Energy from solar panels and electro motor

[Eagle9]

As known, the Earth receives that solar radiation and its value approximately is equal to 1.4 W/m² (data taken from here: http://en.wikipedia.org/wiki/Solar_radiation ). Now imagine that there is some electric motor placed in the Earth’s orbit and it rotates the rod with some certain mass (let’s say 100 000 kg) and at some constant velocity (for example 2 revolutions/sec)

Can we calculate what amount of solar panels (I mean their total area in square meters) will be needed for rotating this rod by means of this electro motor?

 

[russ_watters]

None: once you start rotating something in space, there is nothing to stop it from rotating.

 

[sophiecentaur]

BTW, that's KW and not W of received solar power. Quite a significant difference.

 

[Eagle9]

russ_watters

None: once you start rotating something in space, there is nothing to stop it from rotating.

Yes, I know this, but we will need the energy to force the motionless rod to begin rotating and reach the certain speed, and I want to know-how is it possible to calculate this….

sophiecentaur

BTW, that's KW and not W of received solar power. Quite a significant difference.

Of course, I know that Watt and Volt are absolutely different things and I speak about Watt (Power). So, how can we calculate how much solar arrays (their area and mass) will be needed in space for rotating the rod?

 

[Fish4Fun]

Eagle9,

Ultimately the answer is, "Virtually any size solar panel will do the trick." Why? because you only need a finite amount of energy to initiate the rotation, and you did not specify any time period over which the acceleration from static to 2 RPS needed to occur. To calculate the energy required you would use:

E = 1/2 * m * (wL)^2

I will leave solving that to you, the question was about the size of the solar panel, so let's assume you decide on a 1.5V 100mA panel and motor. Let's further assume that the total energy required to reach 2 RPS is 10kJ (I just picked a number), we would then be able to calculate how long it would take to achieve the final speed,

10,000J = 1.5 * 0.1A * T ==> 66,666 seconds (18.52 Hours).

This obviously assumes 100% efficiency in the conversion of the electricity to mechanical energy. But what I am more curious about is how you intend to get a 100,000kg mass into orbit!

Anyway, in theory, you could use a 1V source with only 1uA of current and still reach the same speed, it would just take a lot longer.

Fish

 

[Eagle9]

Fish4Fun

Ultimately the answer is, "Virtually any size solar panel will do the trick." Why? because you only need a finite amount of energy to initiate the rotation, and you did not specify any time period over which the acceleration from static to 2 RPS needed to occur.

This surprised me, however it seems to me that you are right, but still-I do not think that “any size” solar panel will be suitable, because if the solar panel is too little (that is producing too little electricity) then the electro motor simply will not be able to move the rod, right? So, apparently some minimal size for solar panels is still necessary….

RPS

Is that revolutions/sec?

To calculate the energy required you would use:

E = 1/2 * m * (wL)^2

m-is mass of the rod (kilograms), w-is total power (in Watts) of solar panels, L is length of the rod, right? And for receiving the correct results I need to use unified units, right? I mean that will use only metric system (or only imperial, however in Europe metric system is used), more precisely SI system (meters, kilograms, second and etc.). In which unit will I receive the results? In Joules?

I will leave solving that to you, the question was about the size of the solar panel, so let's assume you decide on a 1.5V 100mA panel and motor.

Well……is this the property of the electric current that generally is produced by one solar panel? Did you pick these values randomly?

Let's further assume that the total energy required to reach 2 RPS is 10kJ (I just picked a number), we would then be able to calculate how long it would take to achieve the final speed,

10,000J = 1.5 * 0.1A * T ==> 66,666 seconds (18.52 Hours).

Interesting formula, could you please tell me what do physicists call it? As I know some formulas is Physics have got their authors and names also….

This obviously assumes 100% efficiency in the conversion of the electricity to mechanical energy.

As I know the electric motors have got very high coefficient of efficiency, 90 % or so, right?

But what I am more curious about is how you intend to get a 100,000kg mass into orbit!

