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4 quadrants and 8 zones of pump operation

Hi everyone

I am researching about complete characteristics of Pump. The diagram (Knapp complete characteristic diagram) has 4 quadrants and 8 zones totally . Some zones are not clear and I could not comprehend them completely. Is it possible for you to have discussion about the diagram? The main article (which is written by R. T. Knapp) is attached.

Have a look at P. 684 and Fig. 2. The first question of mine is that how can 'Positive Head with Negative Discharge' possible? Negative Discharge means that the generated head (of the pump) can not overcome head at discharge of the pump, which results in negative flow (From discharge to suction side of the pump). 

What do you think about this?

H.

     

  • Hello Hoshi,

    Are you using Bentley HAMMER? It sounds like this is more of a theory question, unrelated to Bentley software.

    It might be better to seek other scholarly articles on the subject, but below is my initial interpretation. If we think of the Head as the increase in HGL relative to the pump orientation, your example holds true. I've inserted a diagram below in an attempt to help illustrate my thoughts.

    In the case of a pump whose impeller speed is zero ("0 RPM" curve in Fig. 2), reverse flow through the pump could cause a headloss through the pump impeller. Since the flow is reverse to the pump orientation, the drop in HGL from the headloss shows as the HGL on the discharge side being higher than the HGL on the suction side. This is a positive head though, since the HGL goes from high to low in the direction of the pump orientation.

    In the case of a pump whose impeller speed is non-zero (for example "3100 RPM" in Fig. 2), a situation could arise whereby a transient event causes the downstream HGL to increase beyond the shutoff head. The pump impeller may still be spinning, but unable to overcome the downstram head. So, flow passes in reverse through the pump even when the pump is spinning. In this situation, the discharge HGL is once again higher than the suction side HGL, so there is a positive head across the pump (HGL goes from high to low in the direction of the pump orientation.)

    Other contributors may be able to confirm my assessment and/or add more thoughts.

    .


    Regards,

    Jesse Dringoli
    Technical Support Manager, OpenFlows
    Bentley Communities Site Administrator
    Bentley Systems, Inc.

  • Jesse

    Thank you very much firstly.

    As you mentioned, this topic is one of your product Bentley HAMMER's subjects. But I want to know more about it than HAMMER (besides I have some questions of HAMMER e.g. how is pump curve calculated by specific speed. I will ask this later. But I have some theoretical questions now).

    Also as you advise, I am researching this issue from several sources like: Knapp article, Shawn Batterton's book and Hanif Chaudhry's book. But have some problems yet.

    I ask Dr. Walski to contribute this discussion.

    Let me say my thought about 4 quadrants of every centrifugal pump characteristic curves (regarding Below photo). Then if you have another idea, correct my thought:

    A) Up-Right: normal operation of pumps (at Steady condition). Both Q & N are positive. Generated head is positive in this situation.  

    B) Down-Right: In this quadrant, RPM is positive and it means that pump generates positive head, but whereas discharge head is beyond suction head (because of a transient event which has a high pressure), flow moves from Dis side to Suc side of pump and pump can not transfer fluid from Suc. to Dis. side, in spite of positive Head. generated head is positive in this situation.It was described by Jesse.

    C) Down-Left: Pump behave like turbine exactly. Rotation of pump in quadrant B. is decreasing until 0 rpm. then it rotates in reverse and becomes turbine. Generated head is NEGATIVE, because pump operates in reverse. Am I right?

    D) up-Left: Regarding condition C, Because of  transient event, flow moves in positive direction but impeller still rotates in negative direction. Generated head is NEGATIVE. Am I right?

    Please let me know your ideas about above descriptions. After it we can discuss on Knapp's Diagram as shown below:

  • This is a bit beyond my expertise, but I may be able to add some additional input. It seems to me that an important consideration is whether the "head" being discussed is oriented in the direction of flow or in the direction of the pump orientation.

    I also found Fig. 6 in your PDF to be a bit more helpful to visualize compared to Fig. 7.

