12 replies to this topic

[NoobMi]

My task is to upsize a centrifugal pump due to increased flow rate requirement. For this, i will need to calculate the differential head required.

Info:
Existing pump flow rate = 10m3/hr
Existing Pump calculated head req. = 20m
New Pump flow rate required = 20m3/hr
New Pump head required = ?.

For this case, can i just use 20m*4, since P is proportional to velocity^2? (assuming turbulent conditons).

[Zauberberg]

The head requirement will increase for the amount of additional pressure drop arising from the increased flow, all other things being the same (source and receiver pressure, number of valves/fittings, elevation profile, etc.).

 

If you intend to use the same pump, this might not be possible - check what is the maximum impeller diameter that can be fit into the existing pump casing, and if this is sufficient to provide you with the required flow. If yes, then check the motor and see if it can pull that many Amps you need for 20 m³/hr.

[NoobMi]

The head requirement will increase for the amount of additional pressure drop arising from the increased flow, all other things being the same (source and receiver pressure, number of valves/fittings, elevation profile, etc.).

 

If you intend to use the same pump, this might not be possible - check what is the maximum impeller diameter that can be fit into the existing pump casing, and if this is sufficient to provide you with the required flow. If yes, then check the motor and see if it can pull that many Amps you need for 20 m³/hr.

 

Thanks for the input.

I do understand that re-using the same pump may not be possible. Therefore i would need to change a new pump. And for a new pump, i will need to provide the head required to the vendor.

So what i meant was given all things remain constant, except for the change in flow rate, would the increase in head requirement be proportional to (change in velocity)^2.

 

I have actually calculated the pressure drop across the line and got like 40m head required (increase from 25m head requirement) when the flow rate increased by 2 times. This calculated value is much lower if i consider P proportional to v^2. Therefore not sure if my calculations is accurate or not

Edited by NoobMi, 11 November 2013 - 07:46 AM.

[Zauberberg]

What does pressure head mean? It is the pressure drop of flowing fluid, expressed in terms of height of the equivalent liquid column. That is your additional head required.

 

The relationship between fluid velocity (or velocity²) and total head requirement is not directly proportional because there are other additive factors in the equation which contribute to the overall head requirements for the pump. These additive factors are elevation profile, source and receiver pressure - and these are independent from the flowing velocity.

[fallah]

NoobMi,

 

Simply you can extend the system curve beyond the existing pump curve till to reach the flowrate around 20 m³/hr. Corrresponding head of this point coud be reported to the vendor as new required head...

[breizh]

Hi ,

To add to the posts above,

 

The system curve is a parabola :TDH = a+b*Q^2   ; a = difference altimetry ( destination-source) ; b :coefficient related to head loss ;

Q :flow rate  .

 

Manipulating the equations , you can get access to TDH (20m3/h) vs TDH (10 m3/h) , knowing  a .

(TDH(20) - a)/(TDH(10) -a)  =( 2*Q1/Q1)^2 =4 .

 

Hope this helps

Breizh

[S.AHMAD]

the equation by Beizh is  allso known as system performance curve....

[gregga]

why don't you just solve the bernoulli equation for your system to calculate the head requirement?  you will need to provide flow rate and head to a pump vendor (along with physical properties) to a pump vendor.

[NoobMi]

thanks for the information.

im currently finding the minor losses and i have a question. if my pump is pumping fluid from tank to a height of 10m elevation and back to the tank, do i need to include the losses for the flow from the top back to tank?

[fallah]

NoobMi,

 

Please upload a simple sketch of the system. But, in general to calculate the pump head you should include the frictional loss of piping system...

[NoobMi]

Hi fallah,

 

Please find the sketch in the attachment as requested. Appreciate your advice on this.

Attached Files

•  Minor Loss.xls   47.5KB   34 downloads

[fallah]

NoobMi,

 

Pump head is equal to differential static head (static head from end of the discharge line till liquid surface in the tank as per the sketch) plus corresponding head of the friction loss in whole suction/discharge lines might rarely be more than 10 m in steady state conditions based on the sketch; but the pump shall be capable of generating the head more than 10 m to be able to circulate the fluid in the cycle at start up conditions...

[S.AHMAD]

1. Based on the diagram given, let say the return line inlet to the suction tank is point-2 and the liquid surface in the inlet tank is point-1. The pressure at respective and elevation are P2,Z2 and P1, Z1 respctively.The pump head equation should be:

 

Pump head = (P2 + Z2) - (P1 + Z1) + (Frictional Loss piping and the heat exchanger.)

 

(In the above equation, the terms are in the same unit e.g. m of liquid)

 

As you can see that P1 = P2 but Z1 is not equal to Z2 and Z1 depends on tke tank liquid level.

 

Assuming the frictional loss is neligible the pump head required is reduced to:

 

Pump head = Z2 - Z1

 

2. However as fallah said, make sure that the pump can pump the 10 m elevation differential at start-up since the discharge line is empty at start-up. However if the line is already full of liquid at start-up, the the differential pump elevation at point-2 and point-1 is the pump head required.

 

3. Of course in real world, frictional loss is normally significant

Edited by S.AHMAD, 17 November 2013 - 08:16 PM.