11 replies to this topic

[sai56]

In reciprocating pump as piston moves forward, liquid gets compressed. Hence work input is required. This work is equal to change in pressure multiplied by volumetric flow rate. When liquid pressure equals set pressure of the valve , discharge valve opens and liquid starts flowing out.
From this point onward till piston reaches TDC, what is the power required?
Is it only to overcome frictional losses or power required to push piston against reaction force on piston?

[latexman]

 Capture.JPG   34.4KB   0 downloads

 

Brake HorsePower

U.S. Gallons Per Minute

Pounds per Square Inch

EFFiciency

 

 

[snickster]

In reciprocating pump as piston moves forward, liquid gets compressed. Hence work input is required. This work is equal to change in pressure multiplied by volumetric flow rate. When liquid pressure equals set pressure of the valve , discharge valve opens and liquid starts flowing out.

 

Another way to look at it is that the entire PD pumping process is a constant volume process with an incompressible liquid.  Fluid enters the cylinder at upstream pressure to fill the cylinder volume.  The piston then compresses the fluid causing the fluid to reach downstream pressure but the volume change is very negligible and slightly adds to the P(delta V) work and this work is counted for in the efficiency factor.  Once the pressure reaches discharge pressure the dischage valve opens which is more of a check valve that opens when pressure on upstream is greater than downstream, there is really no set pressure per se.  From that point the ideal work is developed by the piston pushing the volume out of the cylinder at constant pressure against the downstream friction forces.  The governing equations are:

 

Work Pump Wp = Energy out (Hout) - Energy in (Hin)

 

Hin = P1V1 + KE1 + H1  Where KE (kinetic energy) and H (potential energy) differences are zero

Hout=P2V2+ KE2 + H2

 

Wp = P1V1 - P2V2  Consider that V1=V2 then

 

Power Pump = ((P1-P2) V)/Time = (dP )(Q))/t  in units of ft-lb/sec with P in lb/ft2 and Q in ft3/sec.  With P in lb/in2 and Q in GPM and Power in horsepower, the equation becomes:

 

Power = (P(144)(Q))/(7.48(60)(550)) = PQ/1714  Ideal water power

 

Power Actual = (P(144))/(1714(eff))  where eff = pump efficiency

 

Another way to look at it is Work = delta (PV) = P dV + V dp per calculus relationship,  but since dV = 0 then

 

Work = V (dP) = V (P1-P2) which gives same results when working out derivation of power equation as above.

 

 

From this point onward till piston reaches TDC, what is the power required?

 

It is the entire power output of the pump as shown above considering that the pump is pumping against a differential pressure of outlet pressure minus input pressure.

 

 

Is it only to overcome frictional losses or power required to push piston against reaction force on piston?

 

Yes the power is only to overcome frictional forces as it pushes the fluid out the cylinder and into the pipe.  The reaction forces on the piston is from the friction forces in the pipe downstream.  This is what the piston is pushing against.

Edited by snickster, 02 March 2023 - 07:19 PM.

[breizh]

Hi,

You may find some interest reading this PPT presentation.

Breizh

[Bobby Strain]

You surely recognize that the piston moves at varying speed. The simple relationship describes the complete cycle.

 

Bobby

[sai56]

Dear Snickster ,
Thnks for your nicely explained reply. My further question is about viscous fluid. If the fluid is very viscous then can we safely ignore the frictional work required while calculating total power for the pump. I do not know how to attach a file here. I have created a small picture of power demand as the piston moves from TDC to BDC and back to TDC. But not able to post it.
Only frictional work is required when piston starts moving from TDC to BDC I. e during suction cycle.
Peak work is required when piston starts compressing the fluid. As the piston starts moving back from BDC to TDC. Pressure rises almost instantaneously and discharge valve opens. This happens almost within 1 degree of crank movement. So for furthe movement from 181 degree to 360 degree, driver needs to provide only frictional power.
So we need to size electric motor for this spike that occurred for a second. Is it possible to provide flywheel to dampen this spike and average out electric motor load?
Thnks in advance.

Attached Files

•  Motor power demand.PNG   30.69KB   0 downloads

Edited by sai56, 04 March 2023 - 08:31 AM.

[latexman]

 

I do not know how to attach a file here.

 

EDIT your post, click USE FULL EDITOR, under the text window is the file attachment functions.

[snickster]

I am not sure what you are trying to accomplish.  Are you designing the pump?  Why are you getting so deep into the details of what is going on related to energy/power consumption during each stage of the cycle?

 

I am a mechanical design engineer and am more used to sizing a pump, so using the design criteria and equations I determine the required flow of the pump at pressure.  This gives an ideal water horsepower and the actual horsepower if you multiply by pump efficiency at this operating point.  Manufactures provide pump curves that show efficiency and what range of horsepower required for design differential pressure and flowrate. Then I select the pump and prepare a data sheet for puchase for a given project.

 

If you are getting into the specific detail design of the pump then I am not a pump expert.  However looking at your diagram I have the following comments.

 

If the fluid is viscous then it would cause the frictional loss in the piping to be greater, and possibly cause a reduction in efficiency of pump so why would you ignore it?

