HP to spin the fan in a VW type 1 air cooled engine?

[scotty c]

the vw type 1 engine comes with flaps that close when the engine is cold. i know the flow through the fan doesn't completely stop but for the sake off argument let's assume that is does. the fan is spinning but there is no flow. what power is required to maintain the pressure differential between intake and output sides of the fan?

 

[billy_joule]

Here's an experiment you can do at home: Block the nozzle of your vacuum cleaner, what happens to the motor RPM? So what must be happening to the load on the fan?

 

[scotty c]

Block the nozzle of your vacuum cleaner, what happens to the motor RPM? So what must be happening to the load on the fan?

ok, the flow stops. the fan speed increases. the fan speed reaches equilibrium. the air in the vicinity of the fan starts to get really really hot and then plastic fan and the plastic housing melt all because the fan is still sucking up power.

 

[billy_joule]

So what's the answer to my second question?
I don't think your initial question has a reasonable analytical answer. Maybe via CFD but my experience of CFD software is that modelling a blocked fan at constant rpm and determining shaft torque would be incredibly difficult/produce inaccurate results. An experiment would be easier.

 

[scotty c]

after doing some more research, here's what I've found. the blocked flow power depends on the shape, number and angle of the blades (not speed). do the calcs and then throw it out the window because the only way to know for sure is to experiment. so much for a "reasonable analytical answer". i started going to fan manufacturer web sites and over and over again found what they call a fan characteristic performance curve that is independent of the speed. the power at zero flow through the fan (vw type 1 fan is a backward inclined flat blade variant) is typically about 50 percent of peak power at that speed. the range was 60% to 40% with one chart running down to about 20%. i figure the vw fan isn't designed for low power consumption at low flow so in the discussion I'm having with others at thesamba vw forum I'm going to go with 50% of peak power. there is a guy who has worlds of experience and knows a good bit about these old vws who posts on the site all the time. most of what he says is fantastic but then every now and then he just blows something like his claim of zero power at zero flow.

 

[scotty c]

So what's the answer to my second question?

sorry, forgot to answer this. in the case of the vacuum. let's see. the motor is set up to run at a particular power. the fan speeds up until that power is still being utilized. so fan load remains the same.

this vacuum plugged analogy just isn't right though. the vw fan isn't going to speed up when it's flow is stopped.

i'm not done with this topic yet though. i have flow data for the fan from 1000 to 10000 rpm. i don't have pressure data. so I'm struggling with a way of estimating what that pressure is. if i can get estimated pressure combined with actual flow then i can (i think) get a pretty accurate estimated power consumption for the fan. my real goal is to get the power. this zero flow thing is just nit picking with that other samba poster.

 

[billy_joule]

so much for a "reasonable analytical answer".

Yep, that's what I said

I don't think your initial question has a reasonable analytical answer.

most of what he says is fantastic but then every now and then he just blows something like his claim of zero power at zero flow.

Yes, that's the simple answer for an ideal fan. There's no flow so no work is done on the air so no power is consumed.

Fan Power = Change in pressure * volumetric flow rate
http://www.engineeringtoolbox.com/fans-efficiency-power-consumption-d_197.html

sorry, forgot to answer this. in the case of the vacuum. let's see. the motor is set up to run at a particular power. the fan speeds up until that power is still being utilized. so fan load remains the same.
this vacuum plugged analogy just isn't right though. the vw fan isn't going to speed up when it's flow is stopped.

It's not perfect but it clearly shows that blocking the flow path reduces the required shaft torque
In the case of the vacuum, the reduction in torque load will lead to an increase in RPM to maintain about the same power level, in the case of the VW, where RPM is fixed by the idle air control valve, reducing the torque load will result in a proportional reduction in power consumption via
P = Tω (Power = Torque * rotational velocity)
That is, your air control valve will close a tiny amount as less air will be required to maintain the idle RPM.

You could set up a simple experiment on your VW, disconnect your idle air valve, turn off all auxiliaries, open and close the fan flaps and note the change in RPM due to the change in load. Now, try to replicate that RPM change with a known, measurable load, this may involve switching your headlights (or on and measuring the current flow, which, along with alternator efficiency will give you the torque load. Or maybe have the car in gear, on blocks, measure brake line pressure required for the RPM change, calc brake torque load from that etc There's probably better ways, I don't know much about cars.

 

[scotty c]

disconnect your idle air valve

we are on the same page but I'm on the front and you're on the back.

link to photo of the vw type 1 fan
http://www.justkampers.com/113-119-...eetle-1970-on-vw-t2-bay-1600cc-1970-1979.html
2 links to the fan in action.
link to my personal motor running (more or less)

link to high power version of same type of motor running

 

[billy_joule]

we are on the same page but I'm on the front and you're on the back.

OK. best of luck.

 

[OldYat47]

The bottom line is that the fan requires less torque when the flaps are blocked. How much depends on many things, mostly how much leakage there is in the flaps. I don't like to use power in cases like this because it clouds the issue. Power is velocity dependent. For example, if the torque required drops from 10 to 1 (avoiding units) at 1,000 RPM the power reduction is 1/2 the power reduction if torque drops from 10 to 1 at 2,000 RPM. Same drop in torque, different power.