Well, it is not critical actually, I think that Space Elevator will enable this in future

Anyway, in theory, you could use a 1V source with only 1uA of current and still reach the same speed, it would just take a lot longer.

This made me to think a bit…….in that formula (10,000J = 1.5 * 0.1A * T) I see volts is multiplied to amperes and this generally is equal to Watts (Power), right? A*V=W Thus I remember…

 

[Fish4Fun]

Yes, Watts = Joules/Second, so by multiplying Watts * Time you get Joules (frequently converted/referred to as Watt-Hour or Kilo-Watt-Hour).

Obviously in E = 1/2 * m * (wL)^2 you would want to use SI units to get Joules.

E = Energy
m = Mass
w = "Omega" or Angular Velocity in Radians per Second
L = Length

You can get a more detailed derivation here: http://en.wikipedia.org/wiki/Moment_of_inertia

The above formula for rotational inertia is for a simple "light rod" rotating around one end, you would need to adjust this formula for your actual rotating body.

Well……is this the property of the electric current that generally is produced by one solar panel? Did you pick these values randomly?

There are lots of "experimenter" solar cells with roughly these specs.

[/quote]
but still-I do not think that “any size” solar panel will be suitable, because if the solar panel is too little (that is producing too little electricity) then the electro motor simply will not be able to move the rod, right? So, apparently some minimal size for solar panels is still necessary
[/quote]

Yes, in theory "any size" will do. The "theory" assumes you have a lossless DC motor perfectly coupled to your solar panel. In reality, 1.5Vdc, motors are quite common; combining a 1.5V PV cell to a 1.5Vdc motor seemed like an intuitive place to "start" in demonstrating that a "small" motor would eventually get the job done.

Interesting formula, could you please tell me what do physicists call it?

The definition of power wrt electricity is:

P = IE

P = Power in Watts
I = Current in Amps
E = Voltage in Volts

As stated above, Watts = Joules/Second by definition, so multiplying Watts * Seconds = Joules. In this:

10,000J = 1.5V * 0.1A * T

I am simply solving for T = Time in Seconds.

Fish

 

[Eagle9]

Fish4Fun

w = "Omega" or Angular Velocity in Radians per Second

And if I “my” rod makes 2 revolutions per second how can I calculate how many "Omega" is it?

You can get a more detailed derivation here: http://en.wikipedia.org/wiki/Moment_of_inertia

The above formula for rotational inertia is for a simple "light rod" rotating around one end, you would need to adjust this formula for your actual rotating body.

Do I need to do it? Your formula E = 1/2 * m * (wL)^2 is enough for my situation, right? I will specify one thing: the rod is balanced, that is there are the same masses at rod’s both sides (let’s say 50 000 kg and 50 000 kg to each sides):

So, that formula (in red color) is suitable, right?

 

[vk6kro]

Wouldn't the motor also rotate, but in the opposite direction?

If this was to be minimized, then the motor would have to be a lot heavier than the rod or attached to something that was.

 

[Eagle9]

Wouldn't the motor also rotate, but in the opposite direction?

If this was to be minimized, then the motor would have to be a lot heavier than the rod or attached to something that was.

Of course, rod’s rotation generally forces the motor to rotate to opposite direction, but this undesired rotation can be canceled by means of second rod rotating at opposite direction, so it is not problem

 

[sophiecentaur]

russ_watters


sophiecentaur

Of course, I know that Watt and Volt are absolutely different things and I speak about Watt (Power). So, how can we calculate how much solar arrays (their area and mass) will be needed in space for rotating the rod?

You may know the difference between Volts and Watts but I didn't suggest that. I merely pointed out that your value for the Solar Constant was 1/1000 of the accepted one. The difference between kW and W is highly relevant here!

 

[Eagle9]

You may know the difference between Volts and Watts but I didn't suggest that. I merely pointed out that your value for the Solar Constant was 1/1000 of the accepted one. The difference between kW and W is highly relevant here!

Yes, you are right, I should have written 1400 W/m^2