Need help with a fractional cube-square problem, sort of.

[UncleMorgaan]

Hi, all.


I'm dancing with some numbers for air fans (impellers).


I need three answers.

The only hard numbers I have are that with a fan 6 ft in diameter, an 8 mph breeze can be generated with an input of 6.71 hp.

The speed of the fan doesn't matter, since it will simply increase until it uses all the available horsepower and there are no RPM issues so far.


According to the rules. if you cube the power, you will double the wind speed.


So 302.11 hp would generate a 16 mph breeze.

However: I only have 25 hp available, and I'm trying to figure out what the wind speed would be with only that relatively modest increase in power. That's answer #1.


Alternatively, if I doubled the size of the fan to 12 ft in diameter, the power would only have to be squared to keep the same 8 mph breeze.


And, conversely, if I reduced the diameter of the fan to 3 ft, I would only need the square root of the original 6.71 hp to keep the same 8 mph breeze.

And that would be 2.59 hp. for the original 8 mph breeze with the half-size fan.


Then, doubling the breeze to 16 mph would only take the cube of 2.59 hp, which is 17.37 hp.

Close...but...


The next question is what breeze would result with the 3 ft diameter fan if it was spun with 25 hp? That's answer #2


And then, at last (long last!) the Ultimate Answer that I seek:

If the original 6 ft diameter fan was reduced to 4 ft in diameter and the power input was increased to 25 hp, what breeze would result? That's answer #3.


I know there is some way of handling fractional cube-square calculations, if that's what this is, but the math is far beyond my meager skills at the moment.


Any help with this would be much appreciated, even if it's just a signpost pointing to where I might find out how to work all this out for myself.


Thanks, all.

UncleMorgan

 

[Dr.Peterson]

Hi, all.

I'm dancing with some numbers for air fans (impellers).

I need three answers.

The only hard numbers I have are that with a fan 6 ft in diameter, an 8 mph breeze can be generated with an input of 6.71 hp.

The speed of the fan doesn't matter, since it will simply increase until it uses all the available horsepower and there are no RPM issues so far.

According to the rules. if you cube the power, you will double the wind speed.

So 302.11 hp would generate a 16 mph breeze.

However: I only have 25 hp available, and I'm trying to figure out what the wind speed would be with only that relatively modest increase in power. That's answer #1.

Alternatively, if I doubled the size of the fan to 12 ft in diameter, the power would only have to be squared to keep the same 8 mph breeze.

And, conversely, if I reduced the diameter of the fan to 3 ft, I would only need the square root of the original 6.71 hp to keep the same 8 mph breeze.

And that would be 2.59 hp. for the original 8 mph breeze with the half-size fan.

Then, doubling the breeze to 16 mph would only take the cube of 2.59 hp, which is 17.37 hp.

Close...but...

The next question is what breeze would result with the 3 ft diameter fan if it was spun with 25 hp? That's answer #2

And then, at last (long last!) the Ultimate Answer that I seek:

If the original 6 ft diameter fan was reduced to 4 ft in diameter and the power input was increased to 25 hp, what breeze would result? That's answer #3.

I know there is some way of handling fractional cube-square calculations, if that's what this is, but the math is far beyond my meager skills at the moment.

Before I try to help, I need to be sure of the formulas. I suspect you may be confusing some things, such as "squaring" vs. "doubling". For example, although I'm having trouble finding this information about fans, I find that for a wind turbine, the power (output) is proportional to the square of the diameter and to the cube of the wind speed; so doubling the diameter would multiply the power by 4 (the square of 2) -- it wouldn't square the power. And the power is proportional to the cube of the wind speed. I have no idea whether these are also true of a fan, but it gives me reason to want a good source.

Can you show me a source for your facts?

 

[UncleMorgaan]

Before I try to help, I need to be sure of the formulas. I suspect you may be confusing some things, such as "squaring" vs. "doubling". For example, although I'm having trouble finding this information about fans, I find that for a wind turbine, the power (output) is proportional to the square of the diameter and to the cube of the wind speed; so doubling the diameter would multiply the power by 4 (the square of 2) -- it wouldn't square the power. And the power is proportional to the cube of the wind speed. I have no idea whether these are also true of a fan, but it gives me reason to want a good source.

Can you show me a source for your facts?

Thank you for considering this, Dr. Peterson. I really appreciate it.

At the moment, I am considering fans and wind turbines as reversible mechanisms, like some motors and generators. I could be wrong there, but it's a starting point.

Basically, I was thinking that (for a given fan diameter) when the breeze (output wind speed) doubles, the powered required to double it would increase by the cube of two--that is, increase 8 times over. The reverse (for wind turbines) would be:

'The wind power increases with the cube of the wind speed. In other words: doubling the wind speed gives eight times the wind power." https://home.uni-leipzig.de/energy/ef/15.htm

So given a breeze of 8 mph and a power input of 6.71hp, doubling the breeze to 16 mph would require eight times the hp, or 53.68 hp.

At this point, the fan is (still) 6 ft in diameter.

Cutting the fan diameter in half (to 3 ft.) would then reduce the power requirement for the 16mph breeze by a factor of four.

That is essentially the opposite of increasing the size of a wind turbine rotor:

"If you double the rotor diameter, you get an area which is four times larger (two squared). This means that you also get four times as much power output from the rotor." http://drømstørre.dk/wp-content/wind/miller/windpower web/en/tour/wtrb/size.htm (Paragraph 3)

So that gives a 3 ft diameter fan putting out a 16 mph breeze from an input of 14.65 hp.

So, bumping the fan diameter up "just a bit" would raise the power requirement to 25 hp. That "bit" would require increasing the swept area by 1.706 times to increase the power input from 14.65 hp to 25 hp.

The swept area for a 36" diameter fan is 1017.876 sq. ins.

Times1.706 = 1736.496 sq ins.

Dividing by pi gives a [radius squared] of 552.744 sq. ins.

Which is a radius of 23.510 ins.

That gives a diameter of 47.021 ins.

And that's very close to the 4 ft. diameter and 25 hp combo that was being considered as optimal.

The 16mph breeze is nice to know, also.

Of course, this is all hypothetical. The real-world factors just aren't available yet. Coefficients of efficiency, drag vs lift, and all that stuff

The application is essentially an improved propeller for smaller and quieter airboats.

So I think I have wandered to the answer I was originally looking for.

I probably would not have done so without your reply to help straighten out my approach. Certainly not as quickly and easily, anyway.

Now all I have to do is look wise and tell my friend to try an impeller 4 feet in diameter--to start!

Thank you very much!