increasing the V/F ratio
increasing the V/F ratio
(OP)
Dear experts,
We have ID fan for cement plant operation, designed for 1070 rpm and required power is 1593 kw(data is taken from fan name plate).Calculated torque requirement for this fan is approximately 14280 NM . This fan is driven by 6.6kv, 60 Hz, 1600kw, 164amp and 1196 rpm motor, Calculated torque this motor can provide is 12775 NM and this motor is equipped with vfd power flex 7000. We have an issue that when operator want to increase the speed above than 950 rpm, motor reaches up to rated motor current 164 amp while power is 1294 kw. I think it is due to torque requirement of fan on certain process parameters is above than torque that motor can provide. Please see the attached screen shots of vfd for motor model and other parameters.
I have concluded that motor is under rated for this fan, I want to get a more power from same motor by increasing the V/F ratio so that fan speed can go above than 950rpm. Please advise me in this regards that this practice is ok or not and if it is ok up to which level I can increase the v/f characteristics.
Thanks,
Fiaz ahmed
We have ID fan for cement plant operation, designed for 1070 rpm and required power is 1593 kw(data is taken from fan name plate).Calculated torque requirement for this fan is approximately 14280 NM . This fan is driven by 6.6kv, 60 Hz, 1600kw, 164amp and 1196 rpm motor, Calculated torque this motor can provide is 12775 NM and this motor is equipped with vfd power flex 7000. We have an issue that when operator want to increase the speed above than 950 rpm, motor reaches up to rated motor current 164 amp while power is 1294 kw. I think it is due to torque requirement of fan on certain process parameters is above than torque that motor can provide. Please see the attached screen shots of vfd for motor model and other parameters.
I have concluded that motor is under rated for this fan, I want to get a more power from same motor by increasing the V/F ratio so that fan speed can go above than 950rpm. Please advise me in this regards that this practice is ok or not and if it is ok up to which level I can increase the v/f characteristics.
Thanks,
Fiaz ahmed





RE: increasing the V/F ratio
Also, is this a scalar or a vector drive? Vector drives can have their own ideas about V/f while scalar drives are more "obedient".
Gunnar Englund
www.gke.org
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Half full - Half empty? I don't mind. It's what in it that counts.
RE: increasing the V/F ratio
Scroll down in the attachment. It looks as if the 60 Hz drive is hitting the current limit at about 47 Hz.
Bill
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"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
1294kW / 950rpm * 1196rpm = 1629kW.
Motor is 1600kW, so 950rpm is the fastest speed your motor can turn the fan. You need a bigger motor to turn the fan faster, there is no way around it.
Well, if you have temperature monitoring in the motor (RTD sensors) and the ambient is reasonably cool you could always push the motor until the motor is running at its rated temperature (or say 90% or 95% of rated) and hope you get away with it without burning out the motor. This will require more current, so your VFD needs to be capable of sourcing this higher current. It's not really the proper way to do it, but if the motor is rated 1600kW in a 50*C ambient then it could produce more power when running in a 20*C ambient. Remember, the motor is really limited by the temperature rating of its insulation.
RE: increasing the V/F ratio
Bio Mass Counters. That is food for thought...
Gunnar Englund
www.gke.org
--------------------------------------
Half full - Half empty? I don't mind. It's what in it that counts.
RE: increasing the V/F ratio
In THEORY, you could change the input voltage to be higher and run the VFD at higher than rated motor speed, keeping the V/Hz ratio constant. We do that all the time on LV motors. The PF7000 drive will be rated for 7200V and probably 200A, so in theory you could squeeze another 9% speed out of it. But this is much much easier to do at LV. If you cannot tap your transformer to put out 7200V and/or there are other things connected to the same transformer that would be adversely affected, it would likely be cheaper to replace the motor than to replace the transformer and all of the other associated equipment just to get 9% more speed.
I like Lionel's idea better, but before you do anything, double check the output current capacity of the drive. It SHOULD be 200A, but may require changes in sensors to get more than it was programmed to do. It's a Current Source Inverter, so everything it does is based on accurate current sensing and control.
"Will work for (the memory of) salami"
RE: increasing the V/F ratio
RE: increasing the V/F ratio
RE: increasing the V/F ratio
But i want to go for v/f change option because our plant management is considering risk in this option.
