April 22 2008

Components and Performance Simulation.

The motors I will be using are Siemens 1PV5135-4WS28. These are fine liquid cooled induction machines - very rugged and reliable. I already mentioned that power inverters will be EVI-200 - new development of Dutch company EVISOL (Dec. 2009 update: actually inverter choice will be different or heavily modified. Here is why). The power lithium polymer battery is made by Kokam. This battery will be taken care of by a BMS (Battery Management System) developed by Metric Mind Engineering. All power components are liquid cooled - motors inverters and battery.

I wanted to estimate vehicle performance with selected components. Here are few more numbers needed for the initial analysis. Most numbers I know about; others (like diff ratio) is my best guess and subject to fine tuning later.

More details about the battery (single cell data - from manufacturer's spec sheet):

The battery model is being worked on and once settled, performance analysis results will be posted here.

Particular projections I am interested to find out are:

- Acceleration rate (vehicle speed as a function of time)
- Motor shaft rotation speed as a function of time
- Battery power as a function of vehicle speed
- Energy consumption as a function of speed
- Driving range
- Max gradeability with chosen gearing.

Any other by-products of simulation gaining useful insight and allowing to optimize the setup while it's still on the paper. For instance I can optimize for sure the final differential ratio based on simulation only, before actually going shopping for diffs.

Here are photos of the power components

Siemens 1PV5135-4WS28 motor
EVISOL EVI-200 power inverter
Kokam 40Ah lithium polymer power cell

The vehicle performance will be simulated in Matlab environment. The vehicle model is developed by EVISOL, I only wrote scripts to format outputs and generate plots of the parameters I'm interested in.

Top level vehicle software model. This diagram may not tell you much, but you will see the below on this page results of simulation done with this model

At high inverter DC input voltage maximum motor torque is 400 Nm (conservative value). For simulation purposes I will "cheat" - rather than modifying software to deal with two 400Nm motors it will think I have one twice as heavy 800 Nm capable motor. Mathematically this is correct because 800Nm gets down to the  wheels in either case, and I suspect if I'd physically have such a motor, the performance outcome after installing it would be the same as with two 400 Nm motors.

I did two types of simulation: max acceleration performance (wide open throttle), and civil driving - established standard US06 driving cycle. US06 represents quite aggressive drive pattern to evaluate vehicle performance on typical urban route: initial maneuvering on 35 MPH streets toward a hypothetical freeway, high speed ride toward destination at about 100-120 km/h (70...75 MPH), coming off freeway and maneuvering at low speeds in stop-and-go traffic again. This is 10 min long ride and can be repeated over and over in the software to determine max. vehicle range. Plots below represent outcome of initial simulation. Later, when I will have more accurate Kokam battery model, I might fine tune input parameters to get more accurate results, but these are close to reality too. For simulator I restricted motors top RPM to 8,000 (yes, motors can spin faster, but I don't plan them to, and also I don't plan to drive much above 120 km/h. So I chose final differential ratio to keep motors RPM in comfortable zone for continuous driving while maximizing torque at the wheels. This number is 5,500 to 6,000 RPM. The diff ratio available in this range (5.7:1) the motor RPM will be 5,700 @120km/h, about to be ideal. So this is the ratio I used in the simulation.

Let's examine one plot at the time. First - top acceleration test:

Vehicle speed vs. time. Yellow line is speed demand by throttle and pink line is what the vehicle can actually do. It will take 4.3 sec to reach 96km/h (60 MPH) and 6.2 sec to 120 km/h (75 MPH). On 12th second the motors RPM will reach 8,000 and the speed at that time will be 168 km/h (105 MPH). Plenty for how I drive.

Zoomed in part of the plot above. Low speed performance. Not that I'm a fan of braking my neck snapping it every time traffic light turns green, but I curious to find out what the car is capable of. Looks like about 1.6 sec to 40 km/h (25 MPH), 2.2 sec to 56 km/h (35 mph) and confirming 4.3 sec to 96 km/h (60 mph).

