Pump Power Consumption 3

[rusika]

Hello,

I am trying to figure out what happens to electrical power consumption
for a centrifugal pump when one throttles a downstream valve.

Mechanical Power = Torque x RPM
Electrical Power = V x I

My understanding is that when you throttle downstream to reduce the
flow you create backpressure thereby increasing the operating Torque
of the pump. Since Torque and current are directly related (e.g. Force
= current x B x Length of wire) this would mean that current would
increase and per P = V x I (since system voltage is constant) power
consumption increases. But from the mechanical formula it's possible
for the Torque to increase without overall power consumption
increasing. What am I missing?

 

[rcrowl]

Centrifugal Pump Water horse power= Q x H x S.G./3960, where Q = flow, H = head, S.G. = Specific gravity. Every centrifugal pump has a curve and reference to a typical curve will show that horsepower is reduced by throttling the discharge because the product of Q and H is being reduced and the Q reduces several times more than H for a typical throttling operation. I am looking at a pump curve where the pressure is 400 psig and flow is 70 gpm. Throttled to 20 gpm this same pump has an output pressure of 408 psig.

 

[rusika]

Hey,

Thanks for answering my question Crowley. Though I understand your answer , I feel it's not the full story. It is true that through throttling the power consumption will decrease per the mechanical equation Power = torque x rpm. As you say, flow drops way more than head increases. My difficulty is in understanding the electrical side of the equation. Electrical power = Voltage x current or so I'm told. Voltage is constant. That leaves current. Current is proportional to torque through F=NIAB (force exerted on a wire in a magnetic field = number of turns x current in wire x area of wire x magnetic field strength). By this equation as head increases, current increases so overall power consumption should increase. I appreciate that this is just a theory issue that is keeping me awake at nights. When I ask the process guys at work they tell me that the power consumption drops when you throttle a pump. The electrical guys when asked reason that the power consumption should increase.

Another minor issue is that the power consumption of a pump starting up spikes, or so I'm told. This is when the head is maximum and the flos is at a minimum. Shouldn't the power for the starting of a pump be very low per your explanation.

Thanks for not leaving me hanging.

 

[rcrowl]

1) I think the electrical guys that believe the power should increase are thinking about a reciprocating pump, where the power will definitely increase when you throttle the discharge.

2) The starting of a centrifugal pump with the discharge throttled is definitely low power, BUT the starting current for a motor is high because the rotor and any drive train components are brought up to speed very quickly.

3) A centrifugal pump has a very low torque requirement (unlike a reciprocating pump) and current does not increase because the pump head increases. The motor is simply meeting the power requirements of the pump and the power is down, so the current is down.

 

[rusika]

Thanks again. Now I know how the power consumption of a pump will vary for different operating regimes. I must admit that I'm still a little fuzzy on the why.

Why exactly does a pump motor draw a lot of current at startup then the current draw falls off? Also, I appreciate that the current and head of a pump must not be linked the way I interpret that they are from the electrical equations I previously listed, but I don't understand in equation form how.

Still, you have confirmed for me that a centrifugal pump draws a lot of current (and therefore power?) when it starts and the power consumed drops for throttled - high pressure low flow situations.

Conversations like these make for killer Saturday nights.
Keep it coming sage.

 

[JJPellin]

I did not see in the replies one basic point. When the flow of the pump is decreased by throttling, the power requirement goes down precisely because the torque requirement goes down. There is no conflict with physics. Less flow = less torqe = less power = less current. With this said, I feel I must add one more comment. There is a class of centrifugal pumps that do not adhere to this formula. In some vertical turbine pumps power goes up as flow decreases. This would be because the head is going up faster than the flow is declining. Thus, the product of QxH is increasing. But for most applications, this is the exception, not the rule. The power spikes when a pump is started because of the in-rush current. It takes a lot of power to overcome momentum and get the pump spinning. Also, as you can see from a speed versus current curve for a motor, the current increases with more slip if the motor is forced to run at a slower speed. During start-up, it takes some period of time for the motor to get up to full speed. During this time, there is more slip and more current is required. If a motor is started up gradually, using a variable frequency drive, this problem is avioded and there is much less current spike at start-up.

