabendo-se que a reduo do acoplamento de 0,66, a velocidade do compressor de 1150rpm e o conjugado nominal de 40Nm. Velocidade nominal do motor: Wn = rpm Velocidade nominal do compressor: Wc = 19,16 rps Potncia nominal do motor: Pc = W Pm = 7,5 cv (Tabela 22). Existe uma condio operacional de motores muito utilizada em processos industriais, notadamente em esteiras rolantes, quando dois ou mais motores funcionam mecanicamente em paralelo. Se dois ou mais motores idnticos so acoplados por um mecanismo qualquer e trabalham mecanicamente em paralelo dividem a carga igualmente. Para isso, necessrio que os motores tenham o mesmo escorregamento, o mesmo nmero de polos e a mesma potncia nominal no eixo. Se dois ou mais motores tm o mesmo nmero de polos, mas diferentes potncias nominais no eixo, normalmente dividem a carga na mesma proporo de suas potncias de sada. 6.3.2.5.2 Tenso nominal As tenses de maior utilizao nas instalaes eltricas industriais so de 220, 380 e 440V. A ligao do motor num determinado circuito depende das tenses nominais mltiplas para as quais foi projetado. Os motores devem trabalhar dentro de limites de desempenho satisfatrio para uma variao de tenso de 10% de sua tenso nominal, desde que a frequncia no varie. 6.3.2.5.3 Corrente nominal aquela solicitada da rede de alimentao pelo motor trabalhando potncia nominal, com a frequncia e tenses nominais. O valor da corrente dado pela equao: Inm = Pnm potncia nominal do motor, em cv; V tenso nominal trifsica, em volts; rendimento do motor; cos fator de potncia sob carga nominal. 6.3.2.5.4 Frequncia nominal aquela fornecida pelo circuito de alimentao e para a qual o motor foi dimensionado. O motor deve trabalhar satisfatoriamente se a frequncia variar dentro de limites de 5% da frequncia nominal, desde que seja mantida a tenso nominal constante. Os motores trifsicos com rotor bobinado, quando ligados numa rede de energia eltrica cuja frequncia diferente da frequncia nominal, apresentam as seguintes particularidades: Motor de 50Hz ligado em 60Hz: a potncia mecnica no varia; a corrente de carga no varia; a corrente de partida diminui em 17%; a velocidade nominal aumenta em 20%, isto , na mesma proporo do aumento da frequncia; a relao entre o conjugado mximo e o conjugado nominal diminui em 17%; a relao entre o conjugado de partida e o conjugado nominal diminui em 17%. Motor de 60Hz ligado em 50Hz: * a potncia aumenta em 20% para motores de 4, 6 e 8 polos; * a corrente de carga no varia; * a velocidade nominal diminui na mesma proporo da reduo da frequncia; * a relao entre o conjugado mximo e o conjugado nominal aumenta; * a relao entre o conjugado de partida e o conjugado nominal aumenta. 6.3.2.5.5 Fator de servio um nmero que pode ser multiplicado pela potncia nominal do motor, a fim de se obter a carga permissvel que o mesmo pode acionar, em regime contnuo, dentro de condies estabelecidas por norma. O fator de servio no est ligado capacidade de sobrecarga prpria dos motores, valor geralmente situado entre 140 e 160% da carga nominal durante perodos curtos. Na realidade, o fator de servio representa uma potncia adicional contnua. 6.3.2.5.6 Perdas hmicas O motor absorve do circuito de alimentao uma determinada potncia que dever ser transmitida ao eixo para o acionamento da carga. Porm, devido a perdas internas em forma de calor gerado pelo aquecimento das bobinas dos enrolamentos e outras, a potncia mecnica de sada do eixo sempre menor do que a potncia de alimentao. Desse fenmeno nasce o conceito de rendimento, cujo valor sempre menor que a unidade. As perdas verificadas num motor eltrico so: perdas joule nas bobinas estatricas: perdas no cobre (Pcu); perdas joule nas bobinas rotricas: perdas no cobre (Pcu); perdas magnticas estatricas: perdas no ferro (Pfe); perdas magnticas rotricas: perdas no ferro (Pfe); perdas por ventilao: (Pv); perdas por atrito dos mancais: perdas mecnicas (Pm). A 94 ilustra o balano das potncias e perdas eltricas envolvidas num motor eltrico. Todo o calor formado no interior do motor deve ser dissipado para o meio exterior atravs da superfcie externa da carcaa, auxiliada, para determinados tipos de motores, por ventiladores acoplados ao eixo. No se deve julgar o aquecimento interno do motor simplesmente medindo-se a temperatura da carcaa, pois isso pode fornecer resultados falsos. Os motores trifsicos ligados a fontes trifsicas desequilibradas sofrem o efeito do componente de sequencia negativa em forma de aquecimento, provocando o aumento das perdas, principalmente as perdas no cobre e reduzindo, assim, a potncia de sada disponvel. Portanto, deve-se procurar manter a tenso entre fases de alimentao dos motores eltricos o mais equilibrada possvel. 6.3.3 Iluminao 6.3.3.1 Introduo At o momento, foi possvel notar que as unidades volt, ampere e watt tm um grande significado para a compreenso de vrios problemas referentes utilizao da eletricidade. Encontramos, igualmente na iluminao, grandezas que correspondem a uma caracterstica especial da luz e das superfcies iluminadas. Trataremos, neste momento, das grandezas luminotcnicas indispensveis para atingir os objetivos deste estudo: fluxo luminoso, rendimento luminoso e iluminamento. 6.3.3.2 Fluxo luminoso As fontes luminosas eltricas absorvem potncia eltrica e fornecem potncia luminosa. A potncia luminosa irradiada em todas as direes por uma fonte luminosa se denomina fluxo luminoso (). O fluxo luminoso medido em lumens (lm). Exemplos: uma lmpada a vapor de mercrio de 250 watts produz 12500 lumens; uma lmpada a vapor de sdio de 250 watts produz 26000 lumens; uma lmpada incandescente de 100 watts produz 1380 lumens. 6.3.3.3 Rendimento luminoso (h) ou eficincia luminosa Na fonte luminosa, somente uma parte da potncia eltrica se transforma em potncia luminosa produzindo tambm calor. A relao entre o fluxo luminoso irradiado e a potncia eltrica consumida se chama rendimento luminoso. O rendimento luminoso de uma fonte luminosa indica quantos lumens se produzem na lmpada por cada watt de potncia eltrica consumida na mesma (unidade = lm/W) dando, portanto, uma medida da economia de consumo de energia eltrica de uma fonte luminosa. O rendimento luminoso depende do tipo e da potncia da lmpada. TABELA 23 Rendimento luminoso. LMPADA CONSUMO DE POTNCIA (W) FLUXO LUMINOSO RENDIMENTO LUMINOSO Incandescente 100W 220V 100 1380 13,8 Fluorescente 65W 78 3800 49 Vapor de mercrio 80W/220V 89 3100 35 Vapor de sdio 60W/220V 81 5000 62 6.3.3.4 Iluminamento ou iluminncia Uma superfcie iluminada v-se clara se o fluxo luminoso incidente grande e a superfcie iluminada pequena. O iluminamento (E) a relao entre o fluxo luminoso e a superfcie iluminada. Sua unidade o lux. Em que: E = iluminamento (lux); = fluxo luminoso (lumens); A = superfcie iluminada. A intensidade de iluminao se mede com o medidor de iluminao (luxmetro). Nas iluminaes de interiores, se distingue entre iluminao geral e iluminao do plano de trabalho. Os exemplos abaixo do uma ideia de ordem de grandeza. luz das estrelas: 0,002 lux; luar: 0,2 lux; iluminao nas ruas: 6 a 12 lux; luz do dia em interiores: 500 a 2000 lux; luz do dia em exteriores: 1000 a 10000 lux; luz do sol direta: 50000 a 100000 lux. 6.3.3.5 Refletncia A NBR 5413, por meio de duas tabelas, possibilita a determinao de valores de iluminncia segundo o tipo de atividade desenvolvida no ambiente, com base em trs variveis: acuidade visual do observador, velocidade e preciso requerida no trabalho e condies de refletncia da tarefa. A Tabela 24 traz valores de iluminncia (mnimo, mdio e mximo) para trs faixas de atividades A, B e C, cada uma subdividida em trs nveis. A Tabela 25, complementa a aplicao da Tabela 24, permitindo ao projetista o clculo ponderado das variveis que determinam a escolha da iluminncia mnima, mdia ou mxima para cada caso. Para sua aplicao, primeiro atribui-se um peso (-1, 0 ou +1) a cada uma das trs caractersticas que aparece na Tabela 25 relativa ao tipo de observador (idade), tarefa visual (velocidade e preciso exigida) e refletncia do fundo da tarefa. Feito isso, somando-se algebricamente os trs valores encontrados, obtm-se o resultado: quando ele for -2 ou -3 pode-se usar a iluminncia mais baixa do grupo; quando for +2 ou +3, usa-se a iluminncia superior; nos demais casos, utiliza-se o valor mdio. Tabela 24 Iluminncias para cada grupo de tarefas visuais. FAIXA ILUMINNCIA LUX TIPO DE ATIVIDADE A Iluminao geral para reas usadas ininterruptamente ou com tarefas visuais simples 20 30 50 reas pblicas com arredores escuros 50 75 100 Orientao simples para permanncia curta 100 150 200 Recintos no utilizados para trabalho contnuo, depsitos B Iluminao geral para reas de trabalho 200 300 500 Tarefas com requisitos visuais limitados, trabalho bruto de maquinaria, auditrios 500 750 1000 Tarefas com requisitos visuais normais, trabalho mdio de maquinaria, escritrio 1000 1500 2000 Tarefas com requisitos especiais, gravao manual, inspeo, indstria de roupas C Iluminao adicional para tarefas visuais difceis 2000 3000 5000 Tarefas visuais exatas e prolongadas, eletrnica de pequeno tamanho 5000 7500 10000 Tarefas visuais muito exatas, montagem de microeletrnica 10000 15000 20000 Tarefas visuais muito especiais, cirurgia. Tabela 25 Fatores determinantes da iluminao adequada. Caracterstica da Tarefa e do Observador PESO -1 0 +1 Idade em anos 40 40 a 65 65 Velocidade e preciso Sem importncia Importante Crtica Refletncia do fundo da tarefa 70% 30% a 70% 30% A NBR 5413 apresenta ainda valores de iluminncias mnimas em lux, para diversos tipos de atividades. 