Fan Airflow vs Power Consuption

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bpk

Senior Member
This maybe more of an engineering / physics question but when a fan such as ID fan or squirrell cage is restricted on the airflow inlet side, such as louvers or dirty filters, does the fan draw less power. If so what about being restricted on the output side ? Thanks in advance
 

topgone

Senior Member
This maybe more of an engineering / physics question but when a fan such as ID fan or squirrell cage is restricted on the airflow inlet side, such as louvers or dirty filters, does the fan draw less power. If so what about being restricted on the output side ? Thanks in advance
With inlet guide vanes shut, the fan draws very little current. The discharge pressure at the fan increases when discharge dampers are closed but still, the fan drive motor current will be very low.
 

kwired

Electron manager
Location
NE Nebraska
Does not matter if inlet or outlet is what is blocked. Same with a centrifigal pump.

The fan or pump is loaded by the volume of media it is moving. Any restriction limits the volume that is available to move.

If you have a boat with a leak in the bottom are you going to work harder by bailing out with one gallon buckets or five gallon buckets? (Assuming you bail same number of bucket per minute in either case) In which case is more work accomplished in same amount of time?

Now lets say you have to pour the bailed water into a restricted opening - you are now limited in how much you can move no matter how fast you can scoop it up and will accomplish less work in same time period.
 

BJ Conner

Senior Member
Location
97006
No Work no current

No Work no current

A fan with closed dampers is not moving air, no moving air= no work. No work means no current.
Fan manufactures make "Fan Curves" that plot power requirement for various flow and differential pressures accross a fan.
http://www.solutionsforair.com/dayton_content/pdfs/FanCurves.pdf

Axial fans can self destruct if you shut of the flow to them while they are running. Read the instructions and warnings before doing that to one.
 

Besoeker

Senior Member
Location
UK
This maybe more of an engineering / physics question but when a fan such as ID fan or squirrell cage is restricted on the airflow inlet side, such as louvers or dirty filters, does the fan draw less power. If so what about being restricted on the output side ? Thanks in advance
It will draw less power.
Power is head (pressure) times flow.
 

Jraef

Moderator
Staff member
Location
San Francisco Bay Area, CA, USA
Occupation
Electrical Engineer
It will draw less power.
Power is head (pressure) times flow.
The Centrifugal Fan and Pump Third Affinity Law states that power changes as the cube of the flow change. That takes place with VFDs, but also dampers or anything that changes the rate of flow, inlet side or outlet side, doesn't matter. Mechanical devices other than VFDs have more losses associated with them, i.e. turbulence, pressure drops etc. (VFDs have losses too, but less so).
 

BJ Conner

Senior Member
Location
97006
Here's another nit.

Here's another nit.

Not quite right.
A cage induction motor takes typically about 30% of full load current even running uncoupled.
The motor will draw no load current + the current to heat the fluid ( water or air) in the pump or fan. How much heat, how fast depends on the physical charesterics of the device.
I have centrifugal pump blow seals by running against a closed valve.
Both fans and pumps can overload if they are being operated in "wrong" section of the pump/fan curve. Inlet vanes on a fan are set to achieve a particular flow at a particular head. IF you mess with the vanes you can overload the fan motor. Works for pumps too.
 

stevenj76

Senior Member
Inlet vanes were used to control duct static pressure before drives became common.

If you think you are going to save juice, then you may be sabotaging whatever the fan is designed to serve.
 

Besoeker

Senior Member
Location
UK
The Centrifugal Fan and Pump Third Affinity Law states that power changes as the cube of the flow change.
Head times flow:

Pi= D*g*H*Q/eta

where:

Pi is the input power required (W)
D is the fluid density (kg/m3)
g is the standard acceleration of gravity (9.80665 m/s2)
H is the energy Head added to the flow (m)
Q is the flow rate (m3/s)
eta is the efficiency of the pump plant as a decimal

Typically, for centrifugal fan or pump on a variable speed drive, flow varies as the square of the speed and the pressure (head) proportionally with speed.
Thus the power varies with the cube of the speed.
 

Rick Christopherson

Senior Member
I believe the discussion is being a little over simplified. There is a difference between blocking the inlet versus blocking the outlet. This can easily be seen with many common pumps/blowers laying around your house or shop. While it is correct that with a blockage there is no total airflow through the system, what will be different is the pressure differential across the pump/blower. It takes work to create a pressure differential.

When the inlet is blocked, there is no available makeup air to enter into the blower. As a result, there is very little pressure differential across the blower. What differential does exist is also from less dense air due to the resulting partial vacuum. The outlet pressure will be atmospheric, and the inlet will be slightly below that.

When the outlet is blocked, there is an abundant supply of makeup air, but no where for it to go. As a result, there is a significant pressure differential across the blower. The inlet will be at atmospheric pressure, and the outlet will be significantly higher.

You can even demonstrate this for yourself with a common vacuum cleaner if you can totally block the exhaust. (Most of them will prevent this though.) Block both inlet and outlet to get a baseline sound of the motor. Then release the inlet and make note of the tone of the motor; it should drop in pitch as the motor loads. Repeat this, but this time release the outlet. The motor tone should remain nearly unchanged.

