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EddyWirbelstrom (Electrical) (OP)
17 Apr 10 5:16
I do not think the file was attached, I will try to attach it again with this posting.
Artisi (Mechanical)
17 Apr 10 20:42
The answer is maybe and then maybe not.
EddyWirbelstrom (Electrical) (OP)
17 Apr 10 20:53
I am using PTW / TMS software to simulate the softstart of a 215kW, 415V, 50 Hz  submersible Flygt pump with an Emotron MSF-450 softstarter in 'squared torque control' mode which provides a constant accelerating torque during start. This avoids the torque spike near the end of starting which can occur with other types of softstarters.

When using the PTW / TMS 'Exponential' template to build a pump model, the input data Synch rpm ( 1500rpm ), Rated Torque ( 1387.23 Nm ), and Moment of Inertia ( 2.5 kg-m^2 ) result in the constants A = 554.8 Nm ( breakaway torque ), B = 61.01, C = 0.000589 and Exponent 2 being automatically generated.
The formula for pump load torque is :
-    T = A                rpm = 0
-    T = B + C * rpm ^2        rpm > 0

How are the constants B and C calculated ?
The attached spreadsheet includes motor and pump toque / speed data.
Which of the following C values is correct ?   
0.00043474 * 1.355817 Nm / Foot-oound = 0.00058942 Nm.
 
BigInch (Petroleum)
18 Apr 10 7:27
I think you're trying to develop a pump torque load curve related to motor speed, but it also seems like you are confusing the motor torque curve with the pump torque curve and trying to generate a pump torque load using the motor equation, but .. anyway the pump power and torque load is calculated as follows,

you'll need to calculate the power required by the system to start the flow beginning at a flowrate of zero and going up to full system flowrate and the torque input to the pump to develop that power required at each flowrate.

for each flowrate 0 to full flowrate
calculate the brake pump power = flowrate * density / eff
then, calculate the corresponding required input torques from those brake power values,

relate torque to pump speed, via the flowrate by assuming flowrate as a linear function of pump speed starting at 0,0 and running to the pump's BEP flowrate at its rated speed.

A, B, C are used to define the torque curve without knowing any of the above hydraulic information, ie the pump is a black box, so that's done simply by relating the required torque to the motor speed.

A, B, C can be found by knowing the shape of the pump torque load to  pump speed graph.  Plot the torque-speed curve in Excel, make a trend line for a parabolic equation and have it generate values for the A, B, C coefficients.

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

EddyWirbelstrom (Electrical) (OP)
18 Apr 10 11:05
Thank you Big Inch for your very helpful posting.
I am an electrical engineer trying to model the torque versus speed curve of a Flygt centrifugal pump from Flygt performance curves.
The two curves I have are :
1)  Pump input power versus NPSHR,
2)  Plot of NPSHR with Head on the y axis and flow on the x axis.
Is Head the same as NPSHR ?
Assuming flow is a linear function of pump speed, can I get values of Head for various speeds from curve 2), and used curve 1) to get values of pump input power at various pump speeds ?
I will upload the Flygt pump performace curves.


 
BigInch (Petroleum)
18 Apr 10 15:17
You have all the curves you need for the pump, but you need to have one for the piping system too.  Red is pump differential head (read it on the scale to the left.  for a flowrate of 240 l/s its about 58 meters), or head that will be added to the system when the pump is running with a given flowrate shown at the bottom of the graph.  That head added to the pressure at the point where the pump is to be located (I gave you a value of 5 meters at zero flow for that, you have to find yours), will give you your system discharge head at the pump discharge flange.  If the flow is 0, pump diff head is about 18 meters, so there its 18+5 = 23 m pump discharge system head.  At 240 l/s, your 5 m will probably be lower due to more loss in the suction piping, so we'll say its 3 meters at 240 l/s, pump diff head is about 58m, so + 3 m = 61 m pump discharge system head.
 
Orange is the NPSHR, or head required for the pump to function without producing cavitation, read on scale to the right. Its an absolute head (and includes atmospheric head of apx 10m, if the fluid is water. if not water then take the atmos pressure and figure the equivalent head.  To convert your suction pressure head to compare against that NPSHR head compatible with the system overlay, add the approx atmospheric equivalent of 10 meters if the fluid is water.  At 240 l/s you would have the 3 m we said above + 10 m = 13 meters.  That's greater than the 6 m required, so that would indicate you will have no cavitation problems at 240 l/s.

Blue is the pump efficiency curve
The system head is the head you need at the pump discharge to move your fluid at any given flowrate.  You must calculate that for each flowrate, or at enough points where you can develop an equation for head as a function of fluid flowrate.  If you consider the pipe system as a blackbox, that equation can be assumed to be equal to a quadradic equation
Hsys = k * Q^2 + C

k is some constant
Q is flowrate L/s
C is the static head, or head required at a very very small flowrate near zero.
Hsys is the required system head.

If you don't have data points for your pipe system for Hsys vs Q, Hsys can also be calculated hydraulically (non-blackbox) using one of several equations for pressure loss in a pipe due to fluid flow, such as Colebrook-White, or Churchill equations, etc.