6.3.3.6 Mtodo dos lumens ou dos rendimentos O mtodo baseado na expresso conhecida: E = F S Em que: E = iluminamento desejado (lux); F = fluxo luminoso total (lumens); S = rea a iluminar (m2). Para uma relativa preciso no clculo de iluminao, muitos fatores devem ser observados na frmula terica: * tipo de iluminao, isto , incandescente, fluorescente ou outras e, alm disso, se direta, indireta, etc.; * reflexes das paredes e do teto; * depreciao do fluxo luminoso com o tempo; * dimenses relativas entre comprimento, largura e altura do ambiente; * perodos de limpeza do ambiente e dos aparelhos de iluminao. 6.3.3.6.1 Coeficiente de utilizao ou ndice de forma Para correo da forma terica, define-se o coeficiente de utilizao como uma funo dos parmetros notveis, como: hu = f (dimenses, reflexes, depreciao do fluxo com o tempo). As dimenses do local e sua relao com a altura do plano de trabalho (0,75m do piso) podem ser definidas por parmetro prprio do local, denominado ndice do local ou ndice de forma. As reflexes do teto e das paredes tambm podem ser definidas, conhecendo o tipo de pintura ou o material de revestimento, posio, tipo e nmero de janelas, portas, etc. Logo: hu = f (k, parmetros conhecidos) Sendo: k ndice do local, que funo das dimenses. As expresses utilizadas para a estimativa do ndice do local so, para qualquer tipo de luminria: . Em que: c = comprimento do ambiente (maior dimenso); l = largura do ambiente (menor dimenso); h = distncia, na vertical, das luminrias ao plano de trabalho. 6.3.3.6.1.1 Expresso geral (tabelas) De maneira geral, pode-se dizer que o fluxo fornecido por: sendo d a depreciao. A depreciao d, devido limpeza do ambiente, pode ser estimada de acordo com o tempo de manuteno, segundo a Tabela 26. Tabela 26 Fator de depreciao d. Perodo de manuteno em horas Tipo de ambiente 2500 5000 7500 Limpo 1,05 1,10 1,15 Normal 1,10 1,20 1,25 Sujo 1,25 1,50 1,75 6.3.3.6.2 Tabelas para determinao do coeficiente de utilizao A pedido, os fabricantes fornecem o fator de utilizao de suas luminrias. Como exemplo, na Tabela 27 est o fator de utilizao da luminria Philips TBS 912 com duas lmpadas fluorescentes de 32W. Tabela 27 Fator de utilizao. K 80 70 50 30 0 50 50 50 50 50 30 30 10 30 10 0 30 10 30 20 10 10 10 10 10 10 0 0,60 0,34 0,32 0,34 0,33 0,32 0,28 0,27 0,25 0,27 0,25 0,23 0,80 0,41 0,39 0,41 0,40 0,38 0,34 0,34 0,31 0,33 0,31 0,30 1,00 0,45 0,42 0,45 0,43 0,42 0,38 0,38 0,35 0,37 0,35 0,34 1,25 0,50 0,46 0,49 0,48 0,46 0,43 0,42 0,40 0,41 0,39 0,42 1,50 0,54 0,49 0,52 0,51 0,49 0,46 0,45 0,42 0,44 0,43 0,42 2,00 0,59 0,51 0,58 0,55 0,53 0,50 0,49 0,48 0,49 0,47 0,46 2,50 0,63 0,55 0,61 0,58 0,55 0,53 0,53 0,51 0,51 0,50 0,49 3,00 0,65 0,57 0,63 0,60 0,56 0,55 0,54 0,52 0,53 0,52 0,50 4,00 0,68 0,59 0,66 0,62 0,58 0,57 0,56 0,55 0,55 0,54 0,52 5,00 0,69 0,60 0,67 0,63 0,59 0,58 0,57 0,56 0,56 0,55 0,53 Nas Tabelas do fator de utilizao, os nmeros 751, 731, 711, etc. representam os ndices de reflexo de tetos, paredes e pisos, nessa ordem. Assim 751, por exemplo, seria na realidade 70% de reflexo do teto, 50% de reflexo de paredes e 10% de reflexo do piso. 6.3.3.6.3 Exemplo de aplicao Deseja-se iluminar um escritrio de 20 metros de comprimento por 10 metros de largura e 3,2 metros de p direito. Sabe-se que o teto branco e as paredes, na maior parte, brancas, com vidros a meia altura em uma delas, com piso de madeira clara. A altura do plano de trabalho de 0,80m. Deve-se utilizar lmpadas fluorescentes, luminria Philips TBS 912, duas lmpadas de 32W por luminria, com os fatores de utilizao apresentados na Tabela 27. O ambiente normal, com manuteno a cada 2500 horas (aproximadamente um ano). Soluo: adota-se o iluminamento do local em 500 lux. O ndice do local ser: Pela Tabela 26, a depreciao para 2500 horas e ambiente normal ser: d = 1,10 Pela Tabela 27, considerando teto 70%, parede 30% e piso 10%, obtm-se: hu = 0,53 Logo, o fluxo total ser: Luminrias fluorescentes com 2 lmpadas de 32W 2 x 2700lm/lmpada = 5400lm/luminria. Quantidade de luminrias = 207547/5400 = 38 luminrias. 6.3.3.6.4 Critrio de espaamento mnimo Para que a rea a ser iluminada tenha distribuio uniforme do fluxo total calculado, preciso respeitar uma distncia mxima entre luminrias fornecida pela expresso: eL 1,5h Sendo: eL distncia entre luminrias; h altura da luminria ao plano de trabalho. A distncia mxima entre luminrias e paredes ser: eLP 0,75h Sendo eLP a distncia entre as luminrias e as paredes. eL = 3,6m eLP = 1,8m A planta seguinte, 95, mostra a posio das luminrias. Nesta unidade conhecemos os transformadores e os principais tipos de motores de aplicao em uma planta industrial. As noes de luminotcnica foram necessrias para o correto dimensionamento da quantidade de lmpadas utilizadas em um ambiente interno. 6.5 Exerccios 1) Um transformador com ncleo de ferro funcionando numa linha de 120V possui 500 espiras no primrio e 100 espiras no secundrio. Calcule a tenso no secundrio. 2) Um transformador para campainha com 240 espiras no primrio e 30 espiras no secundrio retira 0,3A de uma linha de 120V. Calcule a corrente no secundrio. 3) Calcular a sada em kW de um transformador de 5kVA 2400/120V que alimenta a carga nominal com os seguintes fatores de potncia: a) 100%; b) 80%; c) 40%. 4) Um transformador monofsico tem 800 espiras primrias e 60 espiras secundrias. Se a poro secundria for conectada a uma resistncia de carga de 15, determine o valor da tenso primria necessria para produzir uma potncia de 22,5W na carga (presuma que o transformador livre de perdas). 5) Um transformador monofsico de 10kVA e 7200/120V tem uma resistncia no enrolamento do primrio de 12 e no enrolamento do secundrio de 0,0033. Calcule a perda no cobre: a) com carga mxima; b) com meia carga (5kVA); c) com uma carga de 2kVA. 6) Um teste de curto-circuito para avaliao das perdas no cobre com carga mxima d uma leitura de 175W no wattmetro. O transformador submetido ao teste um transformador abaixador de 240/24V que tem uma especificao para a corrente do secundrio com carga mxima de 60A. Se a resistncia do primrio for de 0,7, qual a resistncia do secundrio? 7) A tenso da linha do secundrio de um transformador D-D de 405V e a corrente da linha do secundrio de 35A. Se o transformador tiver uma relao de espiras de 5:1, calcular: a) a tenso de linha do primrio; b) a corrente de fase ou do enrolamento do secundrio; c) a corrente de linha do primrio; d) a corrente de fase ou do enrolamento do primrio. 8) Um sistema trifsico fornece 34,2A a uma tenso de linha de 208V para uma carga equilibrada com 89% de fator de potncia. Calcule a especificao do transformador em kVA. 9) Um motor trifsico de 6HP ligado em D funciona com um fator de potncia de 0,85 com uma eficincia de 80%. Se a tenso da linha for de 220V, qual a corrente de linha? 10) Qual a corrente nominal secundria de um transformador trifsico com potncia nominal de 112,5kVA trabalhando na tenso de 220V? 11) Calcular a potncia nominal de um motor que ser acoplado a uma bomba centrfuga com rendimento de 70% e vazo de 1 m3/s. A altura manomtrica e as perdas nas tubulaes de 10m e destina-se captao de gua potvel. 12) Determinar a potncia nominal de um motor de um elevador com capacidade para 05 pessoas (~350kg) e velocidade nominal de elevao de 1,0m/s. 13) Um motor eltrico est acoplado a um compressor de ar comprimido atravs de um redutor de velocidade na relao de 1:2, ou seja a velocidade do compressor de 875rpm e a do motor 1750rpm. Qual a potncia necessria do motor para acionar o compressor, sabendo-se que seu conjugado nominal de 20 N.m? 14) Sabe-se que um motor de 40cv e 4 polos est trabalhando em sua potncia nominal na tenso de 380V. Qual a corrente solicitada da rede de alimentao se nessas condies o fator de potncia de 85% e o rendimento de 91%? 15) Dimensionar a quantidade de luminrias fluorescentes 2x32W em uma sala de aula com largura de 8m, comprimento de 10m e p direito de 4m. Considerar as mesmas condies do exemplo e iluminamento do local de 300 lux. 6.5.1 Respostas dos exerccios 1) V2 = 24V. 2) I2 = 2,4A. 3) a) P1 = 5 kW; b) P2 = 4 kW; c) P3 = 2 kW. 4) Up = 244,9V. 5) a) PCU = 46,07W; b) PCU = 11,51W; c) PCU = 1,85W. 6) RS = 0,042. 7) a) U1 = 2025V; b) IF = 20,21A; c) I1 = 7A; d) IF = 4,05A. 8) S = 12,32kVA. 9) IL = 17,3A. 10) IL = 295,2A. 11) Pm = 140kW ou 200cv. 12) Pm = 4,9kW ou 7,5cv. 13) Pm = 1,93kW ou 3cv. 14) Inm = 57,83A. 15) 12 luminrias. 6.6 Estudos complementares 6.6.1 Saiba mais Como complemento do estudo de transformadores de energia, recomenda-se estudar o captulo 16 do livro: GUSSOW, Milton. Eletricidade Bsica. So Paulo: Makron Books do Brasil, 1997. Sobre motores eltricos, recomenda-se estudar o captulo 6 do livro: MAMEDE FILHO, Joo. Instalaes Eltricas Industriais. Rio de Janeiro: LTC Editora, 2007. Sobre luminotcnica: GUERRINI, Dlio Pereira. Iluminao: Teoria e Projeto. So Paulo: rica, 2007. 