# Solar Car Calculations



## SummerTime244 (Nov 5, 2015)

Hello everyone,

I am part of a University team and we're building a solar car to participate in Solar Races.

I have a few basic questions about Power calculations : 

So here's the current setup (without the red lines): 










The current setup is PV+MPPT -> Battery -> Motor.

But as you see, the battery has less power than the motor needs, so it will discharge very fast. Additionnaly, the race regulations impose that we start with a Half-empty battery, which means the Battery voltage will be 24V instead of 48 at the start of the race. 

So my question is : Can i wire the PV+MPPT directly to the motor?
This way i can both power the motor and charge the battery. In this case, the battery will be much slower to charge since the Current flowing from PV is split between motor and Battery. 

Another thing i'm failing to comprehend is what really will happen when the motor tries to absorb 80 Amps when the Panels can only provide 23 Amps ?

Could you please enlighten me ?

Thank you


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## Duncan (Dec 8, 2008)

Hi Summer

You are conflating power and energy

Panels are power devices they produce power - watts 

Batteries are more like a fuel tank they contain energy - Joules - 

so a 48v, 48 Ampere hours (Ah) battery has 48 x 48 Watt hours (a commonly used but incorrect unit) or 48 x 40 x 60 x 60 Joules of energy

You battery can provide 48 amps for one hour - or 480 amps for 6 minutes 

DO NOT NOT discharge to 24v - that will kill your battery
Battery charge is NOT proportional to voltage!

A typical lithium 48v battery will have 46v as dead flat and 52v as fully charged

You MUST track amp hours in and out voltage is not good enough!


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## miscrms (Sep 25, 2013)

The battery is basically your buffer between the relatively constant power/current output of the solar, and the fluctuating power/current demand of the motor. You can think of the battery as a water tank. The solar controller will be putting water into the tank, and the motor controller will be taking it out. When the flow out into the motor is greater than the flow in from the solar, the level in the tank will drop. The level will rise when the opposite is true. 

http://en.wikipedia.org/wiki/Hydraulic_analogy

In this analogy the flow rate in/out is current in Amps, and the volume of water in the battery is the stored charge in Coulombs (Amp seconds, or Amp hours / (60*60) ). Voltage is analogous to water pressure. Since the voltage in the battery is relatively constant from full to empty, power in/out in Watts is basically proportional to current in Amps, and Energy in Joules (Watt seconds, or Watt hours / (60*60) ) is proportional to charge in Ah or C (As). 

The mppt solar charger, and motor controller are like flow control devices. They increase or decrease the pressure (voltage) as needed to provide a certain amount of volumetric flow (current) from the solar panels into the battery, and from the battery into the motor.


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## Sunking (Aug 10, 2009)

SummerTime244 said:


> But as you see, the battery has less power than the motor needs, so it will discharge very fast. Additionnaly, the race regulations impose that we start with a Half-empty battery, which means the Battery voltage will be 24V instead of 48 at the start of the race.


That is pure rubbish. Assuming you are talking about using Lead Acid Battery, at 48 volt nominal, 50% SOC or half capacity open circuit voltage is roughly 48.4 volts. 

If using Lithium, the only battery you should be using, 50% SOC is hard to pin down because the discharge curve is fairly flat but around open circuit voltage of around 52 volts. For Lithium you really need a Coulomb counter counting Amps Hours in/out. Do not let the voltage go below 48 volts. 

I have no idea how you came up with 24 volts. 24 volts on a 48 volt battery is a destroyed worthless battery. 



SummerTime244 said:


> So my question is : Can i wire the PV+MPPT directly to the motor?


Well yes but you are overlooking one thing. At the output of the MPPT controller is connected directly to the battery and Motor 



SummerTime244 said:


> This way i can both power the motor and charge the battery. In this case, the battery will be much slower to charge since the Current flowing from PV is split between motor and Battery.


Does not work that way. Current flows from the higher state of energy to a lower state. You have no control where it goes. 




SummerTime244 said:


> Another thing i'm failing to comprehend is what really will happen when the motor tries to absorb 80 Amps when the Panels can only provide 23 Amps ?


 This is where you are lost. Solar panels are current sources, not voltage like a battery. But here is how it works. 

Assume the motor is off, batteries are low, and you have full sun. All power goes into the battery for charging. Lets say it is 20 amps. 

Now we start the motor and it demands 10 amps. The panels are still supplying 20 amps. 10 to the battery, and 10 amps to the motor. So the panels are still charging the batteries and powering the motor.

