we are using to SC IM ( squarril cage induction motor ) in a hydrulic system , with rating of

Vrate = 380V , cosO = 0.86 , Irated = 196A, KW = 110 KW .

in operation both motors are run together , the energy meter shows a p.f of 0.28 lagging and clamp meter show 45 A on each motor , its clear that low P.f due to less load on the motors . we have no option to run a single motor fro opertion as the hydraulic pressure reduce in this case . is there some option to improve the p.f by introducign the cap bank , what is the ratring and exact location to install the bank?

pls expain ion detail if there are some other reson for the cause

regards

asad

# low p.f of IM

Started by tahira, Oct 03 2002 03:05 PM

5 replies to this topic

### #1

Posted 03 October 2002 - 03:05 PM

### #2

Posted 03 October 2002 - 07:51 PM

Hello tahira

Yes the low power factor is because of the low load on the motors. It is very easy to improve the power factor, but the first question to ask is, "do you really want to improve the power factor?" the improvment in power factor will reduce the current drawn from the supply, but will not reduce the power consumed. Therefore the KWHr will not change. If you are penalised for having a poor power factor, then an improvment is worthwhile.

There are two ways to improve the correction. You can use bulk correction at the point of supply, this requires a number of capacitors and a controller to automatically switch in the amount needed to correct the supply, or static correction, where each motor is corrected individually.

With static correction, the correct capacitance is calculated for the motor, and is connected between the starter and the motor so that when the motor is running, the capacitors are connected, and when the motor is not running, the capacitors are disconnected. This is best done using a separate contactor to cswitch the capacitors, but is commonly done by using the same contactor as is used to switch the motor.

There are many look up tables for capacitor selection, but they are rarely correct and always result in under correction. The problem is that there is no magic value that can be applied to a particular motor size. The correct value will vary between motors of the same size by as much as 2.5 to 1. To apply the correct amount of correction, measure the open shaft current or the motor, take 80% of that and that is the amount of capacitive current that you want. To convert that to KVAR, multiply that current by your line voltage (i.e. 400V) multiply by root 3 and divide by 1000.

In you case, I would guess that you would be looking for correction of about 22KVAR

Have a look at our page on power factor corretion at http://www.lmphotonics.com/pwrfact.htm You could also download the Electrical Calculations program and use that to help you calculate the correction required.

Best regards,

Yes the low power factor is because of the low load on the motors. It is very easy to improve the power factor, but the first question to ask is, "do you really want to improve the power factor?" the improvment in power factor will reduce the current drawn from the supply, but will not reduce the power consumed. Therefore the KWHr will not change. If you are penalised for having a poor power factor, then an improvment is worthwhile.

There are two ways to improve the correction. You can use bulk correction at the point of supply, this requires a number of capacitors and a controller to automatically switch in the amount needed to correct the supply, or static correction, where each motor is corrected individually.

With static correction, the correct capacitance is calculated for the motor, and is connected between the starter and the motor so that when the motor is running, the capacitors are connected, and when the motor is not running, the capacitors are disconnected. This is best done using a separate contactor to cswitch the capacitors, but is commonly done by using the same contactor as is used to switch the motor.

There are many look up tables for capacitor selection, but they are rarely correct and always result in under correction. The problem is that there is no magic value that can be applied to a particular motor size. The correct value will vary between motors of the same size by as much as 2.5 to 1. To apply the correct amount of correction, measure the open shaft current or the motor, take 80% of that and that is the amount of capacitive current that you want. To convert that to KVAR, multiply that current by your line voltage (i.e. 400V) multiply by root 3 and divide by 1000.

In you case, I would guess that you would be looking for correction of about 22KVAR

Have a look at our page on power factor corretion at http://www.lmphotonics.com/pwrfact.htm You could also download the Electrical Calculations program and use that to help you calculate the correction required.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #3

Posted 14 October 2002 - 11:37 AM

Thanks marke plz clear the query :

Through table or by using the formula

Q = P( tan01-tan02) i will get the KVAr require is

Q = 110 ( tan cos-1(0.28) - tan cos-1(0.9)) = 312 KVAr

but as you suggest the value is 22KVAr plz help me out in this regards how you calcualte it. the open shaft current in my case is 40A , 80% of which is 32A , is this be the capacitive current for bank , if it so the KVAR = 22KVAr

through seimens hand book

the cap rating for motro not greater than the no load reactive power consumption

here i have from graph Q0/PN = 0.45 for 110KW motor

thus Q0 = 0.45* 110 = 49.5 KVAr

80% of Q0 = 39.6 KVAr

is over rating of cap bank self excite the motor and o/v at motor terminals.

