"affinity laws"... for pd pumps ?!? 2

[electricpete]

We always hear the affinity laws for centrifugal pumps:
Given two speeds N1, we have:
Q1/Q2~N1/N2, DP1/DP2 ~(N1/N2)^2, (Power1/Power2)~(N1/N2)^3

I believe there inherent in the above an assumption that the system flow characteristic curve is unchanged and follows a relationship DP~Q^2. Without assumed system there is no basis for drawing any conclusion. (if we are not moving along that curve, then please tell me how we find the two operating points at which the relationships hold).

Now let's look at a positive displacement pump. Assume piston type or gear type: a fixed volume is trapped and moved for every revolution of the shaft. It seems to me very likely that Q~N.

Now if I hook up that pump to that same system with a fixed characteristic curve DP~Q^2, the DP of pump and DP of system must match, I must have that DP~Q^2 ~ N^2.
Now look at fluid power Power~Q*DP~N*N^2~N^3.

Hmmm, looks very familiar. It looks to me like positive displacement pump also follows the affinity laws.

If I'm right, why are the affinity laws taught as applicable to centrifugal pumps, without mention of pd pumps?

If I'm wrong, what was my error?
Thx in advance.


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[smckennz]

Hi electricpete

1) We are already in agreement that DP~Q^2 for the purposes of preparing a second pump curve. From what you have previously said, you are aware that this relationship does not necessarily hold for a pump/system combination, where Q is the actual flow. This is the classic: " I reduced the pump speed by 20% and now nothing comes out the end of the pipe. According the the pump laws the flow should have reduced by only 20%" I thought this was the concept you are having difficulty with. Obviously not.

2) DP ~ Q^2 is not an assumption, it is a derivation from the pump laws as you have already shown. The pump laws can be proven but are obviously an idealised case. So your question reduces for a request to provide an example in conflict with the pump laws.

You final statement has me a little mystified inasmuch as your DP~Q^2 was derived from the pump laws, and your curves will be derived from the same laws. So the statement appears to answer itself.

The above assumes that all the "Q"s are on pump curves and do not relate to fixed system flows, in accordance with your recommendations.

I have however, found the squiggle key.

Cheers

Steve

 

[25362]

To electricpete, having read the various messages on the issue of (H2/H1)=(Q2/Q1)², I"ll refer to the true/false questions of your message of February 16th.

Since the affinity rules were arrived at by dimensional analysis assuming internal pump hydraulics independent of the Re number, and the pump/impeller set being defined by the constancy of the specific speed, I'd rephrase your first question as follows:

We cannot have (H2/H1)=(Q2/Q1)² or (H2/H1)=(N2/N1)² without (N2/N1)=(Q2/Q1) as a result ofg the constancy of the specific speed for a particular pump. BTW, the last equality, is also valid (with reserves) for rotary PD pumps.

As for the second question, in which it is stated that CP laws, as above, are based on the assumption that the system complies with (H2/H1)=(Q2/Q1)², I beg to disagree on three counts:

First, historically the pump laws were derived from dimensional analyses, in total disregard of the system characterisitcs.

Second, the law applies for centrifugal pumps, no matter what the characteristic pump curves appear to be, mostly at equal efficiencies (id est, small speed changes), and optimally at the BEP locus.

Third, although the system's friction losses are indeed proportional to f.Q², it happens that 'f' is also a function of Q, thus (H2/H1)=(Q2/Q1)^m where m=2 only applies to very special cases in piping systems.

For example, water flowing in sch 40 steel pipe with typical rugosities (taken from tables prepared by the Hydraulic Institute):

gpm nom. size, in Friction, ft/100 ft of pipe m

100 3 2.39
200 3 8.90 1.93
------------------------------------------------------------
100 4 0.624
200 4 2.27 1.91
------------------------------------------------------------
200 6 0.299
400 6 1.09 1.91
------------------------------------------------------------

For the square law (m=2) to apply in piping, the values of Re numbers should be greater than 2,500,000 for 3" pipes, 3,000,000, for 4" pipes, and 4,500,000 for 6" pipes, and so forth. For smooth tubes in which the rugosity ratios are of the order of 0.000001, Re should be larger than 100,000,000 for m=2 to apply!

