OK, how about we knock this one out once and for all?
Option A: Take the easy way out, skip to the final answer:
That chart was posted by GMT360 on the other board. He was a cooling system Engineer at GM who worked extensively on the GMT360 (hence the username) and has given us all more info on the fan and fan clutch of these things than we likely know all put together. Chances are, that chart came from somewhere....
Option B: Let's assume GMT360 was full of beans and come up with our own number independently. I'll provide some data, if anybody else has some they'd like to add we'll throw it into the mix.
My electric fans, the Derale 16927, are rated to consume 50 amps at full speed at 12 volts. That equates to 50*12/746 =
.804 HP. In reality, they've been measured consuming a bit more than that and a good alternator will keep the voltage higher than 12 volts so they probably operate closer to 1 HP, but to keep things as conservative as possible, we'll leave it at
.804 HP. And let's even assume the alternator is 100% efficient at converting mechanical energy to electrical energy so we don't say it's consuming even more power from the engine....
How much air do they move compared with stock? I took some measurements....
Nothing too high tech, just a wind meter placed at various locations. While these measurements aren't accurate down to the last cfm, they're more than "in the ballpark," I'd say they're pretty darn close. If anybody would like to take more accurate measurements, I would welcome your data.
That chart shows the results. It tells us that at full boogie, the stock fan can move air through the radiator at an average speed of
33.375 MPH, while my Derale fans can do the same at
13 MPH. This tells us the stock fan when fully activated moves
2.57 times as much air as my electric fans.
Now it's time to enter into evidence some "Universal Fan Laws"--available in pretty much any Engineering text that talks about fans at all:
OK, so let's say I want to increase the power of my electric fans to match the output of the stock fan. As you can see from above, given the best case scenario of my fan blades keeping the same efficiency, etc, I would need to increase their RPM by 2.57 times in order to increase the CFM 2.57 times as they are directly proportional.
So how much power would that require? Now you go to the equation that really smacks you in the face. The power required is proportional to the RPM
CUBED!!!! Yes,
CUBED!!!!
Ouch, that hurts.
This gives us, (2.57)^3*.804 =
13.65 HP required to spin my fans fast enough to match the output of the stock fan. And that's with many conservative assumptions that keep the number lower than it is in reality. Amazing how we validated GMT360's chart though, isn't it?
For those thinking the RPM Cubed equation seems off, you need to brush up on your fluid dynamics. It's pretty standard fair. Even forgetting fans, it's the same for a simple object moving through the air--the force of air resistance of an object with a given coefficient of drag and frontal area is proportional to the square if its velocity...and since power is force times velocity, the power required to apply that force is proportional to the cube of the velocity. Standard stuff.
So I do agree with the contention a 14 HP (or two 7 HP) electric motors would be quite impossible to package. The conclusion one should draw from these data are that once a certain level of airflow is required, one should simply cross electric fans off the list of possibilities as making electric motors that huge fit between the engine and the radiator becomes impossible. The more CFM you have, the harder each additional CFM is to get--it's exponential. But when you have a 270 HP "fan motor" and can tap into as much of that power as you want with your blade/clutch design of a mechanical fan, if you really need to move that much air mechanical is the only option.
Of course that wasn't the question posed in this thread. The answer to the question being posed--does the mechanical fan really take that much HP?--when you follow the math above, is an obvious "yes."