Now the motor demands 20 amps. All the panels current of 20 amps is going to the motor and the batteries are not charging with 0 amps.

Now the motor demands 50 amps. The panels supply 20 amps, and the batteries supply the missing 30 amps and is discharging.

Now here is the lesson you will have to learn, and you will not like it. You have to limit motor power to equal or be less than the panels can generate. Otherwise you are DISCHARGING. One thing you will learn quickly, your panels never generate rated power. It is something less. And at low angles, a lot less. Panels only produce their maximum power at solar noon assuming the panels are facing directly at the sun. If at an angle, much less. Throw in low angle morning and afternoonof the horizon and poor alignment, you will learn just how ineffective solar really is. 

If you lived in an area with really great sollar insolation like Tuscon AZ, in summer, a 100 watt panel will only generate roughly 500 watt hours of usable energy or 5 SUN HOURS on a Battery System. . That is if the panels are oriented due solar south at the optimum tilt angle. Cannot do that on a vehicle.


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## miscrms (Sep 25, 2013)

Easy there King. Clearly the OP is new to all this and trying to understand.

Building on the water tank idea above, you can get to a simplified mathematical model to start with.

(Initial Volume in the tank) - ((flow rate out - flow rate in) * time) = (Final Volume in the tank)

Rewriting in electrical terms:

(Initial Charge Stored in battery) - ((Current out - Current in) * time) = (Final Charge Stored in battery)

Looking at units, you have:
Ah (Amp hours) - Amps * hours = Ah

From that you can do lots of useful things. First it should be evident that as long as Current out <= Current in, you could essentially drive forever without depleting the battery. If Current out > Current in, you can set final charge stored to 0 and solve for time to get an idea how long the battery will last. You can use this formula with averages over a long period of time, you can use it with increments of varying length to estimate the battery state of charge over time, or you can use it with integrals to model real time battery state of charge.

So that's all relatively straight forward. The hard part is how to model your hardware to give you useful/realistic inputs to that equation.

On the solar side, your current output will vary with a number of factors as Sunking indicated. The sun angle over the course of the day will make a significant difference in the amount of power your panels generate. The solar controller will match the solar voltage to battery voltage, so again current out of the solar controller will be proportional to solar power generation:

Solar current into battery = (Solar electrical power / battery voltage) * Controller efficiency

A good place to start for estimating how much solar power you can generate at a given time on a given day in a given location is this website:
http://pvwatts.nrel.gov/

From there you can come up with a first order estimate of your "current in" term in the equation above. Note again you can use this as an average over the whole day, but you'll also want to model finer time increments to see if you might run out of battery charge sometime during the day. From there you would probably want to do some real world testing with your actual hardware to correlate it to the calculator, and then calibrate the calculator results to better predict performance at other locations / times of year.

The more difficult one to model is probably going to be the motor/vehicle's power consumption. This will depend on several factors:

- the mechanical characteristics of the vehicle (weight, rolling resistance, aerodynamic properties, etc)
- the electro-mechanical properties of the motor/controller
- the specifics of the driving course (speed, elevation changes, stops starts, etc)

As a very preliminary means of estimating vehicle characteristics, you could try playing with this website. There are also probably better tools out there, this is just one I'm familiar with. Its a useful tool, but I would eventually recommend doing the math yourself. Its not that bad, but in the process you may gain some insights into the tradeoffs that will benefit your design. 
http://www.evconvert.com/tools/evcalc/

I would actually ignore most of the EV specific calculations, as they are not well suited to what you are trying to do (and are pretty out of date even for most EVs). You are just looking to get an idea of mechanical power required to maintain various speeds as a function of vehicle characteristics. So all you need is to specify a motor, battery and controller that won't interfere with your calculations. A Warp 9, Zilla 1k, and Exide Orbital batteries are probably a good choice, with 196V system voltage under "adjustments". From there you'll want to customize the vehicle parameters. Key ones will be frontal area, drag coefficient, weights (use combination of initial weight and lbs removed to get final calculated weight where you want it), drive train efficiency, and tire parameters (will take some finagling if you are using non-car tires). By setting motor max rpms and gear ratios you can also use the top speed calculation. As long as the limitation under top speed says "motor rpm" you should be in a case where the specific motor/controller/battery chosen aren't significantly impacting your results. You'll want to ignore the range estimates, as those are specific to the battery chosen which will not be realistic.

