Stop time for a pump after an electrical shut down 1

[Xalii]

I have read a few treads about this but I am not yet clear on the following subject:
For a simple water system (1 pump with an electrical motor, 1 check valve, 1 pipe delivering in 1 tank) is there a formula to calculate the time needed by the pump to stop assuming that we have :
RPM: initial rotation
I: inertia of the pump+motor
L: length of the pipe
Q: flow
and what is the curve looking like ? rather linear?

According to my understanding this information is very important to know if the surge might create cavitation/collapse in a high oint in the pipeline.
I have asked my pump seller to give me this information but he was very evasive...

 

[MikeHalloran]

Since you have included zero useful information about the geometry of your system, we here will probably also seem evasive.

At least your pump seller has the pump curves.
Does he have the system curve?




Mike Halloran
Pembroke Pines, FL, USA

 

[Xalii]

I am not looking for a specific case but a formula for general situation, we could limit the cases with the following specifications:

L: between 500 and 5000
Pipe Diam from 75 to 300 mm (PE or PVC)
RPM: 1450 & 2800
Q: flow from 15 to 350 CMH
Specific speed (Ns) between 20 and 60 (low head or Francis pump type)

I hope it helps..

 

[racookpe1978]

Sorry.

It will stop flowing between 1 second and 10 minutes. 8<)

There are no "simple formulas" or Ipad apps for this type of calculation.

Is your pump discharging into an open pipe dumping into a open tank? Little back resistance in that case. Discharging into a high-back pressure closed system? It will stop turning faster.

 

[Xalii]

That's what I fear... 8<(((
Usual setup is a pipe dumping into an open tank...
Any other suggestion ???

 

[BigInch]

With those size pumps, it will normally be within a few seconds to 30 seconds.
If pumps & motors are huge, have high suction or discharge pressures, no check valves on discharge, attached to long pipelines with high elevation differences, then they will tend to spin longer.

Ignore the fluid in the pipe, not the pump, and figure the change of angular velocity to get an idea of how fast it will change speed. The fluid in the pipe doesn't affect very much in most cases.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[Xalii]

Does it mean that the main factors are the RPM and the diam of the impeller ?

 

[BigInch]

Yes. Diameter of the impeller, since the larger the diameter the higher is the value of rotational inertia and rpm, because the angular momentum is greater as well. The fluid within the pump casing, as well as the impeller, can be a significant contributor.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[PersonalProfile]

I have completed a similar assessment for starting that I propose may be of use.

Subject to assumptions [you would need to confirm they are suitable for your situation]:
The pump is centrifugal [though through suitable adjustment, I propose other types could be assessed].
The system slows as a "block" [i.e. the pump rotation slows down "proportionally" to the pipeline flow / velocity].
I assume you know your system losses as a function of [pipeline] velocity - to use this method you must then express the system losses in relation to the pump rotational speed [w cf. v] and restive torque [T cf. head].
You know that the [pump speed] of operation [w0 = x rads/s] and that the system will stop [w1 = 0 rads/s].
You know the inertia of the system, viz: pump and product in the pipeline.

1. Given: T = Ia
From above you have I and T and hence you can find a.

2. Given a = w/t and dw/dt.

3. Combining 1 and 2: dw/dtI = T, rearranging: dt = dw/IT.
Note the T must be at the step that is being evaluated [see below].

From above you know w0 and w1 and through selecting appropriate dw stepping intervals you can calculate dts that correspond to each of the dws that you selected earlier.
After you step through your dws from w0 to w1, the sum of the dts would be your stopping time.

I currently don't have time to make this as clear as I would like, though hopefully this is of some use.

Regards,
Lyle

 

[Xalii]

Interesting and nice approach, would be a very neat result.
Unfortunately, according to my understanding the "block" assumption is abusive as what I want to calculate is the water hammer, and in your case we assume that the water is in-compressible? Let me know if I am wrong.

 

[BigInch]

Assuming incompressible fluid normally works, especially with water. What doesn't work is assuming incompressibilities when there is 2 phase flow, or fluid column spearation in progress.

With 1-phase flow, waterhammer can generally be conservatively calculated with Joukowsky's equation using a fluid stopping time that is anything less than the pump's spin down time. Calculate pump spin down time using the angular acceleration equations and inertia of pump, motor, gear (if any) and fluid in the casing. Neglecting the fluid in the upstream and downstream pipelines will be conservative. Most ESD valves are designed for 15 seconds, so that's probably the maximum time you want to use in any case. If you want to do any more than arrive at a simple estimate from that method, buy the right tools and invest in a transient analysis capable computer program.





Learn from the mistakes of others. You don't have time to make them all yourself.

 

[GPRnD]

Given the large range of variables you gave in your second post I'm not all that surprised that the pump supplier was reluctant.

You can make some generalizations as other posters have done, but even your "simple" system has a lot of variables, some of which you didn't even cover such as elevation change to the outlet.