 

[stanier]

check out

http://www.mcnallyinstitute.com

 

[Auth]

In my view the equation is a conservation law and just apply for a total process, not a short transient stage, at this stage an extra variable item should be added for reckoning in extra energy consumption which used for kinetic energy increments of liquid and motor, and this item becomes to zero when process becomes stable.

 

[toothless]

All induction motors will have a high in-rush current during the transient.

For most centrifugal pumps, when you throttle the discharge, the discharge pressure goes up a little, the flow goes down a little more so the power consumption decreases. But often it doesn't decrease much because the efficiency of the pump at that throttled point is not as high (you are less efficient). Just because you cut the flow in half doesn't mean the amps will cut in half. You might have moved from 65% efficiency to 40%. the efficiency loss is in flow recirculation in the pump casing.

A blower has the opposite effect. Throttle the outlet and the amps go up. Therefore to control a blower, throttle the inlet, not the outlet.

 

[rusika]

Good work everybody. Now I will task you further. You all seem to have an intuitive understanding of what will happen. Ten points to the one who can tell me why.
1) JJ Pellin: Why does Less Flow = Less Torque as you state. I thought Torque was the Force of the impeller times the radius. Should that not relate to the head rather than flow. How is current related to torque and flow?
2) I don't know if I'm correct. But my understanding of what you are all saying about starting a pump is that the torque is so high that despite the low flow the overall power consumed is great. This torque is so great because the pump is at it's dead head pressure. When the pump starts the fluid moving it's operating curve will lead to it consuming more and more power as the flow increases because the flow changes between two system operating points will be greater (in general) than the pressure changes.

We shall go on till the end. We shall never surrender.

 

[UWH]

rusika,

As rcrowl said, pump power is a function of flow rate, head and density of the fluid. But power can also be expressed as a function of torque. You may recall work = force x distance, and rate of work-done is power. The power consumed by the pump is provided by a motor or engine supplying a force at a certain rotational speed. Power is then the product of torque and rotational speed in a certain amount of time.

So you've got the following forumals ...

Pump HP = Q x H x S.G. / 3960
Mechanically, HP = Torque (lb-ft) x RPM / 5252
Electrically, HP = V x I / 746

From these equations you can see that when flow (Q) is reduced, power and correspondingly, torque, and current are reduced.

Also, as mentioned above, the high inrush current at start-up is a function of the electrical motor characteristics.

hope this helps.

UWH

 

[rusika]

I will use this thread to post those beautiful motor truths as I come across them. This one helps me to understand the high current drawn when a pump is started.

In a motor a coil carries current in a magnetic field. As the coil rotates in the magnetic field, a back emf is generated that tend to counter the emf that supplies the current. When the motor is first turned on, there is no back emf and the current is very large, being limited only by the resistance in the circuit. As the motor begins to rotate, the back emf increases and the current decreases.

 

[Artisi]

The major factor in starting a centrifugal pump is the "breakaway torque" needed to get the pump away from rest. Generally starting torque can be round 15 - 20% of full load torque which then drops to a lower level before ramping up to full load torque as the pump comes up to speed with increasing flow and head. However, depending on the pump type, hydraulics of the system etc once the pump away from rest other factors come into play which govern the starting / switch gear requirements.

I heard of one large pump installation where the breakaway torque was too high for the motor to overcome - so to solve the problem the pump was allowed to rotate backwards from reverse flow to get it away from rest for a few seconds prior to applying power to the drive unit.



Naresuan University
Phitsanulok
Thailand

 

[rusika]

The power consumed by the pump increases and this is directly proportional to the relative density/specific gravity of the liquid being pumped.

The relationship between the flow through the system and the system resistance is given by:

h = KQ²
h = head loss in metres
K = constant
Q = flow in m³/sec.

Affinity laws
1) Capacity changes directly as the change in speed or impeller diameter
2) Head varies as the square of the change in speed or impeller diameter
3) Brake horsepower varies as the cube of the change in speed or impeller diameter.

That's good for today.