    Some comments/responses:

    Re: "B) Down-Right: In this quadrant, RPM is positive and it means that pump generates positive head, but whereas discharge head is beyond suction head (because of a transient event which has a high pressure), flow moves from Dis side to Suc side of pump and pump can not transfer fluid from Suc. to Dis. side, in spite of positive Head. generated head is positive in this situation.It was described by Jesse."

    [Jesse] Although (as seen in Fig. 6 in your PDF), this entire quadrant experiences positive head, if the "head" we talk about is the difference between the discharge and suction side HGL, then it doesn't necessarily mean that the pump is "adding"/"generating" this head. In this quadrant, the impeller speed is positive but the flow is negative. This is indeed described in my previous response where I used the 3100 RPM curve as an example, though the previously mentioned figure was showing quadrants for head and flow, whereas your most recent post is about quadrants of speed and flow.

    Re: "C) Down-Left: Pump behave like turbine exactly. Rotation of pump in quadrant B. is decreasing until 0 rpm. then it rotates in reverse and becomes turbine. Generated head is NEGATIVE, because pump operates in reverse. Am I right?"

    [Jesse] It may depend on the perspective. From the perspective of the pump orientation, this would be a positive "head", because the discharge HGL is higher than the suction side. See solid lines (of constant head) in your Fig. 7 which are all positive. So, although the pump may not necesarily be "generating" head/energy, there is an increase in HGL going from suction to discharge side. Since the speed and flow are negative, this differential may be interpretted as headloss/resistance as the flow moves in reverse through the pump/impeller.

    There is also the element of the applied torque (ex: whether the pump is on or off) to complicate things further. See dashed lines in your Fig. 7.

    Fig. 6 in the PDF you provided is also helpful to visualize the areas of speed, flow, head and torque.

    Re: "D) up-Left: Regarding condition C, Because of  transient event, flow moves in positive direction but impeller still rotates in negative direction. Generated head is NEGATIVE. Am I right?"

    [Jesse] I don't believe that the "head" would necessarily be negative. Again, I believe the interpretation of "head" in this case refers to the difference in HGL from suction to discharge, which may not necessarily be due to energy added from the pump. Note the "zero head" line in the Fig. 7 from your previous reply. There are lines of negative constant head and positive constant head within this quadrant.


    Here is an article in our Wiki regarding the source of the 4-quadrant curves used in Bentley HAMMER, based on the selected Specific Speed. If you can locate a copy of the source documents, that might help in your research.

    Source of the default Specific Speed for pumps and turbines


    Regards,

    Jesse Dringoli
    Technical Support Manager, OpenFlows
    Bentley Communities Site Administrator
    Bentley Systems, Inc.

  • Hello,

    I think it made me confused :), then let's discuss step by step:

    1- At the quadrant I (up-right) of Fig. 7. of Knapp's diagram, pump operates in its normal conditions. Normal condition results positive head, not negative one. Then when H<0 occurs in this quadrant? What I can say about H<0 is that pump discharges fluid to atmosphere. In this case HGL of discharge becomes lower that HGL of suction side (which means H<0). Is it true?

    2- I thought while impeller rotates in normal rotation, then the head (generated head I mean) is positive. But from your explanations, what I understood about sign of Head is, when we talk about H>0 it means discharge HGL is higher than suction HGL. I hope I understood it correctly. Am I right?

    3- At the quadrant IV (Up-Right), both positive and negative heads occurs. I have not any idea about either positive or negative head. When do H>0 and H<0 occurs in this quadrant?

    4- Regarding your Wiki mentioned, let me ask (also it always was a question of mine), how is any characteristic curves calculated by having specific speed? Assume we know specific speed of a pump, full speed of it, head and flow of it at best efficient point and have inertia of both pump and motor. Now how can calculate and draw complete characteristic curves of the pump (like Knapp's diagram)?

    5- The final question is when inertia of pump and inertia of motor and inertia of other equipment are determined (in kg.m^2), How can calculate total inertia? The total inertia is sum of them? Or multiply of the?

    By the way, How can I send message for Dr. Walski and get some help? Is not he there?