 

"only friction work on suction intake": The pump itself does not suck the fluid in, in cases where the fluid is under positive pressure like tank head.  The suction stroke of the pump creates and increase of volume which is filled by the suction liquid under its own postive pressure unless the pumps intake is so small that a negative pressure is produced in the cylinder, but that resulting pressure will be based on how much delta P gives the resulting equillibrium flow to fill the cylinder.  So I don't think the power you have on the suction stroke will exist as provided by the pump but it will be provided the fluid itself.

 

If the fluid is compressible than power would be required to compress it.  Power will be required to compress the liquid also but it would be based on differential pressure, area of piston and distance piston moved.  The pressures are low and the movement of the piston to compress and incompresible liquid would be very small so the work/power must be very small precentage of overall power.

 

Most of the power from fluid flow will be in the discharge stroke after the fluid is compressed.  This occures at constant pressure and approximately the full stroke of the piston in accordance with equations I already posted.  The pressure force of the pump pushing the liquid is balanced by losses in the downstream line including frictional, dynamic KE ,and elevation difference, all additing up to produce a back pressure at the pump.

Edited by snickster, 05 March 2023 - 11:17 PM.

[breizh]

Hi,

Consider this website to support your query:

 https://www.pumpfund...e.htm#download6

 

Not sure about what is your scope of work and your questions. Share with us the big picture.

 

Breizh

[sai56]

Dear Breizh,
I am trying to study pattern of work input for reciprocating pump. During suction stroke it is obvious that work required is only frictional work which is negligible.
As the fluid is incompressible, pressure will rise within a degree of crank movement ( from 180 degree to 360 degree). So electric motor will be loaded maximum during this period. I have checked pressure rise in CFD analysis and noticed that pressure rises within fraction of crank movement.

After the discharge valve opens ( due to pressure developed), again the piston needs to work against frictional resistance offered by discharge system. So for rest of the crank movement till 360 degree position ( I.e. TDC) electric motor will be loaded to minimum value equivalent to frictional losses. Hence I tried to develop artistic representation of demand faced by electric motor in above sketch.
I agree that total work demand will remain fixed as per formula change in pressure multiplied by volumetric flow rate.
But actually motor is facing this resistance only for a short time. Energy consumed will remain same as per formula. But crucial productive demand on motor will be for very short duration.

Hence question is whether it is possible to lower motor rating by providing flywheel. This can help to average out load on motor.

Request your views on this conceptual design.

[Bobby Strain]

Take a look at pumping units in oil production.

 

Bobby

[snickster]

A flywheel is typically used to store energy to even out loads in piston engines like a car where firing of pistons at separated time steps so if you have a flywheel during the time between firing the energy absorbed by the flywheel is released to keep shaft spinning at a basically constant rate.

 

If the pump is driven by an electrical induction motor there is really no discontinuity of power provided by the motor as it is continuously fed by electricity and driven by electromagnetic forces cause by a rotating magnetic field cutting the rotor coils causing a torque on the rotor.  It does this by "Slip" which is the difference between the rotating magnetic field and the rotating rotor speed.  With no load the rotor rotates at approximately magnetic field speed with very little slip, only enough differential movement between the rotor speed and magnetic field rotation field to overcome friction in the pump and motor itself.  As the load increases the slip gets greater to be in proportion to the load so that the motor automatically adjusts to meet the pump load.  This is a self-adjusting process since if the load gets greater at the present torque output the motor will slow down, but if it slows down the relative speed between the rotating magnetic field and the rotor increase causing greater torque and power output until the value of slip just equals what required to satisfy the load, given the the motor has enough rated horsepower otherwise it will stall.

 

So during the suction stroke 180 to 360 / 0 degrees, there is very little load and the slip is very small.  Say you have a 1800 RPM motor which is the speed of rotation of the magnetic field cutting the rotor.  When first started up the rotor is at zero speed so the slip is maximum and the torque is greatest which gets the motor/pump up to speed very fast.  But with little load, the rotor just rotates just slightly less than 1800 Rpm say 1750 RPM after getting up to full speed.  Now on the upstroke 0 to 180 degrees, the maximum power is required for both compressing the fluid and pushing it out of the cylinder.  This causes a greater resistance to turning of the pump so the pump slows down, but as it slows down there is more slip and hence more torque until it slows down just enough that the output torque of the motor equals the input torque required to turn the pump.

 

That being said I don't think there would be a big discontinuity of power appied by the motor like a piston engine, but with such great change in load when pump piston reaches BDC 0 degrees and starts the compression stroke, there could be some jerking of the motion to slow the motor down and increase the slip instantaneously, which may cause some inefficiencies.  I am not a pump expert so I am not sure if a flywheel would be applied in this case but even so I don't think it would cause a decrease in required motor size but just serve to smooth out rotation and possible make pump a little more effiecient because of this but I don't expect it would effect efficiency that much to reduce motor size.

 

Note that I still believe my description of what is happening within the cylinder during each cycle as I described previously is what is approximately happening and don't agree that major load is during compression and minor load is when fluid is being pushed out.  I think it is opposite.

 

 

Edited by snickster, 07 March 2023 - 08:51 PM.