How to Estimate Centrifugal Pump Rundown Time?

[Bambie]

Would the following iterative calculation provide a reasonable estimate of pump rundown time?

The hydraulic system P vs Q relationship is:
PT = PS/QS^2 * QT^2
PS = operating pressure at start of rundown
QS = operating flow at start of rundown
QT = flow at time T
PT = pressure at time T
Flow Energy is expressed:
FE = PT * QT * T
substitute into FE = PS / QS^2 * QT^3 * T
The pump Rotational Energy is:
RE = .5 * I * WT^2
I = moment of inertia
WT = rotational speed at time T
WS = rotational speed at start of rundown
assume WT = WS / QS * QT
substitute into RE = .5 * I * (WS / QS * QT)^2

For a small time interval (T), solve for the loss of Flow Energy (FE), subtract it from Rotational Energy (RE), then solve (RE) for the new (QT), substitute and solve for (FE), then repeat until flows and pressures reach minimum acceptable values and the sum of time increments is rundown time.

 

[Artisi]

Open discharge or discharge NRV, have you included for pump and motor inertia?

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[Bambie]

Artisi,

This pump circulates fluid around a pipe loop with no NRV.

The pump rotational energy includes flywheel, rotor and impellor moments of inertia.

Do you agree with this iterative solution approach?

Would you know how to use integration to solve for time between starting flow and a minimum acceptable flow?

 

[Artisi]

Bambie, way out of my area, therefore cannot make any comment.

It is a capital mistake to theorise before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts. (Sherlock Holmes - A Scandal in Bohemia.)

 

[LittleInch]

It seems to ignore the issue of flow versus pressure in your loop and also that pressure is proportional to rotational speed^2.

Why do you want to know?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Bambie]

LittleInch,

I don't think I have forgotten any essential relationship:

Flow versus Pressure is represented by the exponential relationship:

PT = PS/QS^2 * QT^2

Rotational speed is directly proportional to flow:

WT = WS / QS * QT

Therefore, as the equations indicate, pressure is proportional to the square of both rotational speed and flow.

The control technicians are looking for backup pump response requirements to avoid process interruptions.

 

[LittleInch]

Okay, what does the curve / numbers tell you?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[hydtools]

Divide the I*w by 1/2 the torque at w to get the time for run-down.

Ted

 

[hydtools]

See page 81.

Ted

 

[Bambie]

LittleInch,

The numbers won't tell me much if the model is flawed.

hydtools,

Thanks for the reference, I will check the failure modes to see whether starting torque could reduce rundown time with retarding torque.

Unlike the discussion on page 81, the flow energy is quite large for my application and needs to be included.

 

[LittleInch]

You have a strange situation here as you're in a loop so the pump inlet is actually connected hydraulically to the discharge.

The thing missing here is that as soon as the pump stops actively pumping and starts to run down, the inlet pressure to the pump will start to rise as fast the discharge pressure falls.

This will actually then take less energy out of the pump rundown than would otherwise occur and result in a longer tail off (I think)

Also there is no account of friction or reverse rotational forces from the motor.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[hydtools]

After shutdown the circulating fluid will slow the pump by tending to drive it backwards against the rotary inertia. Flow friction will also work to slow the circulating fluid. So the circulating flow momentum will be resisted by flow friction and pump inertia.

Ted

 

[Bambie]

hydtools and LittleInch,

It sounds as though you would agree with this iterative approach to calculating rundown time if I add an efficiency factor to the direct proportionality between WT and QT.

 

[hydtools]

What if you take this approach? Knowing the power delivered to the motor/pump while only circulating fluid and the rpm, you can calculate the torque delivered to the impeller. Turn off the motor input. The system must dissipate that energy remaining in the system at shutoff. The pump rpm must go from rpm at shutoff to zero. The torque must go from torque at shutoff to zero. Time for that torque now being delivered by the circulating fluid to bring rpm to zero would be = I*WS/Tavg. Tavg = TS/2 TS would be the torque calculated at shutoff. WS = rotation speed at shutoff.

How accurate do you need to be? Is iteration necessary?

Ted

 

[Bambie]

hydtools,

I think your approach is similar to setting KE = PE and solving for T at shutoff conditions, which will give a very short rundown time since it does not account for the non-linear dependency on flow and rpm, which can be simulated with iteration.

 

[hydtools]

Bambie, it is an impulse-momentum solution.
Look into unsteady flow solutions which is where time is part of the solution.

Ted

 

[LittleInch]

There are transient flow programs which do this if they know the characteristics of the pump.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.