Projeto do eVectra - Início

Para começar o meu projeto de conversão do Vectra em elétrico vou partir de alguns parâmetros:

Qual a potência elétrica de pico que eu preciso?

Carros de passageiros requerem 10Kw/125Kg para uma aceleração aceitável.
Carros sport requerem 13Kw/125Kg.
Compactos e sub compactos tendem a 8Kw/125Kg.

10Kw/125Kg => 8Kw/100Kg
13Kw/125Kg => 10,4Kw/100Kg
8Kw/125Kg => 6,74Kw/100Kg.

Vou partir deste número: 8Kw/100Kg.

O vectra pesa 1200Kg.

Tirando as partes desnecessárias:

Motor + Descarga + ar condicionado/aquecedor + radiador + mangueiras + acessórios = 250kg

Total = 950kg

Incluindo as partes elétricas:
Motor => 10CV = 45Kg Siemens 1LA7131
Motor => 15CV = 63Kg Siemens 1LA7135
Inversor WEG 45a 380/480V = 22Kg
Baterias => 37 baterias 24Ah selada da Global = 7kg x 37 = 259kg

Motor + inversor + baterias + acessórios = 45 + 22 + 259 = 325kg

Total = 1276kg

Potencia de pico => 8Kw/100kg * 1276kg => ou 8Kw * 12,7Kg = 102Kw.

Portanto a potencia de pico que eu preciso para uma aceleração aceitável é 102Kw.

Qual a potência que eu tenho neste exato momento?

Não esquecer que esta usando um inversor.

Usando um inversor, consigo aumentar a potencia nominal do motor pois consigo elevar a rotação nominal para 2x, por exemplo.

Então agora ele trabalha com 3600 rpm, ao invés de 1800 rpm.
Calculo da potência nominal operando em 3600 rpm:

               C(kgfm) * N(rpm)                      4,09 * 1755
P(kw) = --------------------- =>  P(kw) = ---------------------- = 7,4 kw
                          974                                        974



               C(kgfm) * N(rpm)                      4,09 *  3555
P(kw) = --------------------- =>  P(kw) = ---------------------- = 15 kw
                            974                                       974


Potencia de pico:
Conjugado máximo => 340% = 4,09 * 3,4 = 14kgfm
Potencia máxima => (14kgfm * 3555 rpm) / 974 = 51kw

Portanto com o motor de 7,5kw a potencia máxima que consigo é 51kw. Pouco.
Com o motor de 11kw, a potencia vai pra 81kw. 

Esta potência ainda esta abaixo de uma conversão para um carro compacto, que seria de 86kw.


De acordo com informações do fabricante Siemens:

O motor é capaz de operar com rotação de até 2x a rotação nominal.
O motor de 2 polos tem a rotação nominal em 3600. Portanto 2x 3600 é 7200rpm.

A diferença entre o 2 polos e 4 polos é basicamente o numero de bobinas, pois os rotores são os mesmos.
Se o 2 polos consegue girar até 7200, numa boa, o de 4 polos também vai girar numa boa. Eles estão balanceados pra isso.

Então, se girarmos o Motor => 15CV = 63Kg Siemens 1LA7135, até 6000 rpm, a potência alcançada será:

Motor kw
11	C(kgfm)	6
	N(rpm)	6000
	P(kw)	36,96099	kw
	P(cv)	50,27933	cv
	Conjugado Máximo		22,2	kgfm
	Conjugado Máximo		370
	Conjugado Nominal		6
	RPM		6000
	Potencia Máxima		136,7556	Kw

A potência de pico desenvolvida fará o Vectra acelerar em alta performance.

Como elevar a rotação do motor de 1800 rpm para 6000 rpm?

Aumentando a frequência de saída do inversor mantendo o mesmo torque.
Pra fazer isso devemos utilizar a função escalar que os inversores tem.
Fazendo isso verificamos que pra uma tensão nominal de 220v e 60hz, precisaríamos chegar em 200hz.
Sendo 200hz o limite para as perdas por magnetismo começarem a causar problemas.
Mas em 200hz, precisaríamos alimentar o motor com 734v.
Devemos então diminuir a tensão nominal do motor em 60hz para 127v. Veja tabela:

Hz	Tensão	rpm	V/F
60	127	1800	2
70	148	2100	2
80	169	2400	2
90	191	2700	2
100	212	3000	2
110	233	3300	2
120	254	3600	2
130	275	3900	2
140	296	4200	2
150	318	4500	2
160	339	4800	2
170	360	5100	2
180	381	5400	2
190	402	5700	2
200	423	6000	2

A corrente de partida de um MIT é 7x, mas isso é quando o motor está operando em linha. Direto na rede trifásica.