6.6.2 Outras sugestes de fontes de informao (links, revistas, filmes, etc.) https://pt.wikipedia.org/wiki/transformador https://pt.wikipedia.org/wiki/motor_eltrico https://pt.wikipedia.org/wiki/luminotecnica https://pt.wikipedia.org/wiki/lampada_fluorescente Glossrio Alta-tenso alternada: valor nominal maior que 69kV. Aterramento: ligao intencional com a terra, realizada por um condutor ou por um conjunto de condutores enterrados no solo, que constituem o eletrodo de aterramento. Este pode ser constitudo por uma simples haste vertical ou por um conjunto de hastes interligadas. Baixa tenso alternada: valor nominal maior que 50V e menor ou igual a 1000V. Capacidade de conduo de corrente de um condutor: a corrente mxima que pode ser por ele conduzida continuamente, em condies especificadas, sem que sua temperatura em regime permanente ultrapasse um valor predeterminado. Capacidade de interrupo: um valor de corrente de interrupo que o dispositivo capaz de interromper, sob uma tenso dada e em condies prescritas de emprego e funcionamento, dadas em normas individuais. A capacidade de interrupo era antigamente chamada de capacidade de ruptura, termo que no deve mais ser usado. O valor da capacidade de interrupo de particular importncia na indicao das caractersticas de disjuntores que so, por definio, dispositivos capazes de interromper correntes de curto-circuito, o que os demais dispositivos de manobra no fazem. Choque eltrico: o efeito patofisiolgico resultante da passagem de uma corrente eltrica, a chamada corrente de choque, atravs do corpo de uma pessoa ou de um animal. Eletrocusso o choque eltrico fatal. Circuito eltrico de uma instalao: conjunto de componentes da instalao alimentados a partir da mesma origem e protegidos contra sobrecorrentes pelos mesmos dispositivos de proteo. Os condutores podem eventualmente no possuir a mesma seo nominal ao longo do circuito, desde que os dispositivos de proteo sejam selecionados para proteger os condutores de menor seo. Comando: uma ao humana ou de dispositivo automtico que modifica o estado ou a condio de determinado equipamento. Contator: dispositivo de manobra (mecnico) de operao no manual, que tem uma nica posio de repouso e capaz de estabelecer (ligar), conduzir e interromper correntes em condies normais do circuito, inclusive sobrecargas de funcionamento previstas. Corrente alternada: aquela em que a direo da corrente varia ciclicamente, ou seja, a que utilizamos nas instalaes eltricas em frequncia de 60Hz. Corrente contnua: aquela em que a direo da corrente permanece constante, como no caso das baterias de carro, pilhas, etc. Corrente de curto-circuito: sobrecorrente que resulta de uma falha, de impedncia insignificante entre condutores energizados que apresentam uma diferena de potencial em funcionamento normal. Corrente de fuga: a corrente de conduo que, devido imperfeio na isolao, percorre um caminho diferente do previsto. Em uma instalao a corrente que flui para a terra ou para elementos condutivos estranhos instalao. Corrente de partida: valor eficaz da corrente absorvida pelo motor eltrico durante a partida, determinado por meio das caractersticas corrente-velocidade. Corrente de projeto: a corrente prevista para ser transportada pelo circuito durante seu funcionamento normal. Corrente nominal: corrente cujo valor especificado pelo fabricante do dispositivo. Essa corrente obtida quando da realizao dos ensaios normalizados. Curto-circuito: uma ligao intencional ou acidental entre dois ou mais pontos de um circuito atravs de uma impedncia desprezvel. Logo, um curto-circuito acidental uma falta direta. Disjuntor: dispositivo de manobra (mecnico) e de proteo, capaz de estabelecer (ligar), conduzir e interromper correntes em condies normais do circuito, assim como estabelecer, conduzir por tempo especificado e interromper correntes em condies anormais especificadas do circuito, tais como as de curto-circuito. Dispositivo eltrico: ligado a um circuito com o objetivo de desempenhar uma ou mais das seguintes funes: manobra, comando, proteo ou seccionamento. Falta eltrica: o contato ou arco acidental entre partes energizadas sob potenciais diferentes, entre partes energizadas e a massa, num circuito ou equipamento eltrico energizado. As faltas so causadas por falhas de isolamento entre as partes, podendo a impedncia entre elas ser considervel ou desprezvel (falta direta). Fusvel encapsulado: fusvel cujo elemento fusvel completamente encerrado num invlucro fechado, o qual capaz de impedir a formao de arco externo e a emisso de gases, chama ou partculas metlicas para o exterior quando da fuso do elemento fusvel, dentro dos limites de sua caracterstica nominal. Instalaes eltricas: conjunto de componentes eltricos, associados e com caractersticas coordenadas entre si, constitudo para utilizao de energia eltrica. Interruptor: chave seca de baixa tenso, de construo e caractersticas eltricas adequadas manobra de circuitos de iluminao em instalaes prediais, de aparelhos eletrodomsticos e luminrias, e aplicaes equivalentes. Isolao: o material isolante utilizado para impedir a circulao de corrente entre partes condutoras (uma isolao de PVC de um cabo de energia ). Isolamento: o conjunto das propriedades adquiridas por um corpo condutor, decorrentes de sua isolao (isolamento para 0,6/1kV, ou seja, tenso mxima admitida entre fase e neutro de 600V e entre fases de 1000V). Manobra: a mudana na cono eltrica de um circuito realizada manual ou automaticamente por dispositivo adequado e destinado a essa finalidade. Massa: (ou parte condutiva exposta) parte condutiva que pode ser tocada e que normalmente no est energizada, mas pode tornar-se em condies de falta, isto , de falha de isolamento. Um invlucro metlico de um equipamento eltrico o exemplo tpico de massa. Mdia tenso: valor nominal maior que 1000V e menor ou igual a 69kV. Proteo: a ao automtica provocada por dispositivos sensveis a determinadas condies anormais que ocorrem num circuito no sentido de evitar danos a pessoas e animais e/ou a um sistema ou equipamento eltrico. Quadros de distribuio: destinam-se a receber energia de uma ou mais alimentaes e distribu-la a um ou mais circuitos, podendo tambm desempenhar funes de proteo, seccionamento, comando e/ou medio. Trata-se de um conceito amplo que abrange quadros de luz, painis de fora, centros de medio e CCMs (Centros de Comandos de Motores), entre outros equipamentos. Rel (eltrico): dispositivo eletromagntico destinado a produzir modificaes sbitas e predeterminadas em um ou mais circuitos eltricos de sada, quando certas condies so satisfeitas no circuito de entrada que controlam o dispositivo. O rel no interrompe o circuito principal, mas sim faz atuar o dispositivo de manobra desse circuito principal. Resistncia de contato: resistncia eltrica entre duas superfcies de contato, unidas em condies especificadas. Esse valor de particular interesse entre peas de contato, onde se destaca o uso de metais de baixa resistncia de contato, que so normalmente produzidos por metais de baixo ndice de oxidao, ou seno ainda, quando duas peas condutoras so colocadas em contato fsico, passando a corrente eltrica de uma superfcie a outra. , por exemplo, o que acontece entre o encaixe de fusveis na base e a pea externa de contato do fusvel, que no pode ser fabricada com materiais que possam apresentar elevada resistncia de contato. Seccionador: dispositivo de manobra (mecnico) que assegura, na posio aberta, uma distncia de isolamento que satisfaz requisitos de segurana especificados. Um seccionador deve ser capaz de fechar ou abrir um circuito, ou quando a corrente estabelecida ou interrompida desprezvel, ou quando no se verifica uma variao significativa na tenso entre terminais de cada um dos seus polos. Um seccionador deve ser capaz tambm de conduzir correntes em condies normais de circuito, e tambm de conduzir, por tempo especificado, as correntes em condies anormais do circuito, tais como as de curto-circuito. Seccionamento: a ao de desligar completamente um equipamento ou circuito de outros equipamentos ou circuitos, provendo afastamentos adequados entre as partes que garantam condies de segurana especificadas. Sobrecarga: a parte da carga existente que excede a plena carga. Sobrecorrente: uma corrente cujo valor excede o valor nominal. um termo que engloba a sobrecarga e o curto-circuito. Referncias AGNCIA NACIONAL DE ENERGIA ELTRICA. Resoluo n 456 de 29 de dezembro de 2000. Disponvel em: https://www.aneel.gov.br/cedoc/res2000456.pdf. Acesso em: 14 dez. 2009. ______. Resoluo n 505, de 26 de novembro de 2001. Disponvel em: https://www.aneel.gov.br/cedoc/res2001505.pdf. Acesso em: 14 dez. 2009. ASSOCIAO BRASILEIRA DE NORMAS TCNICAS. NBR 5413: Iluminncia de interiores. Rio de Janeiro, 1992. ______. NBR 5410: Instalaes eltricas de baixa tenso. Rio de Janeiro, 2004. ______. NBR NM-280: Condutores de cabos isolados (IEC 60228, MOD). Rio de Janeiro, 2002. CAPELLI, Alexandre. Energia Eltrica para Sistemas Automticos da Produo. So Paulo: rica, 2007. CREDER, Hlio. Instalaes Eltricas. 15. ed. Rio de Janeiro: LTC, 2007. GUERRINI, Dlio Pereira. Iluminao: Teoria e Projeto. So Paulo: rica, 2007. GUSSOW, Milton. Eletricidade Bsica. So Paulo: Makron Books do Brasil, 1997. MAMEDE FILHO, Joo. Instalaes Eltricas Industriais. Rio de Janeiro: LTC Editora, 2007. PIRELLI. Manual Pirelli de Instalaes Eltricas. So Paulo: Pini, 2003. SHIMBO, Ioshiaqui; ZANIN, Maria. Eletricidade aplicada engenharia. So Carlos: EdUFSCar, 2008.