I have invented a system curve equation (lime green curve) to show you how they go and superimposed it on your pump curve.  At 240 l/s , lets say that the churchill equation gave 48 m of head loss and we have 3 there now, so 48 -3 = 45 m of head would be needed from the pump.  Oh good, the pump's giving more so the capacity is OK, but then we need a control valve, because the pump is really giving 58 meters differential head, and a sys discharge head of 63 m.  If the flow was 265 l/s, the pump and system curve would match, so we wouldn't need a control valve for that flow.

OK, so assume the lime green curve is the result of your black box piping and regress the coefficients for the Hsys = k * Q^2 + C and then calculate the power and torque.

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

BigInch (Petroleum)
18 Apr 10 18:32
Just realized I didn't attach the colored pump curves.  Here,

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

EddyWirbelstrom (Electrical) (OP)
19 Apr 10 8:55
Many thanks BigInch, your addition of system head curve to the performance curve and excellent comprehensive description has been a great help in my understanding of pump curves.
 
BigInch (Petroleum)
19 Apr 10 11:13
Super.  That was a pretty fast write-up and I wasn't sure myself if it was making total sense.  Glad you could follow along.  Let me/us know if you get to a roadblock.

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

BigInch (Petroleum)
19 Apr 10 11:33
Actually it didn't make total sense.  I said NPSHr varies with fluid density.  It doesn't.  NPSHr values are always the values read from the pump curve and they remain those values regardless of fluid density.  

Its the [b]NPSHa[/a] that changes with density, when it is calculated from suction pressure.  Example:  With water, a suction pressure of 11 psig would be 11 psig + 14.7 atmos pressure =  25.7 psia and 25.7 psia * 144/62.4 = 59.3 ft.  That/those NPSHa values should be always above the corresponding npshR for each flowrate.  For npshA values, the 62.4 density would change for each fluid.  A fluid with a density of one-half that of water would have twice as much npsha.  A fluid with a density twice that of water would have an npshA of 1/2 of water.  Sorry that I managed to mangle the "r" for "a".  

Remember it like this,
NPSHa chAnges; NPSHr neveR.    

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

ione (Mechanical)
19 Apr 10 11:44
BigInch (Petroleum)
19 Apr 10 13:14
Right.  Nothing is as simple as it initially seems.  Its hard enough to get the basics delivered in a "tip", so I was endevoring to simplify as much as possible for a beginner.  Hydrocarbons are known to require less NPSHr than those standard pump npshr curves with a cold water basis, but I think its a matter for experienced engineers to apply.  I don't even reduce NPSHr values for any reason unless there is some other complication that forces me to do reconsider alternatives.  In the meantime, IMO the OP should adhere to conservative practices.  Besides, we still don't know what fluid he has in there anyway, so discussing reducing the NPSHr ... could be premature.

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

BigInch (Petroleum)
19 Apr 10 14:55
One other thing I noticed I didn't mention, or clarify, is that during spin-up, the pump discharge head at any given time should be approximated with the affinity law for head,
Ht = HQt^2/1490^2 * RPMt,

where

the sub Qt is the flow at time t, so
HQt is the head from the pump curve at a flow Qt

RPMt is of course the RPM at the same time t,

the flow by,
Qt = QBEP/1490 * RPMt
 

**********************
"The problem isn't finding the solution, its trying to get to the real question." BigInch
http://virtualpipeline.spaces.live.com/

EddyWirbelstrom (Electrical) (OP)
20 Apr 10 1:31
Thank you BigInch for the additional pump information; it will take time for me to digest it all.  
In the meantime I need to simulate the impact on the electrical system of sofstarting the submersible water pump with a softstarter which controls the voltage applied to the motor during acceleration such that the accelerating torque remains constant providing a linear speed increase, thus avoiding the torque spike which can occur at the end of starting with some types of softstarters.
ie.  Configure starting time and current to ensure that starting current does not result in unacceptable voltage dips at the switchboard or torque spikes at the end of starting.
In the absence of a system head curve I am assuming the 'worst-case' rated rpm operating point ( lowest pump efficiency and highest flow rate ) of the Flygt pump performance curve where the pump shaft input power of 195 kW corresponds to a flow rate of 396 litres / sec.  At synchronous speed ( motor rated speed  = 1480 rpm ) this corresponds to a pump shaft torque of 1241 Nm.
I will assume a break-away torque of 40% * 1241 Nm and a torque immediately after break-away of 5% * 1241 Nm .
(The PTW / TMS software requires torque inputs at synchronous speeds.)

The PTW / TMS pump models the pump torque as :
T = A                rpm = 0
T = B + C * rpm ^ 2        rpm > 0
where :
A = break-away torque in Nm
B = torque immediately after break-away in Nm
C = multiplying factor calculated by PTW / TMS based on input data
This is similar to your 'black box' formula except the units are in Nm.

It is interesting to note that the intersection of the motor torque and pump torque curves is similar to the intersection of the system head ( pump load ) and pump differential pressure in that it is the steady-state operating point.

The attached spreadsheet has the torque versus rpm spreadsheets I have used to build models of the of the pump and motor.   The PTW / TMS plots are also included.
 

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