Quando operamos através de inversor, podemos controlar está corrente de partida. De modo que o que nos interessa é apenas a corrente em torque máximo.

Corrente de partida é a mesma do rotor bloqueado = 7,6x. Ip / In = 7,6.

Não se usa esta corrente para os cálculos, pois esta corrente provoca uma pancada na máquina e danifica os equipamentos. Além de ser uma corrente altíssima. Usa-se partir o motor com corrente menor.

Como usaremos inversor, a corrente de partida é a que quisermos, desde que fornecemos esta corrente ao motor através da aceleração.
Devemos portanto controlar a tensão de partida e consequentemente controlaremos a corrente e o torque:

Nesta curva podemos perceber que quando partimos com tensão 85% menor que a nominal, a corrente também não atinge seu valor de partida, consequentemente o torque também é reduzido.
Outra observação muito importante é com relação a corrente no torque máximo. Podemos ver que o torque cresce bem rápido, quando a rotação se encontra entre 80% e 98% da rotação nominal. E que o torque máximo se dá em torno de 90% e 95% da rotação nominal e neste instante a corrente está em 3x In.
Neste gráfico vemos que o torque máximo atinge 2x o conjugado nominal quando se encontra em 93% da RPM nominal e a corrente atinge um valor um pouco menor que 3x.
Se o MIT estiver em rotação síncrona a corrente será a nominal e o torque será o menor possível, mas ao menor sinal de aumento de carga que force o eixo a diminuir a velocidade de rotação, automaticamente a corrente se elevará em até 3x e o conjugado atingira 2x o conjugado nominal para este motor.

Partindo desta análise que a corrente aumentará no máximo 3x In para se atingir o conjugado máximo, adotarei o inversor que seja capaz de fornecer 3x In.

O MIT de 7,5Kw/10CV da Siemens que eu consegui tem In de 15,5A em 380V. 3x 15,5A = 46,5A.
O modelo de inversor da WEG será o 30/380-480.
De acordo com tabelas do inversor, este modelo fornece Imax de 45A.

O MIT de 11Kw/15CV da Siemens que eu preciso tem In de 22,7 em 380V. 3x 22,7A = 68,1A.
O modelo de inversor da WEG será o 45/380-480.
De acordo com tabelas do inversor, este modelo fornece Imax de 68A.


Vou procurar pelo modelo 45/380-480. Pois já vou estar preparando o veículo para o motor de 11Kw.

Não podemos esquecer que os inversores aguentam até 1,5 a potencia máxima por curto intervalo de tempo.

De acordo com a tabela, a potência dissipada nominal é de 0,90 Kw = 900W. 900/440 = 2,045A.
De acordo com as tabelas a corrente nominal de entrada do inversor 45/380-480 utilizando um motor de 7,5Kw será de 19 + 2 = 21A => Portanto, em teoria o consumo seria 21A. Baterias com capacidade de carga de 24A.  Estariam de bom tamanho. Para 1h de serviço.

Abaixando o peso do Vectra

Tire: 
12kg - Estepe 
3kg - Macaco 
68kg - Banco Traseiro e Passageiro 
3kg - Acabamento das Portas 
5kg - Desmonte o Painel 
4kg - Forração 
36kg - Sistema do Ar quente + Ar Condicionado 
12kg - Vidros c/ Maquina - Traseiros "Tampe c/ Plastico ou Fibra" 
25kg - Capô - Troque por um de Fibra 
= 168kg 

Abaixando a tensão do motor trifásico

Como primeira postagem, gostaria de colocar o que me reanimou a prosseguir com a conversão aqui no Brasil.
Foi este post abaixo, explicando como religar um motor de indução AC para conseguir uma potência de pelo menos 6x maior do que a original.

O post é deste forum aqui:

from: http://forums.aeva.asn.au/forums/forum_posts.asp?TID=1237&PN=9&title=changing-an-induction-motor-voltage

E o autor é este cara aqui:

coulomb  

Ok, I'll have a go at this. 