Monday, December 23, 2019

Government Test Study Guide - 9816 Words

TEST 3 STUDY GUIDE Student: ___________________________________________________________________________ 1. An audit of a government, conducted in accordance with generally accepted auditing standards (GAAS), includes A. A determination of efficiency and effectiveness. B. An examination of financial statements and underlying records for conformance with generally accepted accounting principles (GAAP). C. Tests for compliance with laws and regulations. D. Both B and C. 2. Audits of state and local governments may be performed by A. Independent CPAs. B. State audit agencies. C. Federal grantor agencies. D. All of the above. 3. In the auditor s report the financial statements on which the opinion is being expressed†¦show more content†¦B. Specifies the financial statements that the auditor has examined. C. States that all applicable accounting records were located and examined. *D. States that the examination was made in accordance with generally accepted auditing standards. 13. Government Auditing Standards (GAS) A. Establish the same scope as GAAS, but use wording appropriate to governmental entities instead of business organizations. B. Are set forth in the Federal Government Red Book. C. Establish more extensive standards than those found in GAAS. D. Establish standard wording of auditor s reports on governmental financial statements. 14. In which paragraph of the standard audit report does the auditor communicate to the user that certain combining fund information in the financial statements is not part of the basic financial statements, but that such information has been subjected to auditing procedures and, in his or her opinion, is fairly presented in all material respects in relation to the basic financial statements? A. Explanatory paragraph. B. Scope paragraph. C. Opening paragraph. D. Opinion paragraph. 15. An unqualified audit opinion rendered on a governmental unit s general purpose external financial statements means those statements A. Contain departures from GAAP that may make them misleading. B. Have been audited by an auditor with limited qualifications. C. Have been certified as free fromShow MoreRelatedA Christian Worldview Of Common Core Standards1350 Words   |  6 Pagesstudents must satisfy to continue to the next grade level or graduate from high school, no matter the state or school district. With a set of standards, consistency occurs because all students’ are held to meet the same standards skill in their academic studies. Why, then, are there concerns about the implementation and use of Common Core standards within the Christian community? First, to discern this question, an assessment of concerns Christians have related to the issue of the implementation, and useRead MoreThe Role of Macroeconomic Variables in the Financial Market673 Words   |  3 Pagesstock market will guide investors a long way to make better investment decision. 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Sunday, December 15, 2019

