Line Shaft Drive for Paper Machine

[astroid]

DC Motor running a paper machine line shaft is 185Kw,400V,490A,1150rpm. Normal running data is 380VDC,320A,1000rpm. It was replaced by 167KW,400V,449A,1750rpm. Amps reached to 450 at only 500rpm. Motor drive pulley dia was reduced and rpm could be raised to 800 at full load. A day later someone raised the speed to 1100rpm and found amps reduced to 400. Please explain!

 

[Skogsgurra]

Hi,

Paper machines is what I've done for forty years. So this sounds interesting.

First: Do you have fixed excitation? I assume that you have that.

Second: Condensate in the dryer cylinders (which is known to accumulate after a stop) is dragged up the walls of the cylinders and does brake more and more until you reach something like 500 m/min at 1500 mm cylinder diameter. Then, as speed increases, the water is no more dragged up the walls - instead it just forms a solid ring, which rotates with the cylinder and doesn't consume energy. That makes current consumption go down.

Third: Have condensate been emptied properly?

This seems to be an old machine. Line shafts are not used in many places nowadays. Time for something new? Section drives are very easy to install, maintain and run.



Gunnar Englund

http://www.gke.org

--------------------------------------
100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[itsmoked]

Wow.. Never would ever guess it was water condensate somewhere.. Ah, experience..

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[TomG33]

astroid,
I agree with Gunnar that condensate in a dryer can load the drive up but I think there is something else at play as well here.
You have to watch the speed rating of your motors.
THe nameplate data tells you the various conditions that must be met for the motor to deliver it's rated POWER, namely volts, Armature amps and speed (Field strength).

In broad terms, The motor must be going at it's nameplate speed in order to be delivering it's nameplate power. If you are going at half speed, then it can only deliver up to half the rated power (give or take a little bit) without overloading something.

Lets look at the two different motors you are dealing with.

FIRST THE ORIGINAL MOTOR
Nameplate data is. 185Kw, 400V,490A,1150rpm

First the efficency.

Mechanical Power output 185,000 W 185,000 W
= ------------------------------ = ---------------- = ---------------- = 94.4% Efficient
Electrical power input 400V*490A 196,000 W



You give the machine speeds in terms of MOtor RPM, but really, as far as the paper makers are concerned, it is machine Meters per minute or Feet per minute that matters. I am assuming that when the original motor was operating at 1150 RPM, the paper machine was runing at it's maximum Meters per minute and that at the motor speed you gave as the normal operating speed corresponded to something less then this.

Now Lets look at the voltage on the motor at the "Normal Load" point you give.

With the field at rated field strength, 400 Volts equals 1,150 RPM, therefore
Using a straight ratio calculation, the volts corresponding to 1,000 RPM is ..

400
Volts at 1,000 RPM = -------- * 1,000 = 348 Volts.
1,150

The Data you give says 380 Volts so perhaps the original installation had the
field current slightly above rated. This is not necesarily bad.


Now we will look at the load on the motor at the "Normal Load" point you give,


power input = 380 V @ 320 Amps = 121.6 Kw

Efficiency = 94.4% therefore mechanical power required by the load at "Normal Load" point

is 121.6 KW * 0.944 = 114.8 KW



NOW THE REPLACEMENT MOTOR
Doing some similar calculations for the replacement motor. From the name plate data provided. 167KW,400V,449A,1750rpm

we get the following


Mechanical Power output 167,000 W 167,000 W
= ------------------------------- = ---------------- = -------------- = 92.9% Efficient
Electrical power input 400V*449A 179,600 W

Assuming the first sceneario where the same gearing ratio was used

400
Volts at 1,000 RPM = ----- * 1,000 = 229 Volts.
1,750

Now the paper machine still needs the same mechanical power to run at its originally stated "Normal Load" point
which is 114.8 Kw.

Taking into account the motor efficency of 92.2%
the input electrical power = 114.8 Kw / 0.922 = 124.5 Kw



Current required to provide this power is

124,500 W
----------- = 544 Amps which is about 20% overloaded according to name plate,
229 V

So the reason for your trouble is that your replacement motor has
it's rated speed at 1750 RPm and not at 1000 Where you are running it.

So ...With each drive pully change, you are increasing the motor speed necesary
to get the machine to its operating speed, which is taking the motor volts
up with it. The bigger the voltage on the bottom line of the calculation above
the lower the armature current necessary to drive it.

Ultimately, if you changed the Pully ratio such that the motor speed went up to
1750 RPm at your paper machine operating speed, the armature current calculation
would be


124,500 W
-------- = 311 Amps which is well within your motor rating
400 V


Hope this all helps.

Tom )

 

[astroid]

Thank you for the inputs. We are looking at all aspects.