maximal head

[YuriB]

Can someone please assure me that the no flow point on a head flow diagram for a centrifugal pump correspond indeed to the maximal head ? The reason I ask is because I want to replace HUPA 25-7.0 U 180 with HGPA 25-7.0 U 180. For the both the no flow (maximal head on the diagramm) is 7 m, only the second is 3 times more powerfull and, for instance, if at the head 5m the first produces about 2 m3/h, the second at the same head pumps about 4 m3/h. Is there I am missing something and my more powerfull pump could burst the system (produce the head more than the mentioned 7m), if a valve get closed?

 

[1gibson]

No flow on the curve is the maximum differential head (pressure difference across the pump) but the for maximum discharge head (pressure that comes out of the flange) you must add maximum suction pressure. And consider it at maximum specific gravity.

 

[LittleInch]

1gibson is correct. By the look of the halm website, these are circulation heating pumps. Hence if the rest of the system remains the same, you may not get that much more flow due to the friction going up as square of the velocity (2m3/hr to 4m3/hr needs 4 times more power unless you make the pipes bigger). These are centifugal pumps, not PD pumps so adding a bigger pump may not do what you anticipate.... The thing you may notice is that the steady pressure will increase to force more flow through the system if the rest of the piping remains the same. 7m is only 10 psi in water so I don't think you're in that much danger.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[YuriB]

Thank you for your prompt answers !
Imagine the both pumps to be sucking from an open to atmoshere source into a pipe and then suddenly a valve on the pipe closes. I understand well that the slugs would be very different in the two pipes, but then, the statics on both are indeed 7m ? (although the pumps produce very different thrusts on the liquid) Where is the trick ?

 

[YuriB]

Sorry, I re-post to be more clear.
Imagine the both pumps are pumping liquid from an open to atmoshere source, each into its pipe. Then, suddenly valves on the pipes closes. I understand well that the slugs would be very different in the two pipes, but then, the statics on both are indeed 7m ? (although the pumps produce very different thrusts on the liquid) Where is the trick ?

 

[YuriB]

Starting to answer myself my own qwestion. The speed of rotation of the both pumps is the same. But the impellers should be different - and the pump with the bigger one, logically, has to have a more powerfull motor. But although, when running, the more powerfull pump throws more liquid, when the flow get obstructed (valves close), the bigger impeller pump starts to pass back more liquid - just because there is left (intentionally) around it proportionally more area than around the impeller of the less powerfull pump. Am I right and such is the 2 different pumps' design - although the (working) heads produced are different, the no-flow head is similar ?

 

[bimr]

You seem to be asking about water hammer. If you have a water hammer situation, then the pressure may be higher than the pump dead head.

Water hammer is related to the water velocity in the pipe.

Water hammer is usually designed out of the system by controlling the time that it takes a valve to close, lower fluid velocities, etc.

 

[YuriB]

No, no its not about a slug or hammer. My qwestion was whether a pump with about 2 times greater flow capacity than a second pump will indeed have the same maximal (7 m., in my case) head at no flow (for instance, with the valve at the discharge closed). So follow from the diagrams of the two pumps, but I am not sure, anyway. So, not dynamic pressure (which at the moment of closing are of course different for the pumps), just static pressure.

 

[LittleInch]

Two of us have already said it will have the same head. What more do you want? The impellor of the larger pump is just twice as wide, but runs at the same speed and will have the same profile and diameter so the potential energy at the tip is the same. Only difference is one has a surface area twice as big as the other one. Don't know what else to say.... Oh one has a bigger motor.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[YuriB]

Thank you. I am going to change one at a cite where some idiots installed small diameter pipes. The heater on the AHU thus receiving only about 0.7 m3/h of water (if my calculation is right) instead of the necessary 2 m3/h. Thanks again.

 

[bimr]

Why would you think that the no flow point on a centrifugal pump operating curve would not correspond to the maximal possible head?

To obtain optimum pump service life, select a pump that operates close to the BEP.

Most piping systems are rated substantially above 21 m design pressure.

 

[LittleInch]

YuriB, As I noted above you may find that a larger pump doesn't increase flow by as much as you think. You may need to put in bigger pipes instead or as well. Either that or put in a pump with a higher head output.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[YuriB]

It seems that the system was designed for steel pipes DN 50 but the "installers" put PPR pipe DN 50. Thus, smaller ID, big losses. The half of all the pipes are consealed under the walls - no possibility to redo. But I was also thinking about putting a !0 m pump. Thank you !

 

[Artisi]

please explain DN 50 versus PPR 50 - what is the difference in internal diameter. Have you established the head and the flow at the pump discharge when in operation, forget about the closed valve conditions at this point - all that is needed is facts not guesses or assumptions, from there we maybe able to offer something meaningful.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

DN 50 steel is actually 60mmOD, say 52mm ID, PPR - basically polymer pipe - I'm not familiar, but it's commonly used in heating systems, OD is 50, ID about 33mm. Relative square area is a 1: 2.2, hence frictional ratio about 1: 4 to 5 Any sort of reasonable length and it's not surprising yuriB hasn't got the flow, but what is needs is both bigger pipes where possible and more pressure. A bigger pump but with the same shut off head will make things a little bit better, but not as much as you need.

2m3/hr is a bit less than 1m/sec so you should be able to get that through a 50 OD pipe, but with more pressure.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[YuriB]

In short, I think, they planned the system with steel pipes but finally put instead PPR with the same OD. Hence big losses. There were other mistakes - idiotic - like two pumps working ... towards each other ! (this I had corrected already), in the heat exchanger (water/glycol) 2 flows moving in the same direction (I am going to corect it), and some others.

 

[Artisi]

Head loss thru pipes varies as the diameter ^5
therefore using the number supplied by littleinch for ND50 and PPR50 the head loss thru PPR 50 as compared to ND 50 is;

increase in Hl == 55/33^5 = 12.8 times the head loss thru the 50 PPR pipe / ND 50.

There still isn't enough info from yourself to make any major recommendations.

you need to supply head and flow thru the system with the existing pump/s - the change in elevation from inlet to discharge, the performance curve for the existing pump and for the "new" pump - from here we can calculate some operating conditions for both pumps.
Alternately, employ an engineer experienced in pump systems who can sort it out for you.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

Artisi - For info, why are using the ratio to the power 5. I would use the power 4 for a rough calc. Assuming the flows are 0.7 versus the design of 2, the relative observed friction (proportional to velocity squared) is just over 8. The additional head loss between 55mmId vs 33mm Id is also about 8 when you use the power 4. So the decrease in ID seems to be in proportion to the decrease in flowrate. However it does seem to indicate that yuri would need a differential head of 8 x what he has at the moment in order to get the flow from 0.7 to 2.... There's no substitute for size.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[Artisi]

Why, because head loss is D1/D2^5 not ^4

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

My working of this is that head loss is proportional to velocity squared. Velocity for the same liquid flow rate is proportional to the diameter (radius) squared. Thus HL is proportional to diameter ratio to power 4. Where does the extra power come from? At these sorts of differences (>10), there are other factors involved which make any HL difference not exactly in proportion to diameter ratio, but I'm open to seeing where my logic is awry. I've looked at specific examples before and found that the power 4 sometimes over estimates the pressure loss between ID's.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way