Battery voltage sagging under acceleration. This plot will be updated later as it is based on crude battery model. Still gives an overall picture of what's expected. Why does steep voltage drop ends after 2 seconds of acceleration? The answer will be apparent after looking at the battery current plot. Read on...

Battery current (yellow line) and motor current (pink line) plot. So the battery voltage on the plot above stops sagging after 2 seconds because linear battery current increase stops and levels off. At any time the voltage sag is comprised of two components: voltage drop on the internal resistance and terminal voltage reduction due to reduced SOC (depletion). In a short time SOC drop is noticeable but minimal. Why the battery current increases linearly in first 2 seconds while the motor current peaks at its maximum limit right away? Because at near zero RPM voltage on the motor is near zero as well, but the battery voltage is high, near nominal. The battery supplies only as much power as motor consumes (+ some to cover system losses). The power at ~zero RPM is also ~zero:  from electrical point of view (~0 volts * peak current is 0 Watts) and from mechanical point of view: the power motor makes is torque*rotation speed (which is zero), and the torque is directly proportional to the motor current. At low speeds there is no back EMF yet, so the motor current (and so motor torque) is usually imposed by the inverter. So with fixed motor current (280A rms in this case) motor power is increasing with its RPM increase and the battery power follows. Because battery power is voltage*current and voltage is relatively constant, current keeps increasing as yellow line indicates. So in the motor loop current stay constant and voltage increasing and so is power. In the battery loop voltage is forced constant (by the battery) and current is increasing and so is power. Thus inverter acts as "impedance transformer" constantly matching input to output similar to a conventional isolated autotransformer.   In other words inverter continuously trades increasing motor voltage and current with battery voltage and current, acting like "electronic transmission" that trades torque for speed. This means after 2 seconds motor power, which is product of motor voltage and motor current must be constant too (neglecting losses for the moment). Is it so?

Yes. But since this constant power is torque*rotation speed and speed increases, torque must be decreasing for the power to be maintained. Check this out:

Motor RPM plot.

  Ta-Daaa... The motor torque is decreasing at exactly the rate of rotation speed increase so the product of the two (motor power) stay constant. Of course because there is fixed gear transferring torque to the wheel, the motor RPM plot exactly match vehicle speed plot:  But why this change happens after 2 seconds? The answer has to do with current limit of the inverter and motor back EMF. Any motor as it rotates also acts as a generator producing voltage on its windings. This is back EMF (Electro-Magnetic Force). Because inverter acts like "transformer" motor voltage, multiplied by "transformer ratio" reflects to the battery side and keep increasing. When motor power increases to the maximum (typically limited by inverter power) inverter comes out of current limit: current will drop because motor voltage keeps increasing. That is what happens after 2 seconds of the run - the motor transitioned from constant torque (current limit) mode to the constant power (power limit) mode. Inverter has reached its maximum power and so the motor. Looking at motor RPM plot I can see that in 2 seconds motor reach about 2,500 RPM. This is the RPM at which max power is achieved and the reason that torque must keep dropping if this power is going to be maintained. Also, the motor acts as a motor only so long as you cram amps in its winding, fighting back EMF motor is producing. To put amps in (so generate torque) the current must flow from the battery to the motor, meaning battery voltage has to stay above back EMF. But at some point, when motor spins too fast, its back EMF reach battery voltage, which means no current flow is possible (there is no difference in voltages). So the torque becomes zero. It is not visible on this plot though, perhaps because the motor must spin far faster than 8,000 RPM in order to generate BEMF exceeding 750+ VDC on the battery side. If I lower battery voltage, this additional cliff will move toward lower RPM and you would see horizontal power line start dropping down. In my case I'm glad this RPM point is beyond maximum 8,000 RPM I'll be driving with.

How is the battery doing after 12 seconds of abuse? Let's examine SOC plot.