In this case at 47 hz there are 5170 volts, if I shall increase v/f ratio in such a way that at the same frequency i will get voltage up to 5300 volts then i can get more torque with compare to 5170 volts and hope it will decease the current while on same speed so we can go little above the 950 rpm. my concern here is to know about the v/f ratio that max can be used to avoid high magnetising current.
i can not write English well so sorry for any mistake while communicating on this forum.
fiaz ahmed
RE: increasing the V/F ratio
How old is the motor? Older motors are usually more generous in the design of the stator iron, which gives more margin for changing the V/f ratio. If it is a relatively new motor, and especially if it is a standard mass-produced type, then you will typically find that the design has been optimised so that there is very little excess iron and the core can easily start to saturate.
Lionel's idea is a good one and you can almost certainly raise the current limit a little if you have stator winding RTDs to protect the machine.
RE: increasing the V/F ratio
Yes, you can get more power by increasing the voltage and frequency, but the power comes from the greater speed at the same max current.
My understanding is that when a motor is driven over-voltage to achieve greater power, the drive ratio is lowered.
Try a larger sheave on the fan and the motor will be able to run faster and develop it's full HP and a greater speed.
Direct drive? Trim the impeller blades to lower the torque required at a given speed and then you will be able to use the voltage headroom to spin the fan faster.
You are running at 81% power and 78% voltage. You have to drop the torque requirement so that the motor can spin faster in order to use the surplus power.
More a mechanical problem than an electrical problem.
Bill
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"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
RE: increasing the V/F ratio
I can tell, you have that tee shirt too.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
RE: increasing the V/F ratio
1. increase of voltage or decrease freq of motor parameters in drive.
2. increase the rated current up to 105 %.
3. increase the ramp up time to reduce the dynamic torque requirement when changing the speed from one setpoint to another setpoint. what is expert,s opinion about this.
is there any way to monitor the motor when it will saturate ?
fiaz ahmed
RE: increasing the V/F ratio
Monitor the current. You'll know if changing the V/Hz ratio is working almost as soon as you start it.
RE: increasing the V/F ratio
#2 is the right way to get same result: all u want is MORE current for more torque so simply turn current limit up while watching the thermistors and motor temp. also, changing mechanical ratio on belts won't help; HP is a figure of speech only: your motor/vfd is limited at 47hz due to the torque/amp rating of the motor; more gear ratio simply moves your limit point from 47 to 60 hz - no difference in final fan speed.
#3 is not the issue from your math since the motor is at rated amps already - slower accel wont change that. slowing it down is simply an exersize.
RE: increasing the V/F ratio
I don't agree. HP actually is important. More HP is required to move more air. So, increasing the drive ratio to increase the motor RPM gives more HP and allows the fan to be turned faster to move more air. Remember, you get a torque multiplication when you change the drive ratio.
RE: increasing the V/F ratio
That said, my thinking comes from my 10' dia windmill "fan" I built. I figured torque went up by square of the speed, thus HP went up by the cube of the speed.
Also 20 years of my life as engineer selling, applying, helping design, torque motors, where we could output 1,000#ft of torque - at zero speed, yet mechanical result is 0 HP! Hence my belief HP means nothing. Sorry.
Am I wrong saying if you increase gear ratio so that FLA is reached at 60hz to rather 47hz, mechanical HP will remain the same? Please show how I err....
60/47hz= 1.277 speed increase. Means same load torque on output goes down by same 1.277.... So HP=N*T/5252 or [(N*1.277)*(T//1/277)]/5252= what? same?
Please correct me! Thanks.
RE: increasing the V/F ratio
Round off 47 Hz to 50 Hz and 1500 RPM.
Round off 60 Hz to 60 Hz and 1800 RPM.
HP = Torque times speed times a constant.
Take any torque you want and plug it into 1500 RPM and then into 1800 RPM. You will find that 1800 RPM always gives a higher HP for a given torque than 1500 RPM.
Torque and current are related. The motor is hitting the current limit at about 80% speed, 80% voltage and 80% HP.
A lower ratio will give more torque on the fan for a given torque on the motor. This will allow the motor to use some of the excess voltage and HP and deliver more air before hitting the current limit.
A mechanical problem.
The only electrical solution is to try to justify an over current on the motor. Calling it a change in the V/Hz ratio does not change the fact that you intend to increase the current above the maximum nameplate value.
If this is a direct drive fan, there are methods to reduce the torque demand and let the fan run faster.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
Looking at it as today's operation is using 80% volts but 100% torque would suggest further mech ratio to reduce the 100% torque since we have some spare volts (speed) as you well point out... But perhaps increasing ratio to 90% speed (volts) & 90% torque (amps) may make more sense so motor does not have to overspeed or drive go above nominal 60hz either... then at 90/90% point he is back to same as today operation and has 10% speed and torque to increase some. To go all the way to 100% speed and thus any increase is at constant HP won't help either since now the torque avail begins to go down with any further speed increase over same as he has today... Perhaps the optimum ratio should be calculated if he wants to try a change: since torque goes up by square of speed, and speed goes up linearly, a little spreadsheet could calculate that something like 1/3 of the total 60/47 = 1.27 increased ratio makes best sense?