The other thing that is being overlooked is that air is still being moved from a low to high pressure. Even if the total air flowing through the system remains zero, there is air flowing (and leaking back) through the blower.

By the way, this becomes even more pronounced on a pump, where you can nearly stall the pump with a blocked outlet, and it will run nearly free with a blocked inlet.
 
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Rick Christopherson

Senior Member
How did you get from what I said to this? ↓↓↓↓↓↓↓↓↓↓ :happysad:

Are you suggesting that the equation for Pi in post #10 is wrong?
I wasn't commenting on your posting. Our first postings crossed last night. I'm not totally familiar with your equation, but considering it has "g" in it suggests it is probaly the equation for a lift pump. Since this thread is discussing blowers, then yes, I say your equation was wrong.
 

Besoeker

Senior Member
Location
UK
How did you get from what I said to this? ↓↓↓↓↓↓↓↓↓↓ :happysad:

I wasn't commenting on your posting.
I was commenting on yours.
Our first postings crossed last night. I'm not totally familiar with your equation, but considering it has "g" in it suggests it is probaly the equation for a lift pump. Since this thread is discussing blowers, then yes, I say your equation was wrong.
The equation refers to fluid. It applies to any fluid - including air.
But, if you think it's wrong, maybe you can post one that you believe to be correct.
 

Rick Christopherson

Senior Member
The equation refers to fluid. It applies to any fluid - including air.
But, if you think it's wrong, maybe you can post one that you believe to be correct.
It is a specific equation for the limited application of vertical flow. As the person that posted it, you should have known that. It also doesn't contradict what I said, nor does it have any bearing on what I said.

I hadn't previously looked at your equation very closely because your term "H = Energy Head" was confusing at first glance. H is more commonly called the "Head Height". If you had used that term, it would have been more obvious that you brought the wrong set of equations with you to the discussion. I originally hadn't noticed the "H", but I did notice the "g" (accel of gravity). That's what told me it was a lift equation.

So according to your equation, all of these unit air handlers up in plenums aren't doing any work, and therefore don't expend any power. (H is zero for horizontal flow.) There are an awful lot of utility bills out there that say otherwise. :D

Your equation (even if it did apply here) relates solely to how much power (work over time) it takes to lift a fluid through a height, where the only pressure differential is the weight of the fluid in the column. Your equation fails if the pressure differential is caused by something other than gravity.

I was commenting on yours.
Yes, and as I said, you still haven't contradicted it. All you have to do to see it is correct is to take the pump that your equation is discussing and block the outlet and see how high the amperage rises. It will go through the roof.
 

topgone

Senior Member
. . . . . . . . . . . . . .. . . . . . . . . . . . . . .
I hadn't previously looked at your equation very closely because your term "H = Energy Head" was confusing at first glance. H is more commonly called the "Head Height".
Bes's equation is correct. It's just that you're not used to seeing it written that way, IMO. H is basically just "head"; the total head the pump is going to be handling. It could include friction head, pressure head, and in liquids to include the difference in elevation of the suction point from the discharge point.
If you had used that term, it would have been more obvious that you brought the wrong set of equations with you to the discussion. I originally hadn't noticed the "H", but I did notice the "g" (accel of gravity). That's what told me it was a lift equation.
The "g" appeared in that general equation because Bes was using ρ "density", not specific weight. γ = ρ g

To wrap it up:
The power input Pi
= ρ Q g h/eff ; substituting γ = ρ g in the equation:

= γ Q h/eff ; where γ = the specific weight of the fluid; Q = the flow ; and h = the total head​
 

Rick Christopherson

Senior Member
Bes's equation is correct. It's just that you're not used to seeing it written that way, IMO. H is basically just "head"; the total head the pump is going to be handling. It could include friction head, pressure head, and in liquids to include the difference in elevation of the suction point from the discharge point.

The "g" appeared in that general equation because Bes was using ρ "density", not specific weight. γ = ρ g

To wrap it up:
The power input Pi
= ρ Q g h/eff ; substituting γ = ρ g in the equation:

= γ Q h/eff ; where γ = the specific weight of the fluid; Q = the flow ; and h = the total head​
Would you like to re-evaluate this before I respond in-full? I will give you a hint: "Why should the weight of the fluid matter unless it is a 'lift' equation?"

Oh, and "g" is the conversion from MASS to WEIGHT, not Density to Mass!
 
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Rick Christopherson

Senior Member
Bes's equation is correct. It's just that you're not used to seeing it written that way, IMO. H is basically just "head"; the total head the pump is going to be handling. It could include friction head, pressure head, and in liquids to include the difference in elevation of the suction point from the discharge point.

The "g" appeared in that general equation because Bes was using ρ "density", not specific weight. γ = ρ g

To wrap it up:
The power input Pi
= ρ Q g h/eff ; substituting γ = ρ g in the equation:

= γ Q h/eff ; where γ = the specific weight of the fluid; Q = the flow ; and h = the total head​
I was too slow to edit my previous post, but here is another hint for you....check your "units"!

"Head" has the units of "meters", but your suggestions are that it includes "friction", "pressure", etc. Do these parameters have the units of "meters"? Should all of these parameters be multiplied by "g"?
 
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