Once these are all set, what you are primarily interested in are the results under drag calculations. This will give you the steady state power required to maintain a given speed in the defined vehicle. This will start to give you an idea of the "current out" parameter above relative to various trade offs in weight, aerodynamics, etc. Similar to the solar controller, the motor controller will manage the voltage difference between the battery and the motor, so you can estimate:

Motor current out of battery = (mechanical power required / battery voltage) / controller and motor efficiency

Note you will need to convert mechanical power in hp to electrical power in kW for this equation. Also note the difference in the efficiency term between this equation and solar equation is a result of the direction of current flow. Power out of the solar controller will by definition be less than power in. Power into the motor controller will by definition be higher than power into the motor. Efficiency can not be better than 100% 

You can also set the percent incline on the calculator to see how dramatically power requirements are impacted when the road is not flat. Another handy estimation, is the power requirement for a 5% incline is approximately the same as a 1mph/s acceleration on a flat. Using these numbers you can start to see how the terrain and stops/starts on your course will impact your power/current consumption.

That would be a step toward incremental modeling of specific legs of a presumed route. So much time at a certain speed, at a certain incline and/or rate of acceleration will will result in a certain amount of Ah drawn from the battery. If that time interval occurs at a certain time, on a certain day, in a certain location, a certain amount of Ah will be replaced into the battery by the solar panel. This is the beginning of a full system model, which can be used to assess performance under a variety of conditions and design tradeoffs.

In practice, the driver needs to basically do this calculation in their heads in real time. They need instruments to show them current coming out of the solar into the battery, and current going out of the battery into the motor, and the present amount Ah available in the battery. Then they can adjust speed based on knowledge of the route ahead. If there is a big climb coming up, they will want to make sure they are storing up charge into the battery so they can make it up the hill. If its morning and they have a good amount of charge in the battery they might be ok going a little faster and using some of that charge knowing the that the solar will replenish it later in the day. Etc. 

From my limited knowledge of these sorts of races, those seem to be the main factors. A vehicle that's well designed and suited to the course through proper modeling and testing, and a driver who is very conscientious about managing power flow ie using and storing charge at the right times to maximize performance. Modeling is certainly a key part of that effort, but there is no substitute for real world testing and driver practice. Hopefully modeling will help you rule out the dumbest options, and make the best choices you can with what budget you have 

Hope that helps a bit, and best of luck!

Rob


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## Sunking (Aug 10, 2009)

miscrms said:


> Easy there King.


OK point taken, but I would expect an engineering student to already have a grasp of physics.

Couple things strike me as off. First is only having battery 50% charged up. That eliminates using lead acid as you never want to discharge them more then 50% if you expect any kind of cycle life from them. Not only th eSOC is troublesome but weight and area are problematic using Pb batteries. An experimental solar model demands the the highest energy density battery there is. Something like lithium Cobalt of 160 wh/Kg. A Pb battery is about 50 wh/Kg. Huge difference. 

I would also caution against using amp hours as it is pretty much useless and just an end result. Seen way too many solar battery designs fail because the user used Amp Hours. Watt Hours is much more accurate. Amp Hours go in at a higher voltage than they come out.

So let's say his 1100 watts of solar panels once incidence is accounted for a given day receives 3 Sun Hours. That is 3.3 Kwh generated at the panel terminals, and 2.3 Kwh usable once you take into account all the losses he is going to be around 70% efficiency or 2.3 Kwh usable. How many amp hours is that? Who cares what it is, motors or anything electrical uses power, not Amp Hours. 

Amp Hours is the very last thing you need to know, and is just math result when everything else is done. So if you need say 2.3 Kwh usable in a day, you have to generate it at some voltage. If using a Pb battery at 48 volts nominal you know you need 50 AH plus another 50 AH to stay above 50% SOC or 100 AH. It will weigh in at 250 to 300 pounds 

If you are using Lithium you can discharge to 10% DOD and only need 60 AH at 16S. That battery will weigh in at 40-50 pounds and 1/5 the volume of the Pb battery. Huge difference. Catch is the Lithium battery will cost many times more than Pb. 

Stick with Watt Hours. More accurate, easier, and less prone to errors from conversions. Use nominal battery voltage, not actual voltage which will be a little higher and inject errors. Once you know how much power can generate and store, the math is 5th grade math. If you make $50/day, and need $100/day to survive, you know immediately you are going bankrupt.


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## Duncan (Dec 8, 2008)

I agree with Sunking about Watthours rather than Amphours 

BUT as a university project he should really be using the correct units! 
- Joules and Coulombs NOT Watthours and Amphours

If I was writing (or marking) his paper I would expect the correct units to be used with the "common" units alongside in brackets


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