Nothing exciting. SOC just naturally drops as the battery being discharged. In 12 sec it is a tad above 0.97, or 3%. With about 32Ah estimated battery capacity (at this rate), 0.96Ah is spent during this acceleration. This figure can be cross-checked this way: 0.92Ah is 0.92*60=~55 Amp-minutes or 55*5=275 Amp-"12seconds" (60sec/5=12 sec). So the average current consumption over the course of acceleration must be 275A. Looking at the battery current again, I can see that during first 2 seconds current ramped up from 0 to 300A (150A average) and next 10 seconds stayed at 300A figure. Amazing, but integrated average over 12 sec. turns out to be exactly 275A. Pretty accurate.

How much energy did I spend and how far have I gone in 12 seconds?

The answer is here. Distance is ~0.07*5=0.35km (350m or 1,148 feet) and energy spent is 0.65 kWh. That translates to horrible efficiency of 1.857 kWh/km (2.97kWh/mile)! Don't drive like that, not to mention my tiny 32*740=23,680 kWh battery will last exactly 23,680/1.857=12.75km (7,97 miles) of that kind of driving, reaching 0% SOC. Its cycle life is going to be miserable too.

Let see what civil way of driving is going to look like. Applying US06 driving pattern.

The vehicle speed vs. time in 10 min run look like this.

With fixed ratio of course motor RPM plot is exact scaled copy of the vehicle speed plot.

It is interesting to examine closely section of the drive demand and actual vehicle acceleration. Inaccuracy of simulator of course contributes here, but I can see that while actual vehicle acceleration (pink line) closely follows throttle demand (yellow line), deceleration has some inertia and deceleration rate is lagging behind demand. I know that regen can be set as aggressive as acceleration, so I'm sure in simulator regen is greatly reduced (negative battery current limited) and the vehicle does not slow down exactly as speed demand decrease - it coasts. Will fix that.

Battery voltage. While swings a lot (when regen is applied it rises above 800V), you can see overall trend down as the battery being depleted.
Battery current (yellow) and motor current (pink). I can see that at 110km/h (~70 MPH) battery current is going to average about 25A. Nice. 13mm² (AWG gauge 6) wire as battery "cable" should do it easy.
Motor power. I can see I'm averaging about 20kW at high speeds. It's big car (large frontal area), and being station wagon, I can see that Cd of 0.31 is not being something phenomenal. Thus power consumption
Motor torque. Regen is definitely suppressed somewhere...
Energy consumption. About 2700Wh in 10 min.
Battery power. Very closely follows motor power, which tells me that the system efficiency is high (losses are low).
Distance travelled and energy spent. With average speed about 80km/h in 10 min I travel 80/6=13.3km. Energy spent is 2700Wh, so the efficiency is 2700Wh/13.3km=203Wh/km (324 Wh/mile). Well, with this kind of driving, drop dead range is about 28/0.203=137km (85 miles). OK for now, but my next battery should be 70Ah, not 40Ah...
SOC over the course of first 10 min US06 cycle driving. 7% out. Every next 10 min run like this will deplete battery by more than 7% because voltage declines, but the power demand remains the same, so the current (and so Ah) rate consumption increases.
This is one of most interesting and useful plots - battery power, energy consumption and efficiency as a function of the vehicle speed. I can see these numbers for most common speeds:

- 56km/h (35MPH): 4kW 80Wh/km (128Wh/mile)
- 88km/h (55 MPH): 11.5kW 128 Wh/km (204 Wh/mile)
-120km/h (75 MPH): 24kW and 200Wh/km (320Wh/mile).

What's apparent here: 32km/h (20MPH) increment from 56km/h to 88km/h is far easier than the same increment from 88km/h to 120 km/h. First extra 32km/h cost you 7.5kW more and 76Wh/km more, but another 32km/h increment will come at the expense of 12.5kW more and 116 Wh/km more respectively.

Turn out the power requirement is a cube function of the speed - to double the speed requires 8 times more power!

The moral of the story - slower you drive - further you will get. Especially if you slow down at high speeds. Every
5km/h slower at near 100km/h will save you far more energy than the same 5km/h slowing at about 50km/h speeds.

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