Fan Airflow vs Power Consuption

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jumper

Senior Member
:D:D


Yeah, I did laugh to myself that I went out in the dark driveway at 4AM to grab my meter, clamp and camera to check out a fan.
The last time time my wife caught me doing a wee hour test, around 3:00 AM, I was seeing if using a 240V 1P supply to power a 120V to 277V ballasted fluorescent light fixture would work, she shook her head and said "plumbers do not do this crap".

Might have helped if I had not been standing there in my skivvies and had all the parts and tools lying on the washing machine and the electrical panel cover off.
 

ActionDave

Moderator
Staff member
Location
Durango, CO, 10 h 20 min without traffic from wing
Occupation
wire pulling grunt
You two...running around in the middle of the night, in your underwear, testing electrical equipment because someone on the internet has a question.... I mean, really....
If you had told the woman that has pledged to stay with you, for better or for worse, that this is the kinda stuff you do; do you really think she would have said, "I do"? :happyno:

Purchasing of flowers before coming home tomorrow is a must.:happyyes:
 

Besoeker

Senior Member
Location
UK
The last time time my wife caught me doing a wee hour test, around 3:00 AM, I was seeing if using a 240V 1P supply to power a 120V to 277V ballasted fluorescent light fixture would work, she shook her head and said "plumbers do not do this crap".
How about this as a response (from The Ball of Kirriemuir):

The village plumber, he was there
But felt an awful fool
He'd come thirty leagues or more
And forgot to bring his tool.


Tool in this context....'nuff said.
 

Besoeker

Senior Member
Location
UK
Not going to debate the validity of the formula for the purpose desired, but the answer to your question is to establish a pressure value.
Careful!
You'll have him again asserting that it relates to pumps and vertical lift.

I gave a formula for power. The basic physics is fairly simple.
Power is force times distance divided by time.

One Newton through one metre in one second is one Watt.

One Newton through one metre is one Joule.
 

jumper

Senior Member
You two...running around in the middle of the night, in your underwear, testing electrical equipment because someone on the internet has a question.... I mean, really....
If you had told the woman that has pledged to stay with you, for better or for worse, that this is the kinda stuff you do; do you really think she would have said, "I do"? :happyno:

Purchasing of flowers before coming home tomorrow is a must.:happyyes:
LMAO :lol:
 

Rick Christopherson

Senior Member
Careful!
You'll have him again asserting that it relates to pumps and vertical lift.

I gave a formula for power. The basic physics is fairly simple.
It's bad enough that you didn’t understand the origins of your own equation, but to have yourself and your cronies gloat over your mistake compounds it. So let's take a look at the origin of your equation and you tell me that it still isn't a vertical lift equation.

Let's start with the general equation for power in a pump. If you need confirmation of this, you can find it in your text books or here is a convenient wiki-link.

P = Δp * Q / η , where:

Δp = (your D) = pressure differential
Q = (your Q) = Flow Rate
η = (your eta) = efficiency

Notice the similarity of the general equation to your equation. The only difference is the Δp. I will rewrite your equation first using your variables, and then using standard variables to conform with the standard equation.

Pi= D*g*H*Q/eta = ρ * g * H * Q / η , where:

ρ = (your D) = density

It is easy to see that ρ*g*H = Δp and the units match as well. (kg/m3) * (m/s2) * (m) = (kg/m/s2).

However, it is also quite obvious that the term (ρ*g*H) is the pressure at the bottom of a column of fluid (gas or liquid). The same wiki article confirms this farther on.

So your Δp is the difference in pressure from the bottom of a column to the top of a column, which is in fact, a fluid lift differential. As I stated back in the beginning, your equation (which I never said was not valid) is a "lift equation". It is valid, but does not apply to the discussion we have been engaged in.

Here is an example where your equation fails: You need to size a recirculation pump, where there is no elevation change (H=0), but the pressure differential is the result of frictional losses and other restrictions in the piping. Your Δp is the wrong Δp for this problem.

If you had actually understood the source and nature of the equation that you have grown so accustomed to using, you would have seen that it was limited to a lift equation as soon as I originally questioned it. The fact is, you cited an equation that you didn’t fully understand.

Power is force times distance divided by time.
By the way, in this continuation of your previous quote, you missed the "dot product", which would have told you that when your acceleration (g) is vertical, the distance portion is ONLY the vertical component.

===================================
For convenience, here is your original posting:
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.
 
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Besoeker

Senior Member
Location
UK
It's bad enough that you didn?t understand the origins of your own equation, but to have yourself and your cronies gloat over your mistake compounds it. So let's take a look at the origin of your equation and you tell me that it still isn't a vertical lift equation.

Let's start with the general equation for power in a pump. If you need confirmation of this, you can find it in your text books or here is a convenient wiki-link.

P = Δp * Q / η , where:

Δp = (your D) = pressure differential
Q = (your Q) = Flow Rate
η = (your eta) = efficiency
Good job!!
It is in accordance with my post #5.
Power is head (pressure) times flow for the hydraulic part.
Of course you need to include efficiency for the shaft power.
And then that of the prime mover if you want wire to water.
 

Rick Christopherson

Senior Member
Preface: A new posting came in while I was typing. I have not read it yet.

I made a non-technical mistake, but it is an incorrect reference to Bes's original variables. Δp is the pressure differential, but it is not Bes's "D". Bes's "D" is density, as I correctly referenced later.

Δp = (your D) = pressure differential
 

Rick Christopherson

Senior Member
Good job!!
It is in accordance with my post #5.
Power is head (pressure) times flow for the hydraulic part.
Of course you need to include efficiency for the shaft power.
And then that of the prime mover if you want wire to water.
Yes, but your "pressure" is wrong for the discussion, and you have adamantly denied that it was due to head lift. Are you agreeing that this pressure, as you defined it, is in fact due to "lift", and solely to lift?
 

Besoeker

Senior Member
Location
UK
Yes, but your "pressure" is wrong for the discussion, and you have adamantly denied that it was due to head lift. Are you agreeing that this pressure, as you defined it, is in fact due to "lift", and solely to lift?
Not at all.
If you pump water or any fluid round a horizontal loop there will be a head or pressure difference that might, in some cases, require quite a lot of power.
 

Rick Christopherson

Senior Member
Not at all.
If you pump water or any fluid round a horizontal loop there will be a head or pressure difference that might, in some cases, require quite a lot of power.
I fully agree with this, however, it does not correctly reflect what you have been arguing for the past couple of days.

You have replaced your Δp with the variables for vertical lift (ρ*g*H = Δp), and then called it a universal pressure differential. That is the mistake that I have been contending since you drew me into this argument with posts #12 and #14.

If H=0, and the entire system lies in a horizontal plane, then your equation goes to zero. Your Δp is related solely to a vertical lift, and no other sources of pressure differential are accounted for. Moreover, you can't simply add things like flow resistance to your Head (H) because those terms would also be multiplied by "g".

To cover this mistake, you then later claimed that "g" was needed to find the "mass". We can come back to that coverup, but I think you're going to want to let that one slide.

I got the impression back in the beginning that you might be using "H" as a "garbage varaible" to add all of the various pressure differentials, but this would be incoerrct, if for no other reason, that the units didn't match. H has the units of "meters". Your other Δp's don't have these units.
 

Besoeker

Senior Member
Location
UK
If H=0, and the entire system lies in a horizontal plane, then your equation goes to zero. Your Δp is related solely to a vertical lift,
Nope. Head is head.
It seems that you are unwilling to accept the the the hydraulic power is head or pressure times flow.
That's it in simple terms. Your own post shows that to be so.
Dispute that and you dispute basic physics.
 
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