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How To Read Pump Performance Curve


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#1 Kiran T

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Posted 09 April 2015 - 12:02 AM

Dear all, 

hi....

 

I have pump with specification as under,

 

Capacity - 20 m3/hr

Head - 40 m

RPM - 2900

Imp Dia - 190 mm

 

Vendor is supply this pump with efficiency 50 %, I don't get it how ? 

Also power requirement as per vendor calculation is 6.09 KW, but  my power calculation gives 5.74 KW ...

here i am confuse to cross check the pump and how to read PPC.

 

 

 

Attached Files



#2 breizh

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Posted 09 April 2015 - 12:20 AM

hi ,
You should give info about the liquid ! is it water ? if yes you have to read info related to the blue curve ( 190 mm) , 20m3/h , H=42 M , efficiency :56% and perform calculation : P=Q*Ro*g*H/Eff

Hope this helps

Breizh

#3 ankitg009

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Posted 09 April 2015 - 04:56 AM

hi ,
You should give info about the liquid ! is it water ? if yes you have to read info related to the blue curve ( 190 mm) , 20m3/h , H=42 M , efficiency :56% and perform calculation : P=Q*Ro*g*H/Eff

Hope this helps

Breizh

What does these colored lines indicate? Like you have mentioned for water,read blue curve(190 mm). What does this 190 mm means.?



#4 gegio1960

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Posted 09 April 2015 - 06:32 AM

190 mm is the diameter of the installed impeller 



#5 curious_cat

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Posted 09 April 2015 - 08:08 AM

Using the formula breizh mentioned I get 4.1 kW & reading directly off the second graph in your vendor's document I'm getting 4.2 kW. 

 

Close enough, I guess. 



#6 fallah

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Posted 09 April 2015 - 09:26 AM

Kiran T,

 

Please send also the pump data sheet being filled by vendor. It might the power submitted by vendor is nominal motor power rather than required hydrualic power...



#7 MTumack

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Posted 09 April 2015 - 11:11 AM

Dear all, 

hi....

 

I have pump with specification as under,

 

Capacity - 20 m3/hr

Head - 40 m

RPM - 2900

Imp Dia - 190 mm

 

Vendor is supply this pump with efficiency 50 %, I don't get it how ? 

Also power requirement as per vendor calculation is 6.09 KW, but  my power calculation gives 5.74 KW ...

here i am confuse to cross check the pump and how to read PPC.

 

I think this is totally fine.

 

You calculated your fluid needs 5.74 kW, your vendor is saying your motor needs to supply 6.09 kW.

 

Thats an Efficiency of 94.3%. Thats pretty reasonable mechanical losses when we are talking about a motor spinning, then the bears on the shaft, the shaft coupling to motor (or even perhaps a belt/chain drive system) then moving into your pump shaft being rotated (again having bearing friction) and actuating whatever pump fluid component is being used (be it a piston, gears, centrifugal components) and the friction of those rotating with respect to their bearings / casings.

 

There is no cause for concern at all, in my opinion. I'd be much more concerned if your pump guy told you his pump was 100% mechanically efficient.



#8 Kiran T

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Posted 10 April 2015 - 06:28 AM

hi ,
You should give info about the liquid ! is it water ? if yes you have to read info related to the blue curve ( 190 mm) , 20m3/h , H=42 M , efficiency :56% and perform calculation : P=Q*Ro*g*H/Eff

Hope this helps

Breizh

Dear breizh Bhai,

The pumping liquid is not water,  its crude having specific gravity 1.45. From my calculation power requirement is 5.74 KW, i use same equation as you suggest.  Here i am confuse that what will be effect of density on pump head ? Will it deliver same head (40 m) for specific gravity 1.45 ??  



#9 Kiran T

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Posted 10 April 2015 - 06:37 AM

Kiran T,

 

Please send also the pump data sheet being filled by vendor. It might the power submitted by vendor is nominal motor power rather than required hydrualic power...

Dear fallah,

Tech data sheet of pump provided by vendor is as under,

 

 

TECHNICAL ANNEXURE

LIQUID TO BE  HANDLED

SUITABLE FOR SS-316

QUANTITY

01 NOS

SPECIFIC GRAVITY

1.45

VISCOCITY

7 CP

TEMPARATURE

70 DEG CEL

PUMP MODEL

SCP-SM-25-150

SUCTION SIZE IN MM

40 MM

DISCHARGE SIZE IN MM

25 MM

BID IMPELLER DIA

190 MM

IMPELLER RANGE

145-205 MM

IMPELLER TYPE

SEMI OPEN

HEAD IN MLC REQD

40 MLC                         

OFFERED :-   40 MLC

CAPACITY  REQUIRED IN M3/HR

20 M3/HR

OFFERED: -   20 M3/HR.           

EFFICIENCY

50%

 

BKW WITH SP. GR of 1.45 at DP

6.09  KW

 

MAX BKW WITH SP. GR of 1.45

7.25  KW

 

MOTOR REQUIRED

12.5 HP / 2900 RPM

MOTOR RECOMMENDED

12.5 HP / 2900 RPM

 

THRUST BEARING NO

3305/SKF

RADIAL BEARING NO

6306/SKF

COUPLING

RATHI(LOVE JOY )MAKE TYPE :- FLEXIBLE/SPACER

MECH.SEAL

GLAND PACKING (NOT IN OUR SCOPE)

 

MATERIAL OF CONSTRUCTION:

CASING

SS-316 (JACKETED)

IMPELLER

SS-316

SLEEVE

SS-316

BACK PLATE

SS-316

SHAFT

SS-316

COUPLING

CAST IRON

BASE PLATE

M.S. FABRICATED



#10 breizh

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Posted 10 April 2015 - 06:58 AM

Hi ,
Consider reading this document to support your work.

Breizh

#11 fallah

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Posted 10 April 2015 - 07:07 AM

Kiran T,

 

Vedor is right to propose 6.09 kW, because with considering the effect of specific gracity of 1.45 on pump power, 4.2 kW from the curve of pumping water (the frist performance curve you attached) has been multiplied by 1.45 that will result in 6.09 kW (4.2*1.45=6.09)



#12 Kiran T

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Posted 11 April 2015 - 04:59 AM

Thank you

fallah and breizh bhai.



#13 P T R Gupta

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Posted 11 April 2015 - 08:03 AM

In the pump curve, what does head mean? Is it pressure head alone or is it combination of pressure head and velocity head?? I am asking this because, when we say pump operates at intersection of system curve and pump curve, head on system curve represents both pressure head and velocity head. Kindly clarify me on this.



#14 fallah

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Posted 11 April 2015 - 11:00 AM

Gupta,

 

Actually it's mostly pressure head because the velocity head, even though the size of discharge nozzle/line of the pump is usually lower than those of the pump suction, would be supposed to be offset across the pump...


Edited by fallah, 11 April 2015 - 11:02 AM.


#15 P T R Gupta

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Posted 11 April 2015 - 12:58 PM

Dear Fallah

thanks for your kind reply. But how does it justify that intersection of system curve and pump curve is the operating point. System head includes both pressure head and velocity head. How come pump head include only pressure head?



#16 fallah

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Posted 11 April 2015 - 02:09 PM

Gupta,

 

In fact, the operating point is where the total differential head (static head, pressure head and velocity head) between discharge line and suction line is equal to the total differential head (static head, pressure head and velocity head) produced by the pump which is located inbetween. Then, as you can see both system differential head and pump differential head included differential velocity head; but as i mentioned due to its low value would be ignored to be included in total head. Obviously if in a case, its value would be considerable, it could be come into play for total differential head calculation for system, for pump or for both items...






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