Short time Overload on a transformer while starting a motor 2

[varri79]

Dear Folks,

I am working on a modification project (brown field, where new loads are going to be added in the existing power system.

Existing system:
Existing Transformer capacity is 300KVA, 11KV/415V, 417A, z= 5%,
Existing LV switchgear is 630A (Both breaker & bus bars).
Existing base load =143KW (248A).

New loads:
75KW- a single motor
15KW – lighting + small power
Total new load = 90KW.

I need to verify the existing switchboard and transformer capacities are adequate for these newly proposed loads (90KW).

After adding new loads, the total load will be

90+143=233KW or 290KVA (PF- 0.8)

It seems that new peak demand (290KVA) is with in the transformer capacity (300KVA) I know this is not good for the transformer in long run.

Now my worry is starting the 75KW motor on the existing base load.

It means when I start this 75kw motor, the starting current (with direct on line starter) will be around 6 times (130A x 6) = 783A

Hence the current on transformer, is the motor starting current plus base load current

= 783A + 248A = 1031A,

which is more than 150% of transformer full load current.

Is it safe to run the transformer in this condition? What is the allowed short time overload current (percentage) as per IEC?

I would greatly appreciate if some one advice me with the code requirements and percentage short time overload allowed.

Thanks in advance,

Varri

 

[Skogsgurra]

Code is one thing. I do not know what your local code says.

But, I would not worry from a thermal perspective. A motor start is usually over in a few seconds. If the load has a large inertie, it may take ten or twenty seconds - but then you certainly need a soft starter.

The only situation that thermal considerations may be important is if you are starting the motor very frequently. But usually the motor gets toasted before the transformer in such a situation.

The problem is practical. How much will the secondary voltage drop during a start? Will your connected equipment ride through the voltage dip?

I think it will. Using your numbers, the transformer load wil be 150 % during a start. Volatge regulation is 5 %. That means your voltage will drop around 7.5 % during the start. That is what you have to consider. Will the other loads work or will they brown out or just drop out?

Modern switchers will not be a problem. Lamps will blink. Contactors and relays may drop, but usually not. Most are designed to stay in down to 85 % of nominal voltage. Motors may slow down a little, but not much.

It is only if you are running motors in brake (regenerating) mode and use a thyristor bridge that you may experience problems. And then only if you are running close to maximum speed.

I don't think you will have any problems at all.


Gunnar Englund

http://www.gke.org

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

 

[7anoter4]

I totally agree with Skogsgurra: You don't need to be worry about the trfx overload.
A soft starting for this motor would be better.
But if nothing will be change the voltage drops are still ok.
Let say motor rated voltage is 400 V
Let say MV will be 95% that means 11*.95=10.45 KV
Existing base load= 178.3 KVA
New [15 KW,0.8 pf]=16.7
Total [before starting 75 KW motor] =195 KVA=65% from 300 KVA
Let say from trfx to switchgear 50 m 2*3*240 sqr.mm cables
Let say from switchgear to motor terminals 100 m 3*70 sqr.mm cable
If MV is 11 kV indeed the switchgear bus bar voltage will be [at motor start IST=6*Irated]: 373 V= 93% from motor rated voltage [permissible up to 97% of rated frequency IEC 34-1 standard].
If MV is 10.45 kV only the switchgear bus bar voltage will be [at motor start Ist =6*Irated]: 358 V = approx.90% still permissible for 100% rated frequency.
The total voltage drop up to motor terminals when starting will be 19.5% permissible according to IEC 34-12.
Regards

 

[varri79]

Hi Skogsgurra, I would not worry if the control supply for existing motors taken from a secured source (UPS) but it is taken from its own bus hence there is a chance of dropping of other running motor contactors during starting of this 75kw motor.

Next one, the total starting load is not 150% of transformer rating, but it is 246%.
{As Current = 783A + 248A = 1031A,
1031 A*415 V*1.732= 740.3KVA , 740.3 KVA/300 KVA = 246%}.


I went though IEC codes, finally I found it in IEC60354.
As per IEC 60354, Clause 1.4.3 at Table, the recommends short time emergency loading is 200%.

In this case, I cannot start motor on the 300KVA transformer.
Pl advice the consequences if I recommend to start this motor (75KW with Direct Online) on the existing 300KVA transformer, with base load of 143KW.