1. For ages, we've known that it is possible to push an induction motor past its nominal power point, simply by loading it more than its nominal power. The controller can increase the mechanical load by increasing the "slip". We call this "overcurrenting" the motor, because it requires more current than the windings can continuously handle. This is OK as long as there is sufficient cooling, and the motor isn't overcurrented for more than short bursts, e.g. for acceleration in an EV. This is also OK for short hill climbing, but not long hills or mountains; they can overload the motor for long periods of time and cause overheating. This is not the subject of this thread; I put it here for context. 

2. The other way of getting more power out of a motor is by increasing the voltage. But most motors are 415 V already; without more than a 600 V pack and a very expensive controller we can't increase the voltage, and it's probably not safe to go beyond 440 VAC since the motor insulation isn't designed for it. But we can reconfigure the motor so that its nominalvoltage is less than 415 V, and put up to 415 VAC on it anyway. We can reconfigure the motor's nominal voltage in several ways: 

2a. Rewinding. That basically means chop out the old wire, and start again. This is beyond most people's garages. A figure of AU$800 has been mentioned for this process. 

2b. Reconnecting. If we are lucky enough to have the windings split into two or more pieces, and there are 9 or more connectors in the terminal box, then we can rearrange the connections to achieve lower nominal voltage. Some American motors are like this, so you might be able to configure for 220 VAC as well as 440 VAC, for example. One special case: some Australian motors can be found (or specified if you are buying new) with 415 V star windings. This is the "S" option for ABB motors. In that case, with the standard 6 wire terminal box, you can rejumper your motor to be 230 V delta. This is a fixed voltage reduction of sqrt(3) = 1.732. 

2c. Rewiring. That's what this thread is about. Some, but as we have seen not all, motors have windings in series that can fairly easily (in the home garage) be wired in parallel. Depending on how many similar windings are in series, we might even have a choice as to what factor the motor becomes undervoltaged: half, third, perhaps even smaller fractions. The big advantage of this method over 2a is that it's cheap and easy, but if the motor isn't suitable for rewiring, then 2a is the only real option for changing the motor's nominal voltage. 2b is even better, since we don't even have to open the motor, but few motors available to Australian converters are likely to have this option. 

3. So changing the nominal voltage of the motor is a prerequisite to overvoltaging. But it has some important effects. Firstly, the nominal power of the motor is increased. This is because when we increase the voltage beyond the nominal voltage, the speed will increase beyond the nominal speed, but the motor current remains largely unchanged. The bulk of the motor current depends on the load torque, so if we keep the load torque the same and double the voltage, we double the speed and hence double the power. 

4. But we can't leave the motor at double voltage and nominal torque continuously, because the iron losses increase dramatically with speed. That's why I said "largely unchanged" in point 3 above. But we can reduce the continuous torque by a certain amount to compensate for this. Weber calculated that the derated increase in continuous power is about the 0.7th power of the voltage increase. 2^0.7 ~= 1.6, which is 80% of 2, so we can derate the continuous torque by 20%. 
With double the speed and 80% of the torque, we have 160% of the original nominal power, and the motor can run at this power level continuously. Note: this is theory right now, yet to be tested. 

5. Let's consider an example to make this clear. In a Barina class vehicle, it might take 16 kW to maintain 100 km/h on the level. Without rewiring, you'd need an 18.5 kW nominal motor to be able to drive at freeway speeds for as long as your battery lasts. You'd probably get away with a 15 kW motor unless you had a really big battery pack, as the thermal time constant for a motor is about an hour, so it might take an hour for the 15 kW motor to overheat, by which time you've travelled 100 km and drained your pack. Now let's consider an 11 kW nominal motor, 415 V delta (can't get the easy 1.732 voltage reduction). Let's say its breakdown torque is a little over 3x nominal, so this is a 33 kW peak power motor. Let's say this motor is suitable for voltage halving. Let's also say it's a 4-pole motor, so its nominal speed is a little under 1500 RPM; for simplicity I'll call this 1500 RPM. By rewiring, you end up with a 208V 11 kW 1500 RPM motor. So we can tell the controller that this is now a 160% * 11 kW = 17.5 kW continuous motor, with a nominal speed of 3000 RPM. We can still overcurrent this motor by 3x (note: this is theory again). So at peak we should be able to get 3x original nominal torque at 2x original nominal speed, for an overall 6x peak power, or 11 * 3 * 2 = 66 kW. That's 32% more than the peak power of a Barina class ICE. 