For flows across staggered tubes m=1.8; if these tubes are finned, m=1.68; for tubular helices, m=1.75. All this is beside various lower 'm' values for slurries such as paper stock.

Therefore, it appears to me that, in the majority of cases, the intersections of the system curve with the pump curves (i.e., operating points), would not correspond to the "square" rule.

I have other comments regarding the exercises we both did with pump curves answering to certain types of equations, but I feel it is time I should rest my case.

 

[electricpete]

It sounds like we are coming into closer agreement. Thanks for your patience.

One thing I believe we have reached agreement on: The earlier contention that pump laws can exist in a vacuum without contemplating a system characteristic relationship I believe has been proven false. (Pump curves can exist in a vacuum, but we cannot relate them by the pump laws unless we consider a system characterstic form).

Steve you said "DP ~ Q^2 is not an assumption, it is a derivation from the pump laws as you have already shown"

Now I have to delve into the semantics. Saying DP~Q^2 is derived from pump laws (rather than pump laws are based upon assumption that DP~Q^2) seems like backward logic to me.

Do we agree on the following equivalent statements:
1 - Pump laws cannot be true unless we consider / apply / contemplate a system characteristic curve with DP~Q^2.
2 - Pump laws are true ONLY IF system DP~Q^2 is true.

If the previous statements 1 and 2 agreed on, then I'm not sure why anyone would disagree with the what I consider another equivalent statement:
3 - System DP~Q^2 is an assumption of the pump laws.

25352

You show that not all systems obey DP~Q^2. That is agreed. Therefore we cannot predict DP and Q as function of speed in these systems using pump laws.

You mention a derivation based on specific speed. I am not that familiar with the subject. It seems irrelevant to my question if we accept statements 1, 2, 3 (do you?).

If 1,2,3 are not true, then Steve or 25 please provide an example of application of pump laws which does not in some way rely on an assumption that system DP~Q^2

Interesting discussion by the way. I know it seems like we are going around in circles. But still interesting

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[25362]

I said before I have one item still to be reviewed, and that is the pump curve itself.

Take, for example, a pump with a drooping curve (regularly found with low-specific-head power-efficient pumps) as its characteristic having two flow rates for one head.

This would be a curve of the type h=a+bQ+cQ², where a (=KN²), b and c are constants for a given speed, describing a parabola having its axis parallel to the "h" axis, and its apex at Q=-b/2c. If this apex is way to the right of shutoff, the curve will show a detectable droop. Upon applying the square law, increasing speeds would move the apex to the right, farther away from shutoff.

Pump affinity rules still hold for such a pump, but how can we speak of an effective dependence or correspondence with a given system ?

Of course, a system is needed for a pump to work on, but the pump laws apply w/o the need of assuming the presence of a particular system.

In a similar manner available NPSH for a given system can be estimated without a pump being even installed, but, of course, to physically verify it, one has to have the pump.

All these seems rather philosophical. Electricpete, I'm looking forward to reading your comments.

 

[electricpete]

"Pump affinity rules still hold for such a pump, but how can we speak of an effective dependence or correspondence with a given system ?

Of course, a system is needed for a pump to work on, but the pump laws apply w/o the need of assuming the presence of a particular system."

I say we need BOTH of two 2 things for affinity laws to apply
1 - Pump with the proper characteristics. This may be positive displacement pump or example B above (DP=k1*N^2 – k2*Q^2). Perhaps there are other forms that will work but certainly not every conceivable pump curve will be able to behave according to affinity laws (example A and example C did not).
AND
2 - Assumed quadratic form of system curve.

I think your recent e-mail focus on the first aspect. Yes, there are certain characteristics of the pump required for pump laws to apply.

I think you cannot deny that #2 is also a firm prerequisite for the pump laws to apply. You have said it yourself about 10 messages ago:

"If the system's curve isn't quadratic, for example H2/H1=(Q2/Q1)m, with m not equal 2, it will intercept the pump (quasi-parallel) curves at points (one on each pump curve) not interconnected by the similarity rule."