 

[sethoflagos]

rusika,

Going back to your original point, when a centrifugal pump is throttled back, eventually you reach a condition where the liquid contents of the pump are simply spinning around in the volute. No fluid requires pressurising since the fluid at the outer wall of the volute is already at shut-in head pressure maintained by centrifugal forces. In the absence of friction, conservation of rotational kinetic energy maintains everything spinning like a top with no need for external energy input, no net force on the impeller vanes, no torque, and hence no load on the motor.

Of course, there are some fairly substantial frictional forces, hence there is some current drawn in the shut in condition.

As you start to open the discharge valve, a small volume V is allowed to escape from the volute periphery, the remaining contents move radially outward along the vanes, and an identical volume V enters system at the suction port to replace it. This tends to slow the rotation since liquids nearer the impeller centre have initially less velocity than the liquid further out that they are replacing, and a decelerating force is exerted on the impeller vanes. This tries to break synchronisation between the constant speed rotating fields of the motor and rotor. Phase difference between supply voltage and current rotates to compensate and power is drawn. The additional torque is now transmitted down to the impeller and via the vanes to restore liquid velocities and replace the lost rotational kinetic energy.
The nett energy required is VdP as mentioned above.

Obviously the more the discharge valve is opened, the more liquid requires acceleration from zero(-ish)to volute periphery speed.

Even as the pressure head drops off, the flowrate is increasing at an even faster rate (and frictional forces more so), hence power increases with flowrate all the way to full run-out.

As stated previously, getting everything spinning in the first place requires considerable torque, hence the high start-up current.

I'm sorry, I've glossed over the electrical side a little (not my field) but essentially that's my understanding of what's going on. Hope it helps.

 

[sarf363]

Hello All

Does any body know how can i calculate the pressure drop across filter if the fluid is the engine exhaust gas.

Thanks

 

[rusika]

Ok ok ok ok... I think I've got it! Please let this be the answer.
1) Power = VI = Torque x RPM
2) When you throttle a pump, as sethoflagos stated, the impeller still spins but it has to accelerate a smaller volume of fluid. F = ma. Mass goes down so force goes down, so current decreases as practice shows.
3) When current decreases the motor has to spin faster because of the following: Kirchoff's rule (sum of voltage changes in a closed circuit = 0). V - IR (motor current x resistance) - BKw (this term is the back voltage produced in the motor winding by induction and increases as the motor spins faster). This relationship means that as current goes down the speed has to go up.
4) Here's the conclusion that I draw that seems counter-intuitive and may or may not be correct. When you throttle a pump the impeller actually rotates faster - rpm up, torque down. When you open the throttling valve the rpm actually goes down, though the flowrate through the pump goes up. It was always intuitive to me that the flow was related to rpm and pressure to torque. Actually they are the opposite.
5) When a pump is first started the torque requirements are maximum, because the motor has to bring fluid from rest to the right speed. So the current spikes. Then the current drops when the motor is turning. If you dead head a pump completely, same thing happens. The difference is that torque x rpm is zero so there should be no power drawn. In this case all the power that is drawn is converted to heat.
6) Two things that sent me in the wrong direction: i) I was told by my boss that flowrate has a bigger impact on power consumption than pressure. This is true, but I was equating flowrate with rpm, which is not true (I think). ii) The motors I was interested in are single speed motors. I didn't understand how the above sited equations made sense if the motor speed was truly constant. If that was true than by the above the current and therefore power drawn would also have to be constant. What seems to happen is that induction motors do allow some slip between the stator magnetic field and the rotor field up to ~ 20%. So I guess this means that the speed can vary by 20% on a single speed pump?

Setho, this is where your answer has led me. What do you think?
Regards,
Rusi

 

[AUChEinAL]

What about the rare event of dead-heading a centrifugal pump? Would you expect the amps to increase or remain at no load?

 

[BigInch]

Already answered... see rusika #5.

Deadhead power consumption equals spinning frictional and internal recirculation loss levels only. Have a look at your pump's power curve at 0 flow. For a lot of centrifugals, its about 30% of BEP power.

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com