    Thank. H.

  • I discussed briefly with Dr. Walski and he did not have anything to add.

    This is quite an advanced topic. Can you provide some background on why you are doing this research? If it is for academic curiosity, you might be better of tracking down the basic research reports and papers. If it is for a particular real system, you may want to hire a consultant who is an expert in this area.

    "1- At the quadrant I (up-right) of Fig. 7. of Knapp's diagram, pump operates in its normal conditions. Normal condition results positive head, not negative one. Then when H<0 occurs in this quadrant? What I can say about H<0 is that pump discharges fluid to atmosphere. In this case HGL of discharge becomes lower that HGL of suction side (which means H<0). Is it true? "

    [Jesse] Take a look at Fig. 6 - there are three "Sectors" within that quadrant, abnormal turbine, energy dissipation and normal pump operation. Normal pump operation is in the area where both the head and torque are positive.

    "2- I thought while impeller rotates in normal rotation, then the head (generated head I mean) is positive. But from your explanations, what I understood about sign of Head is, when we talk about H>0 it means discharge HGL is higher than suction HGL. I hope I understood it correctly. Am I right?"

    [Jesse] Yes, that is my interpretation.

    "3- At the quadrant IV (Up-Right), both positive and negative heads occurs. I have not any idea about either positive or negative head. When do H>0 and H<0 occurs in this quadrant?"

    [Jesse] The upper-right quadrant is quadrant #1. See Fig 6 on the previous page, which more clearly identifies the areas within this quadrant where positive or negative head occur. In quadrant IV, both torque and head are always positive.

    "4- Regarding your Wiki mentioned, let me ask (also it always was a question of mine), how is any characteristic curves calculated by having specific speed? Assume we know specific speed of a pump, full speed of it, head and flow of it at best efficient point and have inertia of both pump and motor. Now how can calculate and draw complete characteristic curves of the pump (like Knapp's diagram)?"

    [Jesse] You can use some basic information like what you mentioned to estimate the specific speed of a pump. See wiki article here:

    Estimating the Specific Speed of a Pump or Turbine (Solution 500000074795)

    However, the complete four-quadrant characteristics cannot be computed with that information alone. The specific speed is linked to a full four-quadrant curve, stored in a text file. You can read more about this in the HAMMER Help. So, you would typically estimate the specific speed and pick the closest one from the list available in HAMMER. If this is not satisfactory, you can enter your own four-quadrant pump curve in the text file. However, the information necessary to do so is often very difficult to come by. You might start by contacting the pump manufacturer.

    "5- The final question is when inertia of pump and inertia of motor and inertia of other equipment are determined (in kg.m^2), How can calculate total inertia? The total inertia is sum of them? Or multiply of the?"

    [Jesse] From the HAMMER help topic "Pump Inertia":

    The value of inertia you enter in Bentley HAMMER V8i must be the sum of all components of the particular pump which continue to rotate and are directly connected to the impeller, as follows:

    • Motor inertia—typically available from motor manufacturers directly, since this parameter is used to design the motor. The pump vendor can also provide this information.
    • Pump impeller inertia—typically available from the pump manufacturers’ sales or engineering group, since inertia is used to design the pump.
    • Shaft inertia—the shaft’s inertia is sometimes provided as a combined figure with the impeller. If not, it can either be calculated directly or ignored. Entering a lower figure for the total inertia yields conservative results because flow in the model changes faster than in the real system; therefore, transients will likely be overestimated.
    • Flywheel inertia—some pumps are equipped with a flywheel to add inertia and slow the rate of change of their rotational speed (and the corresponding change in fluid flow) when power is added or removed suddenly.
    • Transmission inertia—some pumps are equipped with a transmission, which allows operators to control the amount of torque transmitted from the motor to the pump impeller. Depending on the type of transmission, it may have a significant inertia from the friction plates and the mechanism used to connect or separate them.


    Regards,

    Jesse Dringoli
    Technical Support Manager, OpenFlows
    Bentley Communities Site Administrator
    Bentley Systems, Inc.