Moscow and Grand National Capital Free Essays

1. How did Paris become the model for the Grand National capital city? Paris becomes a model for the Grand National capital city because of its impressive public buildings, but mostly because of the work of Haussmann. Before the transformation, Paris was extremely congested, dirty and unpleasant. We will write a custom essay sample on Moscow and Grand National Capital or any similar topic only for you Order Now It was not unified and it had a lack of an effective transportation system. After Haussmann’s work, Paris becomes a model not only within France but also internationally because of its ideal environment and power. 2. How did Haussmann transform the city in just 2 decades? What was his contribution to the economic development of Paris? Haussmann carved out boulevards through the dense city, and opened up the city. He sold land and helped finance construction of new apartment buildings. Sewer systems and clean water was brought into the city, and space was made for both circulation traffic and armies for control. He even created public restrooms for the city, along with parks and the green areas of the city. Because of Haussmann’s urbanization, Paris grew economically because it became a place that reflected the power of the state and a place of exporting. . Why were the advantages of the Boulevard system? What were the disadvantages? The Boulevard system created a sense of orientation for the city as well as dividing it evenly. It made transportation easier and organized the city in a way. Some disadvantages might be that some people don’t like a repetitive look of the boulevards and that it could create traffic. 4. What was the social de mocratic vision for Paris after World War II? What were its advantages and what were its disadvantages? The social democratic vision for Paris after World War II included things such as national healthcare, free education systems, family allowances, pensions and unemployment insurance. The development of the subways and better transportation systems were some advantages of the changes after WWII. Paris also builds new modern buildings, such as the subsidized housing buildings. However even though the housing was built for the white French people, immigrants later started to move in which caused social problems and separations 5. What are some of the problems that have emerged in recent years in Paris as a result of neo-liberalization and market orientation? What are some of the benefits? Some of the problems that have emerged recently in Paris are that there is still separation between the social classes in the suburbs and then in the city. Paris has been moving towards the more global standard of international development as opposed to a highly regulated system. Consumer choice has become more popular and this allows businesses in Paris to have more freedom in the market. How to cite Moscow and Grand National Capital, Papers

Friday, December 6, 2019

The History Of The Atom Essay Research free essay sample

The History Of The Atom Essay, Research Paper THE HISTORY OF THE ATOM I am making my undertaking on the history of the atom. The atom was non merely thought up nightlong it took 1000s of old ages of research and experimenting, but was eventually finalized bty Dalton. Dalton was funny about many things but most of all he wondered how substances combined. He discovered that if you combine two substance, you got a new wholly different substance. After old ages of experimenting he came up with the jurisprudence of multiple proportions. From this Dalton concluded that all affair was made out of atoms which where indivisible.. They could be separated from each other or combined to organize new substances. Dalton besides assumed that if two elements combined to do one compound, that compound consisted of one atom of each component. All this was merely and premise by Dalton and he didn # 8217 ; Ts have scientific plenty equipment to verify this, subsequently scientists found out that some of his premises where incorrect. We will write a custom essay sample on The History Of The Atom Essay Research or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Like the atoms is ga s are non near together and the atom is far from indivisible. Dalton was known as the great experimenter and he based most of his work on other peoples decisions. But he did come up with many historical alterations theories like the atomic theory The different parts of the atom like the negatron, proton and neutron where found by different people. The negatron was discovered by scientists who # 8217 ; s chief involvements where electricity instead so chemistry. The proton was besides discovered by a group of scientists. A proton a individual unit of positive charge and is 1840 times heavier so an negatron. The proton is what remains when a H atom is stripped of an negatron. In 1932 Sir James Chadwick confirmed yet another subatomic atom, the neutron. Neutrons are subatomic atoms with no charge. But there mass is about equal to the proton. In decision an atom is made of three parts negatrons, protons and the neutron, all of these are every bit of import and a necessity to finish an atom.

Friday, November 29, 2019

Police States and Territories of India and Civil Services Examination free essay sample

IPS officers are recruited from the state police cadres and from the rigorous Civil Services Examination conducted by Union Public Service Commission every year. [7] Due to an ongoing shortage of police officers in India, the Ministry of Home Affairs proposed the creation of an Indian Police Service Limited Competitive Examination to be conducted by UPSC. [8] The Civil Services Examination has a three stage competitive selection process. At stage one, there is an objective type examination called the preliminary exam. This is a qualifying examination. It consists of a General Studies paper and an aptitude test. Only the candidates who pass this can appear for the Main Examination which consists of nine papers. Each candidate has to select an optional subject (one paper) and to take six General Studies papers, an Essay, an English language paper and a regional language paper. This is followed by an interview. After selection for the IPS, candidates are allocated to a cadre. We will write a custom essay sample on Police: States and Territories of India and Civil Services Examination or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page There is one cadre in each Indian state, with the exception of three joint cadres: Assam-Meghalaya, Manipur-Tripura, and Arunachal Pradesh-Goa-Mizoram-Union Territories (AGMUT). Two-thirds of the strength of every cadre is filled directly by IPS officers and the remaining are promoted from the respective states cadre officers. State Police Services (India) Indian Police Service and Law enforcement in India The State Police Services , simply known as State Police or SPS are police services under the control of respective state governments of the States and territories of India. The candidates selected for the SPS are usually posted as Deputy Superintendent of Police or Assistant Commissioner of Police once their probationary period ends. On prescribed satisfactory service in the SPS, the officers are nominated to the Indian Police Service. [1] The recruitment to this service is done by the respective State Governments, usually through State Public Service Commissions. The recruitment to SPS is also on the same pattern as that for IPS. The only difference is that the recruitment of SPS officers is made by the State Public Service Commission concerned through usually a combined competitive examination for State Civil Services which in the case of IPS is through the All-India Civil Services Examination conducted by Union Public Service Commission . However, in case of Union Territory, recruitment to SPS is also made by Union Public Service Commission through the Civil Services Examination as one of the Group-B Services. State police requirement in large States like Tamil Nadu, Uttar Pradesh, Madhya Pradesh, Bihar, Rajasthan, Maharashtra and Andhra Pradesh is greater as compared to the smaller States like Himachal Pradesh, Punjab, Haryana, Kerala, Sikkim and north-eastern States. After selection, which again is based upon the choice of service and merit of a candidate, selected candidates are required to undergo vigorous probationary training before getting posted as ACP or DSP.

Monday, November 25, 2019

Types of Fixed Price Contracts

Types of Fixed Price Contracts Fixed price contracts are a bit self-explanatory. You propose a single price to accomplish the work being sought. Once the project is complete the government customer pays you the agreed to price. Your cost to complete the work does not factor into how much you are paid. Types of Fixed Price Contracts Firm Fixed Price or FFP contracts have detailed requirements and a price for the work. The price is negotiated before the contract is finalized and does not vary even if the contractor needs to expend more or less resources than planned. Firm fixed price contracts require the contractor to manage the costs of the work in order to make a profit. If more work than planned is required then the contractor may lose money on the contract. Fixed Price Contract with Incentive Firm Target (FPIF) contract is a firm fixed price type contract (as compared to a cost reimbursable). The fee can vary depending on whether the contract comes in above or below planned cost. These contracts do contain a ceiling price to limit the government’s exposure to cost overruns. Fixed price with economic price adjustment contracts are fixed price contracts but they contain a provision to account for contingencies and changing costs. An example is the contract may contain an adjustment for an annual salary increase. Computing Fixed Price Fixed price contracts can be lucrative or cause a big loss to a company. Computing the proposed fixed price follows a similar to cost plus contract pricing. Study the request for proposals carefully determining the scope of work to be completed, labor categories of personnel needed and materials to be procured. A conservative approach to scoping the work (resulting a higher proposed cost) is preferred to offset the risk level of the work taking more effort and money than planned. However, if you propose too high a price you could lose the contract by not being competitive. Start computing the fixed price you will propose by creating a general work breakdown structure (WBS) for the project. Using the work breakdown structure you can estimate the number of labor hours by labor category needed to complete each phase of the project. Add in the materials, travel and other direct costs to the labor (priced at your labor rates) to get the proposed contract cost. Add fringe, overhead and general administrative rates to the appropriate costs to get the proposed project cost. Fee is then added to the planned cost to obtain the final fixed price you will propose. When deciding the fee take careful account of the amount of risk you have in the project not going at least as well as planned. Any risk of cost overruns should be factored into the fee. If you feel confident you can complete the work in the proposed costs then you can reduce your fee to be more competitive. For example, if the contract is to provide mowing services on base then you can estimate the amount of labor that will be required fairly accurately since the amount of mowing is well defined. If the contract is to develop a new, renewable fuel type for tanks then your risk of incurring more costs than planned is much greater. Fee rates can range from a couple of percent to 15% depending on the risk level. Note that the government and your competitors are also computing the project risk level and the related fee so be reasonable and realistic in your computations. Proposing the Fixed Price Here is where the couple of fixed price contracts come into play. When finalizing the price you will propose know the fee type required in the request for proposals. If an economic adjustment is allowed then you will need to propose what this percentage will be for each year of the contract. This is also called the escalation. Modify the computed fixed price to match the request for proposals and submit your winning proposal.