RE: increasing the V/F ratio
I agree also with your comments on the ratio.
The optimum ratio will be one where the motor is running at rated speed and rated current.
Sort of like shifting to a lower gear when climbing a hill.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
thanks so much for explaining well to solve the issue mechanically, again thanks.
i am still stuck on v/f ratio increase because of current decrease in dol motors when supply voltage increas in plant, this decrease is upto limited increase in supply voltage. when voltage increase due to any reason while frequency remain same that mean increase in v/f ratio. some years back, i increased the supply voltage up to 3% by tapchanger for experimental purpose and observe the current trend of the important motors of plant and i found decrease of current in all motors except the lightly loaded motors. Due to this reason i am thinking maybe i can get some extra kw. Please help me to understand if there are some drawbacks in this option.
I am interested in this option as there is no major modifications required and 100kw power increas may fulfill process requirements.
fiaz ahmed
RE: increasing the V/F ratio
The motor when running at 950rpm is effectively a 1271kW motor, not a 1600kW motor. 1600kW x 950rpm/1196rpm
The motor is running at full load when operating at 950rpm, so it can be assumed the fan is requiring about 1271kW to operate at 950 rpm.
The desired rpm of the fan is 1070rpm. This is 1.13 times faster than 950rpm.
The fan power requirement increases by the cube of the speed change. So, 1271kW * 1.13^3 = 1834kW. This is the power the fan would require to reach 1070rpm. It's not very likely you'll get there using a V/Hz change or even by overloading the motor.
Now back calculate the speed increase possible by changing the drive ratio so the motor is 100% loaded at rated speed. The ratio = cube-root(1600kW/1271kW) = 1.08. So, you can increase the fan speed to 950rpm x 1.08 = 1026rpm by changing the drive ratio to get the motor running 1196rpm when the fan is running 1026rpm.
RE: increasing the V/F ratio
Regarding HP=k*M*N relation, it's not valid for entire speed-torque curve and final operating torque is imposed by load not by motor. In IM motor, torque-speed it's a non-liniar relation; motor reach FLA at 945rpm because is undervoltage, developed torque is higher than nominal but still below to needed value to reach 1070rpm.
RE: increasing the V/F ratio
The simple fact is that this motor is under sized for direct driving this fan. My take would be that someone saw the fan data 1593kW @ 1070rpm and simply picked a 1600kw @ 1196rpm motor thinking they were matching the motor to the fan. In reality, the motor chosen should have been 1800kW @ 1196rpm to properly match with the fan rating plate. A rather expensive mistake to make when the system can not reach the required operating point.
Now, to make matters worse, the fan appears to be running over the design rating. The fan should only be using 1097kW @ 945rpm if it was loaded to match the name plate. Since it is drawing 1294kW @ 950rpm, this means the fan requires a 2100kW @ 1196rpm motor to power the new expected load of 1870kW @ 1070rpm.
I really hope you don't need to reach 1070rpm because I doubt that your motor is capable of that.
RE: increasing the V/F ratio
Please help me to understand if there are some drawbacks in this option.
I am interested in this option as there is no major modifications required and 100kw power increas may fulfill process requirements.
Fion, you do not have a v/hz drive. You have a vector drive. Your vector drive as wired and mechanically attached ALREADY HAS 20% EXTRA VOLTAGE - YOU ARE NOT VOLTAGE LIMITED.
You have NO **NO** control over the output voltage - this is a vector drive. It is controlling current, not the voltage.
You can only increase current limit to get more output from your present system.
Nothing else. Wishing will not make it work.
Either your present motor will be able to overload and do the job (not likely), or you can gain a SMALL amount more by tweaking gear ratio, or you need a larger motor. No way around it.
RE: increasing the V/F ratio
fiaz ahmed
RE: increasing the V/F ratio
I hope that fiaz will let us to know why so high torque difference (fan data error or some system problem?).
RE: increasing the V/F ratio
In this case the current is limited to rated current and this effectively limits the torque . At rated current the motor is generally developing rated torque.
I think that Lionel knows this.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: increasing the V/F ratio
RE: increasing the V/F ratio
The torque vs speed curve of a motor on a VFD with a current limit isn't the same as the full-voltage torque vs speed curve. In theory, the torque curve on a VFD set to current limit to FLA is rated torque at all speeds. Also, the motor breakdown torque might be 2.5 times the rated torque, but the motor won't be able to produce breakdown torque if the VFD can't supply the current required to produce breakdown torque.