I am not avoiding the possibility of soft starter, but I am more interested at Direct on line starter, as my client requires a cost effective solution.


Hi 7anoter4, you have presented with the figures of voltage drop. ThankS for your info. It’s fantastic.
However, the voltage drop at motor terminals during starting is coming around 83%, (during dynamic starting condition). I did motor starting study in a software and I found the voltage drop coming around 83%.

your advices are welcome, and thanks again.

 

[varri79]

Hi 7anoter4 (Electrical),

COuld you please explain me how did you obtain the starting voltage drop of 373V=90%?

What is the cable impedance you consider? and what is the impedance of motor (at the time of starting) you considered?.

I would appreciate your explanation.

Thanks

 

[7anoter4]

Hi varri79
According to IEC 34-12 up to 200 hp 19.5% voltage drop from motor rated voltage at start it is ok . So 83% is only 17% that means it is ok indeed.
For transformer overload –as the motor starting time will be less than 1 sec I think is still ok. If I suppose it is a very heavy motor –cast-iron TEFC- of 8 poles[750 rpm ] and according to ABB catalog the inertia moment will be 4.1 kg*m^2.The starting torque is 1.7 *rated torque and the maximum torque is 2.7.The average motor torque will be:
0.45*(Tstart+Tmax) = 1.98*Trated=1916.6 Nm
Trated =968 Nm
Suppose the load inertia moment would be the same [4.1] and the load torque would be equal to Trated [constant all the time].
Accelerating torque is the average motor torque-loadtorque
Tacc=1916.6-968=948.6 Nm ka=acceleration constant ka=pi*freq*no.poles pi=3.14156 freq=50 HZ ka=78.54
Starttime= (JMotor+JLoad)*ka/Tacc Starttime=2*4.1*78.54/948.6=0.7 sec
According to DIN 57532 a transformer up to 630 KVA withstand a full short-ciruit current for 2 sec.
Isc=417/.05= 8340 A
For 1 sec Imax=sqrt (Isc^2*2sec) =sqrt (8.34^2*2) =11.7 kA .So 1.11KA for this xfrm is far from maximum permissible.
Regards

 

[waross]

I know that Canadian and US codes don't apply but they may provide an insite.
1>Under Canadian codes, the largest motor must be calculated at 125%
This would make your added load:
Motor 75kW x 1.25 = 93.75 kW
Transformer 15 kW
Total additional load 108.75 Kw
Total load 108.75 + 143 = 251.75 @ 0.8 PF = 315 kVA.
This would not be allowed as a new installation under Canadian code.
2> However, as an addition to an existing load, the code will allow a demonstrated base load to be used for additional loads and it would probably qualify under this provision.
I have seen peak recording or trend recording instruments installed for several months to demonstrate to the Authority Having Jurisdiction that there is sufficient reserve capacity for additional loads.
Many large installations already have the peak demand indicated on the power bill, even if they are not on demand billing. Even when the demand is not shown on the power invoice, most industrial meters show the greatest peak demand since the last meter reset.
If your calculations are based on connected load rather than measured demand, you may gain spare capacity that you were unaware of by checking your peak demand.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

I agree with you, waross, as 100% load on this transformer is very limited. I had change the software as the former
edition does not permit more than 80% load [for a new installed transformer].
The parameters I took into consideration are as follows:
Transformer
Strf=0.3 MVA
Uk%=5
Ur= 415 V
Xtrf =uk/100*Ur^2/Strf= 0.0287 ohm
Irxfrm = 417 A
Load%= 100%
Pf= 0.85
SINFtr = 0.52678
I(ACTIVE)= 354.757 A
I(REACT)= 219.858
uka [active] = 1.333%
ukr [reactive]= 4.8189 %
xtrf/restrf= 3.6
ukr%=sqrt(1-1/(xtf/restrf)^2)*uk%
uka%=ukr%/(xtrf/restrf)
Feeder:2*3*240 sqr.mm XLPE 1/0.6Kv
spec.resist. =0.103 ohm/km
spec.react. = 0.07033 ohm/km
cable length =0.05 km
resistance = 0.002575 ohm
reactance = 0.00175828 ohm
Motor
Pmot= 75 kW
Urmotor =400 V
Ist/Irmot = 6
Irmot = 140 A
Pfmot= 0.85
Pfstart=0.37
max.drop%=18
Istmot =840 A
Motor cable 3*70 sqr.mm XLPE 1/0.6 KV
spec.resist. = 0.344ohm/km
spec.react. = 0.072759 ohm/km
cable length =0.104 km
resistance = 0.035776ohm
reactance = 0.0075669 ohm

