Endurance and Range Equations for Electric Aircraft....HELP PLEASE!!!

[sligo565]

The following has been a constant thorn in my side for the past few days let me tell you!

I am currently underway with a thesis based study into a conceptual design of a solar powered UAV, but I am stuck on how to calculate the following equations:

Endurance
Range

I think my problem is that seeing that im using Electric Aircraft, what do i do with the SFC and Fuel Weight in these equations....do i make them constants (1)!?

[BMatthews]

I'm guessing that you need this for night time when the "fuel pump" is below the horizon?

I'm also going to guess that your "fuel delivery" will vary based on the angles of the sun through the daylight hours.

I don't know what SFC means but obviously the fuel weight will be 0 or 1 or whatever is needed to keep the overall weight of the craft the same throughout the flight.

[annihilator]

alright, SFC (matthews its specific fuel consumption) is the wrong thing to look at for a solar UAV. it is given in lb/hp/hour and is used to calculate how fast the airplane lightens up as it flies.

Also, your fuel weight is constant and should be listed as the weight of your batteries.

If im thinking right youre probably looking at the bourget range and the classical endurance equations..... Dont.

electric and solar airplanes stay the same weight as they fly, so your flight time is simply (battery energy / power) where battery energy is in watt hours and power is in watts required to fly. if youre doing this as your thesis I assume you know how to calculate required power.

trick is for a solar plane you modify that equation by saying endurance = (batteryenergy/(power-solar_power_input) and range is just endurance*velocity. If your solar input is greater than power used then you get infinite flight times, but then the sun goes down and life $%*& again.

that help?

[sligo565]

you truley are a legend annihilator!!

I indeed have my Power Required...just for FYI purposes it works out to be 2.242kW and my Power Available is 6.5kW so things seem to be looking up....quite literally!!

[sligo565]

I was just working out my equations and I couldnt help notice where you say:

trick is for a solar plane you modify that equation by saying endurance = (batteryenergy/(power-solar_power_input)

Does this mean i subtract my solar input power from my power required....just that im getting a negative hours answer if I do this!!

[BMatthews]

So that implies that you're getting more power from the panel than you need to fly the model. So you can use the extra power to charge the onboard batteries unless I'm missing something here.

Remember that the sun angle hitting the panels will change during the course of the day so you won't be getting full power all day long. You'd need to do some testing of the panels for power output vs angle of source and then integrate the results to find out the total energy obtained during the daylight hours. It'll also be dependent on the seasonal sun angle and latitude where you're flying the UAV concept.

[annihilator]

2.24 KW? how big is it? that is a lot of power! multiple engines? how many M^2 of solar panels?

solar irradiation is only 1.3 KW per m^2 and panels are AT BEST 20% efficient in direct sunlight with expensive materials.... unless you have a new solar invention. so that gives 260W/m^2 meaning your airplane must have over 25m^2 of wing.

i would reccomend if you havent already looked at them the airplanes "solong" and http://www.solarimpulse.com/en/index.php

[HugePanic]

you could just do some experiments with excel.
for each row you can calculate a specific time period, and then do all your power-calculations (input, output, battery) for each time-period in this row.

did you consider that the solar power is veryvery unstable over the day as the light changes??? (angle of earth surface to the sun)

you should also cosider a flight plan without constat power. fly up with motor, and soar it down as a glider, then up again ... and so on...
(here you could take the aircrafts' flying height into you excel sheet)

[sligo565]

Yes i do think I am an idiot....I'm getting myself muddled up with numbers etc.

Basically I have 36m^2 to play with, I found 2 suitable solar cells that meet my criteria:

1) Power = 2.66W
Area = 0.015625m^2
Efficiency = 17.1%

so 36/0.015625 = 2304 cells can be fitted onto my surface....

2304 x 2.66 = 6.13kW

2) Power = 2.76762W
Area = 0.01
Efficiency = 19.5%

so 36/0.01 = 3600 cells can be fitted onto my surface....

3600 x 2.76762 = 9.96kW


Now here comes the idiotic question....Do I now find 17.1% of my first solar cell power and 19.5% of my second solar cell power? This being my actual Power output???

1) 17.1% of 6.13kW = 1.048kW
2) 19.5% of 9.96kW = 1.9422kW


Or do I find out the max Electrical Power available from the sun and use my percentages to find my answers?

[MarkusN]

You estimate the irradiation you get in the situation you are flying (1.37 kw/m3 out of the atmosphere, somewhere around 1 kw/m3 at ground level, less in haze, fog...)
Then you correct for not ideal angle (somewhere with the cosine of the angle of the rays vs. surface vector, probably less at large angles due to some of the energy being reflected.)
times the area of solar cells
times efficiency.

[sligo565]

Thanks Marcus,

Ill be cruising at 65,000 feet so the irrays will be alot stronger there