Through table or by using the formula

Q = P( tan01-tan02) i will get the KVAr require is

Q = 110 ( tan cos-1(0.28) - tan cos-1(0.9)) = 312 KVAr

but as you suggest the value is 22KVAr plz help me out in this regards how you calcualte it. the open shaft current in my case is 40A , 80% of which is 32A , is this be the capacitive current for bank , if it so the KVAR = 22KVAr

through seimens hand book

the cap rating for motro not greater than the no load reactive power consumption

here i have from graph Q0/PN = 0.45 for 110KW motor

thus Q0 = 0.45* 110 = 49.5 KVAr

80% of Q0 = 39.6 KVAr

is over rating of cap bank self excite the motor and o/v at motor terminals.

### #4

Posted 16 October 2002 - 09:13 AM

Hello Tahira

Your calculation is incorrect because at a powerfactor of 0.28, you do not have 110KW of power being consumed by the motor. I suspect that at 0.28, you will be closer to 10KW than 100 KW.

To calculate, you must know the actual power factor and actual power at that power factor.

I would use the open shaft current, take 80 % of that and make the capacitive current equal to that and you won't go wrong. You will not be over corrected!!

Any calculation or chart bast on generalised figures will be incorrect. The trend for these rules of thumb, is to be conservative to prevent over correction. The problem is, that for a motor of this isze, the magnetising current will vary between 20% of rated current to as high as 60% of rated current depending on the design. If you do not know otherwise, you correct as though the motor is 20% only and so you could be applying less than 50% of the required correction.

Best regards,

Your calculation is incorrect because at a powerfactor of 0.28, you do not have 110KW of power being consumed by the motor. I suspect that at 0.28, you will be closer to 10KW than 100 KW.

To calculate, you must know the actual power factor and actual power at that power factor.

I would use the open shaft current, take 80 % of that and make the capacitive current equal to that and you won't go wrong. You will not be over corrected!!

Any calculation or chart bast on generalised figures will be incorrect. The trend for these rules of thumb, is to be conservative to prevent over correction. The problem is, that for a motor of this isze, the magnetising current will vary between 20% of rated current to as high as 60% of rated current depending on the design. If you do not know otherwise, you correct as though the motor is 20% only and so you could be applying less than 50% of the required correction.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

### #5 Guest__*

Posted 23 February 2005 - 10:04 PM

hello

Please clear my doubt in calculating the capacitor value for 6 nos of IM of total 12hp and 24 KW heating element. How to calculate the value of the capacitor? Is there any formula to calculate?

Regards selva

Please clear my doubt in calculating the capacitor value for 6 nos of IM of total 12hp and 24 KW heating element. How to calculate the value of the capacitor? Is there any formula to calculate?

Regards selva

### #6

Posted 24 February 2005 - 08:42 AM

Hello Selva

There is no correction required for the heating element so you can ignore that from the effective load.

The best way to determine the correction required is to measure the magnetising current of the motors and then correct to 80% of that figure.

To measure the magnetising current, operate the motors under open shaft conditions. Do this for each of the motors and add the total open shaft current and this will give you the total magnetising current.

If you multiply the magnetising current by the phase to phase voltage, times root three will give you the VAR of the motors. Multiply this by 0.8 to get the total correction required. You can apply this to each individual motor (80% of mag current for each motor) or as an automatic bulk correction bank. Look at the power factor correction page on our web site for more information.

Best regards,

There is no correction required for the heating element so you can ignore that from the effective load.

The best way to determine the correction required is to measure the magnetising current of the motors and then correct to 80% of that figure.

To measure the magnetising current, operate the motors under open shaft conditions. Do this for each of the motors and add the total open shaft current and this will give you the total magnetising current.

If you multiply the magnetising current by the phase to phase voltage, times root three will give you the VAR of the motors. Multiply this by 0.8 to get the total correction required. You can apply this to each individual motor (80% of mag current for each motor) or as an automatic bulk correction bank. Look at the power factor correction page on our web site for more information.

Best regards,

Mark Empson | administrator

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

Skype Contact = markempson | phone +64 274 363 067

LMPForum | Power Factor | L M Photonics Ltd | Empson family | Advanced Motor Control Ltd | LMP Software | Pressure Transducers | Smart Relay | GSM Control | Mark Empson Website | Soft Starters

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users