Thursday, November 21, 2019

First line Managment Essay Example | Topics and Well Written Essays - 2000 words

First line Managment - Essay Example Various studies have postulated that most executives and first line managers tend to be unaware of the various issues until they come out of hand and become a crisis that is very complicated and hard to resolve. Most of these can be avoided only through learning of the skills that enable the first line managers implement strategies for ongoing employee performance success though on the general it is necessary that companies or the groups have a game plan intact. Amongst the issues affecting performance include the working environments, the attitude of the employees towards the organization and the management in general, employees morale and most importantly the leadership style of the organization as regards to addressing employees grievances (Katzenbach and Smith 1993). 1.1 The manager’s role in identifying performance issues in a team Various studies have indicated that the act of holding employees accountable has proved to be the most effective in ensuring consistency in cl arity for performance expectations. This has been propagated as the most important element necessary for motivating work environment. For the managers, a career defining moment for most managers arises during instances where decisions are to be made as regards to addressing or opting to ignore the performance problems. The first step role the manager in identifying performance issues in a team includes devising parameters for performance. When a manager knows what is expected of his employees, then he is in a better position of determining whether the employees are able to reach their targets or not (Katzenbach and Smith 1993). When the manager sets the parameter s for a desirable behaviour or performance, the important thing to keep in view is on how poor performance or undesirable behaviour has effects on not only the organization, but also the employees and the customers. This plays a significant role in enabling the manager has a clear view of the desirable behaviour thus conseq uently placing him in a better position to analyze the employees’ performance. However, the manager should be very keen while analysing performance so that focus is not given to people issues but instead on their performance. Previous researches have shown that employing a subjective view during performance analysis might lead to confrontations between the manager and the employees. When performance problems are identified early, it plays a significant role in helping both the manager and the employee to reach an amicable solution to most of the problems. The main reason behind is on the manager who is entitled with the responsibility of leading the process and coming up with a solution that will initiate a positive development that will in the end benefit the employees and the organization as a whole. It should always never be forgotten that the main purpose of the process has always been to resolve the problem and be a valued employee. 1.2 How to evaluate individual and tea m performance and behaviours Performance evaluation has proved to be important in providing employers with an opportunity to establish their employees’ behaviour and determine their contribution to the organization which is critical to developing a powerful work team. Numerous

Wednesday, November 20, 2019

Industrial safety engineering (safety in work enviroment) Term Paper

Industrial safety engineering (safety in work enviroment) - Term Paper Example The government regulations on organizational safety and health (OSHA) demand that organizations maintain health and safety standards to ensure that their employees work in a minimum risk working environment. Failing to comply with these regulations attracts legal penalties that often are consequential to the organization. One of the superior ways of maintaining health and safety within an organization is educating the employees on the need for safety in a working environment and the minimum safety level requirements demanded in the organization. The management should thus be aware of how to induce effective training programs and how to align the employees towards the managerial objectives as far as health and safety is concerned within the work place. In reality, employee training is a key strategy to minimization of accidents, illnesses and death in work environments. Safety and health are two closely related terms yet very distinct in an industrial setting, hence the need to separa te them. Goetsch (2008), an industrial and safety professional, define safety in an organization as the ability to keep the employees away from the accident-causing situations that might cause damage, injury or even kill the employee. For instance, in an engineering firm, engineers have to operate electrically driven machines such as conveyor belts and grinders. When such machines come into contact with human beings, they are likely to cause accidents, some of which may be fatal. As such, it is crucial to safeguard the life of an employee from such machines to avoid unnecessary accidents in an organization. On the other hand, health maintenance refers to the ability to protect employees from disease causing environments. In industrial organizations, say a manufacturing plant, employees are exposed to smoke that may cause lung diseases. In this light, organizations have a core responsibility to avoid accidents and industrial related diseases within the employee population. In the Uni ted States the Organizational Safety and Health Act (OSHA) provides the standards that industrial organizations should comply with. The OSH Act provides that each business organization has a general duty to ensure that their employees are safe and free work environment hazards. Each year, the federal government, through their safety supervisors, conduct random safety inspections to observe whether organization have applied safety programs that comply with the OSHA standards. Over 50, 000 organizations are scrutinized and failure of compliance attracts fines, penalties and even temporary or permanent closure of organizations. One requirement of the OSH Act is that the employees have the â€Å"right-to-know† the health risks and conditions of the organization before they are assigned to their duties (United States Department of Labor, 2013). Resultantly, the act recommends that every organization implement an employee training program to ensure that employees are aware of the d angers that they risk in their working conditions. In essence, the implementing a health and safety training program is a compulsory requirement for every organization. Putting the legal side of organizational health and safety aside, a healthy working environment has many economic benefits to the organization. Burton (2008) points out to the high financial cost that organizations pay for failing to implement standard work environment conditions. First, unhealthy employees are likely to absent from

Monday, November 18, 2019

Tourism Trend Research Paper Example | Topics and Well Written Essays - 1000 words - 1

Tourism Trend - Research Paper Example However, in recent years, Florida has been able to develop a novel niche in tourism business called Cruise Tourism. The paper focuses on this new upcoming tourism trend that is making rapid progress in Florida now. In recent years, Florida has become a major tourist destination for variety of reasons. Florida theme parks such as Busch Gardens at Tampa Bay, Magic Kingdom, SeaWorld and Typhoon Lagoon at Orlando, Epcot in Disney World offers a unique experience to the tourists. Most of the tourists make a dedicated visit to some of the theme parks in Florida. Sun shines in Tampa Bay, for almost 360 days in a year, lures tourists to come over there. Tampas only water park, Adventure Island spread in 30 acres also supplements tourists’ experiences that they never want to forget (Henthorn, 2014). It is widely known that the US economy suffered immensely due to financial crisis during 2008-2009; accordingly, American spending on leisure travel also declined. In view of the depressed economic scenario, the tourism industry had to extend Cruise tourism is going to be a future trend in Florida as disposable incomes of the people increases. As per Satchell (2013), Florida is the only state that has some of the busiest cruise ports that include Port Everglades, PortMiami, and Port Canaveral. These three ports in Florida generate around 36 percent of the total $19.6 billion business estimated nationwide. It is important to note that in 2012 almost 60% of the cruise tourists began their voyage from Florida registering a 2.6 percent increase over 2011. In 2011, 13.5 million passengers took advantage of cruise ship vacation. Port Miami and Port Everglades have become top cruise ship organizing port in Florida. Port Everglades can boast of having the worlds largest cruise ship – a 225,000-ton Royal Caribbean mega ship called the Oasis of