So, we are in full agreement that pump laws doen't work when pump is used in a non-quadratic system, but we can't agree that quadratic system is a prerequisite for pump laws. It is somewhat philosophical and maybe I am being semantic, but I still don't have any comprehension of how anyone believes the pump laws are not based on assumption of quadratic system.

I go back to a repeating question which will silence my objection. Can you give me one example application of the pump laws which does NOT in some way assume a quadratic-form system?

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[25362]

I think I'm being misinterpreted. When I say "If the system's curve isn't quadratic, for example H2/H1=(Q2/Q1)^m, with m not equal 2, it will intercept the pump (quasi-parallel) curves at points (one on each pump curve) not interconnected by the similarity rule", I'm really saying the pump quadratic law applies to pump curves, but not necessarily to system curves and "operating points".

The fact that all kind of pump curves can be drawn following the square rule, whilst not all system curves need to answer to a square "H/Q" relation, is -to my grasping- a proof of the independence of the pump affinity laws from any system characteristic.

 

[electricpete]

So, we are not in agreement that the pump laws can not apply if the pump is used in a non-quadratic system?

I thought it had already been well established.A

PumpQ = System Q
Pump DP = System DP
Let's use non-quadratic system:
SystemDP = K*Q^M, m not equal 2.

Change the speed. If pump laws were true:
Q2 = Q1*N2/N1
DP2 = DP1*(N2/N1)^2 [Equation 1]

Substitute into equation 1 values for DP1 and DP2 based on system relationships (DP1= K*Q1^M, DP2=K*Q2^M_:
K*Q2^M = K*Q1^M*(N2/N1)^2 [Equation 2]

Substitute into equation 2 Q2 = Q1*N2/N1
K*(Q1*N2/N1)^M = K*Q1^M*(N2/N1)^2

Cancel out the K's
(Q1*N2/N1)^M = Q1^M*(N2/N1)^2

Distribute the power of M over the items on LHS
Q1^M*(N2/N1)^M = Q1^M*(N2/N1)^2

Cancel out Q1^M
(N2/N1)^M = (N2/N1)^2

But we started out by saying M (the exponent of our system characteristic) was not 2!

We see that the pump-laws are self-consistent ONLY IF M=2. (Only if we consider a quadratic system)

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[25362]

I think I found a conceptual error in your presentation,

This is OK:

Pump Q1,Q2 = System Q1,Q2: Operating points.
Pump DP = System DP
Let's use non-quadratic system:
SystemDP = K*Q^M, m not equal 2.

The following is not OK:

As I see it, the conceptual error appears in assuming "a priori" that both points, [DP1,Q1] and [DP2,Q2] follow the pump affinity laws. The new point for the pump law to apply should be [DP3,Q3]. Thus your Equation 1 changes as follows:

Change the speed. If pump laws were true:
Q3=Q1*N2/N1
DP3=DP1*(N2/N1)^2 [Equation 1]

One has to show that Q3=Q2 and DP3=DP2

And this could be done only when the system's
DP2/DP1=(Q2/Q1)^m, and m=2.
Since our pre-condition was that m is not 2, the point defined by [DP3,Q3], is not equal to the point [DP2,Q2].

Do you follow my thinking ?

 

[electricpete]

No, I do not follow your logic.

The following equations are meaningless to me:
Q3=Q1*N2/N1
DP3=DP1*(N2/N1)^2

What is the speed at which Q3 and DP3 are measured?
We have a speed N2 but no associated DP or Q... then what does speed N2 have to do with anything? Can you formulate this as a complete problem statement.

In general one will intepret Q~N to mean
Q2/Q1 = N2/N1.

In general one will intepret DP~N^2 to mean
DP2/DP1 = (N2/N1)^2

Have I misunderstood the meaning of proportionality? Is there another definition for proportionality that I should be aware of?

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[25362]

There are two curves, one at N1 and the other at N2, cut by one system curve at two "operating" points: (H1,Q1) and (H2,Q2). There is a third point (H2',Q2') -if you don't like (H3,Q3)- on curve N2, resulting from applying the curve affinity laws to (H1,Q1).