But there's more: at 1500 RPM and below, we know that it can take 100% of its original continuous torque. The original motor's nominal torque dropped away at 1/f, so at 3000 RPM, the original motor had half nominal torque. After rewiring, it has 3/4 nominal torque at 3000 RPM. 

In summary: by rewiring for half voltage, we've gone from an 11 kW continuous 33 kW peak motor, to a 17.5 kW continuous, 66 kW peak motor. You might notice that I've chosen the numbers so that a rather weak motor, not capable of continuous freeway driving, and only 2/3 of the original ICE's peak power becomes capable of continuous freeway driving, and more peak power than the original ICE. That's an impressive improvement. We can only get away with this because and assuming that EV power demand is very peaky, with the peaks of short duration. 

How about considering what a 5hp ac motor which would normally be too small to sufficiently move a light vehicle be applying the above parallel vs series process?

Assuming by light you mean Barina class, you have the right idea, but the effect is not quite as dramatic as that. 

I'm fairly certain that the best voltage reduction we can hope for in EV sized induction motors would be half. But let's suppose we found a 5 kW motor that we could quarter the voltage of. From Weber's equation, we can expect about a 4^0.7 ~= 2.6x improvement in continuous power, and a 4x * 3x = 12x (!) improvement in peak power. So that's 13 kW continuous, and 60 kW peak. The peak power is at 4x the original nominal speed, so it should be a 4-pole motor, so that peak power is a little under 6000 RPM. (Extra balancing will likely be required, and possibly also better bearings.) Whether this would be satisfactory or not depends on your needs. 13 kW will not allow continuous freeway speeds, but it might allow continuous 80 km/h driving, and certainly continuous stop/start traffic. So that would be great; but: can we quarter the nominal voltage? 

Edits: use Weber's equation rather than chart; redid examples with this.
Continuing the big summary post above. Now we know that rewiring for lower voltage is good; what are the ways that this can be done? 

6. Many EV-sized vehicles come with "concentric" windings. I'm pretty sure that this is an attempt to distribute the magnetic flux in a more sinusoidal way. Unfortunately for rewirers, this means that a lot of opportunity for rewiring is removed. In a two pole motor, these windings can span 18, 16, and 14 slots in a typical 36-slot motor. For 4-pole, the numbers are typically 12, 10, and 8 slots. The resistance of the windings varies according to how many slots are spanned, so paralleling these concentric windings would probably cause problems with circulating currents. This is not yet confirmed. If it was safe to do, and could readily be done, this would allow a 1/3 voltage rewire. However, when I attempted to do this, I found that it was very difficult to find the places where the wire changed from one pair of slots to the next (the "crossover wires"). This problem is made much worse by the way the windings are arranged; generally, one pair of windings is very much behind the other overlapping pairs, and this makes it virtually impossible to find all the crossover wires. 

7. Despite that, most EV sized induction motors have two real poles. This is obvious for 2-pole motors, and at least some 4-pole motors use "consequent poles", meaning that the direction that the windings are connected forces these "virtual" or "consequent" poles to form. In the case of my 7.5 kW motor, these poles were actually connected in parallel. No voltage reducing rewiring is possible in cases like that. We need to find windings that are in series, and connect them in parallel instead. The paralleling in this case was obvious from the "Y" connections around the outside of the windings. 

In the case of Electrocycle's 11kW motor, we didn't see any such Y points, so we assume that the poles are wired in series. In this case, it should be possible to do a fairly easy half voltage rewiring, and the varnished cambric sleeving should pretty much highlight the places were wires need to be cut and reconnected. 

8. To prove that this works, we need to know the load we can put on a motor, both to see what load it can handle continuously without overheating, and what peak load it can handle. Two main ways of doing this are setting up a dynamometer (like ACmotor is doing), and driving an actual vehicle and logging as much data as possible. The latter has the problem that when you do get to the point where you've overloaded the motor, you might cause a traffic hazard, and/or have to pull over to the side of the road and let things cool off for a long time. The problem with the former is that things like thermal constants may vary greatly from 2kW class motors to 11 kW class motors.