Saturday, November 16, 2019

The Comprehensive And Flat Tax System Economics Essay

The Comprehensive And Flat Tax System Economics Essay The comprehensive income tax system in its ideal form enjoys a good reputation due to the structure which is based on the ability to pay principle. In general this principle points out that taxpayers with the same low level of income are taxed at the same rate and the ones with higher income are taxed heavier. 4 What makes this taxation system even more attractive to countries is that due to the equalization of capital and labour income, there is no interest in income shifting. This would be a so called tax arbitrage profit. That means it is more profitable to transform labour income into capital income and pay less taxes due to the different tax burden between labour and capital income. 4 2.1.2 Disadvantages of the comprehensive tax system 4 2.2 Flat Tax 5 2.2.1 Advantages of the flat tax system 5 2.2.2 Disadvantage of the flat tax system 5 Introduction In this Paper I will discuss the so called dual income tax. This system is considered to be one of the most auspicious taxation alternatives to existing taxation systems. This paper is build up in three big chapters. Firstly I will describe the environment of the actual taxation systems. On the one hand there is the system which is used in almost all European countries, the so called comprehensive personal income tax. And on the other hand the flat tax system, which can be found in the Este-European countries. Secondly I will point out more accurately the development of the Dual income tax system with its mayor advantages and disadvantages. Finally I will draw my personal conclusion to the dual income taxation system, its possibilities and future evolution. The major objectives of the taxation system in a country are clearly to earn as much tax affordable to obtain a solid and running state. This aim is collateral with the idea that everybody pays the amount of taxes one has to pay according to the ability to pay principle. Therefore leave as little as possible chances to cheat. Aligned with this aim the system should have two additional targets, fairness and simplicity. Fairness as a basis requires a system to be fair in the way that people with same income should be taxed in the same way. Simplicity should make taxpaying as easy and efficient as possible. This means in an ideal world there would be no exceptions or special tax treatments. Through globalization, easier and faster communication, many people start to find more and more possibilities to optimize their tax expenditures. This, for example is possible by investing in foreign countries causing harm to the own. As a consequence of missing tax income and loosing the possibility to keep up the states standards, a change in form of a reform is unavoidable. There are as many possibilities as there are countries on earth to reform a taxation system, by raising / lowering taxes or changing the whole system to become more attractive again. I will now present 3 exemplary systems. The comprehensive tax system The comprehensive income tax system also known by other synonyms as global income tax, unitary income tax or synthetic income tax is the most used taxation system in western European countries. It has got its name due to the fact that all income types are seen as a one and therefore are added together and taxed as one whole income. In Germany it was installed in the late 1950, and since then has been subjected of many changes from its original form. It was seen as the ideal tax system in Europe because in its original form it could align fully with the ability to pay principle and to both tasks of simplicity and fairness. This method is composed as a system which adds together all the taxpayers income (from labor, capital, rent and business) in a single measure and taxes it with a single progressive tax. Labour income is usually defined as income earned from activities as an employed individual. Capital income can take a variety of forms such as dividends, interests, income from real estate and to add further complications capital income is not always from a single organization, but can come from pension funds, life insurance companies, corporations, banks etc. Business income is income earned out of entrepreneurial activities. Normally the minimum subsistence level remains tax free. Income surpassing this minimum level is typically taxed on a progressive scale. For example in Germany it starts with a tax rate of 15% and ends up as high as 50%. Losses of one income type can be offset with gains from other sources. One tax rate will then be applied on the taxpayers total income (Synthetic taxation system). The idea behind the comprehensive tax scale is the assumption that someone with higher income is able not only in absolute but also in relative numbers to pay a higher share of his earnings to the government. This is by the majority of the most western cultures felt as a fair distribution of the tax burden also called tax fairness. This is the reason why there is a progressive tax table lying behind this structure. 2.1.1 Advantages of the comprehensive tax system The comprehensive income tax system in its ideal form enjoys a good reputation due to the structure which is based on the ability to pay principle. In general this principle points out that taxpayers with the same low level of income are taxed at the same rate and the ones with higher income are taxed heavier. What makes this taxation system even more attractive to countries is that due to the equalization of capital and labour income, there is no interest in income shifting. This would be a so called tax arbitrage profit. That means it is more profitable to transform labour income into capital income and pay less taxes due to the different tax burden between labour and capital income. 2.1.2 Disadvantages of the comprehensive tax system Although this taxation system seems to be clear and easy, when installed and worked with there are always specialties and exceptions. These treatments were implemented during many years and destroyed the purity of the comprehensive system. The special treatments came along with a high amount of administrative costs and a loss of the simplicity. The possibility of the capital income to become negative and be cleared with the labour income is just one example. This even can be seen in the tax declaration. Hardly anybody is capable to hand in a correct tax declaration without concerning a tax accountant. All countries with the comprehensive taxation system implied special tax treatment for a particular type of income (e.g. pensions, capital gains or lottery wins) to grant a tax relieve. But there is no systematic reason that for example lottery wins stay completely income tax free in Germany. Theoretically the State should levy taxes to generate income for funding its public tasks, as production public goods (eg. Schools, streets). It should not use the taxation system to channel the spending and investment behaviour of its citizens and tax payers. By doing so the systematic foundation, the justification and the acceptance of the comprehensive taxation system gets lost. The maximum rate should in no case surpass the 50% level because it is perceived as unfair from the tax payers when more than half of the money earned has to be transferred to the government. Some decades ago Sweden had a maximum tax rate of 90% with the consequence that people had no incentive anymore to generate income and tried to transfer their income stream to foreign countries. This effect is described as a so called Laffer curve. (It is used to illustrate the concept of  Taxable Income Elasticity  (that  taxable income  will change in response to changes in the rate of taxation). The curve is constructed by  thought experiment. First, the amount of tax revenue raised at the extreme tax rates of 0% and 100% is considered. It is clear that a 0% tax rate raises no revenue, but the Laffer curve hypothesis is that a 100% tax rate will also generate no revenue because at such a rate there is no longer any incentive for a rational taxpayer to earn any income, thus the revenue raised will be 100% of nothing. If both a 0% rate and 100% rate of taxation generate no revenue, it follows that  there must exist  a rate in between where tax revenue would be a maximu m. The Laffer curve is typically represented as a stylized graph which starts at 0% tax, zero revenue, rises to a maximum rate of revenue raised at an intermediate rate of taxation and then falls again to zero revenue at a 100% tax rate. One potential result of the Laffer curve is that increasing tax rates beyond a certain point will become counterproductive for raising further tax revenue because of diminishing returns.) 2.2 Flat Tax An alternative to the just described global tax system is the so called flat tax system. Herewith a flat proportional taxation for all net income types, capital, labour and other income is installed. This taxation system does not consider the taxpayers ability to pay taxes but sets a flat level for all income types. Some east European countries (Russia and Slovakia) have installed this taxation system. Russia replaced its progressive taxation system with a single flat tax rate of 13%. (Under a pure flat tax without deductions, companies could simply, every period, make a single payment to the government covering the flat tax liabilities of their employees and the taxes owed on their business income.  For example, suppose that in a given year, ACME earns a profit of 3 million, pays 2 million in salaries, and spends an added 1 million on other expenses the IRS deems to be taxable income, such as stock options, bonuses, and certain executive privileges. Given a flat rate of 15%, ACME would then owe the IRS (3M + 2M + 1M) x0.15 = 900,000. This payment would, in one fell swoop, settle the tax liabilities of ACMEs employees as well as taxes it owed by being a firm. Most employees throughout the economy would never need to interact with the revenue authorities, as all tax owed on wages, interest, dividends, royalties, etc. would be withheld at the source.) 2.2.1 Advantages of the flat tax system Aim of this taxation method is to gain a very simplified and transparent Tax system. Those who perceive the comprehensive taxation system with its progressive effects as unfair often are in favor of a flat tax system and claim it to be fairer to apply the same tax rate to all income sources and amounts. Because there is only one tax rate for all income, the administration costs are definitely lower and there is a balance between lower tax and lower cost. Additionally every taxpayer can easily calculate his taxes due which are usually not the case in the very complex progressive taxation system. After Russia installed the flat tax the real revenues from its Personal Income Tax rose by 25.2%, 24.6% and 15.2% in the first three years. 