The "operating points", belong to, and are indeed related by, the system's curve. But it is wrong to assume -a priori- that they are also related by the pump's affinity law square equation.

The "correct" third point on pump curve (N2), called (H2',Q2'), follows H2'/H1=(Q2'/Q1)²=(N2/N1)² by the pump laws.

The object of the exercise is to show that "operating" point (H2,Q2) coincides with "pump" point (H2',Q2'). This can only be shown to be the case when H2/H1=(Q2/Q1)² on the system curve. Otherwise they'll not coincide.

Did I make myself clear ?

 

[electricpete]

Let me say what you said in my own words:

Start with one system curve and two pump curves (at speed N1 and N2). Creates operating points 1 and 2.

Apply the affinity laws to point 1 and we come up with [ficticious] point 3.

Point 3 is equal to point 2 IF AND ONLY IF the system curve follows quadratic relationship.

Repeat. Point 3 has no physical significance if the system does not follow quadratic relationship.

So what you’re telling me is the purpose of the pump laws is to predict operating points which have NOT PHYSICAL SIGNFICANCE unless the system is quadratic (in which case they will be the operating point). But in spite of this you don’t think that the pump laws assume the system is quadratic?

Tell me why on earth do we need a law to predict a ficticious point with no physical significance. Give me an example of a real-world problem that needs to be solved where we are concerned about this ficticious point.


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[25362]

Point 3 is quite real and indeed has a physical and useful significance, because it is a point that exists on pump curve No. 2. Dismissing it as fictitious, would be equivalent to dismissing the whole curve No. 2 as irrelevant and unreal.

An example: whenever the system needs a different flow rate as in point 3 -as when commissioning a flow control valve- while using pump curve No. 2, after a change in speed, it will encounter point 3 defining the new [H3,Q3] conditions.

 

[electricpete]

There are an infinite number of points on curve #2. Point #3 has no special significance other than it is the one which can be associated with point #1 assuming a square-law characteristic.

In your example, it sounds like you are controlling flow by a control valve. Yes, if flow #3 happens to be the target flow, the valve will drive it there. But again, there is no special reason to prefer point #3 over any other point.

The only reason we ever conceieved of point #3 is that it is predicted by pump laws for change in speed N1->N2 given previous point #1. It has no other significance. Sure you can attach a magical significance by inventing a demand for exactly this flow. But you could also invent a demand for any other flow on the curve.

Step back and take a deep breath and re-read our last few messages. Are you really serious or just yanking my chain?

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[25362]

To electricpete, I did just as you advised, and came to the same conclusion.

The same principles govern both, the idealized affinity laws for centrifugal pumps (CP), and the friction drops for systems on Newtonian fluids. As for the square law for pumps and the relation between H and Q on flow systems one sees that both can be arrived at by dimensional analyses.

The important difference I found is that for CP affinity laws, head is normally independent of the Re number, meaning we have rough and fully developed turbulent flow (where the Darcy friction factor doesn't change any more as a function of Re), while "system heads" are normally dependent on Re, from laminar to turbulent flow, and the H=Q^m relationship can thus change from m=1 to m=2.

"New" characteristic pump curves of various geometrical contours can be constructed from points on "existing" curves at differing speeds, and this is a practical, "real" not imaginary, procedure.

[H,Q] parabolic curves can vary, and are in fact being varied, by factors such as control valves. I sincerely thought the example you asked from us could include the FCVs.

The main conclusion I see is that the pump affinity laws and the system curves, while stemming from the same hydraulic concepts, are, however, independent from each other.

Please don't be disappointed. As for myself I feel honoured from getting your considered attention. The governing premise is that all of us are seeking the truth in this discussion. And I still think my viewpoints on this subject are not wrong.

 

[electricpete]

I appreciate your time and patience on this as well.

“New characteristic pump curves of various geometrical contours can be constructed from points on existing curves at differing speeds, and this is a practical, real not imaginary, procedure.”