Paying someone to perform a transient analysis using the correct software is probably your best bet particularly if you want to look for something such as cavitation at specific point in the pipeline.


If all you want to do is calculate the pump rundown time, you can use the specific system curve and pump curve (or 4 quadrant curve) to do an integration. I get asked this a few times a year, although more often it is about pumps running in reverse when the system fails. Since this way is system specific it doesn't lend itself to the general formula you are seeking though.

 

[BigInch]

Stop reverse running by installing a check valve and be done with that.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[LittleInch]

I've read this a few times and I'm struggling to work out what it is exactly you need to find out as you seem to be mixing transient pipeline issues with pump inertial issues. When a pump is deprived of its motive power (i.e. you turn the electricity off), the pump normally becomes a dead head unit as the downstream pressure (head) doesn't decay as fast as the decay of the pump as it is slowing down. This is why you put in a check valve to stop fluid flow going back through the pump even as it is slowing down.

Of course the decay of your system is dependant on many things - elevation, pipe size, velocity, fluid, valves closing downstream etc etc - i.e. far too many variables to work out manually. Ditto your pump decay curve is different for each pump. If you ask the vendor for a dead head pump decay curve he might have that, but what tends to happen is that the pipeline transient program, if it's clever enough, includes this in the transient simulation.

It could happen that during the slowdown at times your pump might still be pumping some fluid and at times not. All system dependant.

You might be able to figure an initial guess out using energy balance, i.e. your pump has a certain amount of kinetic energy at the point you turn it off, you can find out from the pump curve what the energy (power) is at no flow. Iterate every 5 seconds and subtract the energy the pump needs to dead head from the kinetic energy it has - work out the rpm and then repeat, gradually lowering the amount of energy the pump needs to dead head as it will now be spinning slower (which you know) based on the affinity laws of speed being proportion to head squared...

Try it and see what the curve looks like...

My motto: Learn something new every day

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

 

[GPRnD]

They do normally fit a check valve.

However in the case of safety related nuclear pumps they want to know what will happen if the check valve fails open. Its just part of the FMEA they need to do.

 

[bimr]

It is usually assumed in Hazop evaluations that a check valve will fail. If a check valve failure is a problem, then you need an automatically operated valve that will close on pump failure.

It is a general recommendation that a quick closing check valve be used. Such a valve will close immediately eliminating the possibility of reverse flow and water hammer.

http://www.valmatic.com/silentcheck.html

There is no need to do any pump shut down or water hammer evaluation.

If you want to know how long it takes to shut down the pump, run a simple empirical test. Shut off the pump and time the shut down.

 

[BigInch]

That's not exactly true as the faster the fluid comes to a stop, be it from a check valve or ESD valve, waterhammer pressures (both highs and lows) are greater. With a soft/slow closing check, pressures can be somewhat reduced.

Of course the most severe reverse flow transient cases are generated with standard disc-check failures, mostly because they do close so fast.

Learn from the mistakes of others. You don't have time to make them all yourself.

 

[bimr]

Fast closing means the check valve closes prior to reverse flow but the check valves do not slam shut.

 

[Xalii]

Thanks to all for your valuable inputs. For precision, my aim is to demystify a bit the water hammer phenomena and to develop a spread sheet able to calculate simple cases, where the project doesn't worth to ask a special transient analysis.
Basically in simple conditions, the end section of the pipe will have a reduced surge on a length of ts*c/2 (ts: time to stop the pump, c surge velocity) cf EQ 7.4 in the attached doc.
Thus I can add that I will assume that the check valve is working (no reverse flow) and that the stopping time will be shorter than the return period of the surge, so that it will not depend on the pipe but only on the pump.
Then to my understanding I can use the following:
First we have T/I=dw/dt
T: torque on the shaft
I: inertia of the impeller, rotor and water
w: speed (rad/S)
Then Pmec=T*w=rho*g*h*Q/eff => T=rho*g*h*Q/eff*w
h: head
Q: flow
eff: efficiency of the pump
On this we have that
h(w)=hin/(w/win)^2
Q(w)=hin/(w/win)
And we can assume that h= a1Q^2+b1Q+c1 and eff = a2Q^2+b2Q+c2

With Q going from Q in to 0, w : from win to 0 and h from hin to hin-vin*c/g (vin*c/g being the surge)

Does it looks right ?
You can have a look at the attached document to better understand what I am looking for.

 

[PersonalProfile]

Your understanding of the method I proposed is correct and although I do not have a strong background in fluids handling based on my interpretation of your original question, I propose it is still relevant [i.e. calculate the stopping time and then calculate the associated impact of water hammer].
Though only you are aware of your situation and can make this assessment.

I omitted to mention earlier, when I completed my assessment for my situation, I could not find a standard solution for the governing equations and so I used the computational method that I described.
Subsequently I consulted with some mathematicians [seriously] that I had access to at the time and they were able to provide me a "standard solution" and hence there was no need for computational methods.

Regards,
Lyle