2.2.2 Disadvantage of the flat tax system Despite all the Pro arguments the problem of the flat taxation system is that hardly any country could keep it as a fully flat income tax system due to too many special tax treatments. Because everybody has to pay the same percentage of his total income the ability of pay performance is not considered or minded. This is the biggest contra argument of the flat tax system. It only seems to be a fair system but leaves only a small to no gap for a fair sharing of the tax burden. For many people it is not convincing why rich people should only pay as much as the poorer. And therefore it can hardly survive. In Addition to fairness another adverse point is the fact of the pure flat tax system that the government loses its power to steer the investing behavior. So if it would be of interest to support the private pension reserves, a deduction of capital income taxes on these payments is not possible. The Dual Income Tax Development of the Dual Income Tax (DIT) system The Dual Income Tax (DIT) is a combination of both of the recent presented tax systems. It is not a plain comprehensive system with a single progressive tax development or a flat tax with only a proportional tax, but a combination of both. It attempts to tax the personal capital income at a uniform (low) proportional tax while maintaining a (higher) progressive rate on the labour income. This taxation system was first introduced in Denmark 1987, other northern countries as Finland, Norway or Sweden followed. Until today the Norwegian system is seen as the most experienced one and is seen as very respected for the consistency with which it was implemented. Until today the system as such had to be subject to changes. Germany introduced the dual income tax system in 2009. Income was taxed according to the global tax system with the progressive taxation method whereas capital income was taxed deducted at source at a rate of 25 %, without exception. A unique aspect of the dual income tax system is the problem of how to distinguish labour from cpital income when the two are mixed together. A typical example is the business man who provides labour and capital with one business. These types of business are tempted to declare their gains of labour as capital income to avoid the higher progressive tax. The Finnish Government countered this problem with a division of dividends paid by unlisted companies into two components. Each half is then treated as either labour or capital income. Advantages of the Dual Income Tax system A country with lower capital income taxation becomes more attractive to foreign investors with a higher local capital income tax. This also has a positive effect on the attractivity of corporate investments. The further positive impact on the economic development is obvious. The generation of capital income tax by collecting it at its source, creates a higher possibility of tax payment in comparison to individual tax declaration. The incentive for capital exports and tax avoidance is reduced by a lower tax rate on capital income. It also diminishes the risk of tax evasion. Neutrality or simplicity means an equal treatment of capital income tax without personalized deductions or any other special handling. The main tax income which is raised for public finance can be brought up alone by the labour income tax. Disadvantages of the Dual Income Tax system Compared to the comprehensive taxation system, the dual income tax system has left the tax arbitrage is theory open. This is the (income shifting between higher-taxed labour income and lower-taxed capital income) to gain the difference between these two tax rates. Nevertheless it can reduce the incentives of capital exports to cheaper capital income neighbor countries. A common mentioned disadvantage of the dual income tax system it the so called problem of the double taxation. This taxation problem does not only exist in this taxation system but is characteristic for this system. Double taxations is a  taxation  principle  referring to income taxes that are paid twice on the same source of earned income. The problem herewith is the question when and how the best way is to tax capital income. The current scheme taxes the first time when earned as labour income. The taxpayer now has got two options, one would be to consume directly and have no further problems. In times of shrinking pension accounts many people prefer to save the money in forms of capital investment eg. shares, bondsà ¢Ã¢â€š ¬Ã‚ ¦ The real dual taxation problem now appears when these capital investments start to pay their interests or dividends. The argument behind this contra argument is that the capital to buy a investment was already taxed. 3.4 The dual income tax system in Germany and its impact In this section I will outline the general impact of the dual income tax system for Germany. In 2009 Germany installed the dual income tax system, keeping its progressive tax for labour income and changing its capital income to a flat taxation rate of 25%. The aim of this change from the comprehensive to the dual income tax system was diverse. First of all the German state wanted to stop the emigration of capital income to abroad countries. By implementing the 25% capital income tax Germany became more attractive to investors from foreign countries like Switzerland due to their higher taxation rate of 35%. As an additional positive effect Germany could hold its own capital income taxes in the country. If Germany would have kept it this way, a major relief of administration on the governmental and in the private sector would have been seen. The taxation on capital income would have been done directly as a deduction at source by the capital organization like banks or pension organizati on, as a standard anonymous fee. This as consequents would have meant a higher and more continuous reliable income flow and a real established flat tax for the capital income tax. But this was not the case, due to a minority, the people whose income tax is below the minimum subsistence level. These people claimed that the taxation of capital income at a rate of 25 % compared to their taxation level from the comprehensive taxation system which is lower than 25 % is unfair and not feasible. This is why the German model of the dual income taxation system could not be installed in its pure way, but with a major amount of adjustments. To give people living at the minimum subsistence level a chance to be taxed by their correct tax rate the capital income tax from banks had to be change from anonymous to personalized. On the one hand this meant a huge effort for banks to change their standard fee to a personalized fee tracing every capital movement on bank accounts. And on the other hand the anticipated administration reduction was not fulfilled. Taxpayers below the minimum subsistence level still have to create a tax declaration with a special request form. The problem with one special treatment is that it never comes alone, and so it is the same in Germany. As one of the few countries left which still has got the church tax as a fixed component of the tax system, the simple neglection as it is foreseen in the flat tax system for capital income was not possible, has to keep the church tax specialties in mind. The church fee in Germany is even dependant the federal state you life in and can vary between 8 9 %. In the old comprehensive system it was an easy task to retain the church taxes, due to the fact that this tax was retained as one tax from the hole income. With the new dual income tax system the labour income part of the church taxation is treated in the same way, by deduction of the entrepreneur. But to keep the principle of being a fair state, of course the capital income part of the church tax hast do be discharged as well. In principle this doesnt seem to be that big of a task but as the capital income is deduced by a capital company as banks even more information exchange is necessary. An aligned problem with the further information exchange is the question if a, lets stay with this example, bank should have the right or need to know ones religious orientation. As a diverse and social alert state Germany added another special treatment to the capital income taxation system. Church tax payers who have capital income then have two options. These would be on the one hand, provide your bank with you religious orientation who then would have to check each capital income payment on the capital income tax 25 % the 5,5 % solidarity surcharge tax and then depending on the Information the 8 9% church taxes. Or on the other hand declare your taxes through the normal tax declaration and being charged retroactively. Especially here you can see how the gap between fair and simple is a thin line which Is hard to meet. Either a taxation system is held flat and small so special tax treatments like church taxes or the regard of the Taxpayers below the minimum subsistence level are not concerned, or the system is tried to be held fair and a huge amount of administrative costs is installed. Another side effect of the dual income tax system is a well known problem that in general equity in personal businesses should be intensified. With the flat tax for capital income in Germany unfortunately the opposite is achieved. If an business owner gives a loan to his own company, his interest incomes are taxed according to the normal progressive tax rate (up to 50%). Giving the money to the bank instead of his own company allows him to make use of the favour of the flat tax rate. This lead to an unforeseen out pulls of money of personal businesses and reinvesting it in the capital market. Here again the question can be asked, which aim is the dual income taxation system following? If the main aim was to ensure a larger capital spend in the local market then the aim was fulfilled, but if a long term effect as supporting the increase of equity in private businesses was targeted then a clearly miss must be admitted. Anyhow the change to the dual income tax system for people with income greater than the minimum subsistence level presented a pleasant relief. As the capital amount of exemption in Germany is currently at 801 euros any capital income higher than 800 euros (20000 euros at an interest rate of 4 %) and a labour income tax higher than 25 % can make a noticeable saving. The problem in the Germany state is that it always tries to care for everybody, this is why a simplified system is hard or even impossible to install in Germany. Conclusion Because the fiscal functionality is not a static system there is in my opinion hardly any tax system which can last all too long. This is also the reason way a change in the German taxation system was needed badly. In my opinion the dual income tax system as it is thought of, in its original form without any special treatments is a very attractive and a realizable taxation system. With its aim to simplify the capital taxation by installing a flat tax which is deducted at the source is a major relief to administrative effort. As a critical discussion Point concerning the dual income tax system, is my question: can a taxation system be seen as fair if a person who owns an X amount of Euros (which is enough for him to live from the interests) should be allowed life a nice and easy life and only having to pay a low capital income Tax. on The contrary the taxpayer working 330 days paying the higher labour tax rate according to a comprehensive tax system? To illustrate this a little more clearly lets make a simple example with numbers. In the one hand we have the rich person with a capital of 5 million Euros who, with a interst rate of 4 %, has got a capital gain of 200.000 Euros. As this is seen as capital income it is only taxed at a low rate of 25 % so, the rich person can keep net 150.000 à ¢Ã¢â‚¬Å¡Ã‚ ¬. He can keep his life standard without working and exaggerated be on holiday the whole year. On the other Hand the person who works hard and has got a labour income of 200.000 Euros, hast to pay the top ta x rate of 50% and can only use the remaining 100.000 Euros. Summary Due to new European taxation systems many countries changed their view on taxation and had no other opportunity than to reform their own system. The dual income tax system, as a so called Scheduler Taxation System, because it has different schedules for the different income types originates from two different taxation models, these are a comprehensive, and flat tax taxation methods. The Main aims of the DIT system is the separation of capital and labour income. The dual income tax system underlays a proportional tax rate on all net income (e.g. Capital, interests, dividends, business income, income from real estate) and a progressive tax rate on labour income. With this background I took a look at the previous described Tax systems. Conclusions of the A major problem of the dual income tax in Germany is the sense of unfairness that is implemented by the general application of this tool. This lead to possible individual tax deductions. A general capital income tax, that is collected right where it is generated from the bank and is discharged anonymously is administrated easily. An advantage of a flat tax rate is ideally a simplification of the existing system. In Germany however in order to avoid unequal treatment individual tax deductions are possible. This counterstrikes the hoped effect. Vereinfachungs effect flat tax vereinfachung. A) keine abzà ¼hge b) depot gebà ¼hrenà ¢Ã¢â€š ¬Ã‚ ¦ etc. Æ’Â   wird einfacher aber immer grà ¶ber (ungerecht)Aufantragà ¢Ã¢â€š ¬Ã‚ ¦. Auch die capitalertrà ¤ge.