I agree we can construct pump curves without any assumption on the system.
I agree during the process of generating pump curves, we will generate points on curve 2 from point on curve 1 using (Q2,DP2) = (Q1*N2/N1, DP1*N2^2/N1^2) (ie the pump laws). When we are done with the curves there is no particular reason to assume those two points remain associated with each other (unless we are making an assumption that the pump will be connected to a square-law system). There is no reason to suggest that the two pump curves are related to each other by the pump laws unless we are assuming a square-law system.

“[H,Q] parabolic curves can vary, and are in fact being varied, by factors such as control valves. I sincerely thought the example you asked from us could include the FCVs.”

I asked for an example where the pump laws could be applied without regard to requirement for system to be quadratic (prerequisite for applying the pump laws). You gave me an example where we have a non-square law piping system, but when we change speed we want to adjust the FCV to make it move to the exact target operating point (3) that it would move to IF connected to a square-law system. It is a very artificial scenario in which the pump laws are only relevant because the target point that you defined is that which would result if the same change in speed were applied to a square-law system. (It is like saying y’’=-m*g is not relevant to falling-body motion because I can create the same y’’ if I apply a total force unrelated to gravity which is numerically equal to m*g - let’s say using rocket propulsion in space). You are changing two parameters (speed and valve position) when the intent of the pump laws clearly is to describe effect of change in speed only. Change in system characteristics is not incorporated into the pump laws as can plainly be seen by the form of the pump laws (Q~N and DP~N^2)

“The main conclusion I see is that the pump affinity laws and the system curves, while stemming from the same hydraulic concepts, are, however, independent from each other. “

Pump curves of varying speeds are independent of the system curves.
Pump laws (Q~N and DP~N^2) can never be mathematically satisfied without having DP~Q^2.

Can you give me a sentence description of what the pump laws mean. We have written the equations but where do we apply them?

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[25362]

As I see it there is nothing wrong in your last message.

One must remember that the affinity laws as described until now by us, are not complete.

The impeller radius r, or diameter D, should be included.
Using (a.p.t.) for "about proportional to" the laws would include:

H (a.p.t.) r²N²
Q (a.p.t.) r³N, and as a result
P (a.p.t.) r⁵N³
T (a.p.t.) r⁵N²

Where P and T are power and torque, respectively.

It may well be to repeat that the affinity laws in this form neglect any effects of Reynolds number and that they are restricted to incompressible flow.

Applications of the affinity laws are common for pump users and makers. Changes of rotating speeds and of impeller diameters are frequently done to modify the pump's output at varying operating conditions.

The affinity laws enable also attaining a variety of specific speeds N_s=NQ^0.5/H^0.75 and specific radiuses, r_s=rH^0.25/Q^0.5 meaning differing pump designs, to approach optimum performance, i.e., the BEP's.

Can I rephrase your last sentence by saying pump laws
(DP (a.p.t.) Q² and DP (a.p.t.) N²) can never be mathematically satisfied without having Q (a.p.t.) N ?

It was not I that tried to tie up the system's and pump's characteristics by the common "square law". I brought in the FCV to show that systems change to adapt to differing process conditions. Only when these changes aren't feasible or economical any more, pump curve modifications are sought after, for example by varying the impeller diameter or the rotating speed.

 

[electricpete]

Wow. 50+ messages in one thread. I think it is now an appropriate time to draw this thread to a close.

Thanks for all the help.

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[smckennz]

Hi electricpete

perhaps you could close by summarising what you have learned through the effort put into this discussion.

My own view of a simplified centrifugal pump is a "G" machine (rotation) in series with an orifice plate(flow path resistance). The "G" portion of the machine is cabable of producing a zero flow head according to Eulers V^2/g. As flow increases, the available (outlet)head varies as the G head less the head loss across an orifice; commonly considered to vary as the square of the flow. Perhaps this is another way of looking at your H~Q^2 proposition.

Cheers

Steve

 

[electricpete]

I don't want to appear ungrateful to the many people who have genuinely tried to offer assistance (especially 25362 who offered a lot of information).

However, I don't really feel like I've learned anything. I have unsuccessfully tried to communicate the basis of my question, which I still believe to be correct. At this point the conversation is going in circles and no useful end in sight.

Thanks again to all who replied.

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