This is software I did long time ago so is difficult now to explain. Here are some patterns:
The xfrm current when motor start will be:
ITst =SQRT((Irxfrm*pftr-Irmot*pfrmot+Istmot*pfst)^2+(Irxfrm*SINFtr-Irmot*SINFmot+Istmot*sinst)
newtanfi=(Irxfrm*SINFtr-Irmot*SINFmot+Istmot*sinst)/(Irxfrm*pftr-Irmot*pfrmot+Istmot*pfst)
angleFi=atan(newtanfi)
Voltage drop through xfrm:
ufi1=uka*cos(angleFi)+ukr*sin(angleFi)
ufi2=ukr*cos(angleFi)+uka*sin(angleFi)
Ureducing factor kU=0.95 or kU=1
dutrf%=100*kU/100*(1+ITst*kU/Irxfrm*ufi1-sqrt(1-ufi1*ITst*kU/Irxfrm*ufi2)^2)
The voltage drop through cable is similar
Regards

 

[varri79]

Hi 7anoter4 (Electrical,

Thank you for your explanation.

" Isc=417/.05= 8340 A
For 1 sec Imax=sqrt (Isc^2*2sec) =sqrt (8.34^2*2) =11.7 kA .So 1.11KA for this xfrm is far from maximum permissible".

Yes the starting current is far from maximum permissible.

But i wonder the starting time is around 1 sec, as it is motor is going to drive a centri pump.

Any way, even if i take 6 secs starting time, it is still safe.

However, I am confused due to IEC (60354) recommendation, about short time overloads.

Is this limit (shown in Table 1) not for motor starting?
Otherwise, what type loads?

Please see the attachment.


Regards

 

[waross]

1, One thing that I did not see in the parameters was a measured actual demand value for the base load.
2, Rather than cyclic or emergency loading, I would consider motor starting to be a transient. I understand cyclic loading to be loading that may persist for up to one hour. Emergency loading would be until repairs may be effected, possibly several days but hopefully sooner.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Waross you are right again! The overload allowed in the attached page [IEC (60354) recommendation] refers to the hours of overload. But there is no reference to the base load and no reference of duration. In DIN VDE 0536 [see attached page no. 531 from ABB Switchgear Manual] is a reference to this. From fig. 12-5 a [ONAN] if the initial load [K1] was 0.6[60%] the transformer may support 200% [K2] for a half an hour.
Regards

 

[varri79]

Thank you 7anoter4 for your valuable inputs.

I got good info from you.

Regards,

 

[dnqhung]

hi 7anoter4,
You said "According to IEC 34-12(IEC 60034-12?) up to 200 hp 19.5% voltage drop from motor rated voltage at start it is ok"
is there any mistake?
IEC 60034-12 doesn't specify any thing about voltage drop for motor starting, it only specifies the torque requirment.
am i wrong?
it took me alot time searchinh IEC standard about Voltage drop for motor starting.
Please give me an advice.

 

[7anoter4]

hi dnghung,
You are right: I didn't found in IEC 60034-12 the permissible voltage drop.
What I found is only start torque, pull-out and maximum torque, as you said.
I don't remember the edition no. or year but I took from this standard the following list:
Up to[hp]: 12.5
30.0%
Up
to[hp]: 50
25.0%
Up to[hp]: 75
19.7%
Up to[hp]: 200
16.3%

upper 9.6%

One may take this from the beginning and check it against calculation.

 

[waross]

I am embarrassed. We haven't considered the very low power factor of the starting motor. The majority of the starting current will be reactive or quadrature current acting at right angles to the real current. We cannot directly add the starting current to the base load current. When we consider the poor power factor of the starting current, the actual transformer current will be less than we have been assuming.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

hi dnghung
I think that in previous edition [34-12] the permissible voltage drop was calculated from the minimum pull-up torque .The minimum required Tu/TN in order to overcome the load resistance torque would be 0.49 [Tu is pull-up torque IEC 34-12 symbol] In this case the old list is still valid. If we should employ this relation:
DU%=(1-SQRT(0.49/minimumTu)*100 we could get this list [approximately]